## December 20, 2009

### This Week’s Finds in Mathematical Physics (Week 287)

#### Posted by John Baez

In week287 of This Week’s Finds, discover Jean-Pierre Marquis and Gonzalo Reyes’ history of categorical logic. Then continue learning about rational homotopy theory! Learn a quick way to construct differential forms starting from any commutative algebra… and see how Sullivan used this to construct “rational differential forms” starting from any topological space.

Also, guess what this is a picture of:

Posted at December 20, 2009 7:49 AM UTC

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### Re: This Week’s Finds in Mathematical Physics (Week 287)

The answer to the question about DGCA^op is surely “derived affine schemes”, no?

Posted by: Aaron Bergman on December 20, 2009 2:53 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Sounds plausible, but according to some folks, the category of derived affine schemes is the opposite of the category of simplicial commutative algebras. As far as I can tell, this is not equivalent to ${\mathrm{DGCA}}^{\mathrm{op}}$. For starters, a DGCA has an underlying cochain complex, while a simplicial commutative algebra has an underlying simplicial abelian group, which is equivalent to a chain complex.

I’m confused about a lot of basic stuff here… my puzzlements could fill a book, but nobody would want to read it.

Posted by: John Baez on December 20, 2009 6:10 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I think the correct answer is “affine dg-schemes”, but derived schemes are nicer objects. I have this vague recollection that in characteristic zero there isn’t much of a difference, but at this point I’ll let the actual experts take over.

Posted by: Aaron Bergman on December 20, 2009 7:24 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

When writing my reply to your question, I was initially going to suggest the possibility that things work better in characteristic zero. After all, this ‘betterness’ is why people like rational homotopy theory: working over the rationals you can make the cochains on a topological space into commutative DGA, while over the integers it’s at best homotopy-commutative, or more precisely, an ${E}_{\infty }$ algebra.

But then I realized I was confused about so many issues that I should better shut up.

Posted by: John Baez on December 20, 2009 8:26 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

For starters, a DGCA has an underlying cochain complex, while a simplicial commutative algebra has an underlying simplicial abelian group, which is equivalent to a chain complex.

In fact, equivalent to a non-negatively graded chain complex. By renaming degree $n$ to degree $-n$ one can identify non-negatively graded chain complexes with non-positively graded cochain complexes.

according to some folks, the category of derived affine schemes is the opposite of the category of simplicial commutative algebras.

The word “derived” is used rather unspecifically. A central motivation for “derived algebraic geometry” has been spaces that are locally modeled not on (co)simplicial rings, but on ring spectra, for instance.

There is the general concept of a generalized scheme defined by a chosen geometry. As soon as that “geometry” is not simply a 1-category, I suppose peope will feel entitled to talk about “derived geometry”.

(Personally I find “derived” not a terribly descriptive term. Speaking of “higher geometry” sounds more natural to me.)

I’d think that under mild technical extra conditions there are natural structures of such “geometries” not only on simplicial rings, but also on cochain DGCAs.

Posted by: Urs Schreiber on December 21, 2009 11:17 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

took me a while to cotton on to derived’ meaning
resolved by’

Posted by: jim stasheff on December 21, 2009 1:50 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

took me a while to cotton on to derived’ meaning resolved by’

And even that is true only with quite a bit of extra qualification added.

It seems that the term “derived” has an origin in contrast with its uage today: it originates from a time when people didn’t understand conceptually what happened when they inverted morphisms and passed to “derived categories” and not what happened when they passed to “derived functors” on these, apart from knowing that they had a prescription wich allowed them to derive some construction from another construction. Isn’t it?

But today, with things propertly understood, “deriving” is really mostly meant has “$\infty$-categorifying”, only that for some reason higher category theory these days is to a large extent done by people who think of themselves not as higher category theorists, so they don’t use that language.

For what it’s worth, I prefer higher geometry over “derived geometry”. For one, this indicates immediately to everyone who knows about categorification that there is not the notion of derived/higher geometry, but many such.

In the same vein, I am surprised to see so much beautiful higher category theory these days be under-sold as “derived algebraic geometry”. Lurie’s notion of space is vastly and vastly more general than derived algebraic geometry. It is nothing less than a full development of Lawvere’s hugely-abstract-nonsense concepts of space and quantity to the full context of $\left(\infty ,1\right)$-topos theory.

Calling this “derived algebraic geometry” after one of its motivating examples is probably just an attempt to trick abstract category theorists into ignoring these developments. And the trick seems to work. (This is meant jokingly.)

Posted by: Urs Schreiber on December 21, 2009 2:21 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Urs wrote:

In the same vein, I am surprised to see so much beautiful higher category theory these days be under-sold as “derived algebraic geometry”.

Shhh! There are still plenty of people scared of ‘higher category theory’ who are eager to learn ‘derived algebraic geometry’. We should only tell them it’s higher category theory after they’ve all learned it.

This reminds me of the joke about the guy who went to an old-fashioned Jewish restaurant on the lower East Side and was surprised that all the waiters were Chinese, but spoke Yiddish.

Calling this “derived algebraic geometry” after one of its motivating examples is probably just an attempt to trick abstract category theorists into ignoring these developments.

Maybe it’s an attempt to trick algebraic geometers into not ignoring them!

Posted by: John Baez on December 21, 2009 10:06 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

John wrote:

For starters, a DGCA has an underlying cochain complex, while a simplicial commutative algebra has an underlying simplicial abelian group, which is equivalent to a chain complex.

Urs wrote:

In fact, equivalent to a non-negatively graded chain complex.

Right. Just so people know: in week287 I defined a cochain complex to be a non-negatively graded chain complex of vector spaces, and in the not-yet-published week288 I define a chain complex to be a non-negatively graded chain complex of vector spaces.

Why? Mainly just to save space.

One always needs to be careful: different people in different contexts mean somewhat different things when they say ‘chain complex’ and ‘cochain complex’. The main difference is their choice of grading. There’s also the question of whether they’re talking about complexes of vector spaces, abelian groups, $R$-modules, etc.

So, in my conversations based on This Week’s Finds, I’ll define these terms the way I just did. But when I chat with random mathematicians in bars, I always tell them how my chain complexes are graded, to avoid any ambiguity.

The term ‘DGCA’ is similarly ambiguous without explanation, but in This Week’s Finds it’ll always mean a commutative monoid in the category of nonnegatively graded cochain complexes. The book by Félix, Halperin and Thomas calls such a thing a ‘cochain algebra’.

Posted by: John Baez on December 21, 2009 9:04 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

John wrote:

So, let’s focus our attention on functions on |Sing(X)| that when restricted to any simplex give polynomials with rational coefficients.

Is it obvious that there are any such functions?

Posted by: Maarten Bergvelt on December 20, 2009 8:58 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

(Simple answer would be: Every constant function from |Sing(X)| to the rationals would be of the kind we are looking for, but that’s obviously of no help :-).

Maybe this helps (I stand a chance to be the most ignorant reader of this edition of TWF, let’s see if I can prove it :-):

We talk about functions on |Sing(X)| - implying that we mean continuous functions (because |Sing(X)| lives in the category of compactly generated Hausdorff topological spaces) with values in a field containing the rational numbers (since we talk about rational homotopy theory).

What does it mean to be continous? If you build your space explicitly with topological simplexes that are glued together, a function is continous iff it is continous on every topological simplex.

If you want to have a function that is continous and is a polynomial with rational coefficients on each simplex, what you build is essentially what is called a spline in numerics. You can always build such a spline by induction on the dimension of the simplices that you define it on, starting with the values that it takes on the 0-simplices (i.e. “points”, you choose one rational number for each “point”).

Posted by: Tim vB on December 21, 2009 11:20 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Tim vB wrote

If you want to have a function that is continous and is a polynomial with rational coefficients on each simplex, what you build is essentially what is called a spline in numerics. You can always build such a spline by induction on the dimension of the simplices that you define it on, starting with the values that it takes on the 0-simplices (i.e. “points”, you choose one rational number for each “point”).

Thanks, that is very helpful. I just did not think to start at 0-simplices, I tried imagining gluing polynomials on two triangles together, which sounded unlikely to work. Shows that I must be the most ignorant reader here!

Posted by: Maarten Bergvelt on December 21, 2009 4:51 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Shows that I must be the most ignorant reader here!

Sorry Maarten. That honor firmly belongs to me :)

Posted by: Eric on December 22, 2009 12:30 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Given the quote at the end of week287, it sounds like you guys vying for the position of ‘most ignorant reader’ are actually showing off!

We live on an island surrounded by a sea of ignorance. As our island of knowledge grows, so does the shore of our ignorance.
- John Wheeler

Posted by: John Baez on December 23, 2009 8:42 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

This necessitates an explanation, since I am very proud of my humility:
First, my shores of ignorance grow at least ten times as fast as my islands of knowledge. They must be some kind of fractal: For everything I understand, I learn about ten things I don’t understand.
This leads to a negative learning-payoff coefficient (sloppy: “the more you learn, the dumber you feel”).
Given this, it is valid to conclude that I tried to show off.
(Note that I use a localized version of the island and shore metaphor, John Wheeler was speaking of humanity, I’m speaking of myself only).

Here is what I had in mind, however:
The path connected component of my island of knowledge that “rational homotopy theory” lives on has a comparativley small volume, with one reason being that there is no continuous path to my mainland, which core is built on topics from theoretical physics. But that will perhaps be remedied when we arrive at the applications (a promise was given, hopes were created).

P.S.: Although I’m too late: Merry Christmas!

Posted by: Tim vB on December 26, 2009 1:07 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Is it obvious that there are any such functions?

The coend formula may be thought of as a constructive way to obtain such functions:

essentially by definition, an element in the function algebra

${\int }^{\left[n\right]\in \Delta }\mathrm{hom}\left({\Delta }_{\mathrm{Top}}^{n},X\right)\cdot {\Omega }^{•}\left({\Delta }_{\mathrm{Top}}^{n}\right)$

is given by specifying on each simplex a suitable function, such that these coincide on the boundaries of adjacent simplices. This is checked to be homotopy-invariant, so we can use simpler simplicial models of our space to produce this. For instance we may choose the simplicial complex coming from a triangulation of the space. With respect to this a function in question is a piecewise smooth one.

Posted by: Urs Schreiber on December 21, 2009 12:04 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Sorry, I seem to miss something obvious here, my understanding of the question was: If we restrict ourselves to polynomials with rational coefficients that are not constant on every connected component, could it happen that it is impossible to fulfill the matching condition on every face shared by adjacent cells?

Is the answer obviously no? Or does this follow from what you said? (If the answer is “yes” I’m obviously too short of categorical knowledge).

Posted by: Tim vB on December 21, 2009 3:23 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Or does this follow from what you said?

What i said was only meant to be an argument for why it is sufficient to look at functions that are piecewise polynomial/smooth/whatever with respect to a given triangulation, even though $\mathrm{Sing}\left(X\right)$ is much bigger than the simplicial complex obtained from a triangulation.

Posted by: Urs Schreiber on December 21, 2009 3:37 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Ok, thanks, but: Since the X we started with was an arbitrary topological space, how does it help to reduce the question from |Sing(X)| to a triangulation of X?

(In case I seem to be a bit stubborn, here’s my excuse: I’ve never heard of rational homotopy theory before I read week 286, it got me interested, I grabbed the Félix/Halperin/Thomas book as time filler for the christmas holidays - any help that gives me a nudge in the right direction is appreciated!).

Posted by: Tim vB on December 21, 2009 4:14 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

It doesn’t particularly help to reduce the problem from $\mid \mathrm{Sing}\left(X\right)\mid$ to a triangulation of $X$. In fact the real meat of Maarten Bergvelt’s question has nothing to do with the original topological space $X$. It solely concerns the simplicial set $\mathrm{Sing}\left(X\right)$ and its geometric realization $\mid \mathrm{Sing}\left(X\right)\mid$. So let’s call that simplicial set $S$.

The question is now: given a simplicial set $S$, how can we construct simplex-wise polynomial functions on its geometric realization $\mid S\mid$?

Urs’ coend formula gives this answer: “by specifying on each simplex a suitable [polynomial] function, such that these coincide on the boundaries of adjacent simplices”.

Unfortunately this does not make it obvious that any such functions exist, apart from constants. And I don’t agree with Urs when he says “This is checked to be homotopy-invariant, so we can use simpler simplicial models of our space to produce this.” A point is homotopy equivalent to a tetrahedron, but their algebras of simplex-wise polynomial functions are not isomorphic. The point has only constant functions on it; the tetrahedron does not. Of course these algebras of functions, while not isomorphic, are ‘homotopy equivalent’ in a subtler sense. But I don’t see how this helps us with our naive task of constructing non-constant polynomial functions on the tetrahedron!

Of course for the tetrahedron it’s pretty easy…

Anyway, you made a lot of progress in your first comment. Starting with a simplicial set, we can choose a rational number for each vertex and then use a systematic algorithm to construct a rational polynomial on each $n$-simplex with the given values at its vertices. For example, we can use linear interpolation.

For a simplicial complex this is really all we need to do. A simplicial complex is a simplicial set $S$ where there’s at most one $n$-simplex with any given set of vertices. In other words, an $n$-simplex is determined by its vertices.

In this case, we easily see that your method gives ‘enough’ simplex-wise polynomial functions on $\mid S\mid$.

What do I mean by ‘enough’? Well, at least enough to separate points — for any two distinct points $x,y\in \mid S\mid$ we’ll get a function $f$ with $f\left(x\right)\ne f\left(y\right)$.

Indeed, we can get enough functions to separate points merely by using your trick with linear interpolation. These simplex-wise rational linear functions generate an subalgebra $A$ of simplex-wise rational polynomial functions.

Conjecture: for a simplicial complex, the algebra $A$ consists of all the simplex-wise rational polynomial functions.

Regardless of whether the conjecture is right, in the case where $\mid S\mid$ is compact, the Stone–Weierstrass theorem shows that the algebra $A$ is dense in the algebra of all continuous real functions on $\mid S\mid$.

And that makes me feel that we’ve got ‘enough’ functions in this case.

But what if $S$ is not a simplicial complex? Then if we use a systematic algorithm to construct a rational polynomial on each $n$-simplex with given values at its vertices, we won’t get enough functions to separate points! The problem: if two $n$-simplices share the same vertices, our function will agree on these two $n$-simplices.

This is why I want someone to prove this:

Conjecture: Suppose we are given an $n$-simplex ${\Delta }^{n}$ and a continuous function $f$ on its boundary which is a rational polynomial on each face. Then $f$ extends to a rational polynomial on all of ${\Delta }^{n}$.

(If you can extend it in one way, then I can extend it in infinitely many, and thus get enough simplex-wise rational polynomial functions on $\mid S\mid$ to separate points, for any simplicial set $S$.)

Posted by: John Baez on December 21, 2009 6:27 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Conjecture: Suppose we are given an n-simplex ${\Delta }^{n}$ and a continuous function f on its boundary which is a rational polynomial on each face. Then f extends to a rational polynomial on all of ${\Delta }^{n}$.

I can show this.

Realize ${\Delta }^{n}$ as $\left\{\left({x}_{0},{x}_{1},\dots ,{x}_{n}\right):{x}_{0}+\cdots +{x}_{n}=1,{x}_{i}\ge 0\right\}$. For $i=0$, $1$, …, $n$, write ${\partial }_{i}\left(\Delta \right)$ for the face of $\Delta$ where ${x}_{i}=0$ and $\partial \left(\Delta \right)=\bigcup {\partial }_{i}\left(\Delta \right)$.

Observe that any polynomial function $f$ on ${\Delta }^{n}$ can be realized as the restriction of a polynomial function $F$ on ${ℝ}^{n+1}$. There is an integer $D\left(f\right)$ such that, if $d\ge D\left(f\right)$, we can take $F$ to be homogenous of degree $d$. (Just multiply any lower degree terms by appropriate powers of $\sum {x}_{i}$.)

Furthermore, for fixed $d$, the representative $F$ is unique. I emphasize here that this means that any two lifts $F$ and $F\prime$ must have the same coefficients on each term, not merely that they are equal as functions. If not, there would be a homogenous degree $d$ polynomial $F-F\prime$ which was $0$ on $\Delta$. Homogeneity then forces $F-F\prime$ to be $0$ on the positive orthant, which means that $F-F\prime$ is the zero polynomial $0$.

For $i=0$, …, $n$, let ${f}_{i}$ be a polynomial function on ${\partial }_{i}\left(\Delta \right)$, so that the ${f}_{i}$ agree on the boundaries. Choose $d$ sufficiently large that each ${f}_{i}$ can be lifted to ${F}_{i}$, a homogenous polynomial of degree $d$ in $\left\{{x}_{k}{\right\}}_{k\ne i}$. Note that ${f}_{i}{\mid }_{{\partial }_{i}\left(\Delta \right)\cap {\partial }_{j}\left(\Delta \right)}$ is lifted by plugging in $0$ for ${x}_{j}$ in ${F}_{i}$.

Using the uniqueness of lifting, we see that the polynomials ${F}_{i}{\mid }_{{x}_{j}=0}$ and ${F}_{j}{\mid }_{{x}_{i}=0}$ are the same. In other words, any monomial which contains neither ${x}_{i}$ nor ${x}_{j}$ appears with the same coefficient in ${F}_{i}$ and ${F}_{j}$. So there is a single polynomial $F$ such that $F{\mid }_{{x}_{i}=0}={F}_{i}$.

***********************************************************

I wrote this argument out in far more detail than it probably deserves. However, I would like to note that questions of this sort can be subtle. Consider the three lines $x=0$, $y=0$ and $x=y$ in ${ℝ}^{2}$. Let $f$ be the function which is $0$ on the first two lines, and equals $x$ on the third line. Then $f$ is continuous and is polynomial when restricted to each line, but is not the restriction of a polynomial to the locus of these three lines.

Posted by: David Speyer on December 21, 2009 9:01 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I think you may have reinvented (higher order) Whitney forms.

Posted by: Eric on December 22, 2009 12:02 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Nice proof, David!

Just in case anyone is having trouble seeing how David pulled the rabbit out of the hat, the key step is this:

Suppose you have a bunch of homogeneous degree-$d$ polynomials ${F}_{0},\dots ,{F}_{n}$, where ${F}_{i}$ is a polynomial in variables ${x}_{0},{x}_{1},\dots ,{x}_{i-1},{x}_{i+1},\dots ,{x}_{n}$. And suppose that any term containing neither ${x}_{i}$ nor ${x}_{j}$ appears with the same coefficient in both ${F}_{i}$ and ${F}_{j}$. Then there’s a single polynomial $F$ such that $F{\mid }_{{x}_{i}=0}={F}_{i}$.

And this is pretty easy if you think about it, but only you treat the cases $n=0,1$ differently from the rest.

Posted by: John Baez on December 21, 2009 9:47 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

And I don’t agree with Urs when he says “This is checked to be homotopy-invariant, so we can use simpler simplicial models of our space to produce this.”

I must be expressing myself badly. After I said this you made the same statement in reply to David Speyer below.

But I see that what I said didn’t quite address what Maarten was looking for.

Posted by: Urs Schreiber on December 22, 2009 10:50 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Thanks, that was enlightening, as always.

You wrote:

Regardless of whether the conjecture is right, in the case where ∣S∣ is compact, the Stone–Weierstrass theorem shows that the algebra A is dense in the algebra of all continuous real functions on ∣S∣.

I wonder why we need |S| to be compact, since we need to apply Stone-Weierstass to every simplex, not to |S| as a whole?

Posted by: Tim vB on December 23, 2009 11:56 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Tim vB wrote:

I wonder why we need $\mid S\mid$ to be compact, since we need to apply Stone-Weierstass to every simplex, not to $\mid S\mid$ as a whole?

It’s all about topologies on function spaces. The Stone–Weierstrass theorem gives conditions under which a collection of functions on a compact Hausdorff space is dense in the algebra of all continuous functions in the topology of uniform convergence.

If we apply it to each simplex of $\mid S\mid$, we’ll see the algebra of bounded functions that are rational polynomials on each simplex is dense in the algebra of all bounded continuous functions in the topology of simplex-wise uniform convergence.

And that’s probably fine for lots of things.

You’ll note the Wikipedia article states a version of Stone–Weierstrass for locally compact Hausdorff space. A version that weakens the assumption of local compactness can be found in this paper by Miguel Carrión-Álvarez.

Posted by: John Baez on December 23, 2009 8:12 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

John Baez wrote:

If we apply it to each simplex of |S|, we’ll see the algebra of bounded functions that are rational polynomials on each simplex is dense in the algebra of all bounded continuous functions in the topology of simplex-wise uniform convergence.

For reasons unkown that was the only topology I had in mind, well…

John Baez wrote:

A version that weakens the assumption of local compactness can be found in this paper by Miguel Carrion-Alvarez.

That’s a really nice paper! Let me just add for the record that - as claimed in paragraph 6 - the GNS construction on algebras of unbounded operators does indeed work without any major revision, if one takes care of the nasty domain problems. An example can be found here:

• Atsushi Inoue: “Tomita-Takesaki Theory in Algebras of Unbounded Operators”, Theorem 1.9.1

Mr. Inoue and his collegues have been working for quite some time on the generalization of concepts from C*-algebras to algebras of unbounded operators, but as far as I know without using much category theory :-(

Posted by: Tim vB on January 6, 2010 11:11 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

It’s voilà rather than voilá.

Posted by: David Corfield on December 20, 2009 9:39 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Voilà! Fixed!

Posted by: John Baez on December 21, 2009 9:16 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

In the TWF it says

There are many examples of such [generalized smooth] spaces, including manifolds with boundary, manifolds with corners, and infinite-dimensional manifolds.

[…]

But here’s a tougher question: how can we generalize differential forms to an arbitrary topological space $X$ ?

Is it really a tougher question, instead of a special case?

Every topological space is canonically a diffeological space, whose plots are the continuous functions from the test domains. And for every diffeological space there is a notion of differential forms on it. And for topological spaces regarded as diffeological spaces it is pretty much what Sullivan describes for simplices:

Let $C$ be a category of test spaces. Might be cartesian spaces, or domains of these, or contractible domains as in your article, or even just simplices. We have a functor ${\Omega }^{•}:{C}^{\mathrm{op}}\to \mathrm{dgAlg}$ that sends each smooth space to its deRham dg-algebra. This ${\Omega }^{•}$ is a standard example of a non-concrete sheaf on $C$. For any sheaf $X$ on $C$ (concrrete or not, i.e. diffeological space or not), the differential forms on $X$ are

${\Omega }^{•}\left(X\right)=\left[X,{\Omega }^{•}\right]$

where the right is as an internal hom of sheaves given by

$\cdots ={\int }_{U\in C}{\mathrm{Hom}}_{\mathrm{Sets}}\left(X\left(U\right),{\Omega }^{•}\left(U\right)\right)$

but which by thinking of ${\Omega }^{•}$ as mapping $C\to {\mathrm{dgAlg}}^{\mathrm{op}}$ we may read equivalently as being not an end, but a coend in ${\mathrm{dgAlg}}^{\mathrm{op}}$

$\cdots ={\int }^{U\in C}X\left(U\right)\cdot {\Omega }^{•}\left(U\right)\phantom{\rule{thinmathspace}{0ex}}.$

So differential forms on diffeological spaces are given by the geometric realization formula that you mention. And Sullivan’s construction is from this perspective nothing but the special case where $C=\Delta$.

Posted by: Urs Schreiber on December 21, 2009 11:34 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Urs wrote:

Is it really a tougher question, instead of a special case?

Every topological space is canonically a diffeological space, whose plots are the continuous functions from the test domains. And for every diffeological space there is a notion of differential forms on it. And for topological spaces regarded as diffeological spaces it is pretty much what Sullivan describes for simplices…

You’re certainly right that Sullivan’s approach isn’t really ‘tougher’. It’s actually simpler. But it’s shockingly radical at first glance.

I’m having trouble understanding the rest of what you’re saying, perhaps because it doesn’t seem quite right to me. We take a topological space $X$ and treat as a smooth space with continuous plots, using simplexes as the domains of our plots. You build the differential forms on this space following a definition like those of Souriau and Chen. I follow the method of Sullivan.

Then I don’t care at all about when one simplex is covered by other simplices. But you do. So I get a vastly bigger algebra of differential forms. For example, your 0-forms will be the continuous functions on $X$. Mine will be the continuous functions on $\mid \mathrm{Sing}\left(X\right)\mid$.

Maybe what I’m saying is that different Grothendieck topologies are at work here.

But maybe you’re just saying that these are both special cases of the same type of construction. And that’s a good point.

But my original point was this: if we take a topological space $X$, regard it as a diffeological space a la Souriau with continuous plots, and build the differential forms following Souriau’s prescription, we get junk — nothing very useful.

On the other hand, Sullivan’s prescription gives something homotopy-invariant.

What do I mean by ‘junk’? I’m not exactly sure. Say we take $X$ to be the real line. Then what algebra of differential forms do we actually get? I leave this as an open question.

And here’s another puzzle that’s been on my mind lately. Start with the algebra of continuous real-valued functions on the real line, and form the module of Kähler differentials. What do we get?

In other words, take formal linear combinations of symbols

$fdg$

where $f$ and $g$ are continuous real-valued functions on the real line, modulo the relations

$\left({f}_{1}+{f}_{2}\right)dg={f}_{1}dg+{f}_{2}dg$

$fd\left({g}_{1}+{g}_{2}\right)=fd{g}_{1}+fd{g}_{2}$

$fd\left(gh\right)=fhdg+fgdh$

and

$fd\left(cg\right)=cfdg$

for $f,g,h$ functions and $c$ a constant.

What do we get?

For example: as a module over the continuous real-valued functions on the real line, what sort of module do we get? A free module? If so, of what rank?

My intuition is that the product rule

$d\left(gh\right)=gdh+hdg$

leads to very weird conclusions when applied to continuous but nondifferentiable functions, like the square root of the absolute value of $x$.

Posted by: John Baez on December 21, 2009 9:24 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Those are some of my all-time favorite relations.

$f\left(dg\right)=\left(dg\right)f.$

However, I also like to have a “1” and I like to have

$fg=gf.$

And we cannot forget

${d}^{2}=0.$

:)

Posted by: Eric on December 22, 2009 12:11 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Ugh. The Wikipedia article has

$d\left(st\right)=s\left(dt\right)+\left(ds\right)t.$

That isn’t necessary for Kähler differentials, is it? We do not need to assume 0-forms and 1-forms commute. I’d be tempted to change the Wikipedia article to

$d\left(st\right)=s\left(dt\right)+t\left(ds\right).$
Posted by: Eric on December 22, 2009 12:17 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Eric wrote:

I’d be tempted to change the Wikipedia article to

$d\left(st\right)=s\left(dt\right)+t\left(ds\right).$

That’s what it already says, and apparently the last change was December 16th, before you came along.

Posted by: John Baez on December 22, 2009 1:19 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Sorry. It was early and I hadn’t had my coffee yet. I wanted to change the wikipedia article, which current has

$d\left(st\right)=s\left(dt\right)+t\left(ds\right)$

to

$d\left(st\right)=s\left(dt\right)+\left(ds\right)t.$

Sorry. Plus, to be clear, I won’t change it because I’m not qualified to determine whether it really needs to be changed. If someone came and confirmed that it should be changed, I might change it, but then and only then. I’m guessing it is fine as is.

Posted by: Eric on December 22, 2009 2:22 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

It’s fine as is, and maybe technically more correct this way, since then we’re only relying on the left $A$-module structure of our Kähler differentials, instead of assuming people know that a left module of a commutative algebra automatically becomes a bimodule.

In less jargonesgue lingo: the audience for this Wikipedia article — budding young algebraists, apparently — will understand that you can multiply $ds$ on the left by $t$, since the article says the Kähler differentials are a ‘module’ for the functions, and that usually means ‘left module’. But these kids may get confused if you start multiplying $ds$ on the right by $t$.

Of course, I sympathize with you: $d\left(st\right)=s\left(dt\right)+\left(ds\right)t$ is morally superior. So kids who don’t know about bimodules need to learn more math… but vita brevis, ars longa.

Posted by: John Baez on December 22, 2009 3:33 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

d(st) = (ds)t + s(dt)
is also a much better way of expressing/ *teaching*
the product rule
much easier to go on to d(stuvw…)

Posted by: jim stasheff on December 22, 2009 1:32 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

The main situation I know in which “0-forms” and “1-forms” don’t commute is when we are taking k-forms valued in endomorphisms of a vector bundle. Or, to be more concrete, when we are considering matrices with differential forms as entries.

But then $d\left(fg\right)=f\left(dg\right)+\left(df\right)g$ is right and $d\left(fg\right)=f\left(dg\right)+g\left(df\right)$ is wrong. In what setting is the reverse true?

Posted by: David Speyer on December 22, 2009 2:02 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I don’t know any situations where $d\left(fg\right)=\left(df\right)g+f\left(dg\right)$ is wrong and $d\left(fg\right)=\left(df\right)g+\left(dg\right)f$ is right, except highly constrained pedagogical situations where you’re talking to people who only know about left modules and you don’t have time to tell them about bimodules.

As far as I know,

$d\left(fg\right)=\left(df\right)g+f\left(dg\right)$

is always morally superior to

$d\left(fg\right)=\left(df\right)g+\left(dg\right)f$

It’s naughty to go around switching the order of letters unless you have a darn good reason!

In particular, if $A$ is a not-necessarily-commutative algebra, the nice notion of a derivation of $A$ valued in $B$ takes $B$ to be an $\left(A,A\right)$-bimodule. Then a derivation of $A$ valued in $B$ is a linear map

$f:A\to B$

such that

$d\left(fg\right)=\left(df\right)g+f\left(dg\right)$

And then there’s still a way to define Kähler differentials such that they’re the ‘universal $\left(A,A\right)$-bimodule equipped with a derivation of $A$ taking values in it’. It looks just like the usual definition except we need to make sure to say

$d\left(fg\right)=\left(df\right)g+f\left(dg\right)$

People use these Kähler differentials in noncommutative geometry. They call them something like ‘the universal first-order differential calculus for $A$’.

Posted by: John Baez on December 22, 2009 3:56 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I bet the version in the _wrong_ order is due to a lack of use of parentheses dfg is ambiguous

Posted by: jim stasheff on December 22, 2009 1:21 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Eric wrote:

$f\left(dg\right)=\left(dg\right)f.$

That’s because I was just defining a left $A$-module for my commutative algebra $A$, so $\left(dg\right)f$ makes no sense. Of course, since $A$ is commutative, this left module can be made into a bimodule by defining $\left(dg\right)f=f\left(dg\right)$, and people working on Kähler differentials always do this. But I know you have a different approach.

However, I also like to have a “1”…

When I say “commutative algebra” in this series of This Week’s Finds, I’ll always assume it’s unital, so it has a $1$.

Of course not everyone assumes their algebras are unital. And annoyingly enough, this ‘not everyone’ includes the Wikipedia article on algebras. So, I’ll need to correct week287 to include the definition of algebra that I’m actually using!

…and I like to have $fg=gf$.

Just to be clear: this has nothing to do with the definition of Kähler differentials; this is just the assumption that our algebra is commutative!

And we cannot forget

${d}^{2}=0.$

:)

I don’t know if that smiley means ‘just kidding’, so I’ll humorlessly observe that Kähler differentials play the role of one-forms for our commutative algebra $A$ — so ${d}^{2}$ doesn’t enter the game at this point.

However, the Kähler differentials are just the grade-1 part of the bigger gadget that I called $\Omega \left(A\right)$ in week287. In $\Omega \left(A\right)$ we have ${d}^{2}=0$, and also the relation you dislike: $fdg=\left(dg\right)f$.

(I find it impossible to talk about whether I ‘like’ or ‘dislike’ an equation of this sort. Whether or not we should assume it depends on what we’re doing.)

Posted by: John Baez on December 22, 2009 1:21 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

The smiley was an expression of happiness. ${d}^{2}=0$ is so beautiful I thought it made everyone smile :)

The list of things I liked was independent of Kahler. I could have included pizza in the list. It is a happy coincidence that many things I like are also part of the definition of Kahler differentials. I especially like universal differential calculus on commutative unital algebras :)

Sorry for the distraction. My original intention was to simply let you know you sparked my interest and I was paying attention :)

Posted by: Eric on December 22, 2009 5:03 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

But maybe you’re just saying that these are both special cases of the same type of construction.

Yes. Souriau uses $C=\mathrm{OpenDomains}$ where Sullivan uses $C=\Delta$. The topology doesn’t play a role.

Posted by: Urs Schreiber on December 22, 2009 7:54 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

What exactly do you mean by ‘The topology doesn’t play a role?’

Sullivan uses presheaves on $\Delta$, so no Grothendieck topology plays a role there. Souriau uses sheaves on $\mathrm{OpenDomains}$, so a Grothendieck topology does play a role there.

To unify them, we can think of presheaves on $\Delta$ as sheaves on $\Delta$ equipped with a dull sort of Grothendieck topology — that’s why I said ‘different Grothendieck topologies are at work here’.

Posted by: John Baez on December 22, 2009 8:46 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

What exactly do you mean by ‘The topology doesn’t play a role?’

The formula for the differential forms does not involve any sheafification. It works for presheaves just as well as for sheaves.

Posted by: Urs Schreiber on December 22, 2009 10:50 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Concerning my comment on the relation between Sullivan’s differential forms and differential forms on (pre)sheaf models for generalized smooth spaces, John writes:

I’m having trouble understanding the rest of what you’re saying,

Okay, I’ll say it in more detail:

Recollection of the context

For $C$ any category of smooth test spaces, we may think of presheaves on $C$ as generalized smooth spaces. Sometimes one finds this a little too general and restricts among all presheaves to those that are sheaves with respect to some topology, and possibly even further to those that are concrete sheaves. The following discussion works for all presheaves just as well as for these special cases.

Whatever the objects of $C$ actually are, that we think of them as smooth test spaces means usually in particular that we have a functor ${\Omega }^{•}:{C}^{\mathrm{op}}\to \mathrm{dgAlg}$ that sends each test spaces to its dg-algebra of differential forms.

Pedestrian definition of diff forms

In the literature on generalized smooth spaces in the form of (pre)sheaves on a test category $C$, a differential $n$-form on a presheaf $X$ is usually defined to be a collection $\left\{{\omega }_{p}\in {\Omega }^{n}\left(U\right)\mid p:U\to X\right\}$ of ordinary differential forms on the domains of the plots, one for each plot, such that this is compatible with pullback of plots and forms along morphisms in $C$.

The collection of forms on $X$ defined this way becomes a dg-algebra by defining all operations plotwise.

End-definition of diff forms

This pedestrian definition is equivalent to the following more high-brow definition:

the assignment ${\Omega }^{n}:U↦{\Omega }^{n}\left(U\right)$ of the set of $n$-forms on test domain $U$ is itself a presheaf on $C$. So for $X$ any presheaf, we may look at the hom-set of presheaves from $X$ to ${\Omega }^{n}$. This is given by the end

${\mathrm{Hom}}_{\left[{C}^{\mathrm{op}},\mathrm{Set}\right]}\left(X,{\Omega }^{n}\right)={\int }_{U\in C}{\mathrm{Hom}}_{\mathrm{Set}}\left(X\left(U\right),{\Omega }^{n}\left(U\right)\right)\phantom{\rule{thinmathspace}{0ex}}.$

This set is canonically isomorphic to the set of pedestrian $n$-forms on $X$, as defined before: an element of the end is for each $U$ in $C$ an element $\left\{{\omega }_{U}\right\}$ in ${\mathrm{Hom}}_{\mathrm{Set}}\left(X\left(U\right),{\Omega }^{n}\left(U\right)\right)$ (hence one $n$-form on $U$ per plot $p:U\to X$) such that for all morphisms $U\to V$ the canonical diagram

$\begin{array}{ccc}*& \stackrel{{\omega }_{U}}{\to }& \left[X\left(U\right),{\Omega }^{n}\left(U\right)\right]\\ {↓}^{{\omega }_{V}}& & ↓\\ \left[X\left(V\right),{\Omega }^{n}\left(V\right)\right]& \to & \left[X\left(V\right),{\Omega }^{n}\left(U\right)\right]\end{array}$

commutes. This commutativity imposes precisely the compatibility of the forms on the plot domains, as before in the pedestrian approach.

Coend-definition of diff forms

Now the observation that I stated in my comment:

the above end was formed in $\mathrm{Set}$. But we may equivalently think of it in ${\mathrm{Ch}}^{•}$ – if we stick to homogeneous degree as I did so far – or in $\mathrm{dgAlg}$ – if we sum over all form degrees. For that, we think of the expression in the integrand as the $\mathrm{Set}$-powering of ${\mathrm{Ch}}^{•}$ or $\mathrm{dgAlg}$: for $S$ any set and $A$ a cochain comlex/dg-algebra, $\left[S,A\right]={A}^{{\oplus }^{\mid S\mid }}$ is the direct sum of as many copies of $A$ as $S$ has elements.

Indeed, taking the end over the powering in this way produces explicitly the cochain complex/dg-algebra structure on the result, which is used in the pedestrian approach: the scalar multiplication, the wedge product and the differential on ${\int }_{U}\left[X\left(U\right),{\Omega }^{n}\left(U\right)\right]$ are all taken plot-wise.

Finally, all I did was to observe that the $\mathrm{Set}$-powering of $\mathrm{dgAlg}$ translates by abstract duality into the copowering of ${\mathrm{dgAlg}}^{\mathrm{op}}$, written $X\left(U\right)\cdot {\Omega }^{n}\left(U\right)$. Moreover, the end in $\mathrm{dgAlg}$ discussed so far is a coend in ${\mathrm{dgAlg}}^{\mathrm{op}}$, obviously, so we find that the above formula is equivalent to the coend

${\int }^{U\in C}X\left(U\right)\cdot {\Omega }^{n}\left(U\right)$

taken in ${\mathrm{dgAlg}}^{\mathrm{op}}$. This is the geometric realization formula.

In particular, for $C=\Delta$ and ${\Omega }^{•}:\Delta \to {\mathrm{dgAlg}}^{\mathrm{op}}$ the functor that assigns polynomial differential forms to simplices, and for for $X=\mathrm{Sing}\left({X}_{\mathrm{Top}}\right)$ the singular simplicial set of a topological space ${X}_{\mathrm{Top}}$, this is Sullivan’s definition of the dg-algebra of forms on ${X}_{\mathrm{Top}}$.

Or so I think. Please let me know if you (still) think that I am mixed up.

Posted by: Urs Schreiber on December 22, 2009 12:40 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Checking the references, I see that the end-description of the Sullivan forms that I mentioned is also given in def 1.20, page 9 of the review that you reference.

Posted by: Urs Schreiber on December 22, 2009 2:33 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Okay, Urs — thanks for the detailed explanation. I think I already understood these facts from your previous remarks, though in a less formal way. But before, I’d been thinking you were asserting a greater similarity between the Souriau theory and the Sullivan theory than this — something more surprising, more mysterious… and more likely to be wrong.

You see, to me the most interesting thing is not the coend formula but the fact that the Sullivan theory of differential forms gives an interesting homotopy invariant of topological spaces — while the Souriau theory of differential forms, applied to topological spaces regarded as diffeological spaces with continuous maps as plots, does not. But your remarks seemed to hint that it does.

Why did I think that? I guess I was trying to read more into your remarks than was visible. I spend much of my time trying to read between the lines of what mathematicians write: often that’s where the most important points are concealed, but sometimes all I see are mirages.

Posted by: John Baez on December 22, 2009 5:48 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

John wrote:

For example: as a module over the continuous real-valued functions on the real line, what sort of module do we get? A free module? If so, of what rank?

My intuition is that the product rule

$d\left(gh\right)=gdh+hdg$

leads to very weird conclusions when applied to continuous but nondifferentiable functions, like the square root of the absolute value of $x$.

I don’t understand what goes wrong with non differentiable functions, but there is another issue that makes the module of Kaehler differentials to be too big: in the relations we are only quotienting by finite sum relations: $d\left({g}_{1}+{g}_{2}\right)=d{g}_{1}+d{g}_{2}$. This has as consequence that if, for instance, $f={a}_{0}+{a}_{1}t+{a}_{2}{t}^{2}+\dots$ is an infinite sum then we will not get $df=f\prime dt$, but $df$ will be linearly independent from $dt$, and I don’t think you will want that. Serre explains in Algebraic Groups and Class Field Theory how to fix this so that you really get $df=f\prime dt$, see Google books, page 19.

Posted by: Maarten Bergvelt on December 22, 2009 9:32 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

If I could stop the world around me, I would stare at this until I understood it. Very neat.

Posted by: Eric on December 22, 2009 12:25 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

If I understand the basic high-level idea of what you are saying here, it means that an enterprising engineer, numerical analyst, or computational physicist could use this as a recipe for developing numerical algorithms where the test domains could be some finite representation of a physical system.

In a way, the test domain determines the physics which has a very “general relativistic” feel to it. But maybe I am totally confused…

Posted by: Eric on December 22, 2009 12:41 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Back in ordinary algebraic topology, one of the important facts was that singular cohomology could be computed from a triangulation. Is there anything similar here? If I look at polynomial differential forms on a triangulation $T$ of a manifold $X$, how is that related to polynomial differential forms on $\mathrm{Sing}\left(X\right)$?

I’ll point out that the degree zero part, polynomial functions on a simplicial complex, is a Stanley-Reisner ring. These are important in commutative algebra, but I hadn’t heard of them showing up in topology yet.

Posted by: David Speyer on December 21, 2009 1:29 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

David wrote:

If I look at polynomial differential forms on a triangulation $T$ of a manifold $X$, how is that related to polynomial differential forms on $\mathrm{Sing}\left(X\right)$?

Unless I’m seriously confused, they will be ‘the same’ in the way that counts in this game.

In other words: take the category of DGCAs (differential graded commutative algebras) and DGCA homomorphisms. Throw in formal inverses to DGCA homomorphisms that induce isomorphisms on cohomology. This gives the ‘homotopy category’ of DGCAs. And the two DGCAs you mention should be isomorphic in this homotopy category.

I believe this because I believe the ‘rational polynomial differential forms’ functor from simplicial sets to DGCAs gives rise to a functor between their homotopy categories, and the two simplicial sets you allude to are weakly equivalent.

I have vague memories that the Stanley–Reisner ring shows up in work on the Halperin–Carlsson conjecture, which says: Suppose the torus ${T}^{n}$ acts freely on a finite CW complex $X$. Then the sum of the Betti numbers of $X$ is at least ${2}^{n}$.

Posted by: John Baez on December 21, 2009 7:39 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

If I look at polynomial differential forms on a triangulation $T$ of a manifold $X$, how is that related to polynomial differential forms on $\mathrm{Sing}\left(X\right)$?

John Baez suggested that

[…] they will be ‘the same’ in the way that counts in this game.

[…]

I believe this because I believe the ‘rational polynomial differential forms’ functor from simplicial sets to DGCAs gives rise to a functor between their homotopy categories, and the two simplicial sets you allude to are weakly equivalent.

Yes, and more is true: the functor

${\Omega }_{\mathrm{Sullivan}}^{•}:\mathrm{SSet}\to {\mathrm{dgAlg}}^{\mathrm{op}}$

is the left adjoint of a Quillen adjunction (with respect to the standard model structure on simplicial sets on the left and the standard model structure on dg-algebras on the right). As such, it preserves weak equivalences between cofibrant objects. In $\mathrm{SSet}$ all objects are cofibrant and the inclusion $T↪\mathrm{Sing}\left(X\right)$ is a weak equivalence.

Accordingly, ${\Omega }_{\mathrm{Sullivan}}^{•}\left(T\right)$ is weakly equivalent to ${\Omega }_{\mathrm{Sullivan}}^{•}\left(\mathrm{Sing}\left(X\right)\right)$, in that the canonical morphism between them is an isomorphism on cohomology.

A bit of discussion along these lines is now also at nLab:ratonal homotopy theory – Forms on topological spaces.

Posted by: Urs Schreiber on December 22, 2009 2:47 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I’m not quite sure, if I understood the part about Kähler-differentials and differential forms right, but did you really claim, that on a smooth manifold the Kähler-differentials (or the differential forms) of the algebra of smooth functions are exactly the smooth 1-forms (resp. higher forms) on this smooth manifolds?

I can’t see why this should be true at the moment. For example in the case ${ℝ}^{n}$, the differential-forms are the free ${C}^{\infty }\left({ℝ}^{n}\right)$-module on the generators $d{x}_{i}$. But the algebra of ${C}^{\infty }\left({ℝ}^{n}\right)$ is certainly not freely generated on the coordinate functions (certainly as a smooth-algebra it is free, but not as an ordinary algebra). In the function algebra on ${ℝ}^{n}$ is freely generated on the projections (namely the polynomial algebra) and therefore we see that the Kähler-differentials are freely generated on the $d{x}_{i}$.

I think in general we have a map from the Kähler-differentials module to the module of 1-forms on this manifold. It’s clear that this map is surjective, but I can’t see at the moment why it should be injective.

Posted by: Thomas Nikolaus on December 22, 2009 1:18 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I may have been wrong in claiming that Kähler differentials for ${C}^{\infty }\left({ℝ}^{n}\right)$ are the same as smooth 1-forms on ${ℝ}^{n}$. You’re right, there’s an obvious surjective map. But why is it injective?

It should suffice to prove this equation between Kähler differentials ${C}^{\infty }\left({ℝ}^{n}\right)$:

$df=\sum _{i=1}^{n}\frac{\partial f}{\partial {x}^{i}}d{x}^{i}$

Of course this is true in the world of smooth 1-forms. But how can we show it for Kähler differentials? Is it really true?

I do know some nonobvious results in this general direction, which led me to make my claim. But I seem to be missing a key step.

For example, the derivations of ${C}^{\infty }\left({ℝ}^{n}\right)$ form a free module of rank $n$ over this algebra, with basis $\frac{\partial }{\partial {x}^{i}}$ This takes a bit of work to show.

If we could show the module of Kähler differentials is dual to the module of derivations, we’d be done.

Posted by: John Baez on December 22, 2009 7:07 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

John:

If we could show the module of Kähler differentials is dual to the module of derivations, we’d be done.

But that will not be true: let $n=1$ and $f={e}^{t}$ then $df\ne fdt$ in the module of Kaehler differentials, right?

Then the basic derivation $d/dt$ acts in the same way on $df$ and on $fdt$, so that the dual of the module of derivations must be some quotient of the Kaehler differentials where $df$ and $fdt$ are identified.

Posted by: Maarten Bergvelt on December 22, 2009 10:14 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

If we could show the module of Kähler differentials is dual to the module of derivations, we’d be done.

Comments from the peanut gallery. You have been warned:

Aside from the point Maarten mentioned, I think a failure of this to be true might be related to the failure of 0-forms and 1-forms to commute in general. When 0-forms and 1-forms commute, I think you have a shot at saying the differentials are dual to derivations.

This is also related to continuum vs finitary stuff. You need a continuum to have derivations in the first place, but you can have Kähler differentials more generally. When 0-forms and 1-forms commute in a nontrivial way, e.g. $f\left(\mathrm{dg}\right)=\left(\mathrm{dg}\right)f\ne 0$, then it implies you are dealing with a continuum underlying space.

This probably boils down to some set theoretic/topos-related issue, e.g. excluded middle or something, that Toby would be an expert on :)

Posted by: Eric on December 23, 2009 12:15 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Eric wrote:

I think a failure of this to be true might be related to the failure of 0-forms and 1-forms to commute in general.

I think most people here will be mystified by your remarks about noncommutativity, since we’re talking about Kähler differentials in algebraic geometry, ordinary differential forms, and Sullivan’s rational differential forms, and in all these contexts the 0-forms and 1-forms commute.

But I just wanted to say: your day will come! In the next Week’s Finds I’ll talk about the commuting cochain problem.

Is there a nice online source of references regarding this problem, e.g. somewhere in the $n$Lab? So far the closest thing I can find is this. Or maybe your paper with Urs has a thorough bibliography — I haven’t checked yet.

Posted by: John Baez on December 26, 2009 2:07 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Thanks John.

Urs talks about this a little bit starting here where he pointed to his page on $\infty$-Lie algrebroids here.

The comment I left here leading to Urs’ response probably left people similarly mystified.

But after reading the nlab page on differential forms on simplices, which points to Sullivan’s paper:

my beauty and “everything is related” radar went off again. The expression from our paper

${d}_{H}H=\left[H,H\right]$

seems to be analogous to Sullivan’s

${d}_{\theta }\theta =\theta \wedge \theta .$

Then, the discussion you just linked to has a comparison of two approaches: one Sullivan’s and one that I can’t claim to be ours because we’re probably not the first to look at it, but it is the viewpoint I support. It all boils down to how you define your product. Do you want graded commutativity at the expense of associativity, or do you want associativity at the expense of graded commutativity. Only with the continuum can you have both. So which is the bigger sin to give up?

My proposal, for what it is worth, begins here. An explicit comparison of what we did versus Sullivan is here.

Anyway, I think the “commutivity problem” is nicely sumamrized in Urs’ table here.

Posted by: Eric Forgy on December 26, 2009 3:39 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Today, Masoud Khalkhali announced the publication of Basic Noncommutative Geometry. I mention it here because I think it is relevant and also because he mentions Kahler differentials on the top of page 4 in the introduction. There is also a cool table relating commutative concepts and the corresponding noncommutative concept on page 5.

I also added some references and a comment to the nLab entry on universal differential envelope, which might be relevant.

Posted by: Eric Forgy on December 27, 2009 3:00 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

There is another issue that makes the module of Kaehler differentials to be too big: in the relations we are only quotienting by finite sum relations: $d\left({g}_{1}+{g}_{2}\right)=d{g}_{1}+d{g}_{2}$. This has as consequence that if, for instance, $f={a}_{0}+{a}_{1}t+{a}_{2}{t}^{2}+\dots$ is an infinite sum then we will not get $df=f\prime dt$, but $df$ will be linearly independent from $dt$, and I don’t think you will want that.

I certainly don’t want that!

I was certainly aware of this danger, but you’re not quite giving a proof that $df$ is linearly independent from $dt$, and I thought maybe I could get around it.

After all, consider a derivation $v$ of the algebra of smooth real-valued functions on the real line. Suppose we know $v\left(t\right)$ — in other words, $v$ applied to the linear function $f\left(t\right)=t$. Then since $v$ is a derivation, we know what $v$ does to any polynomial function of $t$. But suppose that

$f\left(t\right)={a}_{0}+{a}_{1}t+{a}_{2}{t}^{2}+\dots$

is an infinite sum. Then we can never figure out $v\left(f\right)$, since the definition of derivation says nothing about infinite sums. Right?

Wrong! Suppose $f$ is any smooth real-valued function on the real line. Then we can figure out $v\left(f\right)$ as follows.

First of all, we’ll know $v\left(f\right)$ as soon as we know its value at each point ${t}_{0}$.

Secondly, we have

$f\left(t\right)={a}_{0}+{a}_{1}\left(t-{t}_{0}\right)+g\left(t\right)\left(t-{t}_{0}{\right)}^{2}$

for some smooth function $g$. This takes a little work to show. But it’s true, and it implies that

$v\left(f\right)={a}_{1}v\left(t\right)+v\left(g\right)\left(t\right)\left(t-{t}_{0}{\right)}^{2}+2g\left(t\right)v\left(t\right)\left(t-{t}_{0}\right)$

So, at the point ${t}_{0}$, the terms involving the function $g$ go away, and we have

$v\left(f\right)\left({t}_{0}\right)={a}_{1}v\left(t\right)\left({t}_{0}\right)$

So we know $v\left(f\right)$!

This kind of argument also works in ${C}^{\infty }\left({ℝ}^{n}\right)$, but it’s significantly trickier (try it). This leads to the proof that the vector fields

$\frac{\partial }{\partial {x}^{i}}$

form a basis for the module of derivations of ${C}^{\infty }\left({ℝ}^{n}\right)$.

So, I had my hopes. But if…

Serre explains in Algebraic Groups and Class Field Theory how to fix this so that you really get $df=f\prime dt$, see Google books, page 19.

…then I guess there’s really a problem!

I’d like to see a real proof that we can have $df$ linearly independent from $f\prime dt$ in the module of Kähler differentials. Your argument sounds convincing. But I could also have convinced myself that a derivation of the algebra of smooth functions on the real line is not determined by its action on linear functions, using the argument I presented at the beginning of this comment! Unless I had known otherwise, that is.

Posted by: John Baez on December 23, 2009 6:23 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

By the way, what I pedantically called

$v\left(t\right)\left({t}_{0}\right)$

above would more normally be called just

$v\left({t}_{0}\right)$

that is, the value of the vector field at the point $t$. I was using $v\left(t\right)$ to mean ‘the function obtained from applying the derivation $v$ to the function $f$ given by $f\left(t\right)=t$’. So, I was using $v\left(t\right)\left({t}_{0}\right)$ to mean the result of evaluating $v\left(t\right)$ at ${t}_{0}$.

Part of the problem was that in this argument we’re supposed to distinguish between vector fields and derivations, even though — or precisely because — we’re about to prove they’re the same. So, I felt torn here between the algebraist’s way of thinking of the function $t$ as an algebra element, the mathematician’s way of thinking of $f$ as function that evaluates to the number $f\left(t\right)$ at the point $t$, and the physicist’s way of using $f\left(t\right)$ to mean a function, not the function’s value.

This stuff can get really annoying!

Posted by: John Baez on December 23, 2009 6:46 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I have thought about this for a while today and the problem with the dual spaces confuses me.

It is certainly true that smooth vector fields on a smooth manifold $M$ are exactly derivations on the ring ${C}^{\infty }\left(M\right)$, lets denote this module by $\mathrm{Der}{C}^{\infty }\left(M\right)$. Furthermore the universal property of the module ${\Omega }_{K}\left(M\right)$ of Kähler differentials implies that derivations of ${C}^{\infty }\left(M\right)$ are exactly module morphisms

${\Omega }_{K}\left(M\right)\to {C}^{\infty }\left(M\right)$

So we have

$\mathrm{Der}{C}^{\infty }\left(M\right)=\mathrm{Hom}\left({\Omega }_{K}\left(M\right),{C}^{\infty }\left(M\right)\right)=:{\Omega }_{K}\left(M{\right)}^{*}$

Now it is well known that 1-forms on manifold are the same as ${C}^{\infty }\left(M\right)\right)$-linear maps from the module of vector fields to ${C}^{\infty }\left(M\right)$. We denote the module of smooth 1-forms by ${\Omega }^{1}\left(M\right)$ and thus have:

${\Omega }^{1}\left(M\right)=\mathrm{Hom}\left(\mathrm{Der}{C}^{\infty }\left(M\right),{C}^{\infty }\left(M\right)\right)=\mathrm{Der}{C}^{\infty }\left(M{\right)}^{*}={\Omega }_{K}\left(M{\right)}^{**}$

That means that the 1-forms are the bidual of the module of Kähler differentials, and it is (more or less) clear that the canonical map ${\Omega }_{K}\left(M\right)\to {\Omega }^{1}\left(M\right)$ is then just the canonical map from the module ${\Omega }_{K}\left(M\right)$ to its bidual.

As mentioned before, this map is clearly surjective, so it’s a quotient map. Now I don’t think that ${C}^{\infty }\left(M\right)$ is a principal ideal domain so the structure of modules over it could be complicated. But it’s at least clear that for any ${C}^{\infty }\left(M\right)$ module $M$ the canonical map $M\to {M}^{**}$ kills all torsion elements in $M$. (If ${C}^{\infty }\left(M\right)$ is a PID it would exactly divide out the torsion part, because this has a free complement).

Does anyone know enough module-theory to see what happens here (or maybe see an error in my considerations)?

Posted by: Thomas Nikolaus on December 24, 2009 11:05 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Thanks very much for your thought, Thomas! Happy holidays!

Your argument looks correct to me. You seem to have shown that 1-forms are the bidual of the Kähler differentials:

${\Omega }^{1}\left(M\right)=\mathrm{Hom}\left(\mathrm{Der}{C}^{\infty }\left(M\right),{C}^{\infty }\left(M\right)\right)=\mathrm{Der}{C}^{\infty }\left(M{\right)}^{*}={\Omega }_{K}\left(M{\right)}^{**}$

and also that the canonical map from Kähler differentials to 1-forms

$\alpha :{\Omega }_{K}\left(M\right)\to {\Omega }^{1}\left(M\right)$

is surjective. And also, that this canonical map is the usual canonical map from a module to its double dual:

${\Omega }_{K}\left(M\right)\to {\Omega }_{K}\left(M{\right)}^{**}$

So in short, if we haven’t made a mistake somewhere, the canonical map

$\alpha :{\Omega }_{K}\left(M\right)\to {\Omega }_{K}\left(M{\right)}^{**}$

is surjective, and we’re just wondering about its kernel. Is it zero, in which case Kähler differentials are the same as 1-forms? Is it nonzero? Have you or I made a mistake somewhere?

I must admit that I never dreamt of the possibility that $\alpha$ has a kernel here! If it does, that would explain the failure of my intuition. I’m much more used to functional analysis, where the map from a Banach space to its bidual is an injection, thanks to the Hahn–Banach theorem.

Note we also have another fact at our disposal, namely that ${\Omega }^{1}\left(M\right)$ is a projective module over ${C}^{\infty }\left(M\right)$, and finitely generated whenever $M$ can be covered by finitely many open sets with trivial tangent bundle. (This happens for most manifolds of interest.)

Now I don’t think that ${C}^{\infty }\left(M\right)$ is a principal ideal domain so the structure of modules over it could be complicated.

No, it’s not a principal ideal domain.

The ring ${C}^{\infty }\left(M\right)$ has lots of tricky ideals. For example, we can pick a point $p\in M$ and consider the ideal of all functions that vanish to $n$th order at $p$. These ideals aren’t principal when the dimension of $M$ is more than 1. This is nothing surprising: we see this in algebraic geometry.

But we also have a nonzero ideal smaller than all of these, consisting of functions with all derivatives vanishing at $p$. If $M=ℝ$ and $p=0$ this ideal would include the function that equals $\mathrm{exp}\left(-1/x\right)$ for $x\ge 0$ and $0$ for $x\le 0$.

And there’s an even smaller nonzero ideal consisting of functions that vanish in some neighborhood of $p$.

And there are uncountably many even smaller nonzero ideals: pick a Riemannian metric and take the ideal of all functions that vanish inside a ball of radius $r$ around $p$!

And there are ideals vastly more complicated than all of these…

Posted by: John Baez on December 24, 2009 8:36 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Hi John, thanks for your nice explanations about the ideals in ${C}^{\infty }\left(M\right)$ and merry christmas!

“I must admit that I never dreamt of the possibility that α has a kernel here! If it does, that would explain the failure of my intuition. I’m much more used to functional analysis, where the map from a Banach space to its bidual is an injection, thanks to the Hahn–Banach theorem.”

I think there are different effects around here. Maybe I’m wrong, but let me explain what I mean:

First of all for arbitrary vector spaces $V$ the canonical map $\alpha :V\to {V}^{**}$ is always an injection, but this fails for modules (which might be the case here). And I really don’t know how seriously this could fail for a ${C}^{\infty }\left(M\right)$-Module $X$. I’d guess that the kernel of the map $\alpha :X\to {X}^{**}$ are exactly the torsion elements in $X$ but I can’t prove this.

Secondly, in Banach spaces one takes not the whole dual spaces, but only the space of continous linear forms and then the Hahn-Banach theorems ensure, that there are enough of them. But in our case there is no topology around on the modules (maybe that is part of the problem).

Posted by: Thomas Nikolaus on December 25, 2009 9:20 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

John wrote:

This kind of argument also works in ${C}^{\infty }\left({ℝ}^{n}\right)$, but it’s significantly trickier (try it). This leads to the proof that the vector fields

$\frac{\partial }{\partial {x}_{i}}$

form a basis for the module of derivations of ${C}^{\infty }\left({ℝ}^{n}\right)$.

This is not true, I think. The module of Kaehler differentials ${\Omega }_{1}\left(A\right)$ of an algebra $A$, together with the differential $d:A\to {\Omega }_{1}\left(A\right)$ is universal, so for any derivation $X$ on $A$ with value in a module $M$ there is a unique morphism of $A$-modules ${\mu }_{X}:{\Omega }_{1}\left(A\right)\to M$ such that the obvious diagram commutes (I tried to draw it but that didn’t work, and golem is so slow I have to wait everytime 10 minutes to see what it thinks I typed)

Now if we have linear independent elements (over $A$) in ${\Omega }_{1}\left(A\right)$ then we can specify a homomorphism $\mu$ by choosing the value of $\mu$ arbitrarily at the independent elements. So, for instance, if we take $A=C\left[\left[t\right]\right]$, formal power series in $t$, then in ${\Omega }_{1}\left(A\right)$ the elements $\mathrm{dt}$ and ${\mathrm{de}}^{t}$ are indepedent, as Dave Speyer showed over in Math Overflow (or follows from Theorem 87 in Matsumura). Then there is a homomorphism $\mu :{\Omega }_{1}\left(A\right)\to M$ such that $\mu \left(\mathrm{dt}\right)=1,\mu \left({\mathrm{de}}^{t}\right)=2{e}^{t}$. But this means by the diagram that there is a derivation $X:C\left[\left[t\right]\right]\to C\left[\left[t\right]\right]$ such that $X\left(t\right)=1$ and $X\left({e}^{t}\right)=2{e}^{t}$. This shows that the module of derivations of $A$ is certainly not generated by $\frac{\partial }{{\partial }_{t}}$, it will be much bigger. Note that the funny derivation $X$ is not continuous, if we take the powers of the maximal ideals as neighborhoods of 0. So you need, I guess, to take the continous derivations to get a nice and simple module just generated by $\frac{\partial }{{\partial }_{t}}$.

Posted by: Maarten Bergvelt on December 28, 2009 3:32 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

John wrote:

… the vector fields

$\frac{\partial }{\partial {x}^{i}}$

form a basis for the module of derivations of ${C}^{\infty }\left({ℝ}^{n}\right)$.

Maarten wrote:

This is not true, I think.

I really hope it is true! If the module of derivations of ${C}^{\infty }\left(ℝ\right)$ is not generated by $d/dt$, quite a few books will need to be rewritten. For example, this one, and also one I wrote.

Luckily, your argument here does not convince me, because you are talking about derivations of the algebra of formal power series $ℂ\left[\left[t\right]\right]$, not the algebra of smooth functions ${C}^{\infty }\left(ℝ\right)$.

There’s a homomorphism

$p:{C}^{\infty }\left(ℝ\right)\to ℂ\left[\left[t\right]\right]$

given by taking the Taylor series of a smooth function. This homomorphism is onto — a fact not as well known as it should be.

(For example, find a smooth function whose Taylor series is ${\sum }_{n\ge 0}n!{x}^{n}$. This series has zero radius of convergence, but that’s irrelevant — it just means you have to look a bit harder for a function that does the job.)

However, the homomorphism $p$ is not one-to-one, since there are plenty of nonzero smooth functions all of whose derivatives vanish at the origin.

(I mentioned one a while back: the function that’s $0$ for negative $t$ and $\mathrm{exp}\left(-1/t\right)$ for $t\ge 0$. And this is also the key to the puzzle in my previous parenthetical remark!)

So, there’s no instantly obvious way to turn a derivation of $ℂ\left[\left[t\right]\right]$ into a derivation of ${C}^{\infty }\left(ℝ\right)$.

I’m inclined to believe David Speyer’s argument concerning Kähler differentials for the algebra of smooth functions. And, I’m inclined to believe your argument that the module of derivations of $ℂ\left[\left[t\right]\right]$ is not generated by $d/dt$. But I still believe that the module of derivations of ${C}^{\infty }\left(ℝ\right)$ is generated by $d/dt$. Especially because I’ve seen this asserted in many places, and I’ve given a proof here, that nobody has found fault with.

I’m glad we’re getting this stuff straightened out… it’s important.

Posted by: John Baez on December 28, 2009 9:18 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

John:

I’m inclined to believe David Speyer’s argument concerning Kähler differentials for the algebra of smooth functions. And, I’m inclined to believe your argument that the module of derivations of ℂ[[t]] is not generated by d/dt. But I still believe that the module of derivations of C ∞(ℝ) is generated by d/dt.

I just used power series to be explicit, but the argument does not depend on that choice. So let $A={C}^{\infty }\left(R\right)$, and $\Omega ={\Omega }_{1}\left(A\right)$ its module of Kaehler differentials. $\Omega$ still is universal for derivations, and contains $A$-independent elements $dt$ and $d{e}^{t}$. A derivation $X:A\to A$ comes from a unique $A$-module morphism $\mu :\Omega \to A$, via $X\left(a\right)=\mu \left(da\right)$. We can choose $\mu \left(dt\right)=1,\mu \left(d{e}^{t}\right)=$ anything, so that the corresponding derivation satisfies $X\left(t\right)=1$ and $X\left({e}^{t}\right)$ can be anything. Such an $X$ is not of the form $f\left(t\right)d/dt$, for $f\in A$, unless we choose “anything” to be ${e}^{t}$.

I think that people make silently some assumptions about derivations on smooth functions to derive the properties everybody expects and loves. I’ll check your argument in an other part of this thread.

Posted by: Maarten Bergvelt on December 28, 2009 9:50 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Please do check my argument. I’m not assuming any sort of continuity of the derivation. What I’m using is that we know a smooth function $f$ if we know the values $\mu \left(f\right)$ for all possible homomorphisms $\mu :{C}^{\infty }\left(ℝ\right)\to ℝ$ (that is, the values of $f$ at points of the real line). This is not true for formal power series. I think that’s what makes the difference.

By the way, when using TeX on this blog,

$d e^t = e^t d t$

produces the beautiful formula

$d{e}^{t}={e}^{t}dt$

while

$de^t = e^t dt$

produces the ugly

${\mathrm{de}}^{t}={e}^{t}\mathrm{dt}$

It’s one of the ‘features’ added by Jacques Distler: a bunch of letters not separated by spaces reverts to roman font even inside a math expression, so we can easily write something like

$\mathrm{Diff}\left(M\right)$

and have it come out looking cool. But I’ve spent a lot of time behind the scenes fixing people’s differentials on this thread, to make them look nice!

Posted by: John Baez on December 28, 2009 10:41 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Johm:

I am thinking about it. But please check mine, too, it doesn’t seem to depend on any special properties of ${C}^{\infty }$ versus $C\left[\left[t\right]\right]$

I’m not assuming any sort of continuity of the derivation. What I’m using is that we know a smooth function f if we know the values μ(f) for all possible homomorphisms μ: ∞(ℝ)→ℝ (that is, the values of f at points of the real line). This is not true for formal power series. I think that’s what makes the difference.

But how?

Thanks for the TeX tips, I find the n-category cafe very time consuming and frustrating, if you want to look at a preview you have to wait forever to see it, and then there are these random deviations from standard LaTeX. To get $\mathrm{Diff}\left(M\right)$ you should be able to just type \DeclareMathOperator, as you would do in a paper. The n-lab is much easier for me.

Posted by: Maarten Bergvelt on December 29, 2009 1:16 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

To get $\mathrm{Diff}\left(M\right)$ you should be able to just type \DeclareMathOperator, as you would do in a paper. The n-lab is much easier for me.

But the nLab works the same way as the Café does in that respect. They both use itex2mml, so that typing Diff(M) in math mode gives you $\mathrm{Diff}\left(M\right)$ rather than $Diff\left(M\right)$.

Posted by: Mike Shulman on December 29, 2009 3:39 AM | Permalink | PGP Sig | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

John wrote:

I’m not assuming any sort of continuity of the derivation. What I’m using is that we know a smooth function $f$ if we know the values $\mu \left(f\right)$ for all possible homomorphisms $\mu :{C}^{\infty }\left(ℝ\right)\to ℝ$ (that is, the values of $f$ at points of the real line). This is not true for formal power series. I think that’s what makes the difference.

Maarten wrote:

But how?

Well, if you look at my argument you’ll see a couple of reasons why it might not work for formal power series.

First, I’m using the fact that a derivation $v$ is determined once we know the values of $v\left(f\right)\left({t}_{0}\right)$ for all functions $f$ at all points ${t}_{0}$. This isn’t true for formal power series: they’re a local ring so they only have one ‘point’. This is what I was referring to above.

Second, given a smooth function $f$ and a real number ${t}_{0}$, I can always write $f$ as

$f\left(t\right)={a}_{0}+{a}_{1}\left(t-{t}_{0}\right)+g\left(t\right)\left(t-{t}_{0}{\right)}^{2}$

for some smooth function $g$. I don’t know if this works for formal power series. Hmm, maybe it does.

As for your argument, you say:

So, for instance, if we take $A=C\left[\left[t\right]\right]$, formal power series in $t$, then in ${\Omega }_{1}\left(A\right)$ the elements $dt$ and $d{e}^{t}$ are indepedent, as Dave Speyer showed over in Math Overflow (or follows from Theorem 87 in Matsumura). Then there is a homomorphism $\mu :{\Omega }_{1}\left(A\right)\to M$ such that $\mu \left(dt\right)=1,\mu \left(d{e}^{t}\right)=2{e}^{t}$. But this means by the diagram that there is a derivation $X:C\left[\left[t\right]\right]\to C\left[\left[t\right]\right]$ such that $X\left(t\right)=1$ and $X\left({e}^{t}\right)=2{e}^{t}$.

Since you couldn’t draw the diagram I’m not sure, but it seems like you’re skipping a step when you go from the homomorphism $\mu :{\Omega }_{1}\left(A\right)\to M$ to the derivation $X:A\to A$. It seems to me you’d get a derivation $X:A\to M$… and that you’d need some extra work to get a derivation $X:A\to A$.

It’s possible that this ‘extra work’ can be done in the case $A=C\left[\left[t\right]\right]$ but not in the case $A={C}^{\infty }\left(M\right)$.

Thanks for the TeX tips, I find the n-category cafe very time consuming and frustrating, if you want to look at a preview you have to wait forever to see it…

That’s a newish problem — things were working quite nicely until a few weeks ago. I’ve reported it to Jacques Distler, who is the brains behind this operation. So, hopefully that will be fixed.

… and then there are these random deviations from standard LaTeX.

Well, the problem here is that Jacques invented this system mainly for his own convenience, long before other math blogging systems came into being. Now that we have lots of mathematicians attempting to write TeX on the $n$-Café the advantages of idiosyncratic improvements probably pale in comparison to the drawbacks of having to teach everyone how they work!

Perhaps someday I or we may move to some other system — but as usual, it never seems worthwhile to do it today.

If you want to show me some diagrams over on the $n$Lab, that’s fine. The page on Kähler differentials would be a good place.

Posted by: John Baez on December 29, 2009 2:43 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

John:

Since you couldn’t draw the diagram I’m not sure, but it seems like you’re skipping a step when you go from the homomorphism μ:Ω 1(A)→M to the derivation X:A→A. It seems to me you’d get a derivation X:A→M… and that you’d need some extra work to get a derivation X:A→A.

You are just free to take $M=A$, that is what the universality of the module of differentials means. I have typed up the diagram over on the n-nlab on the page on Kaehler differentials. There is no extra work involved. In formulas: if $X:A\to A$ is a derivation, there is a unique $A$-module morphism $\mu :{\Omega }_{1}\left(A\right)\to A$ such that $X\left(a\right)=\mu \left(\mathrm{da}\right)$. So the derivation $X$ and the module morphism $\mu$ determine each other uniquely. Then if you have $A$-linearly independent elements you are free to specify $\mu$ at those independent elements any way you want. This gives the funny derivations you don’t want.

I am pretty amazed that the universal property of differentials is not the way you seem to think of them. ${\Omega }_{1}\left(A\right)$ is something like the coclassifying space for derivations out of $A$: you get any one of those derivations from a map from ${\Omega }_{1}\left(A\right)$. Looking over at the n-lab I see that the universal property of differentials could be mentioned there at various places. I’ll do that tomorrow.

Posted by: Maarten Bergvelt on December 29, 2009 4:11 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I don’t follow everything that is going on here, but I am also skeptical about the step where you turn a derivation $A\to M$ into a derivation $A\to A$. The $M$ I use in my construction is a pretty weird object: Frac of an ultraproduct. In particular, I don’t think there are any $A$-module homs from $M$ to $A$.

Posted by: David Speyer on December 29, 2009 1:53 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I am sorry I brought up $M$, it suffices to consider just derivations $X:A\to A$.

For each such $X$ there is a unique $\mu :{\Omega }_{1}\left(A\right)\to A$ so that $X\left(a\right)=\mu \left(da\right)$. This is the universality of Kaehler differentials.

In particular, if you can show for $A={C}^{\infty }\left(R\right)$ that any derivation is of the form $X=f\left(t\right)d/dt$, then any $A$-module morphism $\mu$ must have $\mu \left(dt\right)=f\left(t\right)$ and $\mu \left(d{e}^{t}\right)=f\left(t\right){e}^{t}$. This seems to contradict the independence of $dt$ and $d{e}^{t}$: you should for a homomorphism $\mu$ be able to choose the values at independent elements arbitrarily.

Posted by: Maarten Bergvelt on December 29, 2009 2:37 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Maarten wrote:

I am pretty amazed that the universal property of differentials is not the way you seem to think of them. ${\Omega }_{1}\left(A\right)$ is something like the coclassifying space for derivations out of $A$: you get any one of those derivations from a map from ${\Omega }_{1}\left(A\right)$.

Amazed? You must have more confidence in me than I do. I’m rarely amazed when I screw up. But let me describe what I think the universal property of ${\Omega }_{1}\left(A\right)$ is, and why I think your argument is flawed. You can correct me.

Let $A$ be a commutative algebra and let $M$ be an $A$-module, which we treat as a bimodule in the usual way. A derivation from $A$ to $M$ is a linear map

$v:A\to M$

such that

$v\left(ab\right)=v\left(a\right)b+av\left(b\right)$

${\Omega }_{1}\left(A\right)$ is an $A$-module with a derivation

$d:A\to {\Omega }_{1}\left(A\right)$

that’s universal in the following sense: if $M$ is any $A$-module and

$v:A\to M$

is any derivation, then

$v=\mu \circ d$

for some unique $A$-module homomorphism

$\mu :{\Omega }_{1}\left(A\right)\to M$

Now let me turn to your argument. Take $A={C}^{\infty }\left(R\right)$. Here ${\Omega }_{1}\left(A\right)$ contains two elements that aren’t equal:

$d{e}^{t}\ne {e}^{t}dt$

So, there exists some module $M$ and some homomorphism $\mu :{\Omega }_{1}\left(A\right)\to M$ such that

$\mu \left(d{e}^{t}\right)\ne \mu \left({e}^{t}dt\right)$

Take $\mu$ to be the identity if you like! So, we obtain a derivation

$v=\mu \circ d$

with the following curious property:

$v\left({e}^{t}\right)\ne {e}^{t}v\left(t\right)$

But this is a derivation $v:A\to M$ for some module $M$. You claim there is a derivation $v:A\to A$ with this curious property. For this to exist, there must be a homomorphism

$\mu :{\Omega }_{1}\left(A\right)\to A$

such that

$\mu \left(d{e}^{t}\right)\ne \mu \left({e}^{t}dt\right)$

This could be harder to find! If ${\Omega }_{1}\left(A\right)$ is a projective $A$-module, and $\omega \ne \omega \prime$ are two distinct elements of ${\Omega }_{1}\left(A\right)$, it’s easy to find a homomorphism

$\mu :{\Omega }_{1}\left(A\right)\to A$

such that

$\mu \left(\omega \right)\ne \mu \left(\omega \prime \right)$

But if ${\Omega }_{1}\left(A\right)$ is not projective, I don’t see to find such a homomorphism, in general.

This was part of the point of my conversation with Thomas Nikolaus: the double dual ${\Omega }_{1}\left(A{\right)}^{**}$ is projective in this example, in fact free, and there’s a homomorphism ${\Omega }_{1}\left(A\right)\to {\Omega }_{1}\left(A{\right)}^{**}$, but this homomorphism has a kernel, and ${e}^{t}dt-d{e}^{t}$ lies in this kernel. I don’t understand the structure of this kernel.

You wrote:

… you should for a homomorphism $\mu$ be able to choose the values at independent elements arbitrarily.

That’s my question: we can do it for a homomorphism $\mu :{\Omega }^{1}\left(A\right)\to M$ for some module $M$, but can we do it for $M=A$?

Posted by: John Baez on December 29, 2009 5:29 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

John:

Amazed? You must have more confidence in me than I do. I’m rarely amazed when I screw up.

I was not suggesting you screwed up, but surprized that you think about differentials in an other way than I do.

Okay, let me try again. The universal property says that $\mathrm{Der}\left(A,A\right)$ is (isomorphic as $A$-module to) ${\mathrm{Hom}}_{A}\left({\Omega }_{1}\left(A\right),A\right)$. So for each homomorphism from ${\Omega }_{1}\left(A\right)$ to $A$ we get a derivation. Now Dave shows (I think, ultrafilters are not something I am tooo confident about) that $dt$ and $d{e}^{t}$ are linearly independent over $A$. So there is certainly a homomorphism ${\mu }_{F}$ from the free submodule $F$ of ${\Omega }_{1}\left(A\right)$ spanned by $dt$ and $d{e}^{t}$ to $A$ that takes arbitrary values at $dt$ and $d{e}^{t}$. Then the question is if we can extend ${\mu }_{F}$ to a homomorphism on all of ${\Omega }_{1}\left(A\right)$. Hm, maybe that is the point.

We have exact sequence

$0\to F\to {\Omega }_{1}\left(A\right)\to Q\to 0$ When is a homomorphism from the free module $F$ to $A$ extendable to a morphism from all of ${\Omega }_{1}\left(A\right)$ to A?

Posted by: Maarten Bergvelt on December 29, 2009 9:59 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Okay, I think I understand what is going on. Let $A$ be the ${C}^{\infty }$ functions on $R$, and let ${\Omega }_{K}\left(A\right)$ be the module of Kaehler differentials and ${\Omega }^{1}\left(A\right)$ the smooth one forms (generated by $dt$). Then it seems that any homomorphism from the Kaehler differentials to a free module $F$ factors through the smooth one forms:

$\mu :{\Omega }_{K}\left(A\right)\to {\Omega }^{1}\left(A\right)\to F$ In particular this forces $df$ and $f\prime dt$ always to have the same image under $\mu$, in contrast to what I claimed earlier.

Then the Kaehler differentials are universal for derivations from $A$ to any old module, while the 1-forms are universal for derivations to nice free $A$-modules (maybe even projective ones). Then it follows that indeed $\mathrm{Der}\left(A,A\right)$ is generated by $d/dt$.

What about formal series? For simplicity let $B=C\left(\left(t\right)\right)$, the field of Laurent series. Then ${\Omega }_{K}\left(B\right)$ is a vector space over a field, and for a morphism $\mu :{\Omega }_{K}\left(B\right)\to B$ we can choose the values $\mu \left(dt\right)=1$ and $\mu \left(d{e}^{t}\right)=0$, for instance. This implies that for formal Laurent series there is a crazy derivation $X:B\to B$ so that $X\left(t\right)=1$ but $X\left({e}^{t}\right)=0$. Such derivations are not continuous, and one could restrict to the continuous derivations. Dual to that are the 1-forms, generated by $dt$. The 1-forms should then be universal for continuous derivations from $B$ to some suitable class of modules I am sort of hazy about.

I hope this makes sense now.

Posted by: Maarten Bergvelt on December 30, 2009 12:02 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Maarten wrote:

I was not suggesting you screwed up, but surprised that you think about differentials in an other way than I do.

Oh, okay. On the internet, you can always count on other people interpreting your words in the most negative way. I’m afraid that’s what I did.

Then the question is if we can extend ${\mu }_{F}$ to a homomorphism on all of ${\Omega }_{1}\left(A\right)$. Hm, maybe that is the point.

Yes, that was my point.

Right now I’m willing to bet 100 bucks that you can’t — in the case $A={C}^{\infty }\left(R\right)$, that is. I’m agnostic about the case $A=C\left[\left[t\right]\right]$.

Posted by: John Baez on December 29, 2009 11:44 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Now let me turn to your argument. Take $A={C}^{\infty }\left(R\right)$. Here ${\Omega }_{1}\left(A\right)$ contains two elements that aren’t equal: $d{e}^{t}\ne {e}^{t}dt.$

I am probably confused, but if

$d\left(fg\right)=\left(df\right)g+f\left(dg\right)$

then

$d\left({t}^{n}\right)=\left(dt\right){t}^{n-1}+t\left(d{t}^{n-1}\right).$

Now, if

$\left(dt\right)t=t\left(dt\right),$

then you can show that

$d{t}^{n}=n{t}^{n-1}dt.$

Unless I’m confused (impossible!), the only way you can have

$d{e}^{t}\ne {e}^{t}dt$

would be if

$\left(dt\right)t\ne t\left(dt\right).$

Would it be helpful to focus on this simpler commutative relation?

I worked this out on page 7 (marked page 5) here.

Posted by: Eric on December 30, 2009 3:32 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Eric:

Unless I’m confused (impossible!), the only way you can have

$d{e}^{t}\ne {e}^{t}dt$

would be if

$\left(dt\right)t\ne t\left(dt\right).$

No, we want to have $\left(dt\right)t=t\left(dt\right)$. We are dealing with commutative rings and modules over them. If $M$ is a left module we give it a right module structure by $mr=rm$. No need for funny commutation relations.

That $d{e}^{t}\ne {e}^{t}dt$ was explained by David Speyer on Math Overflow.

Posted by: Maarten Bergvelt on December 30, 2009 3:50 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Eric: your final conclusion requires a continuity assumption that we do not have; I believe that this is the basic difference between Kähler differentials and ordinary 1-forms.

It should be no surprise that a result requiring continuity fails for Kähler differentials, since these are defined purely algebraically once you specify the algebra of functions to start with. So what is particularly nice is that the double dual does satisfy continuity.

Posted by: Toby Bartels on December 30, 2009 4:22 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Eric wrote:

Unless I’m confused (impossible!), the only way you can have

$d{e}^{t}\ne {e}^{t}dt$

would be if

$\left(dt\right)t\ne t\left(dt\right).$

We’ve figured it all out, here and on Math Overflow. Your comment ignores all the work we’ve done. For starters, we’re talking about Kähler differentials, where

$t\left(dt\right)=\left(dt\right)t$

As you note, this implies

$d{t}^{n}=\left(n-1\right)dt$

but you can’t get from here to

$d{e}^{t}={e}^{t}dt$

without doing some trick like this:

$d{e}^{t}=d\sum _{n\ge 0}\frac{{t}^{n}}{n!}\stackrel{?}{=}\sum _{n\ge 0}d\frac{{t}^{n}}{n!}=\sum _{n\ge 0}\frac{{t}^{n}}{n!}dt={e}^{t}dt$

But this involves infinite sums and passing $d$ through an infinite sum, which is not allowed here since there’s no topology in this game: Kähler differentials know nothing of infinite sums or convergence!

This is why Toby said “your final conclusion requires a continuity assumption that we do not have”.

Indeed, David Speyer showed that

$d{e}^{t}\ne {e}^{t}dt$

in the Kähler differentials for the ring of smooth functions on the real line. However, they’re equal as 1-forms.

Posted by: John Baez on December 30, 2009 4:54 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Your comment ignores all the work we’ve done.

There is a difference between “ignoring” and “not understanding”. I was guilty of the second and not the first. Thanks for explaining :)

Posted by: Eric on December 30, 2009 5:17 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I’m awfully late to this party, and haven’t tried to read this thread with much care. But I was interested by something John wrote:

There’s a homomorphism

$p:{C}^{\infty }\left(ℝ\right)\to ℂ\left[\left[t\right]\right]$

given by taking the Taylor series of a smooth function. This homomorphism is onto — a fact not as well known as it should be.

(For example, find a smooth function whose Taylor series is ${\sum }_{n\ge 0}n!{x}^{n}$. This series has zero radius of convergence, but that’s irrelevant — it just means you have to look a bit harder for a function that does the job.)

However, the homomorphism $p$ is not one-to-one, since there are plenty of nonzero smooth functions all of whose derivatives vanish at the origin.

(I mentioned one a while back: the function that’s 0 for negative $t$ and $\mathrm{exp}\left(-1/t\right)$ for $t\ge 0$. And this is also the key to the puzzle in my previous parenthetical remark!)

I like this little factoid about $p$ being surjective! I didn’t know it, and it does seem like a very handy thing to know.

I’m curious, John, about what you really meant when you said at the end that this is the key to the puzzle. I had to work a bit to find a general argument (which I’ll give below), and it does use “the key” in a way, but as hints go this seems a bit skimpy! Unless you have something much nicer in mind.

The basic idea I had is to use ${C}^{\infty }$ bump functions to death (and the function named in your second parenthetical comment is the key to constructing those). In outline, the argument is this:

1. The map $p$ factors as

${C}^{\infty }\left(ℝ\right)\stackrel{\pi }{\to }{C}^{\infty }\left(I\right)\stackrel{q}{\to }ℂ\left[\left[t\right]\right]$

where $\pi$, the map given by restriction to $I=\left[-1,1\right]$, is surjective by a bump function argument, so all we have to do is prove that the function $q$, which returns the Taylor coefficients at 0 (aka MacLaurin* coefficients), is surjective.

2. The usual topology on the topological vector space ${C}^{\infty }\left(I\right)$ is given by a complete metric. To find an $f\in {C}^{\infty }\left(I\right)$ whose MacLaurin coefficients are ${a}_{0},{a}_{1},{a}_{2},\dots$, it suffices to construct a Cauchy sequence of smooth functions ${f}_{n}$ whose first $n$ coefficients are ${a}_{0},{a}_{1},\dots ,{a}_{n-1}$. For then the limit of the ${f}_{n}$ is a smooth function $f$ with the desired coefficients.
3. In more detail, the topology on ${C}^{\infty }\left(I\right)$ is given by a countable number of seminorms

${\rho }_{n}\left(f\right)=\underset{-1\le x\le 1}{\mathrm{sup}}\mid {f}^{\left(n\right)}\left(x\right)\mid$

from which we can write down a metric

$d\left(f,g\right)=\sum _{k\ge 0}{2}^{-k}\frac{{\rho }_{k}\left(f-g\right)}{1+{\rho }_{k}\left(f-g\right)}$

which is complete. The Cauchy sequence is gotten by putting ${f}_{0}\left(x\right)\equiv {a}_{0}$ and

${f}_{n}={f}_{n-1}+\frac{{a}_{n}{x}^{n}}{n!}{e}_{n}\left(x\right)$

where the factor ${e}_{n}$ is a bump function chosen for its “calming behavior”, in order to make $d\left({f}_{n-1},{f}_{n}\right)<{2}^{1-n}$.

4. Specifically, given $n\ge 0$, choose $\delta >0$ so small that

$\mid x\mid <\delta ⇒\mid \frac{{a}_{n}{x}^{k}}{k!}\mid <{2}^{-n}$

for $1\le k\le n$. Then construct a nonnegative smooth bump function ${e}_{n}$ which takes values between 0 and 1, is supported on $\left[-\delta ,\delta \right]$, and is identically 1 on $\left[-\delta /2,\delta /2\right]$. I think that should do the trick. The point is that this forces

${\rho }_{k}\left({f}_{n-1}-{f}_{n}\right)<{2}^{-n}$

for $0\le k\le n-1$, and this is enough.

So in summary, the germ at 0 of ${f}_{n}$ agrees with the degree $n$ MacLaurin polynomial for the sequence ${a}_{0},{a}_{1},\dots$, and the sequence ${f}_{n}$ is arranged to be Cauchy in the ${C}^{\infty }$ metric.

I suppose the construction of smooth bump functions is pretty well known, but it won’t hurt to recall some basic tricks. For example, given $a, we can construct a smooth function $e\left(x\right)$ with values in $\left[0,1\right]$ that is identically 0 for $x\le a$ and identically 1 for $x\ge b$. Just define $e\left(x\right)$ for $a by applying the function

$\varphi \left(x\right)=\frac{x}{1+x}$

to the function

${e}^{1/\left(a-x\right)}{e}^{1/\left(b-x\right)}$

where the latter function vanishes rapidly as $x$ approaches 0 from the right and blows up rapidly as $x$ approaches 1 from the left. Then one can spline together similar functions so that, for example, given $a, there exists a smooth bump function $\psi \left(x\right)$ valued between 0 and 1 that is identically 1 on $\left[b,c\right]$ and identically 0 outside $\left[a,d\right]$. From there, one can do pretty much anything one likes!

(* One of the unintentionally funniest lines from the movie Good Will Hunting is when the Fields Medalist, examining Will’s work, says “I see you’ve used MacLaurin here”.)

Posted by: Todd Trimble on December 31, 2009 5:40 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

One of the unintentionally funniest lines from the movie Good Will Hunting is when the Fields Medalist, examining Will’s work, says “I see you’ve used MacLaurin here”.

As I recall, all of the work that demonstrates Will's genius in that movie consists of real problems that they got out of some combinatorics text —an elementary text. So if you ever get despondent and want some validation that you're a mathematical genius, just watch that film, and solve the problems before Will does.

Posted by: Toby Bartels on December 31, 2009 6:29 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

What I like about Good Will Hunting is that the bigshot Fields Medalist at MIT has a big office with a plush carpet and lots of beautiful wooden book cases. When I went to MIT, the bigshot Fields Medalist Dan Quillen had a crappy little office with a linoleum floor and yellow-painted concrete block wall just like everyone else!

(I actually enjoyed the movie, but this struck me as funny.)

Posted by: John Baez on December 31, 2009 9:04 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Yes, I recall some of the problems in that movie (like the ones that stumped the brilliant MIT students) were ones that I used to teach my students in elementary linear algebra how to solve. Knowing this has never really helped assuage my occasional despondency!

Like John, though, I otherwise enjoyed a lot about that movie.

Posted by: Todd Trimble on December 31, 2009 9:15 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Like those above, I agree: it's still a good movie.

Posted by: Toby Bartels on December 31, 2009 11:32 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Speaking of movies with math: there’s also the romantic comedy It’s My Turn (1980), starring Jill Clayburgh and Michael Douglas (before he hit it big). Completely unmemorable, except for the scene where she demonstrates the Snake Lemma in class, successfully fending off the skeptical remarks of her obnoxious grad student (played by Daniel Stern) – Clayburgh is smooth and convincing. Chuck Weibel gives this movie as a reference in his book on homological algebra.

Clayburgh’s character is a specialist in the theory of finite simple groups. A bit of trivia I read somewhere is that originally there was a line where she is revealed to be a specialist in the theory of finite simple abelian groups. When they previewed the movie for a test audience, a mathematician who happened to be there laughed out loud – luckily they asked him why, and saved themselves a little bit of embarrassment!

Posted by: Todd Trimble on January 1, 2010 2:52 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Todd wrote:

I’m awfully late to this party, and haven’t tried to read this thread with much care.

Happy New Year!

Yesterday I took a lot of what we developed in this thread and worked it into the current version of week287 — so now it’s drastically different than the original version. I hope my summary there is a bit easier to follow than the pile of comments here. We were pretty confused for a while! Now we’re not.

Still a few puzzles left, though. You’ll see David Speyer’s proof that $d{e}^{x}\ne {e}^{x}dx$ relies on the axiom of choice — or at least a little bit of ‘choice’. Is that really necessary?

I’m curious, John, about what you really meant when you said at the end that this is the key to the puzzle.

I just meant: “Hey, you algebraists — don’t forget bump functions! In the smooth world, unlike the world you love, a function is not determined by its Taylor series!”

With that reminder, one can see how to cook up a smooth function whose Taylor series is

$\sum _{n}{a}_{n}{x}^{n}$

regardless of how rapidly the numbers ${a}_{n}$ grow. Just use

$\sum _{n}{a}_{n}{x}^{n}{f}_{n}\left(x\right)$

where ${f}_{n}$ is a bump function that’s equal to 1 in a neighborhood of 0 and then drops to zero really fast, and stays there. How fast? Fast enough to make the sum converge! But since ${f}_{n}$ equals 1 in a neighborhood of 0, the Taylor series of

$\sum _{n}{a}_{n}{x}^{n}{f}_{n}\left(x\right)$

will be

$\sum _{n}{a}_{n}{x}^{n}$

That’s the basic idea. Of course one should worry that since ${f}_{n}$ ‘drops to zero really fast’, its derivatives will be big — and the higher the derivative, the bigger! If we take a standard ‘unit width’ bump function and squash it horizontally by a factor of $k$ to get something supported in $\left[-1/k,1/k\right]$, its $n$th derivative will get multiplied by ${k}^{n}$.

But a more careful analysis reveals that this doesn’t spoil the plan of attack.

Posted by: John Baez on December 31, 2009 8:50 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

But a more careful analysis reveals…

Ah, yup… I was indeed attempting such an analysis. Otherwise you basically summarized what I wrote.

Posted by: Todd Trimble on December 31, 2009 9:04 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Thanks for pointing me to the updated TWF, which I’m reading. Nitpick: why do you keep saying that a general topological space $X$ is homotopy-equivalent to $\mid \mathrm{Sing}\left(X\right)\mid$, when we all know that’s not true? Are you reluctant to say “weak homotopy equivalent”?

Posted by: Todd Trimble on December 31, 2009 9:49 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Do I keep saying that? I know I keep wanting to avoid talking about ‘weak homotopy equivalence’, since it sounds scary and technical. When writing This Week’s Finds, I go through complicated conniptions trying to explain stuff with the bare minimum of terminology, while not actually saying anything false. Or at the very least, I want experts to instantly know when I’m telling a lie, and nonexpert to not be seriously damaged by my lies.

It’s really hard to write math and have it be enjoyable to read. In particular, I can’t assume readers know the terminology I’ve introduced in all the hundreds of previous ‘Weeks’… nor do I want to pack each issue with definitions. Often the right solution is just to be vague — much as it pains me. Lately I’ve been including links where you can click on a word and see its definition… but too many of these is not a good thing: This Week’s Finds is mainly supposed to be fun, and past a certain point, it’s not fun to keep running off and looking up definitions.

The distinction between homotopy equivalence and weak homotopy equivalence is not something most people will consider fascinating — it’s just a technical nuance experts need to master. So, when discussing homotopy theory, I often restrict attention to ‘nice spaces’, secretly meaning CW complexes — and then maybe stick a footnote somewhere saying what I meant. Lately I thought I could get away with using ‘homotopy type’ to mean ‘object in the homotopy category $\mathrm{Ho}\left(\mathrm{Top}\right)$’ — so instead of saying ‘$X$ is weakly homotopy equivalent to $\mid \mathrm{Sing}\left(X\right)\mid$, I could just say ‘$X$ has the same homotopy type as $\mid \mathrm{Sing}\left(X\right)\mid$’, or maybe ‘$X$ is the same homotopy type as $\mid \mathrm{Sing}\left(X\right)\mid$’. But now you’re making me feel guilty. Am I supposed to say ‘weak homotopy type’? I think I’ve heard objects in $\mathrm{Ho}\left(\mathrm{Top}\right)$ referred to as homotopy types.

Posted by: John Baez on December 31, 2009 10:11 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Do I keep saying that? I know I keep wanting to avoid talking about ‘weak homotopy equivalence’, since it sounds scary and technical. When writing This Week’s Finds, I go through complicated conniptions trying to explain stuff with the bare minimum of terminology, while not actually saying anything false. Or at the very least, I want experts to instantly know when I’m telling a lie, and nonexpert to not be seriously damaged by my lies.

Lately I thought I could get away with using ‘homotopy type’ to mean ‘object in the homotopy category $\mathrm{Ho}\left(\mathrm{Top}\right)$

I’m not sure what is best. I think that if it were me, I might slip in a single instance of ‘(weak)’ and then let elide over future mentions. But that’s me. Another part of me wonders whether “weak homotopy equivalence” (or just “weak equivalence”) is really so much scarier than “homotopy equivalence”. Maybe it is; I don’t know.

I guess in the old days, when people like Milnor were writing papers on spaces having the homotopy type of a CW complex, language was being used in a more fastidious way, but people have by now gotten so used to these nuances that they relax a bit in their language use and avoid long-winded expressions like “weak homotopy type”. And I don’t have any quarrel with that, if the audience is not likely to misunderstand. But the fact that you are writing with non-experts in mind makes these issues a little bit tricky.

Anyway, sorry for making you feel guilty. The TWF is lovely as always – feel free to ignore my pedantic complaints!

Posted by: Todd Trimble on December 31, 2009 10:37 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Todd wrote:

Another part of me wonders whether “weak homotopy equivalence” (or just “weak equivalence”) is really so much scarier than “homotopy equivalence”. Maybe it is; I don’t know.

I try to write each issue on a ‘sliding scale’, where it starts out easy and enjoyable for everyone, and then people gradually leave in disgust. When they’re all gone, I can talk the way I actually think. But the goal is to get to something interesting before they all slip out the back door.

So, when I write the beginning of each section on rational homotopy theory (and I’m writing a bunch now), I imagine physicists reading them… and I imagine something like this:

If I say “homotopy equivalent”, they’ll think “Hmm, some kind of concept of two spaces being ‘the same’ — some concept topologists like.” Maybe with luck they’ll have heard that a circle and an annulus are homotopy equivalent. I don’t expect they can rattle off the definition.

But if I say “weakly homotopy equivalent”, they’ll think “Hmm, something sort of like homotopy equivalent, but not that — something ‘weaker’ and more technical”.

And of course that’s true. So maybe it’s okay if they think that. But I’m also imagining how they might feel when they think this, and I’m imagining it’s something like “Ho hum, now it’s getting technical — fine-grained distinctions are coming in, this is precisely why I didn’t go into math.”

The problem is that the true object of study of homotopy theory was discovered a bit too late! It’s not topological spaces, that’s for sure. And the concept of ‘sameness’ is not homeomorphism or even homotopy equivalence. So the most important concepts are saddled with names that make them sound ‘technical’ and ‘subtle’, when in fact they’re the most fundamental and robust!

Anyway, it’s fun to have a chance to talk about these issues of exposition. And I’ve changed week287 to make it a bit clearer.

Posted by: John Baez on January 1, 2010 12:45 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

the true object of study of homotopy theory was discovered

Really?? What is it?

Posted by: Tom Leinster on January 1, 2010 7:46 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Tom wrote:

Really?? What is it?

Heh, I don’t mean anything deep here, just the usual thing everybody says. Homotopy theory studies ‘homotopy types’.

But unfortunately, because this was realized rather late, you’ll see that Wikipedia gives a definition of ‘homotopy type’ that’s not the one I mean here.

I don’t mean topological spaces where you treat two as the same if they’re homotopy equivalent. I mean topological spaces where you treat two as the same if they’re weakly homotopy equivalent.

And then it becomes clear that you could equally well be studying simplicial sets, or cubical sets, or objects of any of various other equivalent $\left(\infty ,1\right)$-categories.

So, we say topological spaces are just one ‘model’ of homotopy types. Homotopy theory doesn’t really study topological spaces: it uses topological spaces to study homotopy types.

And of course these days most of us hope and believe that for any good definition of $\infty$-category, $\infty$-groupoids are also a model of homotopy types.

I’m not saying all the details have been worked out! You can discover an elementary particle — the muon, say — yet still be puzzled about it. But I think the object of study of homotopy theory was quite firmly fixed in people’s minds by the time Quillen wrote Homotopical Algebra in 1967. He came up with a way to make precise the notion that you can use either simplicial sets or topological spaces to model homotopy types — but of course he did this because people knew it should be true. We’ll keep coming up with better ways to formalize this notion…

Posted by: John Baez on January 1, 2010 6:23 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I would argue that the true object of study of homotopy theory (insofar as it makes sense to use the definite article at all) is not “homotopy types” but rather $\infty$-groupoids. Unless you define the former to mean the latter—but your comments seem more in line with the definition that a homotopy type is an object of the homotopy category of some model for homotopy types (which isn’t the same thing even if $\infty$-groupoids model homotopy types).

Posted by: Mike Shulman on January 1, 2010 8:49 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Well, if someone forced me to answer the question “What is the true object of study of homotopy theory?” then I guess I’d go with Mike and say “$\infty$-groupoids” — though goodness knows what homotopy theory really is, or how one decides whether something is or is not its true object of study.

Then, of course, you have to say what $\infty$-groupoids are, and there’s a story. I still think there’s a big conceptual difference between algebraic and non-algebraic notions of $\infty$-groupoid. There’s no work that I know of saying that the two approaches are going to have to be substantially equivalent. Maybe they are, maybe they’re not — I don’t know, and there seems to be room for some surprises here.

In a sense this is half the Homotopy Hypothesis. In Pursuing Stacks, Grothendieck spoke very explicitly of an algebraic notion of $\infty$-groupoid. If someone proves that topological spaces are the same as $\infty$-groupoids, but using a non-algebraic notion of $\infty$-groupoid, that would be the other half.

Posted by: Tom Leinster on January 1, 2010 10:11 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

If someone proves that topological spaces are the same as ∞-groupoids, but using a non-algebraic notion of ∞-groupoid, that would be the other half.

Aren’t Kan complexes a non-algebraic notion of $\infty$-groupoid? And they’re certainly the same as topological spaces (with weak homotopy).

I still think there’s a big conceptual difference between algebraic and non-algebraic notions of ∞-groupoid. There’s no work that I know of saying that the two approaches are going to have to be substantially equivalent.

Do you consider topological spaces to be “algebraic” or “non-algebraic”? They’re sort of “algebraic” in that paths can be composed (they can be composed using lots of different parametrizations, but that’s okay). Also every object is fibrant but not every object is cofibrant, which is generally a feature of algebraic rather than non-algebraic definitions. But if topological spaces are algebraic and Kan complexes are non-algebraic, then those two are known to be equivalent.

Apart from that, there may be no direct proofs that algebraic = non-algebraic in the $\infty$-case, but there are certainly such proofs at low dimensions. Indisputably algebraic notions of $n$-groupoid suffice to model $n$-types for $n\le 3$, and I’m not aware of anything suggesting that something new would happen at $\infty$. And if you’re willing to use slightly weirder kinds of weak equivalence, there are certainly “algebraic categories” that model all homotopy types, such as small categories with the Thomason model structure. Are those “algebraic” or “non-algebraic”?

What I’m trying to get at is that I don’t see nearly as clear a conceptual distinction between the two approaches.

Posted by: Mike Shulman on January 2, 2010 5:27 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Aren’t Kan complexes a non-algebraic notion of $\infty$-groupoid?

Yes. That’s a good illustration of why using a non-algebraic notion of $\infty$-groupoid dilutes the Homotopy Hypothesis. In the extreme, you might say “I interpret space to mean Kan complex; I interpret $\infty$-groupoid to mean Kan complex; done”.

Do you consider topological spaces to be “algebraic” or “non-algebraic”?

I’m only applying those adjectives to theories of $n$-category (or $n$-groupoid). Topological spaces aren’t a notion of $n$-category or $n$-groupoid.

Maybe I should say more about my point of view on the Homotopy Hypothesis. In its briefest form, it says “spaces are the same as $\infty$-groupoids”. I want to formulate this in a really strong way. People talk a lot about the difficulty of describing the structure formed by all spaces, or all $\infty$-groupoids. But I mean something else.

I want “space” to mean something really topological — an actual topological space, satisfying certain conditions. I want “$\infty$-groupoid” to mean something really algebraic — they should be the algebras for some monad on the category of $\infty$-graphs.

Thus, a proof of the Homotopy Hypothesis would really say that something topological is equivalent to something algebraic.

If someone comes up with a proof that uses, say, a kinda algebraic/combinatorial notion of space (such as simplicial set) or a not-so-algebraic notion of $\infty$-groupoid (such as Tamsamani’s), then it might be a big achievement — but it won’t prove the Hypothesis in its full glory.

What I’m trying to get at is that I don’t see nearly as clear a conceptual distinction between the two approaches.

The question of whether a theory of $n$-categories is algebraic or not — in the sense that I’m using the word — has nothing to do with homotopy theory or model categories. It’s purely about whether the operations in the theory (such as composition) are genuine operations in the sense of universal algebra, or are only determined up to isomorphism etc. Put another way, it’s a question of monadicity.

A detailed explanation of what I mean by “algebraic” is on p.5-6 of my survey of definitions of $n$-category, where I say “the/a” instead of “algebraic/non-algebraic”.

Posted by: Tom Leinster on January 2, 2010 6:13 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Okay, I see your point that Kan-complexes are not an algebraic definition of $\infty$-groupoids. But can’t we simply turn it into an algebraic definition by choosing distinguished fillers for each diagram? Let me explain what I mean a little bit more in detail:

What makes us think of a Kan-complex $X$ as $\infty$-groupoids is that for each horn $\Lambda \left(n,k\right)$ in $X$ (i.e. a map $\Lambda \left(n,k\right)\to X$) there exists a filler $\Delta \left(n\right)$ in $X$ (i.e. an Element of ${X}_{n}$). For example for two composable edges (aka 1-morphisms) we have a horn $\Lambda \left(2,1\right)$ in $X$ and then the third edge of _a_ filler could be thought of as _a_ composite of those two morphisms. Furthermore we have fillers all the way up.

Now the problem that Tom mentions is that those composites are not defined but have to be choosen in a non unique way, though unique up to higher cell. So why don’t we just single out a certain filler and take this filler as our composition? Technically we make the following

Definition: An algebraic simplicial set is a simplicial set $X$ together with a distinguished filler for each horn in $X$. That means for each map $\Lambda \left(n,k\right)\to X$ we have a map $\Delta \left(n\right)\to X$ that renders the obvious diagramm commutative. A map of algebraic simplicial sets is a map that takes distinguished fillers to distinguished fillers. The category of algebraic simplicial sets is denoted by ${\mathrm{sSet}}_{\mathrm{alg}}$.

In particular each algebraic simplicial set is a Kan complex because all the fillers exist. Now we have the canonical forgetful functor ${\mathrm{sSet}}_{\mathrm{alg}}\to \mathrm{sSet}$In order to realize algebraic simplicial sets as algebras for a monad (as Tom mentioned) we should look for a left adjoint to this functor. I claim that such a left adjoint is provided by the fibrant replacement in simplicial sets which is obtained by Quillens small object argument. So lets see how this works:

Take an arbitrary simplicial set ${X}_{0}$ and consider all the horns $\bigsqcup \Lambda \left(n,k\right)\to X$. We have also the map $\bigsqcup \Lambda \left(n,k\right)\to \bigsqcup \Delta \left(n\right)$. Now we take the pushout over this diagram (sorry I didn’t manage to typeset this diagram) and obtain a simplicial set ${X}_{1}$ together with certain choosen fillers for horns,namely those horns $\Lambda \left(n,k\right)\to {X}_{1}$ which factor through ${X}_{0}$. We proceed like this and get a sequence of simplicial sets ${X}_{0}\to {X}_{1}\to {X}_{2}\to {X}_{3}\to \cdots$Now the fibrant replacement ${X}_{\infty }$ is defined to be the colimit over this sequence of simplicial sets. Furthermore we have distinguished fillers for each horn $\Lambda \left(n,k\right)\to {X}_{\infty }$ that factors through a simplicial set ${X}_{k}$ for finite $k$. Now the smallness of $\Lambda \left(n,k\right)$ tells us, that each diagramm factors through a finite step. That means we have fillers for each horn.

Remark: Now that I write it down it seems to me that there might be severals fillers for the same horn. Maybe in going from ${X}_{1}$ to ${X}_{2}$ one should just take the pushout over those horns that don’t factor through ${X}_{1}$ to avoid this.

Unfortunately I can’t write down the rest of the argument now because I don’t have more time at the moment. But I think it is true that the assignment $X\to {X}_{\infty }$ is left adjoint to the forgetful functor (we just took colimits over horns). Now this adjunction establishes a monad for which I hope that the algebraic simplicial sets are algebras. Furthermore one has a model structure on those algebras (see the article of Moerdijk and Berger) and then the forgetfull functor is a Quillen functor and the homotopy hypotheses would be that this is a quillen equivalence, because we already know that simplicial sets are Quillen equivalent to topological spaces.

Sorry for being so imprecise at the end, but it would be very nice to hear some comment of the experts about my ideas. Is there a fundamental mistake in it?

Posted by: Thomas Nikolaus on January 2, 2010 1:41 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Your definition of “algebraic simplicial set,” which I would call an algebraic Kan complex, is absolutely right. For the left adjoint to the forgetful functor you want to use, not Quillen’s small object argument, but Richard Garner’s improved version of it, which (in this case) is basically your idea of “taking the pushout over those horns whose fillers don’t already exist,” or more precisely “over those horns for which fillers haven’t yet been added”. His construction is also definitely monadic, and it’s quite reasonable to expect that if the monad creates a model structure on the category of algebraic Kan complexes, then it would be Quillen equivalent to the usual one on simplicial sets.

Emily Riehl and I have thought about this some, though, and we got stuck on proving that the model structure lifts. In Moerdijk and Berger I assume you’re referring to the general “transfer principle” for cofibrantly generated model structures across adjunctions, first stated explicitly by Crans. This has two hypotheses: the left adjoint $F$ must preserves small objects, which is automatic if the right adjoint preserves filtered colimits, as it seemingly must in this case—but also relative $F\left(J\right)$-cell complexes must be weak equivalences, where $J$ is the original set of generating acyclic cofibrations. This “acyclicity condition” is the hardest to check in practice, especially because pushouts in the category of algebras are generally quite difficult to write down explicitly. Emily and I weren’t able to do it in this case after a fair amount of thought, and some of what we wrote down made us doubt that it might be true. But if you (or anyone else) has ideas about how to prove it, I’d be very happy!

Posted by: Mike Shulman on January 2, 2010 3:59 PM | Permalink | PGP Sig | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Mike, I’m very happy that someone like you and Richard Garner had a similar ideas before and thought about it more properly than me. I have to admit that I only understood part of Richard Garners paper, but paragraph 6 and 7 have been very insightful.

“I assume you’re referring to the general “transfer principle” for cofibrantly generated model structures across adjunctions, first stated explicitly by Crans.”

Yes, that is what I meant. I simply took a version from this paper of Moerdijk and Berger because they used it to equip the categories of algebras for certain operads with a model structrure.

“This has two hypotheses: the left adjoint F must preserves small objects, which is automatic if the right adjoint preserves filtered colimits, as it seemingly must in this case—but also relative F(J)-cell complexes must be weak equivalences, where J is the original set of generating acyclic cofibrations. This “acyclicity condition” is the hardest to check in practice, especially because pushouts in the category of algebras are generally quite difficult to write down explicitly.”

Okay, I see the problems here. I hoped that we could get around this because the unit and counit of the adjunction are already weak equivalences (in the relevant sense) but I didn’t manage to prove this till now. Did I understand right, that your work with Emily Riehl made you think that there is no model structure on algebraic Kan complexes such that this adjunction lifts to a Quillen equivalence?

In this case we could try to get away with an equivalence of simplicially enriched categories, because it’s clear how the simplicially enriched (or $\left(\infty ,1\right)$-category) should look like and the fact that the unit of the adjunction is a weak equivalence is then enough to see that the enriched forgetful functor is then a Dwyer-Kan-equivalence of simplicially enriched categories.

Posted by: Thomas Nikolaus on January 3, 2010 8:56 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Did I understand right, that your work with Emily Riehl made you think that there is no model structure on algebraic Kan complexes such that this adjunction lifts to a Quillen equivalence?

Emily might remember better than I do, but I think all I’m willing to say right now is that I currently don’t see a weight of evidence in either direction. Certainly, if the adjunction creates a model structure on ${\mathrm{Kan}}_{\mathrm{alg}}$ in the usual way, then the adjunction is a Quillen equivalence, since all simplicial sets are cofibrant, all algebraic Kan complexes would be fibrant, and (as you point out) the unit and counit are weak equivalences (the first since it is a fibrant replacement, and the second by 2-out-of-3).

In this case we could try to get away with an equivalence of simplicially enriched categories, because it’s clear how the simplicially enriched (or $\left(\infty ,1\right)$-category) should look like

Hmm… it’s not immediately clear to me how to make ${\mathrm{Kan}}_{\mathrm{alg}}$ simplicially enriched. I guess you could define the mapping space ${\mathrm{Map}}_{\mathrm{alg}}\left(X,Y\right)$ to be the full sub-simplicial set of the ordinary simplicial mapping space $\mathrm{Map}\left(X,Y\right)$ determined by the 0-simplices that are maps of algebraic Kan complexes (and all higher simplices between these), in which case the forgetful functor acts on homs by the inclusion ${\mathrm{Map}}_{\mathrm{alg}}\left(X,Y\right)↪\mathrm{Map}\left(X,Y\right)$ — but I don’t see any reason why that should be a weak equivalence in general. I would hope it to be an equivalence when $X$ is “cofibrant,” but it’s not immediately obvious to me how to prove that without a model structure.

Of course, we could also define ${\mathrm{Map}}_{\mathrm{alg}}\left(X,Y\right)$ to just be the ordinary mapping space $\mathrm{Map}\left(X,Y\right)$. This is not unreasonable, since maps of algebraic Kan complexes are like “strict” functors between algebraic $\infty$-groupoids, while arbitrary simplicial maps between them are like “weak” functors. It then becomes basically tautologous that the forgetful functor is a DK-equivalence onto the full subcategory $\mathrm{Kan}\subset \mathrm{SSet}$—in fact, it’s an equivalence of simplicially enriched categories. In particular, the underlying ordinary category of this “simplicially enriched category of algebraic Kan complexes” is equivalent to $\mathrm{Kan}$, so it’s no longer really a “simplicial enrichment” of ${\mathrm{Kan}}_{\mathrm{alg}}$. It does make some sense to call it an “$\left(\infty ,1\right)$-category of algebraic simplicial $\infty$-groupoids,” but since it’s equivalent (as a simplicially enriched category, not just DK-equivalent) to the ordinary $\left(\infty ,1\right)$-category of non-algebraic simplicial $\infty$-groupoids (namely $\mathrm{Kan}$) it’s not clear that much has been gained.

Posted by: Mike Shulman on January 3, 2010 9:15 PM | Permalink | PGP Sig | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Mike wrote:

I guess you could define the mapping space Map alg$\left(X,Y\right)$ to be the full sub-simplicial set of the ordinary simplicial mapping space Map$\left(X,Y\right)$ determined by the 0-simplices that are maps of algebraic Kan complexes (and all higher simplices between these), in which case the forgetful functor acts on homs by the inclusion Map alg$\left(X,Y\right)\to$Map(X,Y) - but I don’t see any reason why that should be a weak equivalence in general

I also think that this is not a weak equivalence in general. For the simple reason you mentioned that this are only strict functors. But I’d expect that this enrichment extends to an simplical model category if algebraic Kan complexes have a model structure. But I had more in mind to

define Map alg$\left(X,Y\right)$ to just be the ordinary mapping space Map$\left(X,Y\right)$.

This is not as stupid as one could think at a first glance. First of all it’s true that the forgetful functor $U:\mathrm{AlgKan}\to \mathrm{Kan}$ is then an equivalence of 1-categories which extends to an equivalence of simplicially enriched categories. But there is no explicit invers to this functor. The Garner-fibrant-replacement functor $U:\mathrm{Kan}\to \mathrm{AlgKan}$ is not an 1-invers to this functor. It is only a DK-equivalence of simplicially enriched categories (and here this is not so tautologically.) Even better, the adjoint pair $\left(U,F\right)$ forms an $\left(\infty ,1\right)$-adjoint equivalence of $\left(\infty ,1\right)$-categories (for adjunctions of $\left(\infty ,1\right)$-categories see Higher Topos Theory chapter 5.2.2.) So the equivalence is as near as possible to a Quillen equivalence as possible without having model structures.

Furthermore if we had a model structure on algebraic Kan complexes the weak equivalence would be the weak equivalences of the underlying Kan complexes. So it’s immediately clear that the simplical localization (also calles $\left(\infty ,1\right)$-Homotopy category) of this homotopical category is just the category of Kan complexes. Equvivalently one could say that the functor $\stackrel{˜}{U}:\mathrm{AlgKan}\to {\mathrm{Kan}}_{\left(\infty ,1\right)}$ is a localization functor.

It does make some sense to call it an “(∞,1)-category of algebraic simplicial ∞-groupoids,” but since it’s equivalent (as a simplicially enriched category, not just DK-equivalent) to the ordinary (∞,1)-category of non-algebraic simplicial ∞-groupoids (namely Kan) it’s not clear that much has been gained.

I agree that formally not that much has been gained, but maybe it helps to take Kan complexes as a definition of $\infty$-groupoid. And what is the additional gain of a model structure on algebraic Kan complexes? This would just provide a cofibrant replacement $\stackrel{^}{X}$ for each algebraic Kan complex $X$ such that the groupoid of strict functors $\stackrel{^}{X}\to Y$ (meaning the simplicial set mentioned at the beginning) is equivalent to the groupoid of weak functors $X\to Y$(meaning the simplicial set of morphisms between the underlying Kan complexes of $X$ and $Y$).

Posted by: Thomas Nikolaus on January 4, 2010 3:25 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I’d expect that this enrichment extends to an simplical model category if algebraic Kan complexes have a model structure.

I wouldn’t know what to expect. Note that Garner’s fibrant replacement functor is not a simplicially enriched functor.

And what is the additional gain of a model structure on algebraic Kan complexes?

Yes, cofibrant replacements would be useful. It would also provide a simplicial model for $\infty$-groupoids in which all objects are fibrant. All objects being fibrant is, I think, one of the main technical advantages of topological spaces over simplicial sets.

Posted by: Mike Shulman on January 4, 2010 4:47 PM | Permalink | PGP Sig | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Mike wrote:

I wouldn’t know what to expect. Note that Garner’s fibrant replacement functor is not a simplicially enriched functo

Yes, you are of course totally right. I thought somehow that the Garner replacement could easily be extended to a simplicial functor. And even in this case what I wrote doesn’t make so much sense. When I have a little bit more time at the end of the week I think about this category again and let you know when I find out something more.

All objects being fibrant is, I think, one of the main technical advantages of topological spaces over simplicial sets.

Hm, you mean that this is an advantage in calculating homotopy limits? I would imagine that one could just as well say that in Topological spaces not everything is cofibrant. And isn’t taking the subcategory of CW-complexes (or cell-complexes) the same as taking the subcategory of Kan-complexes? Or is CW-complexes a better behaved subcategory?

Posted by: Thomas Nikolaus on January 5, 2010 10:45 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I didn’t mean that simplicial sets don’t have plenty of their own advantages over topological spaces! I wouldn’t undertake to say that either is better than the other in any absolute sense. Having all objects fibrant is indeed nice for calculating homotopy limits, and also for other purposes, while having all objects cofibrant is nice for homotopy colimits and other purposes. And yes, the category of CW complexes is, in some ways, analogous to the category of Kan complexes, in that both are the subcategory of fibrant-cofibrant objects in some model category, and neither have good 1-categorical properties by themselves, although I think calling them “the same” is a bit strong. In the ideal world, we’d have a model for $\infty$-groupoids in which all objects were both fibrant and cofibrant, but as far as I know no one has ever come up with one.

Posted by: Mike Shulman on January 6, 2010 4:17 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

In the ideal world, we’d have a model for ∞-groupoids in which all objects were both fibrant and cofibrant, but as far as I know no one has ever come up with one.

Just for the record one might mention here that for a model category with all objects cofibrant, the subcategory of fibrant objects is, while no longer in general a model category, still, tautologically, a Brown category of fibrant objects, a type of homotopical category equipped with slightly less extra structure and tools than a model category, but still strong enough to support a rich theory, at least strong enough to admit good computations of homotopy categories.

Not the least, using this tool Brown was able to study – and elegantly so – the homotopy categories of $\infty$-groupoid valued $\infty$-stacks and of spectra valued $\infty$-stacks.

Once on this blog here I made some post on how one might find canonically a simplicial enrichment of Brown categories-of-fibrant-objects along the lines of simplicially enriching model categories using framings, such as to go beyond the homotopy category and construct the $\left(\infty ,1\right)$-category presented by the Brown category. Thomas and I had some ideas that seemed like they would work out, but at some point I didn’t follow this further.

Posted by: Urs Schreiber on January 6, 2010 11:50 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

one might mention here that for a model category with all objects cofibrant, the subcategory of fibrant objects is … a Brown category of fibrant objects

What does all objects being cofibrant have to do with that? I thought that was true for any model category whatsoever.

Posted by: Mike Shulman on January 6, 2010 3:51 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Yes, that’s true for any model category.

Posted by: Thomas Nikolaus on January 6, 2010 4:27 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

one might mention here that for a model category with all objects cofibrant, the subcategory of fibrant objects is … a Brown category of fibrant objects

What does all objects being cofibrant have to do with that? I thought that was true for any model category whatsoever.

It’s of course true for every model category, but I said this in reply to your remark that

the subcategory of fibrant-cofibrant objects in some model category, [does not] have good 1-categorical properties by [itself],

Because if already all objects are cofibrant, then the sub-category of fibrant objects, is the subcategory of fibrant-cofibrant objects, and my remark was to point out that this has still the nice structure of a Brown category.

Hence in the case that all objects already happen to be cofibrant, this gives a “reasonably good” 1-categorical structure on the subcategory of fibrant-cofibrant objects.

Posted by: Urs Schreiber on January 6, 2010 5:17 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Ah, okay. Well, yes, a category of fibrant objects is better than nothing, but it still doesn’t even have, for instance, finite limits and colimits. That’s the sort of 1-categorical structure I was thinking of.

Posted by: Mike Shulman on January 6, 2010 8:53 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Thomas, I’ll just reply to this in general terms.

Okay, I see your point that Kan-complexes are not an algebraic definition of $\infty$-groupoids. But can’t we simply turn it into an algebraic definition by choosing distinguished fillers for each diagram?

That’s exactly the kind of thing that one would hope to be able to do. And in other contexts, this kind of thing has been done successfully.

What I wanted to emphasize was the a priori difference between the two types of theory of $n$-groupoids, algebraic and non-algebraic. Those of a topological bent tend to gravitate towards the non-algebraic type, motivated by thoughts of non-unique horn fillers etc. Others like the idea that there is a purely algebraic theory of $n$-groupoids, like the theory of 1-groupoids or categories. Both conceptions are valid. Each comes with its own valuable intuitions.

In the long run the two may turn out to be “equivalent”. But at present we don’t even know how to formulate that properly, let alone prove it. That’s why I’m emphasizing the distinction. We don’t know enough yet to be confident that they’re the same.

Posted by: Tom Leinster on January 3, 2010 1:32 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Tom said:
“That’s exactly the kind of thing that one would hope to be able to do. And in other contexts, this kind of thing has been done successfully.”

In which contexts has this been done? And do you agree that the definition of algebraic simplicial set I gave (or better as Mike suggested: algebraic Kan complex) _is_ in fact an algebraic definition of $\infty$-groupoid?

Tom said:
“In the long run the two may turn out to be “equivalent”. But at present we don’t even know how to formulate that properly, let alone prove it. That’s why I’m emphasizing the distinction. We don’t know enough yet to be confident that they’re the same.”

I don’t know what you exactly mean by that, but I think we know very well what has to be done here. The category of $\infty$-groupoids will form an $\left(\infty ,1\right)$-category in the end (that means an $\infty$-category where all morphisms for $n\le 2$ are weakly invertible). There are several models for $\left(\infty ,1\right)$-categories around (simplicial categories, Segal categories, complete Segal spaces, quasi-categories) which are all equivalent in a certain, very precise sense. For example a model category is a very clever way to produce a $\left(\infty ,1\right)$-category and a Quillen equivalence then encodes the data of a $\left(\infty ,1\right)$-adjoint equivalence. So, if we equip our category of algebraic $\infty$-groupoids with a model structure and show it’s Quillen equivalent to the model category of topological spaces we’re done.

Posted by: Thomas Nikolaus on January 3, 2010 8:36 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Thomas wrote:

In which contexts has this been done?

(‘This’ was showing that a non-algebraic definition of some higher categorical structure was equivalent to an algebraic one, by making choices of composites etc.)

One context where it’s been done is monoidal categories, which can be defined in both algebraic and non-algebraic ways. See, for instance, Chapter 3 of my book.

do you agree that the definition of algebraic simplicial set I gave […] is in fact an algebraic definition of $\infty$-groupoid?

I agree that it’s an algebraic theory: the category of algebraic simplicial sets is monadic over simplicial sets, and it’s also the category of models in Set for a finite limit theory. Whether it deserves to be called a theory of $\infty$-groupoids is another question. It’s a subjective judgement, and I haven’t thought hard enough to have an opinion.

Regarding the rest of your comment: there’s a famous difficulty in comparing any two proposed definitions of $n$-category (or $n$-groupoid, etc). See, for instance, the paragraph at the bottom of page 2 of this, or (for a closely related but not identical point) pages 27–8 of this version of Grothendieck’s Pursuing Stacks.

In brief, the difficulty is that if you propose a definition of $n$-category, and I propose one too, and we wish to decide whether our definitions are equal, then what we should probably do is ask whether the $\left(n+1\right)$-category of Thomas-$n$-categories is equivalent to the $\left(n+1\right)$-category of Tom-$n$-categories — but whose definition of $\left(n+1\right)$-category do we use?

So this has nothing, in fact, to do with algebraic and non-algebraic. It’s a more general difficulty.

There are dirty solutions. For example, suppose that you have a definition of ‘equivalence’ of Thomas-$n$-categories, and I, similarly, have a definition of ‘equivalence’ of Tom-$n$-categories. Then we might try to:

1. find a way of turning any Thomas-$n$-category into a Tom-$n$-category
2. find a way of turning any Tom-$n$-category into a Thomas-$n$-category
3. show that when you start with a Thomas-$n$-category, turn it into a Tom-$n$-category, and turn that into a Thomas-$n$-category, what you end up with is equivalent to what you started with
4. similarly the other way round.

Doing that might well be very hard (and for some pairs of proposed definitions of $n$-category, no one’s managed it yet). Even so, it’s only a tiny fragment of what it would mean for your and my theories of $n$-category to be essentially the same. In fact, all it says is that there’s a one-to-one correspondence between equivalence classes of Thomas-$n$-categories and equivalence classes of Tom-$n$-categories. We’ve said nothing about $n$-functors, $n$-transformations, etc., or for that matter associated notions such as bimodules etc. between $n$-categories.

Another dirty solution — though slightly less dirty than the previous one — is to use model categories. We can agree on what a model category is. Maybe you have a way of making your $n$-categories (or $n$-groupoids) into a model category, and I have a way of making mine into a model category. We can then ask whether those model categories are Quillen equivalent. But again, while we puny humans might be finding this extremely difficult, it’s still only a tiny fragment of what it means for two theories of $n$-categories to be essentially the same.

That’s why I used the word ‘properly’ when I wrote this:

In the long run the two may turn out to be “equivalent”. But at present we don’t even know how to formulate that properly

There are diluted formulations such as the two I mentioned. Proving even the diluted formulations can be a big achievement. This is why, I think, some people are losing sight of the fact that they are diluted. We haven’t got anything like a full understanding of what it means for two theories of $n$-categories to be equivalent.

Posted by: Tom Leinster on January 3, 2010 11:10 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

In brief, the difficulty is that if you propose a definition of $n$-category, and I propose one too, and we wish to decide whether our definitions are equal, then what we should probably do is ask whether the $\left(n+1\right)$-category of Thomas-$n$-categories is equivalent to the $\left(n+1\right)$-category of Tom-$n$-categories — but whose definition of $\left(n+1\right)$-category do we use?

This is quite true for $n$-categories, but $n$-groupoids can be expected to form only an $\left(n+1,1\right)$-category, and we do have a pretty well-developed theory of $\left(\infty ,1\right)$-categories, hence also of $\left(n+1,1\right)$-categories. In particular, a model category is generally regarded as a presentation of an $\left(\infty ,1\right)$-category; do you disagree with that? If one grants that, then Thomas’ argument that “we know quite well what has to be done” is exactly right.

That also means that if we show that two definitions of $n$-category form Quillen equivalent model categories, we’ve shown in particular that they form equivalent $\left(\infty ,1\right)$-categories. I wouldn’t call this “only a tiny fragment” of what it means for two theories to be the same; it means in particular that the notions of functor and of invertible $k$-transfor for all $k\ge 1$ are the same. Of course noninvertible transfors are also important, as are bimodules and so on, but I think this is a noticeable fraction of everything we’d like.

There’s even a possible way to use model categories to “bootstrap” our way up to showing an equivalence of $\left(n+1\right)$-categories. If the model categories of Tom- and Thomas-$n$-categories are closed monoidal, then they present closed monoidal $\left(\infty ,1\right)$-categories, whose internal-homs tell you about higher transfors. Thus, if the Quillen equivalence is a closed monoidal Quillen equivalence, we can conclude that the equivalence between the $\left(\infty ,1\right)$-categories of Tom- and Thomas-$n$-categories in fact extends to an equivalence of the hom-$n$-categories in our putative $\left(n+1\right)$-categories of $n$-categories. Now if both notions of $n$-category have the (clearly desirable) property that a category enriched over $n$-categories induces an $\left(n+1\right)$-category, then we get a Tom-$\left(n+1\right)$-category of Tom-$n$-categories and likewise for Thomas’ definition. We can then hope to apply the monoidal Quillen equivalence to get a Tom-$\left(n+1\right)$-category of Thomas-$n$-categories and vice versa, and finally use the equivalence between hom-$n$-categories to show that the Tom-$\left(n+1\right)$-category of Tom-$n$-categories is equivalent to the Tom-$\left(n+1\right)$-category of Thomas-$n$-categories, and vice versa.

Of course, not many definitions of $n$-categories are known to form closed monoidal model categories, and many of them probably don’t. But a similar game can probably be played with a less strict notion of “closed monoidal $\left(\infty ,1\right)$-category,” such as for instance a closed monoidal quasicategory, and a notion of “weak enrichment.” There are lots of details that haven’t been worked out, but I think it’s an exaggeration to say that “We haven’t got anything like a full understanding of what it means for two theories of $n$-categories to be equivalent.”

Posted by: Mike Shulman on January 4, 2010 2:51 AM | Permalink | PGP Sig | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

1. Conventional wisdom holds that $\left(\infty ,1\right)$-categories should form an $\left(\infty ,2\right)$-category
2. Model categories are morally wrong.
Posted by: Tom Leinster on January 5, 2010 2:32 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

1. Conventional wisdom holds that (∞,1)-categories should form an (∞,2)-category

And they do. For instance as described here.

But Mike recalled that Thomas remarked that the discussion was about comparing $\infty$-groupoids=$\left(\infty ,0\right)$-categories which form an $\left(\infty ,1\right)$-category, in fact a presentable $\left(\infty ,1\right)$-category.

It so happens that an equivalence of presentable $\left(\infty ,1\right)$-categories may be modeled by a zig-zag of Quillen equivalences of combinatorial model categories.

2. Model categories are morally wrong.

This is a theorem, not a question of moral.

Therefore for checking if two different definitions of $\infty$-groupoids are equivalent it is indeed sufficient to exhibit a Quillen equivalence of suitable model categories.

Posted by: Urs Schreiber on January 5, 2010 3:18 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Urs said basically what I would. In particular, I don’t understand what you mean by “model categories are morally wrong.” Model categories are perfectly good presentations of $\left(\infty ,1\right)$-categories, and Quillen equivalent ones present equivalent $\left(\infty ,1\right)$-categories. Granted, not all $\left(\infty ,1\right)$-categories can be presented by a model category, and for some things you want to do with $\left(\infty ,1\right)$-categories, a presentation in terms of a model category is not helpful. But for purposes of saying that two $\left(\infty ,1\right)$-categories are equivalent, having a Quillen equivalence of model categories is perfectly good.

Posted by: Mike Shulman on January 5, 2010 10:25 PM | Permalink | PGP Sig | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Another aspect is that as long as we are interested just in deciding if two $\left(\infty ,n\right)$-categories are equivalent, we do not actually need the $\left(\infty ,n+1\right)$-category of all of them:

since the equivalence consists only of invertible morphisms, it is perfectly sufficient to have the $\left(\infty ,1\right)$-catgory of all $\left(\infty ,n\right)$-categories. If two are equivalent in there, they are also equivalent in the full thing. (In fact, for that it is already sufficient to know the $\infty$-groupoid of all $\left(\infty ,n\right)$-categories.)

For this reason for instance the construction of $\left(\infty ,1\right)$-categories of $\left(\infty ,n\right)$-categories here goes a long way.

And I think this is the reason why $\left(\infty ,1\right)$-categories – and hence their standard model: model categories – play such a pivotal role in all of higher category theory: they are the smallest structures that

a) support a category theory (limits, adjunctions, Grothendieck construction, etc.)

b) while at the same time knowing everything about higher equivalences.

To my mind the main problem with all the existing algebraic definitions of higher categories is that for them point a) tends to be entirely unknown. I think one reason why for instance the cobordism theory was solved in terms of geometric $\left(\infty ,n\right)$-categories and not say in terms of operadically defined Batanin-style higher categories is that for the former there exists “category theory” (limits, adjunctions, Grothendieck construction), while for the latter there does not (at least it’t not known).

I can see why one would want to see algebraic higher category theory. But one will then have to face the question why after all the years that people banged their head against this, no algebraic higher category theory has actually come into existence.

I think André Joyal has been emphasizing for decades that $\left(\infty ,1\right)$-categories ($\simeq$ quasicategories) are good, because “there is category theory for $\left(\infty ,1\right)$-categories”.

That this message was finally picked up by homotopy theorists and algebraic geometers instead of by category theorists is mayby surprising, but not the fault of the homotopy theorists.

Posted by: Urs Schreiber on January 6, 2010 12:13 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Urs, I agree that if two $\left(\infty ,n\right)$-categories live in the same $\left(\infty ,n+1\right)$-category, that to compare the two for equivalence it is enough to compare them in the underlying $\left(\infty ,1\right)$-category, and this is, as you pointed out, a basis to the inductive definitions of $\left(\infty ,n\right)$-categories. On the other hand, if you compare two THEORIES of $\left(\infty ,n\right)$-categories, then this means comparing them with the whole machinery of appropriate functors, transformations etc. rather than comparing two objects in the same surrounding world.

Namely, while the collection of (any sensible sort of) $\left(\infty ,n\right)$-categories likely could be organized into a Quillen category of models, such a model structure presents some information about some $\left(\infty ,1\right)$-category of $\left(\infty ,n\right)$-categories. There is a forgetful (partial groupoidification) functor from $\left(\infty ,n+1\right)$-categories to $\left(\infty ,1\right)$-categories. Thus there may be two substantially different refinements of the $\left(\mathrm{infty},1\right)$-category of $\left(\infty ,n\right)$-categories to a $\left(\infty ,n+1\right)$-category of such. Quillen model categories can certainly not see these differences as they work in the world after forgetting. This is one argument in favour that the categories of models are not appropriate here and why it is important to create collections, as Tom put shortly that the (infinity-) 1-categories form 2-categories (your remark that this is a theorem is a construction which holds in one setup, for each definition/setup there may be in principle even more than one such sensible construction, or no sensible one).

In general, the model categories, while old, standard and successful, have indeed some principal deficiences for ultimate goals. First of all there is no developed notion of higher model categories (say a model 2-category should “have an underlying $\left(\infty ,2\right)$-category” (that is they should be a presentation of some such categories), and model categories should organize into a special model 2-category of model categories). Second the generality is just intermediate: not all $\left(\infty ,1\right)$-categories come as the simplicial nerves of the categories of cofibrant-fibrant objetcs of some model categories; sometimes one can pass to the homotopy category and do derived functors when one has less than a model category and (Grothendieck and Heller derivators and, more funamentally,) recent Cisinski’s “categories derivables” capture much greater generality; next the internalization of model categories (say in a topos, as shown by Durov), that is, of model stacks, if one wants to include natural examples, leads inevitably to the need, (without much harm in usage!), of (non-symmetric) weakenings of the model stack axioms.

Posted by: Zoran Skoda on January 6, 2010 6:00 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Hi Zoran,

Namely, while the collection of (any sensible sort of) (∞,n)-categories likely could be organized into a Quillen category of models, such a model structure presents some information about some (∞,1)-category of (∞,n)-categories.

Yes, and this $\left(\infty ,1\right)$-category of $\left(\infty ,n\right)$-categories is enough to compare two $\left(\infty ,n\right)$ categories up to equivalence. For example in the initial discussion the question was whether diffrent models für $\infty$-groupoids are equivalent or not.

It’s true that this doesn’t tell you how to construct the $\left(\infty ,n+1\right)$ category of $\left(\infty ,n\right)$-categories.

Thus there may be two substantially different refinements of the (infty,1)-category of (∞,n)-categories to a (∞,n+1)-category of such.

The model structure normally encodes the the $\left(\infty ,1\right)$-core of your higher category. That means we forget all higher, non-inertible cells and not that we throw in inverses to them. (Think of quasi-categories, simplicially enriched categories etc…)

Posted by: Thomas Nikolaus on January 6, 2010 6:25 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

It’s true that if all you have is a model category of $\left(\infty ,n\right)$-categories, it presents only an $\left(\infty ,1\right)$-category and nothing about the $\left(\infty ,n+1\right)$-category of such. But as I pointed out up here, there are model-categorical methods that can be used to get at the noninvertible higher transfors.

a model 2-category should “have an underlying (∞,2)-category” (that is they should be a presentation of some such categories), and model categories should organize into a special model 2-category of model categories

I think the second of those things is very unlikely to ever happen. A lot of people have thought about it, but AKAIK no one has any idea how to do it. However, there are already perfectly good notions of model 2-category in the first sense, namely model categories enriched over any monoidal model structure for $\left(\infty ,1\right)$-categories, such as quasicategories or complete Segal spaces. For instance, a model category that is enriched over the quasicategory model structure “presents” an $\left(\infty ,2\right)$-category: its full subcategory of fibrant-cofibrant objects is enriched over quasicategories, which is certainly one sensible notion of $\left(\infty ,2\right)$-category.

(By the way, there is a third, also completely different, notion of “model 2-category” which perhaps might be better called a “locally-model 2-category,” in which the hom-categories are model categories.)

My point is that closed-monoidal and enriched model categories can give you lots more information than ordinary model categories. I already said that model categories have flaws, but when they do apply they can give you lots of information.

Posted by: Mike Shulman on January 6, 2010 9:08 PM | Permalink | PGP Sig | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

not all (∞,1)-categories come as the simplicial nerves of the categories of cofibrant-fibrant objetcs of some model categories;

Yes, but all presentable $\left(\infty ,1\right)$-categories do. In fact they come from combinatorial model categories.

By the remark here on that page, effectively all model categories may be assumed to be equivalent to a combinatorial one. So model category theory should be thought of as modelling presentable $\left(\infty ,1\right)$-categories.

In fact, I think one can read Lurie’s book backwards, nicely: the very last three theorems in the appendix recall Dugger’s theorem about how every combinaotrial model categories is a left Bousfield localization of one of simplicial presheaves.

Now just look for the intrinsic, model-indendent version of this statement, and find Higher Topos Theory.

Posted by: Urs Schreiber on January 7, 2010 12:16 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

But as I pointed out up here, there are model-categorical methods that can be used to get at the noninvertible higher transfors.

Indeed, notably the Joyal model structure on simplicial sets is naturally enriched as a model category over itself and hence does present a structure that should model the $\left(\infty ,2\right)$-category of all $\left(\infty ,1\right)$-categories.

However, for this I suppose one will have to concede that at the present stage there is little to substantiate that this is not a “dirty model” for an $\left(\infty ,2\right)$-category: because there is no well-developed theory (as far as I am aware at least) of how Joyal-enriched model categories do model intrinsically defined $\left(\infty ,2\right)$-categories in the way that we do know precisely how Quillen-$\mathrm{SSet}$-enriched model categories do model the intrinsic notion of $\left(\infty ,1\right)$-category.

I would be surprised if this theory won’t exist, but at the present it doesn’t, really. What do you say?

Posted by: Urs Schreiber on January 7, 2010 12:50 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

because there is no well-developed theory (as far as I am aware at least) of how Joyal-enriched model categories do model intrinsically defined (∞,2)-categories in the way that we do know precisely how Quillen-SSet-enriched model categories do model the intrinsic notion of (∞,1)-category.

Hm, I don’t think that all $\left(\infty ,2\right)$ categories could be modelled by Joyal-sSet- enriched categories (meaning quasi-category (strictly) enriched categories). In contrast to the fact that all $\left(\infty ,1\right)$-categories could be modelled by Quillen-sSet-strictly enriched categories (meaning Kan-complex enriched). So, to model $\left(\infty ,2\right)$-categories (also called $\infty$-bicategories) we should also consider categories weakly enriched over $\left(\infty ,1\right)$-categories (for example like Rezk does). Isn’t that true?

But because for each Joyal-enriched model category, the subcategory of fibrant-cofibrant objects is strictly enriched over quasi-categories, I don’t expect that those categories play such an imortant role for $\left(\infty ,2\right)$-categories as simplicial model categories do for $\left(\infty ,1\right)$-categories.

Despite this general considerations, as Urs pointed out, we do expect that Joyal enriched model categories provide important examples of $\left(\infty ,2\right)$-categories, right?

Posted by: Thomas Nikolaus on January 7, 2010 1:39 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I actually would expect that all $\left(\infty ,2\right)$-categories could be modeled by categories strictly enriched over quasicategories. My intuition is that this should follow from the same sort of “strictification” which shows that any $\left(\infty ,1\right)$-category can be modeled by a category strictly enriched over Kan complexes, or that any ${A}_{\infty }$-space is equivalent to a strict topological monoid. In general it seems as though you can always make at least “one level” of composition strict. But, as Urs says, we don’t yet have a formal comparison between quasicategory-enriched categories and some other notion of $\left(\infty ,2\right)$-category.

Posted by: Mike Shulman on January 7, 2010 3:47 PM | Permalink | PGP Sig | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Looks like you are right. In $\left(\infty ,2\right)$-Categories and the Goodwillie Calculus I Jacob Lurie compares several models for $\left(\infty ,2\right)$-categories among which are especially Joyal-sSet-enriched categories.

But thats very good, because we know how to produce a quasicategory enriched category out of a Joyal-simplicially enriched model category (namely by taking the full subcategory of fibrant-cofibrant objects).

Posted by: Thomas Nikolaus on January 7, 2010 8:44 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Thomas wrote:

In (∞,2)-Categories and the Goodwillie Calculus I Jacob Lurie compares several models for (∞,2)-categories among which are especially Joyal-sSet-enriched categories.

Thanks, Thomas, I forgot about that. So I take back my statement that Joyal-enriched model categories are presently not fully understood as models for $\left(\infty ,2\right)$-categories. They are.

I added a quick note on this to the Lab here:

Posted by: Urs Schreiber on January 8, 2010 3:30 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Good points all. I’m not sure, though, that people have banged their heads for years against developing algebraic higher category theory — I don’t know whether people have really been trying that hard. (Although maybe they have been, and I just don’t know about it since nothing has ever come of it.) And actually, now it seems that there are people making progress on it; possibly what we really needed was impetus from successful applications of non-algebraic higher category theory to make it worth the effort to develop the algebraic version.

Posted by: Mike Shulman on January 6, 2010 4:51 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

And actually, now it seems that there are people making progress on it;

Could you make this more specific? I’d be really interested. Which references are you thinking of?

When I asked Todd Trimble about this here, he kindly started providing very interesting details on his personal web (see towards the bottom of the page).

Posted by: Urs Schreiber on January 6, 2010 5:26 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Here’s one.

Posted by: Mike Shulman on January 6, 2010 9:32 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Mike wrote:

There are lots of details that haven’t been worked out, but I think it’s an exaggeration to say that “We haven’t got anything like a full understanding of what it means for two theories of $n$-categories to be equivalent.”

I hope we don’t spend too much energy arguing over whether “we haven’t got anything like a full understanding” of this issue.

Is a cup that’s 1/3, or 2/3, or 3/4 full “not anything like” a full cup? Where do we draw the line? Do we even need to know? Personally I’d rather know exactly how much milk is in the cup.

Luckily your conversation is shedding a lot of light on that! And I think the conflict between the way topologists and category theorists think of these issues is very fruitful, as long as it keeps pushing people to prove more theorems.

What I’d really like is a theory of $\infty$-groupoids that allows me to treat them simply as topological spaces if that’s what I want, but also enable me to unpack all their Postnikov invariants and see them in great detail as various kinds of ‘-ators’, the way we can already do for homotopy 2-types and 3-types. In other words: easily switch back and forth from ‘topological’ to ‘category-theoretic’ intuitions. Kan complexes are somewhere in between the two extremes.

And, since wishes are cheap, I’d also like something similar for $\left(\infty ,n\right)$-categories. For example, I wish I could treat an $\left(\infty ,1\right)$-category as something very similar to a topological space, but with ‘one-way paths’ — paths you can’t necessarily reverse. Directed algebraic topology is supposed to tackle this kind of issue, but I don’t think they’ve come up with a model of $\left(\infty ,1\right)$-categories yet.

Posted by: John Baez on January 4, 2010 9:21 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Directed algebraic topology is supposed to tackle this kind of issue, but I don’t think they’ve come up with a model of $\left(\infty ,1\right)$-categories yet.

Are “they” trying? My impression has mostly been that people who study DAT are more interested in constructing fundamental (maybe $\left(\infty ,1\right)$-) categories from directed spaces, rather than (in the other direction) showing that $\left(\infty ,1\right)$-categories can be modeled by directed spaces.

I’ve thought a tiny bit about how such a thing might go, but of the various notions of “directed space” none ever seems to be quite right, especially regarding how products should behave.

Posted by: Mike Shulman on January 4, 2010 10:47 PM | Permalink | PGP Sig | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Mike wrote:

Are “they” trying?

I don’t know! You may know more about “them” than I do.

My impression has mostly been that people who study DAT are more interested in constructing fundamental (maybe (∞,1)-) categories from directed spaces, rather than (in the other direction) showing that (∞,1)-categories can be modeled by directed spaces.

Well at least that’s a start. But eventually the concept of $\left(\infty ,1\right)$-category could provide a useful ‘anchor’. There are lots of concepts of directed space floating around, and people seem to like arguing about which one is ‘right’. I’ve never understood the ground rules in these arguments. But someone could argue their concept is ‘right’ when they get some sort of equivalence between their directed spaces and $\left(\infty ,1\right)$-categories.

Ain’t it weird? On the one hand we have this notion of ‘homotopy type’ defined topologically, and everyone is struggling to find nice concepts of $\infty$-groupoid and check that these concepts match the notion of homotopy type.

And on the other, we have this concept of $\left(\infty ,1\right)$-category, and everyone should be struggling to find nice notions of ‘directed homotopy type’ defined topologically, and checking that these notions match the concept of $\left(\infty ,1\right)$-category.

But maybe nobody is.

And of course these two problems couldn’t possibly be related.

Posted by: John Baez on January 4, 2010 11:16 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Concerning people who work on establishing a directed homotopy hypothesis – along the lines indicated for instance at $\left(n,r\right)$-category – homotopy theory – John wrote:

But maybe nobody is [working on this].

Somebody is. When I asked about this a few weeks back on the CatTheory mailing list I received a private reply from somebody telling me that he is working on it with somebody else and that he is “pretty sure that things will work out”.

I am not sure if these somebodys want me to reveal their names in connection with this claim. But I think at least one of them reads this blog.

Posted by: Urs Schreiber on January 5, 2010 11:02 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I certainly agree that the homotopy hypothesis seems to be notably easier, and less interesting, for non-algebraic notions of $n$-category. I think saying that “topological spaces are not a notion of $n$-groupoid” is too strong, though, since they do model all homotopy types—but it’s certainly reasonable to restrict the question to notions of $n$-groupoid that are algebraic in some precise sense.

It does seem to me that you’re being a bit inconsistent with your meaning of “algebraic.” The explanation you refer to says nothing about monadicity; it just says that we have “the” composite rather than “a” composite. This is perfectly true for topological spaces, as I said: any two paths or homotopies certainly have a specified composite, given by traversing one and then the other at double-speed. It’s when using “algebraic” in this sense that I don’t see as clear a distinction.

On the other hand, mere monadicity over $\infty$-graphs doesn’t seem to me to be quite enough to reduce things to the “truly algebraic” notions. The category of 1-categories is certainly monadic over $\infty$-graphs, and 1-categories model all homotopy types via Thomason’s model structure, but we presumably don’t consider them to be a “truly algebraic” notion of $\infty$-groupoid. It seems to me we also need to say something about how the “homotopy theory” of a notion of $\infty$-groupoid must be derived from the $\infty$-graph structure, in order to have a “truly algebraic” notion.

Posted by: Mike Shulman on January 2, 2010 4:10 PM | Permalink | PGP Sig | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I think the word “$n$-groupoid” has different mental associations for you and me, Mike. For you it seems to have various associations to do with topology. For me it doesn’t — at least not immediately. Of course it’s true that there are deep, interesting, etc., relationships between groupoids and topology. But as far as I’m concerned that’s not the primary nature of $n$-groupoids.

I view $n$-groupoids as simply an abstract structure, just like $n$-categories. Various connections can be made between them and other parts of mathematics, including topology. But I think there’s a danger here. The success of category theory has depended on the fact that it has its own independent existence, and its own language, independent of any other part of mathematics. If it had adopted, say, the language of algebraic topology (where it was born) then the connections between category theory and logic, computer science, analysis, physics, etc etc, may never have been realized.

So, I’m a bit worried that in the excitement about connections between higher category theory and homotopy theory, people are getting carried away and downplaying the independent nature of higher category theory — trying to absorb it into homotopy theory. This strikes me as dangerous.

I think saying that “topological spaces are not a notion of $n$-groupoid” is too strong

I stand by what I wrote, for the reasons above. (There’s another reason too: many topological spaces are of the type that homotopy theory usually likes to ignore — e.g. many spaces that arise as spectra, in any sense of the word other than the homotopical one.)

It does seem to me that you’re being a bit inconsistent with your meaning of “algebraic” […] for topological spaces, as I said: any two paths or homotopies certainly have a specified composite, given by doing one and then the other at double-speed

I disagree. You can specify a composite, but it’s not intrinsically specified.

The topological space itself doesn’t tell you to specify that rule of composition. You could specify some other rule (e.g. do the first one at $p$-speed and the second at $q$-speed, for any $p,q$ with $1/p+1/q=1$), and there’s nothing in the topological space saying that one is better than the other.

In any case, as I said, I was using the words “algebraic” or “non-algebraic” to describe theories of $n$-category and $n$-groupoid, and as far as I’m concerned that doesn’t include topological spaces.

mere monadicity over $\infty$-graphs doesn’t seem quite enough to reduce things to the “truly algebraic” notions. The category of 1-categories is certainly monadic over $\infty$-graphs.

Right, and it’s a perfectly respectable algebraic theory. It is, of course, a terrible notion of $\infty$-groupoid! And you could write down a billion other monads on $\infty$-Graph whose algebras are nothing like what we imagine $\infty$-categories or $\infty$-groupoids to be.

It seems to me we also need to say something about how the “homotopy theory” of a notion of $\infty$-groupoid must be derived from the $\infty$-graph structure, in order to have a “truly algebraic” notion.

Again, I disagree. As I’m using the word “algebraic”, at least, whether a theory is algebraic has nothing whatsoever to do with how well it fits any criteria (homotopical or otherwise) for being a decent theory of $\infty$-groupoids.

I’m puzzled by this exchange, because I know you know very well what “algebraic” means in its various standard senses, both formal and informal, and I thought there was widespread understanding of the “a/the” distinction — even if there are those, such as Ross Street, who don’t think it’s as significant as I do. Maybe it’s not as widespread as I thought.

So maybe you could tell me something about your general intuitions. What kind of thing is an “$n$-groupoid” for you? Would you use the word “algebraic” to describe theories of $n$-groupoid or $n$-category? If so, how would you use it?

Posted by: Tom Leinster on January 2, 2010 10:50 PM | Permalink | Reply to this

### Algebraicity of definitions of n-category

I think the word “$n$-groupoid” has different mental associations for you and me, Mike. For you it seems to have various associations to do with topology.

I don’t think so, although I can see how you got that impression. I agree that $n$-groupoids are simply an abstract structure, like $n$-categories. It just so happens that (I believe) one of the many equivalent ways to define them is by using topological spaces.

Of course, when you have two or more equivalent characterizations of the same thing, there’s always room for argument about which is the “definition” and which is the “theorem.” But I believe a reasonable “category-theoretic” attitude is to say that when you have two equivalent categories (or higher categories), either one is a perfectly good definition of the “same” thing. If they look very different a priori, that just means each has a chance of furnishing unexpected useful insights about the other. I’d rather focus on that aspect, instead of arguing about which is the “right” definition. (I’ve expressed elsewhere the same opinion about “material” versus “structural” set theory.) I certainly agree that higher category theory shouldn’t be absorbed into homotopy theory—but I also think higher category theorists can profit by understanding and making use of all the techniques that homotopy theorists have developed.

So, if we agree that Kan complexes are one possible definition of $\infty$-groupoid, then since the $\left(\infty ,1\right)$-category of Kan complexes is equivalent to the $\left(\infty ,1\right)$-category of (say) CW complexes, I think it must follow that CW complexes are also a possible definition of $\infty$-groupoid.

(I am a bit more inclined that John is to emphasize the distinction between homotopy equivalence and weak homotopy equivalence, or between the homotopy category of all topological spaces and that of CW complexes. I’ll try to start saying “CW complex” instead of “topological space” when that’s what I mean.)

What kind of thing is an “$n$-groupoid” for you? Would you use the word “algebraic” to describe theories of $n$-groupoid or $n$-category?

I think there are several uses of the word “algebraic” here that may be getting conflated. I feel that I have a clear intuitive notion of “weak $n$-category.” This intuitive picture of an $n$-category is “algebraic” in the intuitive sense that there is the specified composite of morphisms. However, this the is meant in the higher-categorical sense of the “generalized the”, which means that “the” composite only needs to be specified up to unique specified equivalence (where of course the equivalence is only “uniquely specified” up to uniquely specified equivalence, etc.). According to this philosophy, we can speak of “the” product of two objects in a category, because it is specified up to unique specified isomorphism.

Now, I also agree that it makes some sense to separate the proposed definitions of $n$-category into two classes. I believe that all of them are trying to capture the above “higher-categorically-algebraic” notion, but some of them are “1-categorically algebraic” while others are not. In the “1-algebraic” definitions, “the” composite of morphisms is uniquely specified up to equality, rather than just up to equivalence. By contrast, in the “non-1-algebraic” definitions we merely define what it means to have “a” composite, hoping to then prove that such composites are in fact uniquely specified up to equivalence. I guess you like to give these more formal definitions, such as “1-algebraic” = “monadic over a presheaf category” (or maybe over globular sets), and “non-1-algebraic” = “full subcategory of a presheaf category.”

I would expect, however, that in any correct 1-algebraic definition, we could modify “the” specified composite of any pair of morphisms along a specified equivalence to obtain a different (but equivalent) $n$-category. So “the” composite is still not really determined up to more than unique equivalence; we’ve just chosen a particular one, out of all the equally valid composites, to call “the” composite. Moreover, as you noted here, we can often use a “nerve” construction to convert a 1-algebraic definition into an equivalent non-1-algebraic one, in the sense that the category of algebras for any suitably nice monad on a presheaf category is equivalent to a full subcategory of some other presheaf category.

On the other hand, in any non-1-algebraic definition, we are free to choose “a” particular composite of each pair of morphisms and call it “the” composite, thereby obtaining a “1-algebraic” notion such as that of “algebraic Kan complex.” We can also make this precise in a general way: if the existence of composites is defined by lifting (or “fibrancy”) conditions in some presheaf category ${\mathrm{Set}}^{{C}^{\mathrm{op}}}$, then Garner’s general construction provides a monad $R$ on ${\mathrm{Set}}^{{C}^{\mathrm{op}}}$ (the “fibrant replacement” monad for the corresponding natural WFS) such that a presheaf satisfies the lifting conditions if and only if it admits some $R$-algebra structure (generally non-unique). If we then regard $R$-algebras as a 1-algebraic notion of $n$-category, then $R$-algebra maps can be regarded as “strict functors,” and arbitrary maps in ${\mathrm{Set}}^{{C}^{\mathrm{op}}}$ between $R$-algebras as “weak functors;” thus the category of $R$-algebras and weak functors is equivalent to the original category of “fibrant objects” and maps between them. Hence we can also often convert a non-1-algebraic definition into an equivalent 1-algebraic one.

In conclusion, while there are undoubted technical differences in how we define and work with the 1-algebraic and non-1-algebraic definitions, I find it “obvious” that they are both trying to capture the intuitive “$n$-algebraic” notion. In particular, if they don’t turn out to be the same, then one or both of them is wrong. For example, it might turn out, for some proposed non-1-algebraic definition of $n$-category, that composites are not determined up to unique specified equivalence—but I would regard this as merely evidence of a mistake in that proposed definition.

for topological spaces, as I said: any two paths or homotopies certainly have a specified composite, given by doing one and then the other at double-speed

I disagree. You can specify a composite, but it’s not intrinsically specified.

The topological space itself doesn’t tell you to specify that rule of composition. You could specify some other rule (e.g. do the first one at $p$-speed and the second at $q$-speed, for any $p$, $q$ with $1/p+1/q=1$), and there’s nothing in the topological space saying that one is better than the other.

That’s quite true. But it’s equally true of Todd’s definition of weak $n$-category, which (I hope you agree?) is 1-algebraic, but in which any two morphisms are equipped with as many composites as there are ways to parametrize the gluing of two intervals. In fact, hardly any 1-algebraic definition of $n$-category has only one specified composite. For instance, in a bicategory, two morphisms $f:x\to y$ and $g:y\to z$ have a specified composite $\left(g\circ f\right):x\to z$, but we can also regard $\left(g\circ \left({1}_{y}\circ f\right)\right)$ as a composite of $g$ and $f$, as well as $\left(g\circ f\right)\circ \left(\left({1}_{x}\circ {1}_{x}\right)\circ {1}_{x}\right)$, and so on—in fact, $f$ and $g$ have infinitely many specified composites. The same is true for most globular operads.

Of course, CW complexes (or topological spaces) are not “1-algebraic” in the precise sense of being monadic over a presheaf category. (Although they’re closer than one might naively think—all CW complexes are sequential spaces, which form a reflective subcategory of subsequential spaces, which in turn is a “Grothendieck quasitopos” and hence a reflective subcategory of a presheaf category. Of course the presheaf category in question is not an intuitively sensible “shape category” such as simplicies, globes, or cubes.) But I think there’s a good case to be made that they are a “1-algebraic” definition of $\infty$-groupoid in the loose intuitive sense of having composites specified up to equality.

mere monadicity over $\infty$-graphs doesn’t seem quite enough to reduce things to the “truly algebraic” notions. The category of 1-categories is certainly monadic over $\infty$-graphs.

Right, and it’s a perfectly respectable algebraic theory. It is, of course, a terrible notion of $\infty$-groupoid!

No, actually, it’s a perfectly good notion of $\infty$-groupoid. If you define a functor between 1-categories to be an “equivalence” when it induces a homotopy equivalence of geometric realizations of nerves, then the resulting simplicial localization of $\mathrm{Cat}$ is equivalent to that of $\mathrm{Top}$ or $\mathrm{SSet}$. This is what Thomason’s model structure says.

Maybe you think this is cheating because these “equivalences” are not the “natural” notion of “equivalence” between 1-categories, and you have to formally invert them or restrict to fibrant-cofibrant objects in order to get the $\left(\infty ,1\right)$-category you want. But the point is that a “definition” of $n$-category is never complete after giving only a 1-category of things you call “$n$-categories.” You always have to specify, additionally, how to get the “correct” or “fully weak” notions of “functor”, “transformation”, and so on. For non-1-algebraic definitions, it often happens that the morphisms in the naturally defined 1-category are a good enough (i.e. “fully weak”) notion of functor, but this is not usually true for 1-algebraic definitions. In particular, to get a “fully weak” notion of functor for the 1-algebraic definitions of $n$-category that use globular operads, you have to either restrict to fibrant-cofibrant objects or invert some “weak equivalences” that are not in general already invertible.

So if “(1-)algebraic” just means “monadic over $\infty$-graphs,” then 1-categories with Thomason’s model structure are just as good a “1-algebraic” definition of $\infty$-groupoid as anything else is. But they aren’t “1-algebraic” in the intuitive sense of having composites of $n$-morphisms specified up to equality, which is why I said that you need to say more in order to give a precise meaning to “1-algebraic definition of $n$-category” that captures the intuitive notion (or, at least, that captures my intuitive notion).

Posted by: Mike Shulman on January 4, 2010 1:57 AM | Permalink | PGP Sig | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Tom Leinster wrote:

I’m a bit worried that in the excitement about connections between higher category theory and homotopy theory, people are getting carried away and downplaying the independent nature of higher category theory — trying to absorb it into homotopy theory. This strikes me as dangerous.

But it’s the other way round! The excitement is about how homotoy theory has been fully absorbed into higher category theory as a special case.

Category theory is independent as ever. But it has been enriched by a powerful toolset for a certain special case that was originally developed under the name of homotopy theory. What appears dangerous to me is downplaying the existence of the applicability of this toolset to higher category theory.

Everybody should embrace the $n$POV on homotopy theory.

Posted by: Urs Schreiber on January 4, 2010 10:40 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Urs wrote:

Everybody should embrace the nPOV on homotopy theory.

That’s exactly the over the top sort of comment you acknowledge got cat theory into trouble. More moderately;

Many would find it useful to be aware of the nPOV on homotopy theory.

Posted by: jim stasheff on January 4, 2010 1:57 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

What’s an \infty-graph?
Posted by: Walt on January 6, 2010 8:05 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

An $\infty$-graph is another name for a globular set.

Posted by: Mike Shulman on January 6, 2010 9:11 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Multiple terminology inhibits comprehension
and infinity as a modifier seems to be losing
clear connotation.

Posted by: jim stasheff on January 7, 2010 2:16 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Multiple terminology inhibits comprehension and infinity as a modifier seems to be losing clear connotation.

Generally this is of course very true, but in the special case we are talking about here the prefixes “$\left(\infty ,n\right)$” are not random qualifiers but refer specifically to the notion of $\left(n,r\right)$-category, which makes sense for $n=\infty$.

Now, the definition of “$\left(n,r\right)$-category” is not a precise definition of one specific object, but rather a general “design specification”. Parts of the discussion we are having here is about how one goes about deciding whether two different realizations of this “design specification” are equivalent.

So it’s not absolutely specific. But much more specific than a random use of the prefix $\infty$. And importantly, it is quite widely used by now.

Posted by: Urs Schreiber on January 7, 2010 3:22 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Mike wrote:

I would argue that the true object of study of homotopy theory (insofar as it makes sense to use the definite article at all) is not “homotopy types” but rather $\infty$-groupoids.

Of course I believe that someday everyone will realize that homotopy types are $\infty$-groupoids. But I was talking about history, not the future.

Posted by: John Baez on January 1, 2010 10:59 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

But unfortunately, because this was realized rather late, you’ll see that Wikipedia gives a definition of ‘homotopy type’ that’s not the one I mean here.

Luckily, nobody would link to Wikipedia if an $n$Lab entry is around: $n$Lab: homotopy $n$-type.

(I am joking. But the $n$Lab entry gets it right. It deserves to be expaned, though.)

Posted by: Urs Schreiber on January 4, 2010 10:57 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

No, just say for nice spaces with perhaps a footnote or link

Posted by: jim stasheff on January 1, 2010 2:46 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

The argument I gave can also be used to show this much about Kähler differentials:

Suppose $f$ is a smooth real-valued function on the real line and let $d$ denote the Kähler differential — not the usual $d$ in the theory of differential forms!

Then for any point ${t}_{0}$, there is a smooth function $g$ such that

$f\left(t\right)={a}_{0}+{a}_{1}\left(t-{t}_{0}\right)+g\left(t\right)\left(t-{t}_{0}{\right)}^{2}$

So,

$df={a}_{1}dt+\left(t-{t}_{0}{\right)}^{2}dg+2\left(t-{t}_{0}\right)gdt$

So, while I haven’t shown that $df$ is some function times $dt$, I’ve shown it modulo the ideal generated by $\left(t-{t}_{0}{\right)}^{2}$.

Perhaps this is the best I can do. And now that I look, it seems this is closely related to what Serre does. Is that right, Maarten? All this talk of ‘local uniformizers’ makes it sound like Serre is fixing the ill-behaved Kähler differentials ‘at one point’. It would be nice to have a way to fix them globally all at once.

My usual way of tackling this subject is to take the dual of the module of derivations. If we do that starting with the algebra of smooth functions on the real line, we get the usual 1-forms straightaway. For some reason in week287 I thought I could get away with using Kähler differentials instead.

Posted by: John Baez on December 23, 2009 8:58 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I have a proof that d(ex) is not ex dx in the Kahler differentials. I wrote it up at mathoverflow.

http://mathoverflow.net/questions/6074/kahler-differentials-and-ordinary-differentials/9723#9723

Posted by: David Speyer on December 25, 2009 4:13 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I notice that your proof uses a free ultrafilter on a countable set. Now I want to consider whether the Kähler differentials of smooth functions and the smooth differential 1-forms might be isomorphic in some classically false dream universe.

Posted by: Toby Bartels on December 25, 2009 7:15 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

The link for reference #5 is broken.

Posted by: Toby Bartels on December 24, 2009 4:02 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Fixed!

Posted by: John Baez on December 24, 2009 4:19 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

The email version has it correct: http://www.numdam.org/item?id=PMIHES_1977__47__269_0.

Posted by: Toby Bartels on December 24, 2009 4:20 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Hah! Beat you!

Posted by: John Baez on December 24, 2009 4:29 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

You define the standard $n$-simplex as the space of nowhere negative solutions to

${x}_{0}+{x}_{1}+\cdots +{x}_{n}=0,$

which makes it a point. You want

${x}_{0}+{x}_{1}+\cdots +{x}_{n}=1$

Posted by: Toby Bartels on December 24, 2009 4:07 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Happy Festivus!

Posted by: John Baez on December 24, 2009 4:23 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I'm sensitive to this example since this is one of those things where I have a firm opinion as to which is the right way and which is the wrong way. So I was glad to see you do it the right way, but that got me looking closely enough to notice the typo.

(The wrong way is

${x}_{1}+\cdots +{x}_{n}\le 1,$

of course.)

Posted by: Toby Bartels on December 24, 2009 4:35 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Why is that the wrong way? Just curious.

Posted by: Tom E on December 28, 2009 10:45 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Why is that the wrong way?

It is a matter of æsthetics, of course, since the two definitions are equivalent as topological spaces (homeomorphic), even equivalent as smooth manifolds (diffeomorphic), even equivalent as convex spaces (is there a fancy word for that?). But they are not equivalent as metric spaces, and the inequality in $n$ coordinates introduces an asymmetry in the shape that I find repugnant.

On the other hand, the inequality does equip the $n$-simplex with $n$ coordinates, as befits an $n$-dimensional space. On the other other hand, the equation in $n+1$ coordinates simply equips it with $n+1$ different coordinate systems, each with $n$ coordinates, each obtained by dropping one of the $n+1$ coordinates. So that's still OK, I think.

Posted by: Toby Bartels on December 29, 2009 7:06 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

For what it’s worth, I had some diagrams that help demonstrate this “aesthetic” issue in my dissertation. Have a look at page 71.

Posted by: Eric on December 29, 2009 8:42 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Yes, thank you. (That's on pages 62 and 63 of internal numbering, if anybody's lost.)

Posted by: Toby Bartels on December 29, 2009 10:09 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Ah yes, the origin is a distinguished point in the definition via inequality.

Posted by: Tom E on December 29, 2009 2:45 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Would it be possible for someone to please summarize what was learned here? It seems neat, but was a bit over my head. Who knew that $d$ couldn’t pass through infinite sums? :)

Posted by: Eric on December 30, 2009 8:26 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Quoth Eric:

Would it be possible for someone to please summarize

I’ll do my best. Hopefully an expert will step in to point out any screw-ups.

First, what do we mean by $d{e}^{t}\ne {e}^{t}dt$ in the Kähler differentials? It’s much harder for two Kähler differentials to be equal than for two $1$-forms to be equal. With $1$-forms, the operator $d$ is the exterior derivative—for clarity, let’s call it ${d}_{\mathrm{ext}}$—which obeys a lot more constraints than the different operator $d$ in the Kähler differentials (let’s call that ${d}_{\mathrm{Kah}}$).

In particular, ${d}_{\mathrm{ext}}$ is required to act smoothly as well linearly and in accordance with the Leibniz law. By contrast, ${d}_{\mathrm{Kah}}$ only needs to act linearly and Liebnizially; it knows nothing about smoothness. Using only linearity and Leibniz, you can work out the differential of a polynomial function $P\left(x\right)$ of $x$ in terms of $dx$. But you need (at least) continuity to extend that from polynomial functions to arbitrary ${C}^{\infty }$ functions. (Do you need smoothness?)

So in the ${C}^{\infty }$-module of Kähler differentials, the function ${e}^{t}$ must have a Käher derivative ${d}_{\mathrm{Kah}}{e}^{t}$. There must also exist a Kähler differential that comes from multiplying ${d}_{\mathrm{Kah}}t$ by ${e}^{t}$, viz. ${e}^{t}{d}_{\mathrm{Kah}}t$. But you can’t prove these things are equal using only linearity and Leibniz and so, given the rules of Kähler differentials, they aren’t equal: the symbols $d{e}^{t}$ and ${e}^{t}dt$ refer to different objects.

That’s basically it. It’s easier for typographically different expressions denoting $1$-forms to denote the same object than it is for typographically different expressions denoting Kähler differentials to do so; there are more of the latter.

Posted by: Tim Silverman on December 30, 2009 1:54 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Tim wrote:

But you need (at least) continuity to extend that from polynomial functions to arbitrary ${C}^{\infty }$ functions. (Do you need smoothness?)

Continuity is enough. And this works on ${ℝ}^{n}$. Let’s put a topology on ${C}^{\infty }\left({ℝ}^{n}\right)$ where a sequence (or more general ‘net’) ${f}_{\alpha }$ converges to $f$ if on any compact subset, $D{f}_{\alpha }$ converges uniformly to $Df$, where $D$ is any operator like this:

$D={\partial }_{{x}_{1}}^{{k}_{1}}\cdots {\partial }_{{x}_{n}}^{{k}_{n}}$

and ${\partial }_{{x}_{i}}$ is short for $\frac{\partial }{\partial {x}_{i}}$.

In this case we say ‘${f}_{\alpha }$ and all its derivatives converge locally uniformly to those of $f$’.

In this topology, the polynomials on ${ℝ}^{n}$ are dense in ${C}^{\infty }\left({ℝ}^{n}\right)$. And in this topology, the partial derivative operators

${\partial }_{{x}_{i}}:{C}^{\infty }\left({ℝ}^{n}\right)\to {C}^{\infty }\left({ℝ}^{n}\right)$

are continuous. So, with some extra work, we see any derivation of ${C}^{\infty }\left({ℝ}^{n}\right)$ is continuous in this topology. And with a little more work, you can put a very similar topology on the differential forms on ${ℝ}^{n}$, and then the $d$ operator is continuous in this topology. So, in this topology you can compute $df$ by writing $f={\mathrm{lim}}_{\alpha }{f}_{\alpha }$ where ${f}_{\alpha }$ are polynomials, and doing

$df=d\underset{\alpha }{\mathrm{lim}}{f}_{\alpha }=\underset{\alpha }{\mathrm{lim}}d{f}_{\alpha }$

I’m not sure this fully answers your question, but maybe it helps.

Posted by: John Baez on December 31, 2009 6:23 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Tim wrote:

But you need (at least) continuity to extend that from polynomial functions to arbitrary ${C}^{\infty }$ functions. (Do you need smoothness?)

Continuity is enough. And this works on ${ℝ}^{n}$. Let’s put a topology on ${C}^{\infty }\left({ℝ}^{n}\right)$ where a sequence (or more general ‘net’) ${f}_{\alpha }$ converges to $f$ if on any compact subset, $D{f}_{\alpha }$ converges uniformly to $Df$, where $D$ is any operator like this:

$D={\partial }_{{x}_{1}}^{{k}_{1}}\cdots {\partial }_{{x}_{n}}^{{k}_{n}}$

and ${\partial }_{{x}_{i}}$ is short for $\frac{\partial }{\partial {x}_{i}}$.

In this case we say ‘${f}_{\alpha }$ and all its derivatives converge locally uniformly to those of $f$’.

In this topology, the polynomials on ${ℝ}^{n}$ are dense in ${C}^{\infty }\left({ℝ}^{n}\right)$. And in this topology, the partial derivative operators

${\partial }_{{x}_{i}}:{C}^{\infty }\left({ℝ}^{n}\right)\to {C}^{\infty }\left({ℝ}^{n}\right)$

are continuous. So, with some extra work, we see any derivation of ${C}^{\infty }\left({ℝ}^{n}\right)$ is continuous in this topology. And with a little more work, you can put a very similar topology on the differential forms on ${ℝ}^{n}$, and then the $d$ operator is continuous in this topology. So, in this topology you can compute $df$ by writing $f={\mathrm{lim}}_{\alpha }{f}_{\alpha }$ where ${f}_{\alpha }$ are polynomials, and doing

$df=d\underset{\alpha }{\mathrm{lim}}{f}_{\alpha }=\underset{\alpha }{\mathrm{lim}}d{f}_{\alpha }$

I’m not sure this fully answers your question, but maybe it helps.

Posted by: John Baez on December 31, 2009 6:33 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Eric,

It is a matter of the definition of Kaehler differentials: if $A$ is a commutative ring, then you look at the module ${\Omega }_{K}\left(A\right)$ over $A$ generated by $da$, subject to relations $d\left(ab\right)=\left(da\right)b+adb$ and $d\left(a+b\right)=da+db$. In particular there are only finite sums in the module of Kaehler differentials, and the usual $df={f}^{\prime }dt$ works only if $f$ is a polynomial in $t$, say, if $A$ is ${C}^{\infty }\left(R\right)$ or power series.

The point of Kaehler differentials is a universality property: if $X:A\to M$ is any derivation from $A$ to a module $M$, then there is a unique $A$-module morphism $\mu :{\Omega }_{K}\left(A\right)\to M$ such that $X\left(a\right)=\mu \left(da\right)$, and conversely, for each $A$-module morphism $\mu :{\Omega }_{K}\left(A\right)\to M$ there is a derivation from $A$ to $M$, satisfying that relation.

Now the discussion was about the relation between these algebraic Kaehler differentials, and the usual one forms for $A$, were we do have $df={f}^{\prime }dt$. Let us call ${\Omega }^{1}\left(A\right)$ the usual 1-forms. Then we concluded that in case $A={C}^{\infty }\left(R\right)$ any $A$-module morphism $\mu :{\Omega }_{K}\left(A\right)\to M$ factors through the ordinary 1-forms:

$\mu :{\Omega }_{K}\left(A\right)\to {\Omega }^{1}\left(A\right)\to M$

in case $M$ is free, in particular if $M=A$. This is equivalent to the fact that all derivations of the ${C}^{\infty }$ functions are of the form $f\left(t\right)d/dt$. This is not true for power series, there there are crazy derivations such that $X\left(t\right)=1$, but $X\left({e}^{t}\right)=0$.

There is also a relation between the Kaehler differentials, ordinary 1-forms and the double dual of the Kaehler differentials, but maybe somebody else can summarize that.

Posted by: Maarten Bergvelt on December 30, 2009 2:20 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Would it be possible for someone to please summarize what was learned here?

Would it be possible for that someone to please copy his summary into Kähler differential?

Probably not. So Eric, since you asked the question and got the answer, could you archive it in that entry?

Posted by: Urs Schreiber on December 30, 2009 3:58 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Urs:

Would it be possible for someone to please summarize what was learned here?

Would it be possible for that someone to please copy his summary into Kähler differential?

I’ll try and add it later today.

Posted by: Maarten Bergvelt on December 30, 2009 4:25 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Eric wrote:

Would it be possible for someone to please summarize what was learned here?

For what it’s worth, this whole discussion arose when Maarten spotted a mistake in This Week’s Finds — namely, the place where I said Kähler differentials for the algebra of smooth functions are the same as 1-forms. Now it’s all completely clear. So, now I’m ready to correct that mistake and add the information uncovered to the Addenda of week287. I’ll do it in the next hour or two. As always, people are welcome to lift information from This Week’s Finds and put it in the nLab, though I’d appreciate a reference and link.

Posted by: John Baez on December 30, 2009 5:58 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Would it be possible for someone to please summarize what was learned here?

Kähler differentials are what you get if you start with the ring of smooth functions on the real line (or on some other manifold) and consider what you get as ‘differentials’ of these functions if you only allow yourself to use the purely finitary algebraic properties of differentials. In contrast, $1$-forms are what you get as differentials of these functions if you allow yourself to use all the properties of differentials. We were confused about how different these are.

One thing that we have have learnt is that every Kähler differential has a corresponding 1-form and every $1$-form arises in this way (which should be obvious, at least on the real line), but distinct Kähler differentials may lead to equal $1$-forms (which we might expect but which was not quite as easy to prove). In particular, $\mathrm{d}\left({\mathrm{e}}^{t}\right)$ and ${\mathrm{e}}^{t}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t$ are distinct as Kähler differentials, even though they are equal as $1$-forms.

We have also learnt that there is still a nice algebraic way to get $1$-forms out of the ring of smooth functions (at least on the real line): we form the space of Kähler differentials and then take the double dual (as a module over the ring of smooth functions). That is the point that I found somewhat surprising; I knew that something like this was supposed to be true, but it wasn't clear to me how. Then the operation in the previous paragraph (from Kähler differentials to $1$-forms) may be understood as the inclusion of a module into its double dual (although this ‘inclusion’ is not injective, which is possible since the ground ring is not a field).

And now I have a question: If we impose the finitary algebraic relations obeyed by Kähler differentials and additionally impose a requirement that $\mathrm{d}$ may pass through infinite sums, then do we get $1$-forms? This is a tricky question, since there are a few different ways even to make the last requirement precise (too strict and even $1$-forms will not satisfy it).

Posted by: Toby Bartels on December 30, 2009 4:55 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Toby wrote:

We have also learnt that there is still a nice algebraic way to get 1-forms out of the ring of smooth functions (at least on the real line)…

In fact this ‘nice way’, which I call $\Omega \left(A\right)$ in week287, works for any smooth manifold.

I needed to check this fact when I was writing Gauge Fields, Knots, and Gravity, because that book uses this ‘nice way’ to define differential forms! I figured out how it works for the real line, but I got stuck for ${ℝ}^{n}$. (From then on, the proof is easy.) The problem is generalizing the ‘Taylor series trick’ which lets us take any smooth function $f$ and any point ${x}_{0}$ and write $f\left(x\right)$ as the sum of a constant, a linear function, and a smooth function times something that’s quadratic in $\left(x-{x}_{0}\right)$. This trick makes it obvious that as soon as we know what a derivation does to the linear functions, we know what it does to all functions! So there aren’t any weird derivations that don’t come from vector fields.

I was trying to generalize this trick using multivariable Taylor series, but it wasn’t working. So, I asked Jim Dolan for help, and he proved it. He used a multivariable version of ‘Hadamard quotients’. I describe the 1-variable version in week287, but not the trickier multivariable version. So, I challenge the world to find this multivariable version, which makes it obvious that every derivation of smooth functions on ${ℝ}^{n}$ comes from a smooth vector field.

Interestingly, Hadamard quotients play a big role in the work of Anders Kock and Gonzalo Reyes on synthetic differential geometry! Kock lectured on this in Ottawa conference on smooth structures in logic, category theory and physics. He considered only the one-variable case, but defined a ‘Fermat theory’ to be an algebraic theory that has Hadamard quotients built in. Here are the abstracts of his talks:

Anders Kock: Part 1: Kaehler differentials for Fermat theories

A Fermat theory is an algebraic theory $T$ (in the sense of Lawvere) with a certain property, making “partial differentiation” an intrinsic (canonical) structure in it. Examples are the theory of commutative rings, and also the theory of smooth functions. There is a notion of a module over a Fermat algebra, and of a derivation into such. For a given algebra $A$, there is a universal module ${\Omega }_{A}$ receiving a derivation from it. This is the module of Kaehler differentials. We describe in particular the Kaehler differentials of a free algebra $S\left(V\right)$, which turns out to be the tensor product $S\left(V\right)\otimes V$.

Anders Kock: Part 2: Synthetic meaning of Kaehler differentials.

We describe the context of synthetic differential geometry, and some particularly simple topos models for it, built out of Fermat theories. In this context, the objects (“spaces”) M carry a reflexive symmetrive “neighbour”-relation, in terms of which the notion of differential form can be described in simple combinatorial terms. There exists in such a topos an object $\Omega$ such that differential 1-forms on M are in bijective correspondence with maps $M\to \Omega$. A certain subspace of $\Omega$ similarly classifies closed differential 1-forms, and is therefore a kind of Eilenberg-Mac Lane space $K\left(R,1\right)$.

I’m sure the reappearance of Hadamard quotients in the notion of ‘Fermat theory’ is no coincidence. All this is very fascinating, and it reminds me of your question: could there be a topos in which David Speyer’s proof that Kähler differentials are different from ordinary 1-forms fails?

And now I have a question: If we impose the finitary algebraic relations obeyed by Kähler differentials and additionally impose a requirement that d may pass through infinite sums, then do we get 1-forms? This is a tricky question, since there are a few different ways even to make the last requirement precise (too strict and even 1-forms will not satisfy it).

This is indeed a bit tricky. I think the right way to approach it is this.

We know the canonical map from Kähler differentials ${\Omega }_{K}^{1}\left(A\right)$ to their double dual — the ‘ordinary’ 1-forms ${\Omega }^{1}\left(A\right)$ — is also the surjection we get thanks to the universal property of the Kähler differentials. So, we have two very nice ways of understanding this map. So, there could be a chance of giving a nice description of its kernel. And this kernel consists of the ‘extra relations’ you are wondering about.

So, instead of starting by trying to prove these extra relations all take the form of ‘passing $d$ through an infinite sum’, you might start by simply working out some nice description of these relations!

I think this comment is a clue. And I think this clue is actually related to intuitionistic logic.

Posted by: John Baez on December 30, 2009 11:41 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

When I started reading the first paragraphs about linear tricks, my mind raced immediately to

Then I quickly got lost, but thought I’d share the link in case anyone else hadn’t seen it. Especially since you mention some other related work by Kock.

Posted by: Eric on December 31, 2009 1:17 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

That sounds quite interesting what Anders Kock told there. Did he especially talk about ${C}^{\infty }$-Rings and gave a definition of module and derivations over it (the ingredients to give the universal property of Kaehler-Differentials)?

I’d really like to know whether this can be done, because ${C}^{\infty }$-rings solve some of the problems we have with the ring of smooth functions on a manifold. For instance one problem which is in my opinion the reason for the trouble with Kaehler differentials is that the rings ${C}^{\infty }\left({ℝ}^{n}\right)$ are not free on $n$-generators. I already mentioned this in my first post here, where I raised the question that Kähler-Differentials and 1-forms might be different. In the world of ${C}^{\infty }$-rings we also have a version of ${C}^{\infty }\left({ℝ}^{n}\right)$ (which probably led to the name “${C}^{\infty }$-ring”). And here it is true that this is the free ${C}^{\infty }$-ring on $n$-generators. Now think of algebraic geometry, where the fact that the coordinate ring on ${ℝ}^{n}$ is free on $n$-generators (namely the polynomial ring in $n$ variables) implies that the Kähler-Differentials are also the free module on “d” of this generators. Having a reasonable definition of module and Kaehler-differentials for ${C}^{\infty }$-rings the freeness of ${C}^{\infty }\left({ℝ}^{n}\right)$ should also imply that the module of Kähler-Differentials is free.

So, to subsume what I try to say, the idea is not to give Kähler-Differentials not as an ordinary module over the ordinary ring of smooth functions, but as a ${C}^{\infty }$-module over the ${C}^{\infty }$-ring of smooth functions and that the universal property should then by satisfied in this ${C}^{\infty }$-module category.

An indication for a idea like this could be the following observation which I made, while thinking about how the map of a Module to its bidual could fail to be an injection (my first claim was that the kernel are the torsion elements, but this is obviously wrong if the ring is not an intregral domain like ${C}^{\infty }\left({ℝ}^{n}\right)$). John already mentioned that for projective, finitely generated modules $M$ the map $M\to {M}^{**}$ is an isomorphism. And in fact it is true that projective, finitely generated modules are exactly the fully dualizable objects in the module category over a commutative ring.

Now lets assume that we have a module $M$ for which we know, that ${M}^{**}$ is projective and finitely generated. Think of $M$ as the module of Kähler-differentials over a smooth finite dimensional, compact manifold. Finitely generated, projective modules over ${C}^{\infty }\left(M\right)$ are exactly vector-bundles and we know that 1-forms (i.e. the bidual of Kähler-differentials) are sections into the cotangential bundle which is a vector bundle.

For such an module I claim that ${M}^{**}$ is the universal fully dualizable replacement for $M$. To see this, assume we have a map $f$ from $M$ to a fully dualizable Module $N$. Now we have a unique map ${f}^{**}:{M}^{**}\to {N}^{**}$ rendering the obvious diagramm commutative. But because ${N}^{**}$ is isomorphic to ${N}^{**}$ this is exactly the statement that ${M}^{**}$ is the universal fully dualizable object.

These to things together show that the bidual of the module of Kaehler-differentials for the ring ${C}^{\infty }\left(M\right)$ is the universal fully-dualizbale module with a derivation. Or to reformulate: The universal finite dimensional vector bundle such that the module of sections represents the functor takingderrivations from ${C}^{\infty }\left(M\right)$ into the module of sections of finite dimensional vector bundles.

Posted by: Thomas Nikolaus on January 2, 2010 7:33 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I assume by ‘${C}^{\infty }$-ring’ you mean an algebra of the algebraic theory whose $n$-ary operations are smooth functions from ${ℝ}^{n}$ to $ℝ$. While these are very interesting, Anders Kock took a different approach which doesn’t even rely on the real numbers!

You can see my summary of his talk over here — at our blog entry about the Fields Workshop on Smooth Structures in Logic, Category Theory and Physics.

Posted by: John Baez on January 2, 2010 10:32 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

John replied to Thomas:

I assume by ‘${C}^{\infty }$-ring’ you mean an algebra of the algebraic theory whose $n$-ary operations are smooth functions from ${ℝ}^{n}$ to $ℝ$.

Yes, Thomas means these ${C}^{\infty }$-rings, the basis for all standard Models for Smooth Infinitesimal Analysis.

To expand on the discussion we started here on the $n$Forum:

while Anders Kock may not use these in what you referred to, his general work on synthetic differential geometry finds in particular models in terms of ${C}^{\infty }$-rings. In particular his definition of differential forms in this context applies to models based on ${C}^{\infty }$-rings, as described at infinitesimal singular simplicial complex and differential forms in synthetic differential geometry.

This is not explicitly expressed in terms of derivations. But I believe at the heart of it its the same mechanism that makes both definitions tick: the crucial point is that for ${C}^{\infty }\left(ℝ\right)$ regarded as a ${C}^{\infty }$-ring, the quotient ${C}^{\infty }$-ring ${C}^{\infty }\left(ℝ\right)/⟨{x}^{2}⟩$ is the ring of dual numbers.

This is another way of saying that every smooth function on line may be expanded as $f\left(x\right)=f\left(0\right)+xf\prime \left(0\right)+{x}^{2}h\left(x\right)$ for $h$ smooth. In the geometric picture of synthetic differential geometry this makes the formal dual smooth locus

$D:=\ell \left({C}^{\infty }\left(ℝ/⟨{x}^{2}⟩\right)\right)=\left\{x\in R\mid {x}^{2}=0\right\}$

the standard infinitesimal object with the property, for instance, that for $X$ any smooth space ${X}^{D}$ is its tangent bundle in synthetic incarnation.

Clearly, this is a different, equivalent perspective on the statement that the derivations of ${C}^{\infty }\left(X\right)$ are precisely the vector fields on $X$, proven by expanding every smooth function on ${ℝ}^{n}$ at every point in the above fashion.

Posted by: Urs Schreiber on January 4, 2010 10:10 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Yes, Thomas means these ${C}^{\infty }$-rings, the basis for all standard Models for Smooth Infinitesimal Analysis.

Exactly. And if one goes through the book “Models for smooth infinitesimal analysis” of Moerdijk and Reyes one realizes that they use this Hadamard Lemma at several places to prove important facts (for example that ${C}^{\infty }\left(ℝ\right)$ is free or equally important that ${C}^{\infty }\left(M×N\right)={C}^{\infty }\left(M\right)\otimes {C}^{\infty }\left(N\right),$ where the tensorproduct is just the coproduct in the category of ${C}^{\infty }$-rings.) These are facts which also hold in algebraic geometry for the corresponding objects and make the whole theory work very well. So I’d expect that much nice facts about ${C}^{\infty }$-rings also hold for algebras for an arbitrary fermat theory.

Clearly, this is a different, equivalent perspective on the statement that the derivations of ${C}^{\infty }\left(X\right)$ are precisely the vector fields on X, proven by expanding every smooth function on ${ℝ}^{n}$ at every point in the above fashion.

Yes. But there should also be two such different perspectives on differential forms using some kind of Kaehler differentials in the line of Anders Kock’s work. By the way: The fact that vector fields are exactly derivations of ${C}^{\infty }\left(X\right)$ (of and ordinary module over an ordinary ring) might suggest that we just have to replace the notion of module for an fermat theory but not the notion of derivation (or in some special cases the two notions agree).

Posted by: Thomas Nikolaus on January 4, 2010 12:01 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

But there should also be two such diffrent perspectives on differential forms using some kind of Kaehler differentials in the line of Anders Kock’s work. […] suggest that we just have to replace the notion of module for an fermat theory but not the notion of derrivation (or in some special cases the two notions agree).

I know what you mean. We talked about this before: I had tried some time ago to figure out a good notion of module for a ${C}^{\infty }$-ring. I wanted to give a definition of Chevalley-Eilenberg algebra of an $\infty$-Lie algebroid by saying that it is a certain monoid in cochain complexes of modules over a ${C}^{\infty }$-ring.

I didn’t find a good solution to this problem (which doesn’t mean that it can’t exist) and then at some pointI realized that it is maybe asking the wrong question:

notice how it works in the case that we are discussing:

we start with a manifold $X$, then using $D=𝓁\left({C}^{\infty }\left(ℝ\right)/⟨{x}^{2}⟩\right)$ we form the singular simplicial complex ${X}^{\left({\Delta }_{\mathrm{inf}}^{•}\right)}$ of $X$, a simplicial object in the smooth topos.

Then we take degreewise functions to get the cosimplicial commutative algebra $\left[{X}^{\left({\Delta }_{\mathrm{inf}}^{•}\right)},R\right]$. If we do this externally, we find, using a sequence of theorems by Anders Kock, that the normalized cochain complex of this is the deRham dg-algebra.

Notice that it is only this step of passing from cosimplicial algebras to Moore dg-algebras of (normalized) cochains that introduces the notion or the need for modules. The cosimplical algebra $\left[{X}^{\left({\Delta }_{\mathrm{inf}}^{•}\right)},R\right]$ doesn’t involve the notion of module. But its cochain dg-algebra does. Externally they are equivalent, by the Dold-Kan theorem (externally meaning: not in the topos, i.e. forgetting the smooth structure).

But $\left[{X}^{\left({\Delta }_{\mathrm{inf}}^{•}\right)},R\right]$ makes perfect sense also internally, as a cosimplicial algebra object in the smooth topos. As such it is a ${C}^{\infty }$-incarnation of differential forms. But it doesn’t involve the notion of modules.

In light of this fact at one point I stopped looking for the right notion of ${C}^{\infty }$-modules. But of course I’d be interested to learn of any good proposals. Probably there is a very simple answer, and I am just being dense.

(Back then I tried the notion of co-presheaves on $\mathrm{CartSp}$ with values in modules that respect not all the Lawvere theory structure, but only its restriction to isomorphisms, or something like that.)

Posted by: Urs Schreiber on January 4, 2010 2:12 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

But [X (Δ inf •),R] makes perfect sense also internally, as a cosimplicial algebra object in the smooth topos. As such it is a C ∞-incarnation of differential forms. But it doesn’t involve the notion of modules.

In light of this fact at one point I stopped looking for the right notion of C ∞-modules. But of course I’d be interested to learn of any good proposals. Probably there is a very simple answer, and I am just being dense.

Yes, I know that we could get around the notion of ${C}^{\infty }$-module if we are just interested in differential forms and the way you sketch is very elegant . But there might be other reasons to be interested in ${C}^{\infty }$-modules. For example if we are interested in smooth pull-push operations (aka Fourier-Mukai-transforms) which are an important tool in the algebraic context. So we could take differential forms as a test case for our definition of ${C}^{\infty }$-modules, because we already know how the answer should look like in this case.

Note that we have a discussion on the other blog entry about modules for fermat theories and operads. Maybe you see how this fits toghether or know what Anders Kock has in mind when he talkes about modules for Fermat theories.

Posted by: Thomas Nikolaus on January 4, 2010 3:48 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Yes, I know that we could get around the notion of $C$-module if we are just interested in differential forms and the way you sketch is very elegant . But there might be other reasons to be interested in ${C}^{\infty }$-modules.

Right, so when we talked about this last time, my next reply was that we can express modules often/sometimes in terms of objects of overcategories, and that one can use this here and look at overcategories of internal cosimplicial algebra objects.

By the way, we should be looking at Lurie’s Deformation theory which is all about Kähler differentials and their generalizations.

Notice that he discusses results along the lines I just indicated. For instance corollary 1.74 on page 36.

A good place to start with this is probably example 8.6 on page 25 of Stable $\left(\infty ,1\right)$-categories.

Posted by: Urs Schreiber on January 4, 2010 4:10 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I wrote:

Notice that he discusses results along the lines I just indicated.

Just for the record, I started collecting some material at

This tries to summarize also how Kähler differentials et al fit into the picture (all following JL’s article, of course ).

In particular, notice how the notion of tangent $\left(\infty ,1\right)$-category encodes the notion of modules over arbitrary objects, without ever talking about actions. Instead, it identifies categories of modules over an object exactly with the overcategories of that object – or rather with their stabilization.

In this vein, one can immediately write down a definition of cotangent complexes of ${C}^{\infty }$-rings $A$, as objects in stabilized over-$\left(\infty ,1\right)$-categories over $A$ in, I think, the opposite of the $\left(\infty ,1\right)$-category of cosimplicial algebras internal to the smooth topos.

And I’d think this gives the “right” answer. I had talked about this before at some places. Mainly, what I was missing was the passage to the stabilization. More later, have to run now.

Posted by: Urs Schreiber on January 4, 2010 7:00 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

It would be good if nLab: deformation theory
gave some indication _up front_ of why it’s labeled
deformation theory.

Posted by: jim stasheff on January 5, 2010 2:59 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Jim stasheff wrote:

It would be good if nLab: deformation theory gave some indication up front of why it’s labeled deformation theory.

Thanks for your order. The lab elves are pleased to help. Here you go:

Posted by: Urs Schreiber on January 5, 2010 4:10 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Jim Stasheff asks me to announce that there is now also a link to a new write-up of lecture notes by Yan Soibelman in the References-section at $n$Lab: deformation theory.

More reference have been added by David Corfield.

Posted by: Urs Schreiber on January 7, 2010 3:30 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I’m still catching up on everything that happened here, but I’m making progress…

Now, if you allowed $d$ to pass through the infinite sum then I think you do need to impose the commutative relation

$\left(df\right)g=g\left(df\right)$

to get ordinary 1-forms and $df=f\prime dt$.

As a note that I find interesting (and hope someone else might), since Kahler differentials are generated by $da$, then when they have a basis $d{x}^{\mu }$ (when do Kahler differentials have a basis?), the commutative relation can be expressed as

$\left[d{x}^{\mu },{x}^{\nu }\right]=\sum _{\lambda }\stackrel{⟵}{{C}_{\lambda }^{\mu ,\nu }}d{x}^{\lambda }.$

In this paper, I called these

$\stackrel{⟵}{{C}_{\lambda }^{\mu ,\nu }}$

“left structure constants”, which are distinct from “right structure constants” defined by

$\left[d{x}^{\mu },{x}^{\nu }\right]=\sum _{\lambda }d{x}^{\lambda }\stackrel{⟶}{{C}_{\lambda }^{\mu ,\nu }}.$

Now since the ring is commutative, we have

$d\left({x}^{\mu }{x}^{\nu }\right)=\left(d{x}^{\mu }\right){x}^{\nu }+{x}^{\mu }\left(d{x}^{\nu }\right)=\left(d{x}^{\nu }\right){x}^{\mu }+{x}^{\nu }\left(d{x}^{\mu }\right)=d\left({x}^{\nu }{x}^{\mu }\right),$

resulting in

$\left[d{x}^{\mu },{x}^{\nu }\right]=\left[d{x}^{\nu },{x}^{\mu }\right]$

which means the structure constants (left and right) are symmetric in the upper indices, i.e.

$\stackrel{⟵}{{C}_{\lambda }^{\mu ,\nu }}=\stackrel{⟵}{{C}_{\lambda }^{\nu ,\mu }}$

and

$\stackrel{⟶}{{C}_{\lambda }^{\mu ,\nu }}=\stackrel{⟶}{{C}_{\lambda }^{\nu ,\mu }}.$

We can use these structure constants to define another commutative algebra with product

$•:{\Omega }_{K}\left(A\right)\otimes {\Omega }_{K}\left(A\right)\to {\Omega }_{K}\left(A\right)$

defined on basis elements by

$d{x}^{\mu }•d{x}^{\nu }=\sum _{\lambda }\stackrel{⟵}{{C}_{\lambda }^{\mu ,\nu }}d{x}^{\lambda }.$

So the special case we’ve been talking about here is when these structure constants vanish, but there is a whole new path we could follow by considering nonvanishing structure constants.

Happy New Year!

Posted by: Eric on December 31, 2009 9:15 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Hi,

I will probably take this to my personal nLab page to work things out, but wanted to point something out here in the hopes that someone might find it interesting enough to help turn it into something worth putting on the main nLab.

In this discussion we have considered (commutative) Kahler differentials on the real line ${\Omega }_{K}\left(R\right)$. We can also consider (noncommutative) Kahler differentials on the integers ${\Omega }_{K}\left(Z\right)$.

To do this, we want to turn $Z$ into a directed graph so that we can think of each integer $i$ as sitting over a node ${e}^{i}$ with a directed edge denoted by ${e}^{i}\otimes {e}^{i+1}$ between each consecutive integer.

Let $A$ denote a commutative unital $k$-algebra generated by ${e}^{i}$ with product defined on basis elements by

${e}^{i}{e}^{j}={\delta }_{i,j}{e}^{i},$

where ${\delta }_{i,j}$ is the Kronecker delta. The unit element is given by

$1=\sum _{i}{e}^{i}.$

Let $M$ denote the bimodule generated by ${e}^{i}\otimes {e}^{i+1}$ with left multiplication defined on basis elements via

${e}^{i}\left({e}^{j}\otimes {e}^{j+1}\right)=\left({e}^{i}{e}^{j}\right)\otimes {e}^{j+1}$

and right multiplication defined on basis elements by

$\left({e}^{j}\otimes {e}^{j+1}\right){e}^{i}={e}^{j}\otimes \left({e}^{j+1}{e}^{i}\right).$

Note that for $a\in A$ and $m\in M$ we have

$am\ne ma.$

Define a $k$-linear map

$d:A\to M$

given by

$da=\left[G,a\right],$

where

$G=1\otimes 1=\sum _{i}{e}^{i}\otimes {e}^{i+1}$

and we’ve set

${e}^{i}\otimes {e}^{j}={\delta }_{i+1,j}{e}^{i}\otimes {e}^{i+1}.$

The map $d:A\to M$ is a derivation since

$d\left(ab\right)=\left[G,ab\right]=\left[G,a\right]b+a\left[G,b\right]=\left(da\right)b+a\left(db\right).$

This differential may be recognized as the universal differential when expressed in the more familiar way

$da=\left[G,a\right]=\left(1\otimes 1\right)a-a\left(1\otimes 1\right)=1\otimes a-a\otimes 1.$

Let $t\in A$ be a “coordinate” given by

$t=\sum _{i}i\Delta {e}^{i}.$

It follows that

$dt=\sum _{i}\Delta {e}^{i}\otimes {e}^{i+1},$

any $\alpha \in M$ can be written as

$\alpha ={\alpha }_{t}dt$

for some ${\alpha }_{t}\in A$, and $t$ and $dt$ satisfy the commutative relation

$\left[dt,t\right]=\Delta dt.$

This commutative relation separates these finitary (noncommutative) Kahler differentials from the continuum (commutative) Kahler differentials we’ve been discussing and the “finiteness” is measured by the lack of commutativity. In fact, commuting the differential $dt$ from the left to the right involves a translation in $t$, i.e.

$\left(dt\right)t=\left(t+\Delta \right)dt.$

Observation

I just tried to compute $d\left({t}^{n}\right)$ in this finitary framework and found something interesting (but probably totally well known). By simply using Leibniz plus the commutative relation we get

$\begin{array}{rl}d\left({t}^{n}\right)& =\left(dt\right){t}^{n-1}+td\left({t}^{n-1}\right)\\ & =\left(t+\Delta \right)\left(dt\right){t}^{n-2}+td\left({t}^{n-1}\right)\\ & =\dots \\ & =\left[\sum _{r=0}^{n-1}{t}^{r}\left(t+\Delta {\right)}^{n-r-1}\right]dt.\end{array}$

That term in brackets looks like a convolution, which reminds me of something I recently read on the nLab, but can’t find it now :|

I’m pretty sure my algebra is correct because

$\underset{\Delta \to 0}{\mathrm{lim}}\sum _{r=0}^{n-1}{t}^{r}\left(t+\Delta {\right)}^{n-r-1}=n{t}^{n-1}$

as expected.

I found it! It was a recent comment by Urs:

Concerning that table:

since I wrote it, I understood a few more things. There might be a better story to be told here:

it’s all about taking “algebras of functions on an $\infty$-groupoid”, using pointwise or convolution product.

take a Lie $\infty$-groupoid $A$, let $𝔞\subset A$ be its sub-object of infinitesimal morphisms. Take degreewise functions on this, equipped with the pointwise product. The resulting cosimplicial algebra has as its complex of chains a commutative dg-algebra: the Chevalley-Eilenberg algebra of the ${L}_{\infty }$-algebroid $𝔞$.

But take instead functions on $A$ equipped not with the pointwise, but with the convolution product, i.e. the $\infty$-version of the category algebra. This should be the quantization of the previously mentioned CE-algebra (hence account for the entries labeled “Clifford” in the above table).

Could this be related to my convolution appearing above?

That would be kind of cool. It would support my ages old gut feeling that going from continuum differential geometry to finitary (abstract) differential geometry was akin to quantization, where going in the opposite direction, the “continuum limit” is analogous to the “classical limit”.

Posted by: Eric Forgy on December 31, 2009 3:26 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

What’s ‘$\Delta$’ here, Eric? A number, an operator of some sort…?

Posted by: John Baez on January 1, 2010 6:27 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

In what I wrote above, $A$ is a commutative $k$-algebra generated by the nodes ${e}^{i}$ with $i\in Z$ after turning the integers $Z$ into a directed graph (similar to the way $R$ is a directed space).

I introduced a coordinate

$t=\sum _{i}i\Delta {e}^{i},$

which you can read as saying the value of the coordinate at node $i$ is given by

$t\left(i\right)=i\Delta ,$

where $i\in Z$ and $t\left(i\right),\Delta \in k$.

Conceptually, $\Delta$ is the “distance between nodes” or the “lattice spacing” but we do not have a metric here (yet), so it is a “coordinate distance” between nodes.

In the limit $\Delta \to 0$, I think you can say the “calculus” on $Z$ approaches the calculus on $R$.

Posted by: Eric Forgy on January 2, 2010 1:07 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Okay, thanks. So $\Delta$ is a number — that is, an element of your commutative ring $k$. That’s what I was hoping.

Have you considered taking $k$ to be the ring of polynomials in a formal variable $\Delta$, modulo the relation ${\Delta }^{2}=0$? I.e., taking $k$ to be the dual numbers?

This should make $\Delta$ act like an ‘infinitesimal’. It should simplify your formula for $d\left({t}^{n}\right)$ a lot.

One reason I mention this is that it’s analogous to the ‘semiclassical limit’ in quantum theory, where we set terms proportional to ${\hslash }^{2}$ equal to zero.

I think there is an analogy to quantization in what you’re doing. I don’t fully understand it, but I’d try to think of it in terms of deformation quantization.

Posted by: John Baez on January 2, 2010 7:55 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Have you considered taking $k$ to be the ring of polynomials in a formal variable $\Delta$, modulo the relation ${\Delta }^{2}=0$? I.e., taking $k$ to be the dual numbers?

Hmm…

If ${\Delta }^{2}=0$, then I’m not sure if the coordinate

$t=\sum _{i}i\Delta {e}^{i}$

makes sense because then we’d also have ${t}^{2}=0.$

Once you have specified a directed graph, your freedom to specify $d$ is over, i.e. the graph uniquely determines $d$. So however you define your coordinate $t$, then $dt$ follows and you cannot proclaim

$\left[dt,t\right]=\Delta dt.$

You have to actually compute the commutative relation from knowledge of $t$ and hence $dt$. That is why it is so miraculous (to me) that both exterior and stochastic calculus drop out from a binary tree under different limits.

I think Urs has stated somewhere around here and/or the String Coffee Table something along the lines that synthetic differential geometry (where ${\Delta }^{2}=0\right)$ represents a kind of “correct” continuum limit of what we did in our paper, but I’m not sure how that would work.

By the way, I think that term in $d\left({t}^{n}\right)$ looks a little less scary (not much) when you express it more clearly as a convolution:

$\sum _{r=0}^{n-1}{t}^{r}\left(t+\Delta {\right)}^{n-r-1}=\left[{t}^{x}*\left(t+\Delta {\right)}^{x}\right]\left(n-1\right)$

so that

$d\left({t}^{n}\right)=\left[{t}^{x}*\left(t+\Delta {\right)}^{x}\right]\left(n-1\right)dt.$
Posted by: Eric Forgy on January 3, 2010 1:10 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I am very interested in this approach to the stochastic differential calculus. I’ve read your brief survey, but I haven’t fully understood it yet.

One thing I don’t understand is that you first assume non-commutativity, and then derive the Ito formula. Later you say that in the usual approach 0-forms and 1-forms do commute.

So, is this to say that in the usual approach is consistent with the Ito formula, but if you assume non-commutativity you are required to derive the Ito formula?

Posted by: Tom E on January 3, 2010 3:25 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Hi Tim,

In the usual approach to stochastic calculus, functions and differentials commute at the expense of the Leibniz rule and nilpotency of $d$, i.e. in the usual approach to stochastic calculus $d$ is not a derivation.

What Dimakis et al first showed is that allowing noncommutative of 0-forms and 1-forms, you can derive both quantum mechanics and stochastic calculus in such a way that $d$ is a derivation. It is the noncommutativity of 0-forms and 1-forms that gives rise to the Ito formula in this approach. It is actually very pretty in my opinion.

I wrote a lower brow paper on the discrete version

Financial modeling using discrete stochastic calculus

With discrete calculus, the calculus, i.e. commutative relation, is uniquely determined by a directed graph. Dimakis showed that a particular commutative relation gives rise to stochastic calculus, so Urs and I found that a particular graph, i.e. a simple binary tree leads to the commutation relations

$\left[dx,x\right]=\frac{\left(\Delta x{\right)}^{2}}{\Delta t}dt$ $\left[dx,t\right]=\left[dt,x\right]=\Delta tdx,$ $\left[dt,t\right]=\Delta tdt.$

The neat thing is if you first set $\Delta t=\left(\Delta x{\right)}^{2}$ and then let $\Delta x\to 0$, we get the commutative relations that lead to stochastic calculus.

It was an interesting inverse problem. A graph determines a calculus, so find the graph that corresponds to a given calculus.

I think we’ve really only scratched the surface of what can be done with this. In the paper above, I described this approach as a “meta algorithm”. It is an algorithm for generating algorithms.

Posted by: Eric Forgy on January 3, 2010 5:17 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Ah, so I think you’re saying the usual stochastic calculus is not a differential calculus, but that you can actually recover the Ito formula in a differential calculus setting.

In which case, is the first the “abelianisation” of the second in some sense, which is too harsh an operation to allow the derivation property to be preserved?

Posted by: Tom E on January 3, 2010 5:42 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Ah, so I think you’re saying the usual stochastic calculus is not a differential calculus, but that you can actually recover the Ito formula in a differential calculus setting.

Yeah, I think that is a fair thing to say.

In which case, is the first the “abelianisation” of the second in some sense, which is too harsh an operation to allow the derivation property to be preserved?

Experts in stochastic calculus (which I am not) would probably not describe things this way because there is a pretty large machinery that is usually built up before being able to derive the Ito formula, but I think this a more direct route that ties into a large branch of traditional mathematics.

For a general first order calculus we have

$d\left(fg\right)=\left(df\right)g+f\left(dg\right)$

but it turns out the general commutative relation is given by a symmetric $A$-bilinear map

$\left[df,g\right]=C\left(df,dg\right),$

where

$C:{\Omega }^{1}×{\Omega }^{1}\to {\Omega }^{1}$

so that we can write Leibniz in “left component form” as

$d\left(fg\right)=g\left(df\right)+f\left(dg\right)+C\left(df,dg\right)$

or “right component form” as

$d\left(fg\right)=\left(df\right)g+\left(dg\right)f-C\left(df,dg\right).$

Due to noncommutativity of 0-forms and 1-forms we also have

$aC\left(\alpha ,\beta \right)=C\left(a\alpha ,\beta \right)=C\left(\alpha ,a\beta \right)$

and

$C\left(\alpha ,\beta \right)a=C\left(\alpha a,\beta \right)=C\left(\alpha ,\beta a\right),$

but

$aC\left(\alpha ,\beta \right)\ne C\left(\alpha ,\beta \right)a,$

where $a\in A$ and $\alpha ,\beta \in {\Omega }^{1}\left(A\right)$.

You obtain stochastic calculus by setting

$C\left(dx,dx\right)=dt,C\left(dx,dt\right)=C\left(dt,dt\right)=0$

and you obtain Schrodinger equation by setting

$C\left(dx,dx\right)=idt,C\left(dx,dt\right)=C\left(dt,dt\right)=0.$

You get Kahler differentials (I think) by setting

$C\left(dx,dx\right)=C\left(dx,dt\right)=C\left(dt,dt\right)=0.$

Having said this, I don’t think anyone working in stochastic calculus ever really uses commutativity of 0-forms and 1-forms. Instead, they talk about things like Ito integrals and Stratonovich integrals. However, I think of these as merely different ways to express the product rule. Ito comes from the “left component form” and Stratonovich comes from an average of left and right component forms, i.e.

$d\left(fg\right)=\frac{1}{2}\left[\left(df\right)g+g\left(df\right)\right]+\frac{1}{2}\left[\left(dg\right)f+f\left(dg\right)\right],$

where $C$ cancels out, but these are all just different ways of expressing the same thing.

Posted by: Eric on January 4, 2010 1:57 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Thanks Eric. Could you please explain why $C$ is a form on ${\Omega }^{1}$ when you only ever seem to apply it to things like $df$ for $f\in A$? What is the benefit of considering it as a form on ${\Omega }^{1}$ as opposed to a form on $A$?

Also it’s not obvious to me why $\left[\mathrm{df},g\right]$ *does* extend to bilinear form, but maybe this is a basic result in module theory and my module theory is minimal!

I had a quick look in the Dimakis paper but didn’t see anything relevant.

Posted by: Tom E on January 4, 2010 12:58 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Hi Tom,

If you are interested in this stuff, a really good paper I’ve been enjoying rereading lately is

Differential Calculi on Commutative Algebras
H. C. Baehr, A. Dimakis, F. Müller-Hoissen

Have a look at Section 3 (page 8). Since $A$ is commutative

$d\left(fg\right)=d\left(gf\right)⇒\left[df,g\right]=\left[dg,f\right],$

which they claim means that the commutator depends $A$-bilinearly on $df$ and $dg$, i.e.

$\left[df,g\right]=C\left(df,dg\right).$

Now, since ${\Omega }^{1}\left(A\right)$ is generated by $dA$, any element of ${\Omega }^{1}\left(A\right)$ can be written as (possibly of sum of) $\alpha =adf$ for $a,f\in A$ so that

$C\left(\alpha ,\beta \right)=abC\left(df,dg\right).$

I started some notes here:

Noncommutative Stochastic Calculus

Feel free to leave any comments there. It is like a public notebook.

Posted by: Eric Forgy on January 4, 2010 3:28 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Brilliant! I’d skimmed the paper but hadn’t noticed that. I also missed the crucial point that ${\Omega }^{1}$ is generated by $dA$. I’ll try to get some time to read it thoroughly and leave notes in your notebook.

Posted by: Tom on January 4, 2010 3:43 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Okay, I massively rewrote week287 to fix it and explain most of what we learned in our conversations here. Here’s an extract. I’m too lazy to make the typography look nice here — it looks a bit better over there.

The smooth real-valued functions

f: X -> R

on a manifold X form an algebra over the real numbers. In other words: you can add and multiply them, and multiply them by real numbers, and a bunch of familiar identities hold, which are the axioms for an algebra. Moreover, this algebra is commutative:

fg = gf

Starting from this commutative algebra, or any commutative algebra A, we can define differential forms as follows. First let’s define vector fields in a purely algebraic way. Since the job of a vector field is to differentiate functions, people call them “derivations” in this algebraic approach. A “derivation of A” is a linear map

v: A -> A

which obeys the product rule

v(ab) = v(a)b + av(b)

Let Der(A) be the set of derivations of A. This is a “module” of the algebra A, since we can multiply a derivation by a guy in A and get a new derivation. (This part works only because A is commutative.)

Next let’s define 1-forms. Since the job of these is to eat vector fields and spit out functions, let’s define a “1-form” to be a linear map

w: Der(A) -> A

which is actually a module homomorphism, meaning

w(fv) = f w(v)

whenever f is in A. Let Omega^1(A) be the set of 1-forms. Again, this is a module of A.

Just as you’d expect, there’s a map

d: A -> Omega^1(A)

defined by

(df)(v) = v(f)

and you can check that

d(fg) = (df) g + f dg

So, we’ve got vector fields and 1-forms! It’s a bit tricky, but you can prove that when A is the algebra of smooth real-valued functions on a manifold, the definitions I just gave are equivalent to all the usual ways of defining vector fields and 1-forms. One advantage of working algebraically is that we can generalize. For example, we can take A to consist of *polynomial* functions. We’ll use this feature in a minute.

But what about other differential forms? There’s more to life than 1-forms: there are p-forms for p = 0,1,2,…

To get these, we just form the “exterior algebra” of the module Omega^1(A). You may have seen the exterior algebra of a vector space - if not, it may be hard understanding the stuff I’m explaining now. The exterior algebra of a module over a commutative algebra works the same way! To build it, we run around multiplying and adding guys in A and Omega^1(A), all the while making sure to impose the axioms of an algebra, together with these rules:

f (dg) = (dg) f

(df) (dg) = - (dg) (df)

The stuff we get forms an algebra: the algebra of “differential forms” for A, which I’ll call Omega(A). And when A is the smooth functions on a manifold, these are the usual differential forms that everyone talks about!

And then later, the technical stuff:

We can take *any* DGCA and violently kill C_p for all p > 0, leaving a commutative algebra C_0. We can think of this as a forgetful functor

[DGCAs] -> [commutative algebras]

And this functor has a left adjoint, which freely generates a DGCA starting from a commutative algebra:

[commutative algebras] -> [DGCAs]

Now, I’ve already told you about process that takes a commutative algebra and creates the DGCA. Namely, the process that takes a commutative algebra A and gives the DGCA of differential forms, Omega(A). So, you might think this left adjoint is just that!

I thought so too, when I was first writing this. But it turns out not to be true - at least not always! The left adjoint gives a slightly *different* kind of differential forms for our commutative algebra A. Let’s call these the “Kaehler forms” Omega_K(A).

The Kaehler 1-forms are usually called “Kaehler differentials”. We can can build them as follows: take the A-module generated by symbols

df

for f in A, with these relations:

d(cf) = c df

d(f + g) = df + dg

d(fg) = f dg + (df) g

where f and g are in A and c is in our field.

Let’s call the Kaehler differentials Omega_K^1(A). By how we’ve set things up, they’re blessed with a map

d: A -> Omega_K^1(A)

f |-> df

And this map is a “derivation”, meaning it satisfies these 3 familiar rules for derivatives:

d(cf) = c df

d(f + g) = df + dg

d(fg) = f dg + (df) g

But here’s the cool part: the Kaehler differentials are the *universal* A-module with a derivation. In other words, suppose M is any A-module equipped with a map

v: A -> M

that’s a derivation in the above sense. Then there’s a unique A-module homomorphism

j: Omega_K^1(A) -> M

such that

v = j d

The proof is easy: just define j(df) = v(f) and check that everything works!

Thanks to this universal property, Kaehler differentials are much beloved by algebraists. So, it’s natural to wonder if they’re the same as the 1-forms Omega^1(A) that I explained above!

As it turns out, these 1-forms are the double dual of the Kaehler differentials:

Omega^1(A) = Omega_K^1(A)**

Sometimes we get

Omega^1(A) = Omega_K^1(A)

and this case it’s easy to check that

Omega(A) = Omega_K(A)

But sometimes the 1-forms and the Kaehler differentials are *different*. Let me explain why. It’s technical, but fun if you’re already familiar with some of these ideas.

For starters, let me explain what I mean! We’ve got a commutative algebra A. If we have an A-module M, its “dual” M* is the set of all A-module maps

w: M -> A

The dual becomes a module in its own right by

(gw)(f) = g w(f)

So, we can take the dual of the dual, M**. And then there’s always a module homomorphism

j: M -> M**

given by

j(f)(w) = w(f)

for f in M, w in M*. Sometimes j is an isomorphism: for example, when M is finitely generated and projective. But often it’s not. And that’s where the subleties arise.

If you look back at my definition of 1-forms, it amounted to this:

Omega^1(A) = Der(A)*

The universal property of Kaehler differentials gives us this:

Der(A) = Omega_K^1(A)*

Putting these facts together, we get

Omega^1(A) = Omega_K^1(A)**

So, we always have a module homomorphism

j: Omega_K^1(A) -> Omega^1(A)

This is *both* the map we always get from a module to its double dual, *and* the map we get from the universal property of Kaehler differentials.

Now, here’s the tricky part. This map j is always a surjection. And it will be an *isomorphism* when the Kaehler differentials are a finitely generated projective module. But it won’t *always* be an isomorphism!

For example, when A is the algebra of rational polynomials on a simplex, Omega_K^1(A) is a finitely generated projective module: in fact it’s the free module with one generator dx_i for each independent coordinate. So in this case we actually get an isomorphism

Omega^1(A) = Omega_K^1(A)

and thus

Omega(A) = Omega_K(A)

More generally, this is true whenever A is the algebraic functions on a smooth affine algebraic variety, by the same sort of argument. So in this case, you don’t need to worry about the niggling nuances I’m rubbing your nose in here.

But when A is the algebra of smooth functions on a manifold, the 1-forms are *not* the same as the Kaehler differentials!

Indeed, let A be the algebra of smooth functions on the real line. Then one can show

j: Omega_K^1(A) -> Omega^1(A)

is not one-to-one. In fact, David Speyer showed this after Maarten Bergvelt noticed I was being overoptimistic in assuming otherwise. You can see Speyer’s argument here:

9) David Speyer, Kaehler differentials and ordinary differentials, Math Overflow.

He shows that in Omega_K^1(A), d(e^x) is not equal to e^x dx. The intuition here is simple: showing these guys are equal requires actual calculus, with limits and stuff. But Kaehler differentials are defined purely algebraically, so they don’t know that stuff!

However, turning this idea into a proof takes work. It can’t be as easy as I just made it sound! After all, Omega^1(A) was *also* defined purely algebraically, and in here we *do* have d(e^x) = e^x dx. Indeed, is *why* Speyer’s argument shows that

j: Omega_K^1(A) -> Omega^1(A)

fails to be one-to-one.

So now you should be wondering: how do we know d(e^x) = e^x dx in Omega^1(A)? Since Omega^1(A) is the dual of the derivations, to show

d(e^x) = e^x dx

we just need to check that they agree on all derivations. The hard part is to prove that any derivation of A is of the form

v(f) = g f’

for some g in A, where f’ is the usual derivative of f. Then we have

d(e^x)(v) = v(e^x) = g e^x = e^x v(x) = (e^x dx)(v)

so we’re done!

(Here x is the usual function by that name on the real line - you know, the one that equals x at the point x. Sorry - that sounds really stupid! But anyway, the derivative of x is 1, so v(x) = g.)

So here’s the hard part. Say we have a derivation v of the algebra A of smooth functions on the real line. Why is there a function g such that

v(f) = g f’

for all functions f? As you can guess from my parenthetical remark, we should try

g = v(x)

So, let’s prove

v(f) = v(x) f’

We just need to check they’re equal at any point x_0. So, let’s use a kind of Taylor series trick:

f(x) = f(x_0) + (x - x_0) f’(x_0) + (x - x_0)^2 h(x)

Here it’s utterly crucial that h is a smooth function on the real line. Check that yourself!!! Then, apply the derivation v and use the three rules that derivations obey:

v(f)(x) = v(x) f’(x_0) + 2(x - x_0) v(x) h(x) + (x - x_0)^2 v(h)

Then evaluate both sides at x = x_0. A bunch of stuff goes away:

v(f)(x_0) = v(x) f’(x_0)

Since this was true for any point x_0, we indeed have

v(f) = v(x) f’

as desired.

Sneaky, huh? The argument looked “purely algebraic” - but only because we could pack all the calculus into the utterly crucial bit that I made you check for yourself. By the way, this utterly crucial bit uses the theory of “Hadamard quotients”: if f is smooth function on the real line then

(f(x) - f(y))/(x - y)

extends to a smooth function on the plane if we define it to be the derivative of f when x = y.

A fancier version of this argument works for R^n. This in turn gives the usual proof that derivations of the algebra A of smooth functions on a manifold X are the same as smooth vector fields. And that, in turn, guarantees that Omega(A) as defined algebraically matches the ordinary concept of differential forms on X. The Kaehler forms are different, but as we’ve seen, there’s a surjection of DGCAs

j: Omega_K(A) -> Omega(A)

sending any function f in Kaehler land to the same function f in ordinary differential form land.

So that’s the story! It’s a bit technical, but if we didn’t occasionally enjoy being dragged through the mud of technical details, we wouldn’t like math. I think even more details will become available here:

10) nLab, Kähler differential.

Posted by: John Baez on December 30, 2009 11:01 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Very nice! I had sort of tuned out this whole discussion, but the summary was very comprehensible and drew me back in.

What would you all say is the “right” way to think about the difference between “ordinary” 1-forms and “Kähler” 1-forms? Is one of them “more correct” than the other? Does it have to do with whether we think of our “spaces” as more “smooth” or as more “algebraic”? (Sorry if this was answered somewhere in the long discussion already…)

Posted by: Mike Shulman on December 31, 2009 12:55 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Mike wrote:

Very nice! I had sort of tuned out this whole discussion, but the summary was very comprehensible and drew me back in.

Thanks! I sort of blew the whole day writing it, but it’s all clear in my mind now and it probably won’t be next week. So I guess it’s worthwhile… but it seems more worthwhile if someone actually learns something from it!

Does it have to do with whether we think of our “spaces” as more “smooth” or as more “algebraic”?

I don’t know if I’m ready to draw any big conclusions. For that, I might need to look at some more examples, and I probably don’t want to now.

Anyway… algebraists love Kähler differentials. But for smooth affine algebraic varieties they’re isomorphic to ordinary 1-forms, so that case doesn’t prove much. For nonsmooth algebraic varieties I think this isomorphism breaks down. Anyone who has carefully read Hartshorne could probably explain what goes wrong and what conclusions we should draw. But for smooth manifolds it’s clearly the ‘ordinary’ 1-forms, the dual of vector fields, that we want to use. If we didn’t have

$d{e}^{x}={e}^{x}dx$

all the TAs here at Riverside would be out of work!

I think some experts on logic should check to see if David Speyer really needed the axiom of choice to prove that

$d{e}^{x}\ne {e}^{x}dx$

for Kähler differentials. If he did, that would be very interesting. I’d be sort of shocked.

Posted by: John Baez on December 31, 2009 1:19 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I think some experts on logic should check to see if David Speyer really needed the axiom of choice

Not pretending to be an expert on logic …

He certainly didn't need the full axiom of choice; he used only the weak ultrafilter principle: the existence of a free ultrafilter on $ℕ$. The full ultrafilter principle is much stronger, stating that ever proper filter on every set (such as the filter of cofinite subsets of $ℕ$) is contained in some ultrafilter. And even that is not quite as strong as the full axiom of choice.

Actually, now that I check again, Speyer also uses the existence of a transcendence basis for a field extension of $ℝ$. In full generality, that is probably equivalent to the axiom of choice, although this was a special case. My intuition, however, is that this choice is not necessary; at this stage we've got our hands on what we want and are only checking that ${\mathrm{e}}^{t}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t$ and $\mathrm{d}\left({\mathrm{e}}^{t}\right)$ come out differently.

Posted by: Toby Bartels on December 31, 2009 6:28 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I’m in the process of transferring material to Kahler differential.

In John’s TWF he defines things like

$d\left(fg\right)=\left(df\right)g+f\left(dg\right)$

and

$d\left(f+g\right)=df+dg.$

However, the first line above means we are really working with bimodules. In that case, I think it is important to specify that we also want

$\left(df\right)g=g\left(df\right)$

in order to make things work out.

I keep getting stuck on this because the bimodules I am most interested in (and I hope the nLab remains general enough to accommodate them) are bimodules over a commutative unital ring, but the important and defining property is

$\left(df\right)g\ne g\left(df\right).$

Should we make

$\left(df\right)g=g\left(df\right)$

explicit or should we instead define Kahler differentials like wikipedia does as a left module

$d\left(fg\right)=f\left(dg\right)+g\left(df\right)?$

My preference, and what I’ve done on the nLab, is to define three relations for Kahler differentials

$d\left(fg\right)=\left(df\right)g+f\left(dg\right),$ $d\left(f+g\right)=df+dg,$ $\left(df\right)g=g\left(df\right).$

Then we can more easily refer to the general noncommutative case as the page currently does.

What do you think?

Posted by: Eric on December 31, 2009 7:53 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

My preference, and what I’ve done on the nLab, is to […]

What do you think?

I agree with that.

Since Kähler differentials are commutative, we may have to figure out where the noncommutative case really belongs; your noncommutativity seems rather orthogonal to Kähler differentials' discontinuity. But it belongs somewhere.

Posted by: Toby Bartels on December 31, 2009 2:51 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Eric: it’s perfectly fine to consider $fdg\ne \left(dg\right)f$, but then you’re not talking about Kähler differentials. You’re talking about some quotient of the the universal differential envelope. For example, a first-order differential calculus (see the definition on page 1).

These other structures are very interesting. But it’ll only confuse people to talk about them in the section on Kähler differentials.

In John’s TWF he defines things like

$d\left(fg\right)=\left(df\right)g+f\left(dg\right)$

and

$d\left(f+g\right)=df+dg$

However, the first line above means we are really working with bimodules.

Yes, because ‘everyone knows’ that given a left module $M$ of a commutative algebra $A$, it automatically becomes a bimodule by defining $ma=am$. People who work on commutative algebra take this for granted.

Of course you don’t have to do this — other bimodule structures can also be useful. But it’s the default assumption, so I figured that drawing attention to it would be more distracting than helpful. Of course I did think about you when writing this equation — but I just gritted my teeth and moved on.

Posted by: John Baez on December 31, 2009 6:37 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I didn’t plan on discussing the noncommutative version on the Kahler differential page, but I think it might worth noting there more explicitly that the Kahler differentials are a special case of a general first-order differential calculus (on a commutative algebra) where 0-forms and 1-forms commute.

Differential Calculi on Commutative Algebras
H. C. Baehr, A. Dimakis, F. Müller-Hoissen

It’s been years since I read this paper, which explains the sense of deja vu I was having. I think some of the results you guys have rediscovered are contained in this paper for the more general case and it is pointed out the Kahler differentials are a special case where

$\left(df\right)g=g\left(df\right).$
Posted by: Eric Forgy on January 1, 2010 1:44 PM | Permalink | Reply to this
Weblog: The n-Category Café
Excerpt: A brief introduction to generalized multicategories, in honor of a new draft of a paper about them.
Tracked: January 4, 2010 2:19 AM

### Re: This Week’s Finds in Mathematical Physics (Week 287)

Another dirty solution — though slightly less dirty than the previous one — is to use model categories.

Maybe there was justification for calling model categories a “dirty solution” at some point in the past. But hardly today.

A model category structure is no more dirty as a way to describe an $\left(\infty ,1\right)$-category than, say, a presentation by generators and relations is. In fact, by Dugger’s theorem, we know that at least for combinatorial model categories, it’s precisely the same thing.

Posted by: Urs Schreiber on January 4, 2010 11:18 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I am not sure if we ever said the following explicitly here:

if we do take Kähler differentials of ${C}^{\infty }\left(X\right)$ regarded as a ${C}^{\infty }$-ring as described by Kock-Dubuc and summarized and referenced at Fermat theory, then the definition does coincide with the usual one.

I added a brief remark to this effect now also to the entry on Kähler differentials. But more should be said. It seems Kock-Dubuc don’t say this fully explicitly, but they do discuss in section 3 of their article that the ${C}^{\infty }$-Kähler differentials do reproduce the definition in forms as functions on infinitesimal neighbours of points, which is known to give the right answer

Thanks to Thomas amplifying this.

Posted by: Urs Schreiber on January 14, 2010 9:19 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

And let me maybe add what this means more concretely:

For a ${C}^{\infty }$-ring $R$ the ordinary Module of Kaehler-Differentials can be constructed by taking the free Module on elements $\mathrm{dr}$ and quotienting out the relations $d\left(r\cdot s\right)=r\mathrm{ds}+s\mathrm{dr}$ as mentioned by John in the entry.

But the Module of ${C}^{\infty }$-Kaehler-Differentials for $R$ is constructed by taking the free module on generators $\mathrm{dr}$ but imposing the relations

$d\left(f\left({r}_{1},\dots ,{r}_{n}\right)\right)=\sum _{i=1}^{n}\frac{\partial f}{\partial {x}_{i}}\left({r}_{1},\dots ,{r}_{n}\right){\mathrm{dr}}_{i}$

for each smooth map $f:{ℝ}^{n}\to ℝ$. And this is the module of 1-forms. Thus its clear what the map from (standard) Kähler-Differentials to 1-forms (=${C}^{\infty }$-Kaehler-Differentials) does.

Posted by: Thomas Nikolaus on January 14, 2010 12:15 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 287)

I have spent quite a bit of time working on the entry $n$Lab: Kähler differential.

The motivation was that I had pointed out the general idea of Kähler differentials at MathOverflow to somebody here. But when I then checked the entry, i found it didn’t actually convey the right general picture.

Have a look at what I did, I did invest quite a bit of time in order to bring both the fully general nonsense $n$POV, but lead up to it gently and understandably.

So there is a big “Idea and definition” section now that is meant to explain what is really going on, in the large and in the small.

Then the previous content of the entry, on Kähler differentials over ordinary rings and over smooth rings regarded as ordinary rings, I made subsections in a section titled “Specific definitions”.

I added more subsections to this. A stubby one with a pointer to the ${C}^{\infty }$-ring case that is discussed in detail at the new version of Fermat theory, and then a bit on modules over monoids in general abelian categories, which is what the MO question had been about.

Posted by: Urs Schreiber on January 26, 2010 2:18 AM | Permalink | Reply to this
Read the post Algebraic Models for Higher Categories
Weblog: The n-Category Café
Excerpt: On (oo,1)-categories of algebraic oo-groupoids and algebraic (oo,1)-categories.
Tracked: March 2, 2010 10:02 PM

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