The n-Category Café

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!

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!

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 $\{ (x_0, x_1, \ldots, x_n) : x_0 + \cdots + x_n=1, x_i \ge 0\}$. For $i =0$, $1$, …, $n$, write $\partial_i (\Delta)$ for the face of $\Delta$ where $x_i=0$ and $\partial(\Delta) = \bigcup \partial_i(\Delta)$.

Observe that any polynomial function $f$ on $\Delta^n$ can be realized as the restriction of a polynomial function $F$ on $\mathbb{R}^{n+1}$. There is an integer $D(f)$ such that, if $d \geq D(f)$, 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 $F'$ 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 $F-F'$ which was $0$ on $\Delta$. Homogeneity then forces $F-F'$ to be $0$ on the positive orthant, which means that $F-F'$ is the zero polynomial $0$.

For $i=0$, …, $n$, let $f_i$ be a polynomial function on $\partial_i(\Delta)$, 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 $\{ x_k \}_{k \neq i}$. Note that $f_i|_{\partial_i(\Delta) \cap \partial_j(\Delta)}$ is lifted by plugging in $0$ for $x_j$ in $F_i$.

Using the uniqueness of lifting, we see that the polynomials $F_i|_{x_j=0}$ and $F_j|_{x_i=0}$ 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 $F|_{x_i=0} = F_i$.

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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 $\mathbb{R}^2$. 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.

If you are interested in learning more about this issue, the keyword is “seminormal”.

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 $F|_{x_i = 0} = F_i$.

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

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.

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?

Tim vB wrote:

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?

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 $|S|$, 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.

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.

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

$$d(s t) = s(d t) + t(d s)$$

to

$$d(s t) = s(d t) + (d s)t.$$

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.

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 $d s$ 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 $d s$ on the right by $t$.

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

I don’t know any situations where $d(f g) = (d f) g + f (d g)$ is wrong and $d(f g) = (d f) g + (d g) f$ 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,

$d(f g) = (d f) g + f (d g)$

is always morally superior to

$d(f g) = (d f) g + (d g) f$

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 $(A,A)$-bimodule. Then a derivation of $A$ valued in $B$ is a linear map

$$f : A \to B$$

such that

$$d(f g) = (d f) g + f (d g)$$

And then there’s still a way to define Kähler differentials such that they’re the ‘universal $(A,A)$-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

$$d(f g) = (d f) g + f (d g)$$

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

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:

Infinitesimal computations in topology,

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

$$d_H H = [H,H]$$

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.

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.

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 $C^\infty(M)$, lets denote this module by $Der C^\infty(M)$. Furthermore the universal property of the module $\Omega_K(M)$ of Kähler differentials implies that derivations of $C^\infty(M)$ are exactly module morphisms

$$\Omega_K(M) \to C^\infty(M)$$

So we have

$$Der C^\infty(M) = Hom(\Omega_K(M),C^\infty(M)) =: \Omega_K(M)^*$$

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

$$\Omega^1(M) = Hom(Der C^\infty(M),C^\infty(M)) = Der C^\infty(M)^* = \Omega_K(M)^{**}$$

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 $\Omega_K(M) \to \Omega^1(M)$ is then just the canonical map from the module $\Omega_K(M)$ to its bidual.

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

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:

$$\Omega^1(M) = Hom(Der C^\infty(M),C^\infty(M)) = Der C^\infty(M)^* = \Omega_K(M)^{**}$$

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

$$\alpha: \Omega_K(M) \to \Omega^1(M)$$

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

$$\Omega_K(M) \to \Omega_K(M)^{**}$$

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

$$\alpha : \Omega_K(M) \to \Omega_K(M)^{**}$$

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 $\Omega^1(M)$ is a projective module over $C^\infty(M)$, 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 $C^\infty(M)$ 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 $C^\infty(M)$ 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 $M = \mathbb{R}$ and $p = 0$ this ideal would include the function that equals $exp(-1/x)$ 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…