## April 11, 2007

### Quantization and Cohomology (Week 20)

#### Posted by John Baez

In this week’s class on Quantization and Cohomology, we introduce Chen’s "smooth spaces" which generalize smooth manifolds and provide a more convenient context for differential geometry. These will allow us to define "smooth categories" and study the principal of least action starting with any smooth category $C$ equipped with a smooth functor $S: C \to \mathbb{R}$ describing the ‘action’.

• Week 20 (Apr. 10) - Smooth spaces and smooth categories. The concept of a "category internal to $K$" where $K$ is any category with pullbacks. The category of smooth manifolds does not have pullbacks. Grothendieck’s dictum. Chen’s category of smooth spaces. Examples: the discrete and indiscrete smooth structures on a set. Any convex set or smooth manifold is a smooth space. The product and coproduct of smooth spaces. Any subset of a smooth space becomes a smooth space. Homework: the category of smooth spaces has pullbacks.

Last week’s notes are here; next week’s notes are here.

Posted at April 11, 2007 3:08 AM UTC

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1236

### Smooth spaces

I actually like diffeological terminology better than “smooth space”, but regardless of what one calls the notion, I hope your notes increase general awareness of it. It’s another one of those very good ideas (like the Kurzweil-Henstock integral) that hasn’t quite caught on yet.

I wonder if diffeological/smooth spaces fit into Paul Feit’s general scheme of passing from local to global notions:

P. Feit, Axiomization of passage from “local” structure to “global” object, Mem. Amer. Math. Soc. 101 (1993), no. 485.

Posted by: Michael Kinyon on April 11, 2007 4:26 AM | Permalink | Reply to this

### Re: Smooth spaces

I like the word ‘smooth’ better than ‘diffeological’ because it’s one syllable instead of three, English instead of Greek, and generally less scary. However, I’m not sure what the best category of smooth spaces is — or if there even is one. In addition to the Chen approach being used here (equipping a space with maps from nice test spaces), there’s also the Mostow approach (equipping a space with a sheaf of maps from it to nice test spaces). There are also approaches that combine the two ideas.

If there turns out to be a best approach, ultimately I would like the term ‘smooth spaces’ to be used for that.

I haven’t read Paul Feit’s paper, but smooth spaces as Chen defined them are really just special sheaves on the category of convex sets equipped with a certain Grothendieck topology. Sheaves are very much a ‘local-to-global’ kind of construction.

Why are these sheaves ‘special’? Because of axiom 3, which says they have enough global points to make the global points functor faithful. This axiom is thrown in to make sure that smooth spaces act like sets with extra structure! Keeping it is probably just a sign of my reluctance to fully accept the tao of mathematics. If I dropped it — and also dropped the insistence that a smooth space ‘be a set’ — we’d have a better-behaved category of smooth spaces, namely a topos. As it is, we just have a ‘concrete quasitopos’ — or so I’ve been told.

Posted by: John Baez on April 11, 2007 6:30 PM | Permalink | Reply to this

### Re: Smooth spaces

There are also approaches that combine the two ideas.

St. Stolz and P. Teichner, in their work, talk about smooth functors on categories of paths and encounter essentially the same issues that we are using smooth spaces for.

However, instead of using the concept of a smooth space, they use something like categories fibered over manifolds (think: dual point of view to a stack on the site of manifolds), thus defining every operation on arbitrary smooth families of, for instance, paths.

This can be seen at work for instance on p. 17 of the thesis

Florin Dumitrescu, Superconnections and Parallel Transport

and, more pronounced, in the preprint

S. Stolz & P. Teichner, Super symmetric field theories and integral modular forms

(which, however, you are supposed not to read, or at least not to worry about in detail, since it should be superceded by a new version with slightly different concepts in a short while).

The translation between this point of view and the one of smooth spaces is left as an exercises for the inclined $n$-Café reader.

Posted by: urs on April 11, 2007 7:34 PM | Permalink | Reply to this

### smooth structure on Vect

The translation between this point of view and the one of smooth spaces is left as an exercises for the inclined $n$-Café reader.

Here is what I think the answer is:

we may equip the category $\mathrm{Vect}$ of vector spaces (finite dimensional, say, over some given ground field) with a smooth structure, as follows.

Let’s declare that for any (convex/open) subset $U$ of $\mathbb{R}^n$ a map $\phi : U \to \mathrm{Mor}(\mathrm{Vect})$ is a plot, if and only if $\phi$ is the component map of a smooth morphism of two smooth vector bundles on $U$, i.e. if and only if there exist two smooth vector bundles $p : E \to U$ and $p' : E' \to U$ on $U$ together with a smooth morphism $f : E \to E'$ of vector bundles, fixing the base such that $\phi(u) = f|_{u} : E_u \to E'_u \,.$

Clearly, all constant maps into $\mathrm{Mor}(\mathrm{Vect})$ are plots, by this definition, and if $\phi : U \to \mathrm{Mor}(\mathrm{Vect})$ is a plot by way of the existence of a morphism $\array{ E &\stackrel{f}{\to}& E' \\ p \downarrow \;\; && \;\; \downarrow p' \\ U &=& U }$ then for any smooth $w : V \to U$ also $w \circ \phi : V \to \mathrm{Mor}(\mathrm{Vect})$ is a plot due to the existence of the pullback vector bundles $\array{ w^*E &\stackrel{w^*f}{\to}& w^*E' \\ w^*p \downarrow \;\; && \;\; \downarrow w^*p' \\ V &=& V } \,.$

Using this smooth structure on $\mathrm{Vect}$, and the standard smooth structure on the path groupoid $P_1(X)$ of some smooth manifold $X$ (coming from the standard smooth structure on path space $P X = \mathrm{Hom}_{C^\infty}([0,1],X)$) we may try to characterize smooth functors

Proposition: given any smooth vector bundle $E \to X$ on $X$, with connection $\nabla$, the corresponding parallel transport $\mathrm{tra}_\nabla$ is a smooth functor $\mathrm{tra}_\nabla : P_1(X) \to \mathrm{Vect}$ for this smooth structure on $\mathrm{Vect}$. (I think also the converse is true, but won’t elaborate on that here.)

Let’s unwrap what this means:

a plot into $\mathrm{Mor}(P_1(X))$ is any map $\phi : U \to P X$ such that $\mathrm{ev} \circ \phi : U x [0,1] \stackrel{\phi \times \mathrm{Id}}{\to} P X \times [0,1] \stackrel{\mathrm{ev}}{\to} X$ is an ordinary smooth map.

Let $s : U \stackrel{\phi }{\to} PX \stackrel{\sim}{\to} P X \times \{0\} \stackrel{\mathrm{ev}}{\to} X$ and $t : U \stackrel{\phi }{\to} PX \stackrel{\sim}{\to} P X \times \{1\} \stackrel{\mathrm{ev}}{\to} X$ be the two endpoint evaluations, then the image of our transport functor of the above plot is the morphism of vector bundles $s^* E \stackrel{\phi^*\mathrm{tra}}{\to} t^* E$ over $U$.

The full diagram $\array{ s^*E &\stackrel{\phi^* \mathrm{tra}}{\to}& t^*E \\ s^*p \downarrow \;\; && \;\; \downarrow t^*p \\ U &=& U }$ is essentially the diagram on p. 17 of Florin Dumitrescu’s thesis.

I believe similar reasoning allows to reformulate the notion of smooth functors on bordisms with values in vector spaces in terms of categories fibred over smooth manifolds in terms of Chen-smooth functors.

I am expecting that there should be some general abstract nonsense underlying this, which tells me that the Dumitrescu-Stolz-Teichner concept of smooth is necessarily just another formulation of Chen-smoothness in a suitable sense.

Can anybody comment on this?

Posted by: urs on April 17, 2007 12:56 PM | Permalink | Reply to this

### Re: smooth structure on Vect

I wrote:

I am expecting that […]

Well, there is actually nothing mysterious going on. Here are some more details.

Posted by: urs on April 17, 2007 1:59 PM | Permalink | Reply to this

### Re: smooth structure on Vect

When trying to show that Chen-smooth functors $F : 1\mathrm{Cob}_{\mathrm{Riem}} \to \mathrm{Vect}$ with both $1\mathrm{Cob}_{\mathrm{Riem}}$ and $\mathrm{Vect}$ equipped with Chen-smooth structures as above, are the same as the smooth functors $F : \mathrm{RB}^1 \to V$ as considered in that preprint, I am running into a subtle subtlety that is either irrelevant or a real problem:

on the Chen-smooth side I consider maps $\phi : U \to \mathrm{Mor}(whatever)$ to be plots if there is a $whatever$-bundle over $U$ such that $\phi$ sends any point in $U$ to the fiber above it.

On the other hand, in the other approach we send $U$ to the collection of all $whatever$-bundles over $U$.

Now, if I could pick the bundle associated with a plot as above, then it would seem a triviality to show that both notions of smooth functors coincide.

But I can’t. I feel like this should be easy to handle. But right now I am not sure. More details in the last section here.

Posted by: urs on April 17, 2007 5:35 PM | Permalink | Reply to this

### Re: Smooth spaces

In response to my asking Don Davis’ mail list about a text book version of Chenb spaces, I received a response from Hain (already posted here?) and :
From: Andrew Stacey
Date: Mon, 10 Sep 2007 15:53:47 +0200

You will probably get lots of replies about ‘diffeological spaces’, and in
particular references to Patrick Iglesias’ book (see his homepage at:
http://math.huji.ac.il/~piz/Site/Welcome.html). Another approach is Frolicher
spaces. I’m not sure if there is a book about Frolicher spaces, but the book
“A Convenient Setting for Global Analysis” (Kriegl and Michor, available for
linear Frolicher spaces”. There’s quite a lot of geometry in that book.

I think that Frolicher spaces and diffeological spaces (Chen’s spaces) are one
and the same. Frolicher spaces are specified by giving the smooth curves and
smooth functionals, Chen spaces by giving the plots. However, the smooth
curves completely determine the Frolicher structure, and a plot is completely
determined by its restriction to the smooth curves that factor through it.
Thus both are completely determined by declaring a family of curves to be
“smooth”.

Andrew

Posted by: jim stasheff on September 12, 2007 1:38 AM | Permalink | Reply to this

### Re: Smooth spaces

Hi Jim,

We also discussed diffeological spaces, Patrick Iglesias’s book, Chen smooth spaces, and Frolicher spaces last year - for instance at this post, in which I mentioned Andrew Stacey’s viewpoint.

Posted by: Bruce Bartlett on September 12, 2007 2:34 AM | Permalink | Reply to this

### Re: Smooth spaces

We also discussed [X, Y and Z] last year - for instance at this post,

By the way, if we had that Wiki we’d have in there entries (searchable by keyword, I suppose) on X, Y and Z, summarizing that information and maybe referencing the blog discussion it originated from.

We’d simply answer every question on X, Y and Z by pointing to these entries. And whenever new aspects appear, or corrections are being made, we’d work these into the existing entries on X, Y and Z.

This would be good. :-)

Posted by: Urs Schreiber on September 12, 2007 10:06 AM | Permalink | Reply to this

### Re: Smooth spaces

Michael wrote:

I actually like diffeological terminology better than “smooth space”, but regardless of what one calls the notion, I hope your notes increase general awareness of it.

Alas, the definition of diffeological space used by Patrick Iglesias-Zemmour is a bit different from K. T. Chen’s 1986 definition of ‘differentiable space’, which is what I call a ‘smooth space’. And, this difference of definitions seems to make it harder to treat manifolds with boundary using diffeological spaces.

Chen’s 1986 definition allows the domain $C$ of a plot $\phi: C \to X$ to be any convex set of Euclidean space. So, we can take $C$ to be a half-space:

$\{(x_1,\dots,x_n) \in \mathbb{R}^n : x_1 \ge 0 \}$

Plots where $C$ is a half-space are useful for manifolds with boundary.

Iglesias-Zemmour instead uses a definition that requires $C$ be an open subset of Euclidean space. This rules out half-spaces.

This makes a correct treatment of manifolds with boundary a bit tricky. In fact, I’m not sure it’s possible. I’m very curious.

Iglesias-Zemmour says in the second sentence of his book that the definition of ‘diffeological space’ is equivalent to Chen’s definition. However, he cites Chen’s 1977 paper, which actually gives a different definition than the 1986 paper! I think Chen changed his definition for a good reason.

I believe the issue boils down to this. Suppose

$f: [0,1] \to \mathbb{R}$

is a function such that for every smooth function

$\phi: \mathbb{R} \to [0,1]$

the composite $f \circ \phi$ is smooth. Is $f$ smooth?

This is obvious away from the points $0$ and $1$. But is it true at $0$ and $1$? More precisely, does $f$ have continuous one-sided derivatives of all orders at $0$ and $1$?

Note the derivative $\phi^'(x)$ vanishes if $\phi(x)$ is $0$ or $1$. That’s what makes the question annoyingly tricky. If we were allowed to use plots defined on convex sets, like the identity function

$\phi: [0,1] \to [0,1],$

the question would be easy — in fact trivial!

I should figure this out, but I’m feeling lazy. Maybe someone else would enjoy taking a crack at it.

Posted by: John Baez on April 14, 2007 2:22 AM | Permalink | Reply to this

### Re: Smooth spaces

Hmm, maybe it works! Suppose

$f : [0,1] \to \mathbb{R}$

is a function such that

$f \circ \phi : \mathbb{R} \to \mathbb{R}$

is smooth for every smooth

$\phi: \mathbb{R} \to [0,1].$

Suppose $\phi(x) = 0$. This implies $\phi^'(x) = 0$, so knowing the derivative

${d\over d x} f(\phi(x)) = f^'(\phi(x)) \phi^'(x)$

exists tells us nothing about $f^'(\phi(x)) = f^'(0)$. But, knowing the second derivative

${d^2\over d x^2} f(\phi(x)) = f^{' '}(\phi(x)) \phi^'(x)^2 + f^'(\phi(x)) \phi^{' '}(x) = f^'(\phi(x)) \phi^{' '}(x)$

should tell us that $f^'(\phi(x)) = f^'(0)$ exists, since $\phi^{' '}(x)$ needn’t vanish.

This is sort of hand-wavy, and it’s just a small step towards proving the desired claim. But maybe it works.

Has someone carefully studied manifolds with boundary, manifolds with corners, etcetera using techniques from diffeology? I don’t want to suffer a lot just to prove that diffeological spaces correctly reproduce the standard concepts of ‘smooth map’ for such spaces! With Chen’s definition it’s easy.

Posted by: John Baez on April 14, 2007 2:38 AM | Permalink | Reply to this

### Re: Smooth spaces

I don’t think the two definitions (Chen’s, using convex domains, and Iglesias-Zemmour’s, using open domains) are equivalent, and I think the square-root function from the unit interval to the reals is an example showing the difference. This function appears to be smooth using the natural I-Z smooth structure on [0,1] but not smooth using the natural Chen smooth structure.

Posted by: Dan Christensen on April 16, 2007 11:15 PM | Permalink | Reply to this

### Re: Smooth spaces

Dan wrote:

I don’t think the two definitions (Chen’s, using convex domains, and Iglesias-Zemmour’s, using open domains) are equivalent, and I think the square-root function from the unit interval to the reals is an example showing the difference.

That’s what I’d thought from our previous conversations on this kind of issue.

In the comment above, I’d almost changed my mind.

But now, I realize my remarks were flawed: I was assuming $f''(x)$ converged to a finite value as $x \to 0$. This isn’t true for functions like $f(x) = \sqrt{x}$.

So, I should just check that you’re right. Unless I’m missing some tricks, it suffices to consider 1-dimensional plots and show:

For every smooth function $\phi: \mathbb{R} \to [0,1]$, the function $\sqrt{\phi}: \mathbb{R} \to [0,1]$ is also smooth.

It suffices to consider a point $x \in \mathbb{R}$ with $\phi(x) = 0$, and show the square root of $\phi$ is also smooth at $x$.

Anyone want to help out? I’d like to do this calculation, but I need to do another calculation today — my taxes!

Posted by: John Baez on April 17, 2007 12:01 AM | Permalink | Reply to this

### Re: Smooth spaces

I thought my claim was straightforward to check and I was just about to include a proof, but it turns out that I can only handle the one-sided limits. In fact, I take back my claim. If $f(x) = \sqrt{x}$ and $\phi(x) = x^2$, then $f(\phi(x)) = |x|$ which is not smooth, so $f$ is not smooth in the I-Z category. It feels like I’m being caught by a technicality!

Posted by: Dan Christensen on April 17, 2007 3:31 AM | Permalink | Reply to this

### Re: Smooth spaces

Hmm! It would be pretty odd if the theory of diffeological spaces turned out to match the theory of Chen spaces for sneaky reasons like this! But for now, I guess we should simply try to see how it works for the closed unit interval… is every real-valued function on $[0,1]$ that is smooth in the diffeological sense smooth in Chen’s sense?

Posted by: John Baez on April 17, 2007 7:25 AM | Permalink | Reply to this

### Re: Smooth spaces

I had not grokked the fact that Chen uses arbitrary convex subsets of Euclidean space when I sent my reply to Jim on the AlgTop mailing list. Thanks, John, for pointing out my mistake there.

I still think that the definitions are equivalent. Here is an attempt at a proof for your test case of $[0,1]$. I suspect that once this is done, the proof for an arbitrary convex set will be a reasonably straightforward extension.

We want to prove that if $f : [0,1] \to \mathbb{R}$ takes smooth curves to smooth curves then it is smooth in the usual sense. It is clear that it restricts to a smooth function on the open interval $(0,1)$ so all that needs to be shown is that the one-sided derivatives exist at the end points. Clearly if we can show that $f$ is differentiable at the end points then smoothness follows by induction.

Firstly, an easy $\epsilon/3$ argument shows that if $f : [0,1] \to \mathbb{R}$ is a continuous function which is $C^1$ when restricted to the open interval $(0,1)$ and is such that the derivative extends to a continuous function on the whole interval $[0,1]$ then in fact $f$ is $C^1$ on the closed interval $[0,1]$.

Thus we need to show that if $f : [0,1] \to \mathbb{R}$ is such that it takes smooth curves to smooth curves then the limit of $f'(x)$ as $x \to 0$ (from above) exists.

Let $c : \mathbb{R} \to [0,1]$ be a smooth curve such that in a neighbourhood of the origin, $c(x) = x^2$. By assumption, $f \circ c$ is smooth. Its derivative near $0$ is, by the chain rule[1],

(1)$(f \circ c)'(x) = f'(x^2) 2 x.$

Now $f \circ c$ is a smooth even (near the origin) function and so $(f \circ c)'(0) = 0$. As it is smooth, its second derivative exists at the origin and is, by definition, given by

(2)$(f \circ c)'(0) = \lim_{h \to 0} \frac{(f \circ c)'(h) - (f \circ c)'(0)}{h}.$

Substituting in, we see that

(3)$\lim_{h \to 0} \frac{f'(h^2) 2 h}{h}$

exists. As the function $h \mapsto h^2$ is a homeomorphism of the unit interval to itself, it preserves and reflects limits[2]. We can therefore deduce that the limit

(4)$\lim_{h \to 0^+} f'(h)$

exists and thus $f'$ extends to a continuous function on $[0,1)$. Using a similar argument for the other endpoint we deduce that $f$ is of class $C^1$ on the whole of $[0,1]$.

The geometry underlying this is that we can define tangent vectors using second derivatives of curves providing we know that the first derivatives vanish (I’ve seen this used implicitly a few times). Thus when we cannot use first derivatives, we do have a fall-back position to go to. A derivative is “that-which-acts-on-directions”[3], and whichever of the vast number of equivalent definitions of directions we actually use, we still get the same answer to the question “What is the derivative?”

Does the above convince you? Have I missed something obvious?

Andrew

[1] I’ve just been telling my students how important the chain rule is in differential topology. Nice to have an explicit example to tell them!

[2] With no apologies at all for the deliberate abuse of language.

[3] See, teaching is useful. This was our slogan for last week’s lectures.

Posted by: Andrew Stacey on September 13, 2007 10:11 AM | Permalink | Reply to this

### Re: Smooth spaces

Nice argument, Andrew! And nice post all around! Thanks.

You’re only showing $f$ is $C^1$, but I guess we can then apply the argument to $f^'$ and show $f$ is $C^2$, and so on ad infinitum, concluding that $f$ is $C^\infty$.

It’s eerie how you posted your argument 8 minutes after I posted another proof, due to Anders Kock.

Kock’s argument uses an old result of Whitney, but it’s possible Whitney’s proof uses the same argument you describe. I’ll have to check. The thing I like about your argument is that it hints at a possible strategy for proving a much stronger conjecture:

Conjecture: If $C \subseteq \mathbb{R}^n$ is a convex subset and

$f: C \to \mathbb{R}^n$

is a function such that

$f \circ \phi: \mathbb{R}^n \to \mathbb{R}^n$

is smooth for every smooth function

$\phi: \mathbb{R}^n \to \mathbb{R}^n$

then $f$ is smooth in the usual sense.

I don’t really feel sure this conjecture is true, but I think that if it is, the theory of Chen spaces and the theory of diffeological spaces are equivalent.

It would be nice to know, one way or another, whether these theories are equivalent or not.

Posted by: John Baez on September 13, 2007 10:34 PM | Permalink | Reply to this

### Re: Smooth spaces

John Baez said:

You’re only showing $f$ is $C^1$, but I guess we can then apply the argument to $f'$ and show $f$ is $C^2$, and so on ad infinitum, concluding that $f$ is $C^\infty$.

Yes, I mentioned that smoothness would follow by induction.

As for your conjecture, I do think that it is true and I think that it will follow fairly simply from the case for the half closed interval. Here’s a sketch of a proof. Suppose that $C$ and $f$ are as in your conjecture. The interior of $C$ is diffeomorphic to an open subset of Euclidean space and so $f$ is smooth on that. Thus it remains to check the boundary of $C$. The interior of $C$ is open and dense in $C$ and so we need to show that the derivative of $f$, $D f$, extends to a continuous function on the whole of $C$ (higher derivatives will follow by induction). If it doesn’t extend then we can find a sequence of points in the interior of $C$, tending to a point on the boundary of $C$, such that the limit of $D f$ does not exist along this sequence. By passing to a subsequence if necessary, we can find a smooth curve $\gamma$ with domain $[0,1]$ through this sequence such that $0$ is mapped to our point on the boundary and the rest is mapped into the interior of $C$. But then we apply our above result to $f \circ \gamma$ to find a contradiction.

That’s more sketchy than my previous post! Does it still convince anyone?

I’m pondering Michor’s response to my claim about Frölicher spaces and diffeological spaces. I’m not quite ready to assign my claim of equivalence to the dustbin [trashcan] but it’s looking a little shakey.

Andrew

PS Thanks for sorting out my footnotes in my previous post. I noticed on previewing that \emph{…} didn’t convert but missed the one in the footnote, and forgot about the spacing.

PPS I may have posted 52 minutes before your post with the proof from Anders Kock. It depends on how the time stamps are formed; are they standardised to the same time zone? In any case, the café didn’t tell me about your post until today which may have been down to firefox not refreshing the page, or may have been down to the Sheffield Effect affecting one or other of us.

Posted by: Andrew Stacey on September 14, 2007 10:12 AM | Permalink | Reply to this

### Re: Smooth spaces

I wrote:

I believe the issue boils down to this. Suppose

$f: [0,1] \to \mathbb{R}$

is a function such that for every smooth function

$\phi: \mathbb{R} \to [0,1]$

the composite $f \circ \phi$ is smooth. Is $f$ smooth?

Anders Kock sent me an email saying, in part:

I think the answer is yes; the analogous question for the closed half line $[0,\infty)$ should follow from an old theorem of Whitney that any smooth even function $g:\mathbb{R} \to \mathbb{R}$ is of the form $g(x)=h(x^2)$ for some smooth function $h:\mathbb{R} \to \mathbb{R}$.

So let $f$ be a diffeologically smooth function from the closed half line to $\mathbb{R}$. If you apply the assumption in your question to $\phi(x)=x^2$, $f(x^2)$ is a smooth even function, and the $h$, asserted by Whitney, then extends $f$ smoothly to the whole line.

This argument may be found in my old book on Synthetic Differential Geometry; Whitney’s Theorem is in Duke J. Math 1943.

So, it seems that with the diffeology that $[0,\infty)$ inherits from being a subset of $\mathbb{R}$, a smooth function $f: [0,\infty) \to \mathbb{R}^n$ is the same as what we normally consider to be a smooth function. I haven’t checked, but this makes me vastly more optimistic that the same holds if we replace $[0,\infty)$ by $[0,1]$, as in my comments above.

With more work, perhaps we could show the same sort of thing for any manifold with boundary — namely, that it can be given a diffeology by treating it as a subset of a collared version of itself, and then the diffeologically smooth functions to or from it are the same as what we normally consider as smooth.

It seems more challenging to handle manifolds with corners, like cubes and simplices. But this is very crucial for anyone who wants to use diffeology as a general theory of smooth spaces.

Posted by: John Baez on September 13, 2007 10:03 PM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

I have a precise (I hope) question about smooth spaces.

Let $X$ denote the $x$-axis in $R^2$, let $Y$ denote the $y$-axis, and let $P$ denote the parabola defined by $y=x^2$. Is it true that $X\cap Y$ and $X\cap P$ are isomorphic as smooth spaces? (By intersection, I mean the fiber product over $R^2$.)

I ask because in algebraic geometry, they aren’t. The space $X\cap P$ has nilpotent functions on it which “see” the tangency of the curves $X$ and $P$. I would think that you also wouldn’t want them to be isomorphic in smooth geometry.

Posted by: James on April 11, 2007 1:39 PM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

A smooth point is isomorphic to another smooth point if and only if both have the same collection of plots on them.

But can we have two smooth points with differing collection of plots? I’d think not, since by one of the axioms, every constant map is a plot and all maps to the point are constant, so all maps to the point are necessarily plots.

Maybe if one were to relax that axiom, things would be different.

But I don’t think that the point of diffeology is to work like synthetic differential geometry does. Instead of providing us with a concrete notion of ‘infinitesimal’ it rather provides us with a way to rephrase every potential question about infinitesimals in terms of ordinary differential calculus.

Posted by: urs on April 11, 2007 3:41 PM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

Urs is right; there’s only one smooth structure on a point in this setup. With this smooth structure, every function to or from a point is smooth.

As Urs notes, this uses the 3rd axiom in the definition of ‘smooth space’. This the axiom that makes smooth spaces be merely sets equipped with extra structure. In other words, this axiom guarantees that the forgetful functor from smooth spaces to sets is faithful.

This is deliberately different from algebraic geometry or synthetic differential geometry, where we have many nonisomorphic spaces that have only one point. (Or more precisely, one ‘global point’.)

The difference is that we are not trying to give our smooth spaces ‘infinitesimal fuzz’… if you know what I mean.

It’s possible that in the long run this will be regarded as a mistake, but it’s definitely worth exploring.

Posted by: John Baez on April 11, 2007 8:49 PM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

Thanks, Urs and John. What then are the advantages of Chen’s point of view to that of synthetic differential geometry?

Also, suppose we look at unions instead of intersections in my original question. Is the union $X\cup P$ of the $x$-axis and the parabola isomorphic to the union $X\cup Y$ of the $x$-axis and the $y$-axis?

Posted by: James on April 11, 2007 11:44 PM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

What then are the advantages of Chen’s point of view to that of synthetic differential geometry?

Might be that I simply have not run across the existing “exotic” examples of smooth spaces, but my impression is that smooth spaces are really supposed to be nothing overly fancy or mysterious, but just a very slight generalization of the idea of a manifold, such that it is easy to equip common “infinite dimensional” but otherwise perfectly non-pathological spaces with a smooth structure – in particular spaces of smooth maps between smooth spaces.

This is the only advantage that I have ever seen in applications (which of course might simply be a result of the focus on certain applications that I have. I should read that big book on diffeology some day to enlarge my horizon.)

The axioms of smooth spaces may be read like this (which is just very slightly more general than the idea behind the definition of a manifold): we know what it means to do differential calculus on $\mathbb{R}^n$. So we will simply specify for everything that we want to regard as smooth how to pull it back to $\mathbb{R}^n$s in various ways and then work on these $\mathbb{R}^n$s.

Posted by: urs on April 12, 2007 9:46 AM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

Here is something that appears to me to be a disadvantage to Chen’s approach.

Assuming $X\cup P$ is not isomorphic to $X\cup Y$ in my example above (which I would guess is true since they probably have different tangent spaces at the origin), then $X\cup P$ is not the result of gluing $X$ to $P$ along their intersection. More precisely, consider the diagram

(1)$X\cap P \to X\coprod P$

where the single arrow represents the two obvious embeddings. (I couldn’t get that displayed right.) But this diagram is isomorphic to the diagram

(2)$X\cap Y \to X\coprod Y.$

(Same conventions.) Therefore their coequalizers, which is to say the results of the two gluings, are isomorphic.

Posted by: James on April 12, 2007 11:37 PM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

Indeed $X \cup Y$ is not isomorphic to $X \cup P$ in your example.

I believe that a smooth function

$f : X \cup Y \to A$

is the same as a smooth function on $X$ and a smooth function on $Y$ that happen to agree at the point of intersection (the origin). This is a lowbrow way of saying that $X \cup Y$ is a pushout.

I believe a smooth function

$f: X \cup P \to A$

is not just a smooth function on $X$ and a smooth function on $P$ that happen to agree at the origin. For one thing, the derivatives of these two functions must agree at the origin!

I don’t find this problematic. It’s just the way it is.

However, it does mean you can’t build a smooth manifold from a triangulation by taking the obvious colimit of simplices (viewed as smooth spaces).

Posted by: John Baez on April 14, 2007 2:52 AM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

The difference is that we are not trying to give our smooth spaces ‘infinitesimal fuzz’… if you know what I mean.

It’s possible that in the long run this will be regarded as a mistake, but it’s definitely worth exploring.

Currently I tend to think that it is a virtue.

As you know, for quite a while, under the influence of Kock/Breen-Messing, I was trying to do what we need to in synthetic language (this or that).

Probably it’s all just a matter of taste, but currently I have come back to the point where I tend to like the less fancy smooth world better.

For the reason encoded in the following slogan:

infinitesimal is the same as smooth and functorial.

It seems to me that, when properly combined with categorical notions, Chen-like smoothnes neatly does for us all that we would like to hope for.

As an example: one of the motivations for using synthetic reasoning (and also super-reasoning, by the way) is that we want to be able to say that $T$ is the “infinitesimal interval”, that $\mathrm{Hom}(T,X) \simeq T X$ is the tangent space to $X$ and that $\mathrm{Hom}(T X,\mathbb{R}) \simeq \Omega^1(X)$ is the space of 1-forms on $X$.

The axioms of synthetic differential geometry are geared such that this works.

But consider this: the “infinitesimal interval” should be a special case of a path between two points.

But a path between two points is begging to be regarded not as an object, but as a morphism.

So I’d do this: I take the ordinary point $\{\bullet\}$. Regarded as a 0-category, this is just a point. But now regard it as a 1-category (with only the identity morphism on its single object).

What’s the difference? The difference is of course that now maps from the point into something may have paths between them. And in a moment we will see that, albeit all these paths are finite, only their infinitesimal part really matters.

So, I set $\mathrm{Hom}(\{\bullet\},X) := \mathrm{Hom}_{\mathrm{Cat}}( \{\bullet\},P(X)) \,,$ where $P(X)$ is the strict path $\omega$-groupoid of $X$.

Then this “space” of maps is $\mathrm{Hom}(\{\bullet\},X) \simeq P_1(X)$ the path 1-groupoid of $X$.

We know, indeed, that this behaves like the tangent bundle $T X$ to $X$ in the following sense:

smooth morphisms $\mathrm{Hom}_{\mathrm{Cat}_{C^\infty}} (P_1(X), \Sigma \mathbb{R})$ are in bijection with ordinary 1-forms on $X$ $\mathrm{Hom}_{\mathrm{Cat}_{C^\infty}} (P_1(X), \Sigma \mathbb{R}) \simeq \Omega^1(X) \,.$

1-Forms appear here automatically, even though I never used infinitesimals. Everything is finite.

The magic happens when we combine smoothness with functoriality.

Functoriality implies that maps on paths are determined by maps on sub-paths, which are determined by maps of sub-paths, etc. Magically, if the functor is smooth, this automatically knows about “infinitesimal subpaths”, in the sense that we have the theorem that $\mathrm{Hom}_{\mathrm{Cat}_{C^\infty}} (P_1(X), \Sigma \mathbb{R}) \simeq \Omega^1(X) \,.$

And that’s nice. For one, the theorem goes even further: we may replace $\Sigma \mathbb{R}$ with any Lie groupoid and get “Lie groupoid-valued 1-forms” this way. For single-object Lie groupoids this coincides with ordinary Lie-algebra valued 1-forms.

Also, we might want to modify the domain. The domain here just encodes the structure of connectivity in our space $X$. There are lots of other (smooth) categories which we may want to regard as encoding the information of “points and paths between them”.

Posted by: urs on April 12, 2007 10:46 AM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

Is every subset of Rn a smooth space? Are nonmeasurable sets smooth spaces? That would disturb me.

Posted by: Walt on April 11, 2007 5:48 PM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

Is every subset of $\mathbb{R}^n$ a smooth space? Are nonmeasurable sets smooth spaces? That would disturb me.

A smooth set is a set equipped with an extra structure.

Every set admits at least one (boring) choice of that extra structure: the discrete smooth structure where the only smooth maps from (convex if you like, open if you like) subsets of some $\mathbb{R}^n$ into your set are the constant maps. (see also slide number 5 in John’s lecture notes).

So the question is what kind of “interesting” smooth structures a given set admits.

If you have any smooth space, then any of its subsets naturally inherits a smooth structure: that defined in terms of the collection of all plots into the former that factor through the latter.

In this sense every subset of any manifold (every manifold is a smooth space) naturally inherits a smooth structure.

But potentially this will coincide with the boring discrete smooth structure.

For instance the subset $\{0,1\} \subset \mathbb{R}$ of two disjoint points on the line inherits a smooth structure from that of the line, but it is boring.

I’d guess that for generic non-measurable subsets of $\mathbb{R}^n$ the induced smooth structure will turn out to be the boring discrete one, too.

That depends on how many ordinary smooth maps into $\mathbb{R}^n$ factor through your chosen non-measurable subset. I guess generically only the constant smooth maps do.

Posted by: urs on April 11, 2007 6:17 PM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

Walt wrote:

Is every subset of $\mathbb{R}^n$ a smooth space? Are nonmeasurable sets smooth spaces?

Yes.

That would disturb me.

Perhaps this is an argument for calling them ‘diffeological spaces’ or ‘Chen spaces’ instead of ‘smooth spaces’. The point about these spaces is not that they’re smooth sort of like a manifold. The point is that we can recognize when a map between them is smooth.

Suppose $X$ is any smooth space. Even if $S \subseteq \mathbb{R}$ is some horrible nonmeasurable set, we have no trouble recognizing when a function

$f : X \to S$

is smooth: it’s smooth precisely when its composite with the inclusion

$i: S \to \mathbb{R}$

is smooth. In other words: it’s a smooth real-valued function that just happens to take values in $S$.

Similarly, a function

$f : S \to X$

deserves to be called smooth if its composite with every map

$g: Y \to S$

is smooth, for all smooth spaces $Y$. Note that here we are using the fact that we can tell when a function to $S$ is smooth, to decide when a function from $S$ is smooth.

Finally, there’s nothing special about subsets $S \subset \mathbb{R}$ in what I just said. The same stuff is true for any subset of any smooth space!

Posted by: John Baez on April 12, 2007 10:11 PM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

Perhaps this is an argument for calling them ‘diffeological spaces’ or ‘Chen spaces’ instead of ‘smooth spaces’.

Precisely, and I knew if I waited long enough, you would see it for yourself. My guess is that the area generally known as “nonsmooth analysis” (see the book by Clarke et al or the forthcoming one by Wilfred Schirotzek) has nontrivial intersection with the area we are discussing. On the face of it, they seem quite different, but I suspect a deeper look would reveal overlap.

Posted by: Michael Kinyon on April 12, 2007 11:29 PM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

I have heard that one problem with Chen’s spaces is that one gives up the inverse function theorem. Is this correct?

John sort of addressed my other question, but I still don’t quite understand. If one “imporves” the category of manifolds by embedding it into the category of sheaves of sets, why not then consider all sheaves and not just the ones represented by sets? After all, isn’t this what algebraic geometers do when they talk about algebraic spaces?

Finally, does one have a descent inverse function theorem in synthetic differential geometry? If one does, it may be useful in understanding loop spaces of orbifolds as stacks.

Posted by: Eugene on April 12, 2007 4:44 AM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

I have heard that one problem with Chen’s spaces is that one gives up the inverse function theorem. Is this correct?

I would think so. In fact, I would not even know how to state the inverse function theorem for smooth spaces, since it is not obvious what the rank of the differential of a smooth map should be.

Posted by: urs on April 12, 2007 10:57 AM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

understanding loop spaces of orbifolds as stacks.

What I would tend to do is:

think of the orbifold as a Lie groupoid (the action groupoid of the orbifold group action) and then consider paths in that Lie groupoid (or loops, if desired).

These would be generated from ordinary paths in the space of objects combined with morphisms of the groupoid itself. There is an obvious Chen-smooth structure on this, where a plot into such orbifold-paths is any map that arises as a collection of plots into paths in the space of objects and into the space of morphisms, postcomposed with the formal composition of these.

Posted by: urs on April 12, 2007 11:07 AM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

I agree that your description of paths/loops is what one would get if one thinks of orbifolds as groupoids or if one think of them as stacks (Thinking of them as stacks is a bit better since then this definition is more natural). The problem is that even if my orbifold is an actual manifold, this is not the right “topology” on the space of loops. For one thing, I see no way to do Morse/Floer theory on such a space.

Posted by: Eugene Lerman on April 12, 2007 7:03 PM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

Thinking of them as stacks is a bit better since then this definition is more natural)

Actually, without further thinking, I wouldn’t know how to reformulate my groupoid of “paths with jumps” in stack language. Can you help me?

not the right “topology” on the space of loops

I guess that I am thinking of equipping paths with the compact-open topology of continuous maps.

Could you indicate, for a layman like me, in which sense this is unsatisfactory?

Posted by: urs on April 12, 2007 7:48 PM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

Speaking as an ignorant categorist, I’ll hazard some guesses as to what else one might give up. John Baez said that rumor has it that the category of smooth spaces is a “concrete quasitopos”, which certainly sounds right to me, given the strong family resemblance between the axioms for smooth spaces and the axioms for quasitopological spaces, which I think is the ur-example of quasitopos [and even where ‘quasitopos’ gets its name].

In a typical quasitopos, one has a lot of the Girard exactness conditions for a topos, with the notable exception of one, that every epi is a coequalizer (of its kernel pair). This allows one to prove for example that a morphism which is both mono and epi is an isomorphism. This corollary typically fails in a quasitopos. I’m guessing it fails for smooth spaces in particular. So while smooth spaces are sets with extra structure, they do not behave much like sets in some sense.

If there is some advantage to synthetic differential geometry over smooth spaces, it may be just in the fact that you can reason with the objects in a smooth topos “synthetically”, i.e., pretty much as you reason with sets, provided that you do so “constructively”. (But then eventually, you come back down to earth and ask, “just what are these objects, analytically?” – there’s a rub.)

Posted by: Todd Trimble on April 12, 2007 2:38 PM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

This allows one to prove for example that a morphism which is both mono and epi is an isomorphism. This corollary typically fails in a quasitopos. I’m guessing it fails for smooth spaces in particular.

Yes, I think so.

Here is what I believe to be an example for that:

Take any manifold $X$ and the identity diffeomorphism on that.

Now, think of $X$ as a smooth space in two different ways.

First, in the ordinary sense: the collection $P_X$ of plots includes all ordinary smooth maps from (convex subsets of) $\mathbb{R}^n$ (for all $n$) into $X$.

Second, choose some fixed surjective submersion $\pi : Y \to X$ and allow the collection of plots $P'_X \subset P_X$ to be only the smooth maps from (convex subsets of) $\mathbb{R}^n$ to $X$ that factor through $Y$.

For generic $Y$, say some open cover of $X$ by open subsets and $X$ itself not contractible, $P'_X$ is strictly smaller than $P_X$. Let’s pick such a $Y$.

Then, clearly, the identity on $X$ becomes a smooth map $(\mathrm{Id}) : (X,P'_X) \to (X,P_X)$ which is manifestly epi and mono.

But it is not iso.

At the level of sets, the inverse would of course still be the identity map. But now this fails to be a smooth map $(X,P_X) \to (X,P'_X)$ since (by my wicked construction) not every plot in $P_X$ is a plot in $P'_X$.

Posted by: urs on April 12, 2007 3:07 PM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

Sorry, there is a much easier way to give an example:

Let $X$ be a manifold regarded as a smooth space in the standard way. Let $X'$ be the same manifold, but equipped with the discrete smooth structure.

Then the identity map $X' \to X$ (on the level of sets) is smooth. But the identity map the other way round, $X \to X'$, is not.

Posted by: urs on April 12, 2007 5:50 PM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

To expand a bit on Urs’ examples…

Any set $X$ has a discrete smooth structure. With this smooth structure, every function from $X$ to another smooth space is smooth, and only constant maps into $X$ are smooth.

At the other extreme, any set $X$ has an indiscrete smooth structure. With this smooth structure, every function from another smooth space into $X$ is smooth.

Unless $X$ has $0$ or $1$ points, the identity map from $X_{disc}$ to $X_{indisc}$ is smooth, but it doesn’t have a smooth inverse.

Posted by: John Baez on April 13, 2007 8:21 PM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

To expand a bit on Urs’ examples…

Oh, of course!

And less concretely but more generally: this kind of example works whenever there are two smooth structures on one set with one collection of plots containing the other but being strictly larger.

So, this demonstrates one way in which smooth spaces fail to be a topos. But otherwise it is not a particularly shocking property, is it? I mean, my example above where we have a manifold once with the ordinary smooth structure and once regarded as a discrete set of points and the monic epi going one way failing to be an iso works just as well in the category of ordinary manifolds.

Posted by: urs on April 13, 2007 11:37 PM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

But otherwise it is not a particularly shocking property, is it?

Nope – as you’ve made clear, not shocking at all! Thanks (to you and John).

If there is some advantage to synthetic differential geometry over smooth spaces, it may be just in the fact that you can reason with the objects in a smooth topos “synthetically”, i.e., pretty much as you reason with sets, provided that you do so “constructively”.

In actual fact, I’m far from “sold” on SDG; here I was just reporting on how a proponent might see matters. The jury on Chen spaces v. smooth topos is still very much out as far as I’m concerned.

Your examples show that the embedding of the category of manifolds into Chen spaces preserves certain epis which are not preserved by the embedding into a smooth topos. The embedding into Chen spaces also preserves non-transverse intersections of submanifolds (such as the parabola and the line considered by James here). It may be arguable, then, that because Chen spaces are supported on their points (are sets + structure), that this embedding tends to preserve some behavior in the category of manifolds which from a certain point of view could be seen as “pathological” (cf. the next point made by James here), and that the embeddings into smooth toposes tend to smooth out some of these kinks, while preserving the limits and colimits of manifolds which behave “nicely”. (Again, I’m not absolutely convinced, but it might be worth a pause to consider.)

Posted by: Todd Trimble on April 14, 2007 4:38 AM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 20)

Eugene wrote:

I have heard that one problem with Chen’s spaces is that one gives up the inverse function theorem. Is this correct?

That’s probably true — as Urs points out, it’s not even clear how one would state such a theorem.

Here are two ‘weird’ smooth spaces to expand your intuition:

1. Take any manifold $X$ and let $D$ be any sub-vector-bundle of the tangent bundle of $X$. Say that a map $f: A \to X$ is smooth (where $A$ is any smooth space) iff it’s smooth in the usual sense and $d f$ applied to any tangent vector gives a tangent vector in $D$.

2. Take any topological space $X$ and say that a map $f: A \to X$ is smooth iff it’s continuous with respect to the natural topology on the smooth space $A$.

By the way: if you don’t feel comfortable enough with smooth spaces to enjoy my use of a general smooth space $A$ in the above definitions, just pretend $A$ is a smooth manifold! Then the ‘natural topology’ is just the one you’re used to.

John sort of addressed my other question, but I still don’t quite understand. If one ‘improves’ the category of manifolds by embedding it into the category of sheaves of sets, why not then consider all sheaves and not just the ones represented by sets?

Because some backwards people still think mathematical gadgets should be sets with extra structure! That’s the main reason so far. Urs and I have certain concrete applications of smooth spaces to problems in mathematical physics. These applications don’t seem to require the more general sort of smooth space that’s not a set. So, why bother? It would only make it harder to market our results.

See, it already takes some work to convince physicists that they should be interested in ‘2-category objects in the category of smooth spaces’. I don’t think it’ll help to say ‘oh yeah, and smooth spaces are just sheaves on a certain site’.

I have ideas about some other reasons for restricting to ‘concrete’ smooth spaces, but I haven’t worked out enough details yet to make a good case.

Posted by: John Baez on April 13, 2007 7:41 PM | Permalink | Reply to this
Read the post Quantization and Cohomology (Week 19)
Weblog: The n-Category Café
Excerpt: Finding critical points of the action in a general context: a smooth category equipped with a smooth 'action' functor.
Tracked: April 20, 2007 8:54 PM
Read the post Quantization and Cohomology (Week 21)
Weblog: The n-Category Café
Excerpt: Understanding the 'action' in classical mechanics as a smooth functor --- the case of a path groupoid.
Tracked: May 2, 2007 10:35 PM
Read the post The First Edge of the Cube
Weblog: The n-Category Café
Excerpt: The notion of smooth local i-trivialization of transport n-functors for n=1.
Tracked: May 5, 2007 2:20 AM
Read the post Large Smooth Categories
Weblog: The n-Category Café
Excerpt: Stacks versus categories internal to sheaves.
Tracked: June 8, 2007 1:21 PM

Post a New Comment