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• CommentRowNumber1.
• CommentAuthorEric
• CommentTimeMay 2nd 2010

I’ve been looking a little bit at diffeological spaces and am becoming a fan of the idea. I’ve got some notes on my personal web too.

Quick question…

Given any two smooth spaces $X,Y$, we can form something of a canonical mapping “T” diagram:

$\array{ X & \stackrel{\pi_1}{\leftarrow} & X\times [X,Y] & \stackrel{ev}{\rightarrow} & Y \\ {} & {} & \mathrlap{\;\;\;\scriptsize{\pi_2}}{\downarrow} & {} & {} \\ {} & {} & [X,Y] & {} & {} }$

which is also discussed on transgression.

I drew this with “plots” in mind, so I had arrows like

$\array{ {} & {} & U & {} & {} \\ {} & \swarrow & \downarrow & \searrow & {} \\ X & \stackrel{\pi_1}{\leftarrow} & X\times [X,Y] & \stackrel{ev}{\rightarrow} & Y \\ {} & {} & \mathrlap{\;\;\;\scriptsize{\pi_2}}{\downarrow} & {} & {} \\ {} & {} & [X,Y] & {} & {} }$

with another arrow (not shown) from $U\to [X,Y]$.

This made me think that $U$ was actually the limit of the mapping “T”

$\array{ X & \stackrel{\pi_1}{\leftarrow} & X\times [X,Y] & \stackrel{ev}{\rightarrow} & Y \\ {} & {} & \mathrlap{\;\;\;\scriptsize{\pi_2}}{\downarrow} & {} & {} \\ {} & {} & [X,Y] & {} & {} }$

So, my question is “What is the limit of this diagram?” Is it $\mathbb{R}^\infty$ or something?

Thanks for any words of wisdom.

• CommentRowNumber2.
• CommentAuthorTodd_Trimble
• CommentTimeMay 2nd 2010

If the underlying scheme $D$ of a diagram $F: D \to C$ has an initial object, then the limit of the diagram is $F$ applied to the initial object. So the limit of this diagram is just $X \times [X, Y]$.

• CommentRowNumber3.
• CommentAuthorEric
• CommentTimeMay 2nd 2010
• (edited May 2nd 2010)

Interesting. Thank you Todd. That makes sense.

I was thinking in terms of plots. Since the smooth space $X\times [X,Y]$ is defined in terms of plots $\phi:U\to X\times [X,Y]$ from open subsets $U\subset\mathbb{R}^n$ (but in this case I think $n = \infty$), then it seems that since $X\times [X,Y]$ is the limit of the diagram, then maybe the plots for $X$, $Y$, and $[X,Y]$ all factor through plots for $X\times [X,Y]$ via $\pi_1$, $\pi_2$, and $ev$.

Next, I’m hoping that there is some functor $C:Diff\to Chains$, where $Chains$ is a category of chains on smooth spaces and morphisms are push forwards of chain spaces. If this functor is sufficiently nice, e.g. preserves limits, then I think the “T” diagram becomes

$\array{ C(X) & \stackrel{\pi_1}{\leftarrow} & C(X)\times C([X,Y]) & \stackrel{ev}{\rightarrow} & C(Y) \\ {} & {} & \mathrlap{\;\;\;\scriptsize{\pi_2}}{\downarrow} & {} & {} \\ {} & {} & C([X,Y]) & {} & {} }$

But this is all just my humble attempt at seeing if there is some relation between $C([X,Y])$ and $[C(X),C(Y)]$. I think it would be interesting if

$C([X,Y]) = [C(X),C(Y)],$

but I’m not sure how to make sense out of any of that. If anyone could help shed some light on that, I’d appreciate it.

I’d also obviously be interested in any references that discuss chains on mapping spaces between smooth spaces. It seems that this subject is relatively young in math years, so maybe there is not a reference yet? It seems pretty straightforward, but maybe not enough eyes looking in this direction or something.

• CommentRowNumber4.
• CommentAuthorTodd_Trimble
• CommentTimeMay 2nd 2010
• (edited May 2nd 2010)

If one has an Eilenberg-Zilber theorem for chains, to the effect that there is a canonical isomorphism

$C(X) \otimes C(Z) \cong C(X \times Z)$

so that $C$ is a monoidal functor, then putting $Z = [X, Y]$ one can form a composite map

$C(X) \otimes C([X, Y]) \cong C(X \times [X, Y]) \stackrel{C(ev)}{\to} C(Y)$

and then by the adjunction $(C(X) \otimes -) \dashv [C(X), -]$, this map corresponds to a map

$C([X, Y]) \to [C(X), C(Y)]$

so there is at least some relation between the two. My guess is that this last map is usually not an isomorphism, but maybe this gives you something to work with anyway.

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeMay 2nd 2010
• (edited May 2nd 2010)

I’d also obviously be interested in any references that discuss chains on mapping spaces between smooth spaces.

You mean singular chains? I have thought a bit about singular cochains on diffeological spaces and more general things: here.

• CommentRowNumber6.
• CommentAuthorEric
• CommentTimeMay 2nd 2010

Thanks Todd!

Let me try to rephrase what you said to make sure I understood…

If $C$ is a monoidal functor, then we’ve got a map

$C(X) \otimes C([X, Y]) \to C(Y).$

Then there is an adjunction $(C(X) \otimes -) \dashv [C(X), -]$ giving us a map

$C([X, Y]) \to [C(X), C(Y)]$

but we’re not sure whether this map is invertible.

I would be surprised if this map were not invertible.

Next Steps:

• Find out if $C$ is a monoidal functor for diffeological spaces (I bet it is since $C(X)$ is defined by pushing forward spaces of the form $C(U)$ for $U\subset\mathbb{R}^n$)
• Make sure I understand that adjunction (which I don’t at the moment, but am willing to believe it and move on :))
• Try to see if the map $C([X, Y]) \to [C(X), C(Y)]$ is invertible (which I’m pretty it should be for diffeological spaces. It reminds me of Fourier transform or convolution)
• CommentRowNumber7.
• CommentAuthorEric
• CommentTimeMay 2nd 2010
• (edited May 2nd 2010)

You mean singular chains? I have thought a bit about singular cochains on diffeological spaces and more general things: here

Thanks Urs! But one thing that trips me up on that page and transgression is that you guys jump straight to cohomology. I prefer to stay in the realm of (co)chains for as long as I can and rarely ever feel the need to pass to (co)homology. So I’m interest in seeing what can be learned at the chain and cochain level.

• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeMay 2nd 2010

Well Eric, as I said before, we do not jump straight to cohomology, but describe the oo-groupoids of cochains, coboundaries, etc.

I imagine you want to take $\Delta^p_{Diff}$ to be the standard smooth $n$-simplex and then define the complex of smooth singular simplices in a diffeological space $X$as

$Hom_{DiffSp}(Delta^p_{Diff}, X) \,.$

This manifestly preserves products, hence is monoidal. That passing to free abelian groups on this still preserves products is then the standard lore.

• CommentRowNumber9.
• CommentAuthorEric
• CommentTimeMay 2nd 2010

Well Eric, as I said before, we do not jump straight to cohomology, but describe the oo-groupoids of cochains, coboundaries, etc.

Hi Urs. I’m probably just being dense, but when I went to the page you referred to, the first thing I did was search for the word “chain”. This brought me down to a discussion about singular cohomology. I didn’t see the word chain, cochain, or coboundary anywhere. The content is probably there, but in words I don’t understand.

Also, when I go to transgression, it jumps right into a discussion about cohomology. But I think transgression is not (or maybe “should not”) just about cohomology, but is about (co)chains. Anything (co)homological is just a result of maps commuting with (co)boundary.

Speaking of which, as I read about diffeology, I am tempted to try to reformulate things in terms of (co)boundary. Something whose moral would be, “The category of smooth spaces consists of spaces with (co)boundary and morphisms that commute with (co)boundary.” This would include “finitary smooth spaces”. I’m not sure if diffeology includes “finitary smooth spaces”, but I kind of doubt it.

• CommentRowNumber10.
• CommentAuthorUrs
• CommentTimeMay 2nd 2010

Also, when I go to transgression, it jumps right into a discussion about cohomology. But I think transgression is not (or maybe “should not”) just about cohomology,

This is true. I think Andrew meant to motivate the idea as an operation on cohomology and then later on assumes “nice geometric models” and discusses what the operation there does on cochains. In particular for the case of de Rham cohomology: on differential forms.

This follows the school of thought where cohomology is defined not a priori as equivalence classes of cochains, but by its expected properties (assignment of cohomology groups to spaces satisfying certain conditions). In most cases a cochain model for cohomologies defined this way is available. But not always.

But don’t blame this on “us”. Here on the nLab we went through some pain of discussing cohomology entirely in the un-decategorified sense that you are after.

• CommentRowNumber11.
• CommentAuthorUrs
• CommentTimeMay 2nd 2010

Oh, or maybe this here is a source of some puzzlement:

what is true is that the notion of cochain which is not a cocycle is very much de-emphasized “by us” (in the sense that “by us” is currently being used here). In the vew of cohomology described at cohomology there is no real fundamental notion of cochain at all, while there is a fundamental notion of cocycle.

• CommentRowNumber12.
• CommentAuthorEric
• CommentTimeMay 2nd 2010

PS: I don’t “blame” you guys for anything other than being awesome :) If it weren’t for you guys, I wouldn’t be able to pretend I know anything about this stuff at all.

It is probably standard stuff to think of $C$ as a covariant functor and $\Omega$ as a contravariant functor and is pretty obvious once you start thinking about $Diff$ as a convenient category, but I’m having fun rediscovering this :)

• CommentRowNumber13.
• CommentAuthorEric
• CommentTimeMay 2nd 2010
• (edited May 2nd 2010)

what is true is that the notion of cochain which is not a cocycle is very much de-emphasized “by us”

Yes yes :)This is what I mean by “passing to cohomology”. There is more to life than cocycles. When I say things like “I like to consider things at the (co)chain level”, I mean I care about cochains that are not cocycles too. This could just be because I haven’t been enlightened enough yet though.

By the way, I’m not so sure that I was smart enough to mean what you thought I meant when you said:

I imagine you want to take $\Delta^p_{Diff}$ to be the standard smooth $n$-simplex and then define the complex of smooth singular simplices in a diffeological space $X$ as

$Hom_{DiffSp}(Delta^p_{Diff}, X) \,.$

This manifestly preserves products, hence is monoidal. That passing to free abelian groups on this still preserves products is then the standard lore.

It might turn out to be the same thing, but since diffeological spaces are defined by the plots $\phi:U\to X$ into them, my idea was to use the already-understood chains $C(U)$ (since $U\subset\mathbb{R}^n$) and push them forward to $X$ and glob them all together somehow.

One idea I was playing around with was if the topology $T(U)$ on $U\subset\mathbb{R}^n$ has a basis $B(U)$, then I thought we might be able to think of $C(U)$ as the free abelian group $F(B(U))$ generated by the basis. If that made any sense, I would go ahead and define

$B(X) = \{\phi_* b | b\in B(U), \phi: U\to X\}.$

In other words, the basis for $B(X)$ is the union of all images of all bases for all plots. Then $C(X) = F(B(X))$. Or something…

Would that make any sense? Can we think of the space of chains as the free abelian group generated by the basis of a topology? I’m tempted to consider the basis of “closed” sets as opposed to open sets though.

• CommentRowNumber14.
• CommentAuthorUrs
• CommentTimeMay 2nd 2010
• (edited May 2nd 2010)

This is what I mean by “passing to cohomology”. There is more to life than cocycles.

Ah, now I finally understand what you mean. You mean you want to consider cochains and not just cocycles.

Hmm.

But:

1. cochains cannot be “passed to cohomology” at all.

2. I don’t actually think that there is more to life than cocycles. :-)

The notion of cochain is really something without any intrinsic meaning. BUT, and this is what is happening here, various cocycles in some cohomology theories look as just cochains in others.

Notably differential forms: in de Rham cohomology theory a non-closed differential form appears as just a cochain. But typically if you need it as such, it is wrong to think of it as being a cochain in de Rham cohomology. Better is to think of it as a cocycle in Deligne cohomology or in Hochschild homology, for instance.

It is reasonable to ask natural constructions such as transgression to act not just on cohomology, but on cocycles. But if you ask it to act naturally on cochains, you are asking a general abstract construction to work well with a very specific man-made model. That may work sometimes, but is in general too much to ask for. Better think of your cochains as cocycles and work with these.

• CommentRowNumber15.
• CommentAuthorEric
• CommentTimeMay 3rd 2010
• (edited May 3rd 2010)

Thanks Urs :)

1. cochains cannot be “passed to cohomology” at all.

I know, so when you “pass to cohomology” you leave behind the lonely cochains :)

1. I don’t actually think that there is more to life than cocycles. :-)

This is the answer I was expecting when I said:

This could just be because I haven’t been enlightened enough yet though.

So now the cohomology theory I need depends on the questions I’m asking? That doesn’t sound very pleasing from a physical perspective unless there is a “cohomology theory of everything” :)

Hmm… so something that is not a cocycle in one cohomology theory, e.g. a differential form $\alpha$ with $d\alpha\ne 0$, can be a cocycle in another cohomology theory. It sounds like I have some more homework to do…

Edit: After looking at Deligne cohomology and Hoschild cohomology, I see my chance of understanding it is almost surely 0 :)

• CommentRowNumber16.
• CommentAuthorMike Shulman
• CommentTimeMay 3rd 2010

So now the cohomology theory I need depends on the questions I’m asking?

Surely the answer to a question always depends on what the question is? (-:

The notion of cochain is really something without any intrinsic meaning.

I’m not sure I entirely agree with that. A cochain in a cocomplex doesn’t directly appear in the cohomology groups, but it does affect the homotopy type of the complex. In particular, it does things like kill off its coboundary and prevent that from appearing in the cohomology. So it seems to me that if you want, you could think of a cochain $\alpha$ as “a reason why $d \alpha$ equals zero in cohomology.”