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Stream: learning: questions

Topic: exterior product and tensor of chain complexes

Amar Hadzihasanovic (May 05 2021 at 12:21):

I would like to understand how these facts are connected.

• The exterior product of differential forms takes a $p$-form $\omega$ and a $q$-form $\eta$ and returns a $(p+q)$-form $\omega \land \eta$, such that $\omega \land \eta = (-1)^{pq} \eta \land \omega$.
• The tensor product of chain complexes $K \otimes L$ is generated by a $(p+q)$-chain $x \otimes y$ for each $p$-chain $x$ in $K$ and $q$-chain $y$ in $L$. The tensor product is symmetric with the symmetry defined by $x \otimes y \mapsto (-1)^{pq} y \otimes x$.
• The generalised Stokes theorem establishes a kind of duality between differential $p$-forms on a manifold and singular $p$-chains in the manifold.

Altogether these make me think that the sign $(-1)^{pq}$ in the exterior product is “the same” as the sign $(-1)^{pq}$ in the symmetry on the tensor product of chain complexes. But I am failing to find a source that explains the connection. Can anyone help?

fosco (May 05 2021 at 12:34):

I will just throw a blatantly false/unrelated hot take in order to get notified about answers :smile:

"yadda yadda, Dold-Kan for graded abelian groups: the category of simplicial objects in $\textsf{grAb}$ is equivalent to the category of positively concentrated chain complexes of graded abelian groups"

Tim Hosgood (May 05 2021 at 13:13):

for why this sign is the same, I think you'd find reading about the so-called Koszul sign convention helpful :smile:

Tim Hosgood (May 05 2021 at 13:13):

https://mathoverflow.net/a/359960/73622 is very nice, for example

Tim Hosgood (May 05 2021 at 13:14):

whenever you work with dgla's (or just dga's (or even just graded algebras)), this sign turns up aaaalll the time

Amar Hadzihasanovic (May 05 2021 at 13:48):

Yep, I'm aware of it! (This is not to dismiss your answer -- I like the answer on MO you linked and it's always good to read nice explanations :smile: )

In this particular case I'm thinking that “Stokes duality” between forms and chains suggests a more direct relation, rather than “just” a common topological origin. To be a bit more precise, I'm thinking of something along the lines of “integration of $\omega \land \eta$ on a $(p+q)$-chain $x$ is defined by convolution of integrations of $\omega$ on $x_1$ and $\eta$ on $x_2$ over decompositions of $x$ as $x_1 \otimes x_2$, and that determines $\omega \land \eta$ uniquely” (even though I don't think it makes sense 'literally' as written!)

Amar Hadzihasanovic (May 05 2021 at 13:49):

(In fact now that I've written it like this it looks very co-endy @fosco )

Simon Burton (May 05 2021 at 13:49):

Even before you get to chain complexes you can consider $\mathbb{Z}$-graded vector spaces with the same tensor (forget the boundary operation) and the same symmetry operation $x \otimes y \mapsto (-1)^{pq} y \otimes x$. Exterior product in grade zero is then the same thing as symmetric product in grade 1. So that's how I think of exterior product as a graded thing, it's the symmetrization of odd graded vector spaces...

Amar Hadzihasanovic (May 06 2021 at 13:05):

Since I didn't find an answer, I have tried to figure this out myself... So now I have a few more ideas and new questions.

Let $M, N$ be smooth manifolds, $\omega$ a $p$-form on $M$ and $\eta$ a $q$-form on $N$. Then I have forms $\pi_1^* \omega$ and $\pi_2^*\eta$ on the cartesian product $M \times N$, by pullback along the two projections.

Suppose $c$ is a $p$-chain in $M$ and $d$ a $q$-chain in $N$. The tensor product gives me a $(p+q)$-chain $c \otimes d$ in $M \times N$. Now, I believe we have the equation

$$\int_{c \otimes d} \pi_1^*\omega \land \pi_2^*\eta = (\int_c \omega) (\int_d \eta).$$

I would like to make the bold guess (that I'm not entirely convinced of!) that this equation, together with

$$f^*(\omega \land \eta) = f^*\omega \land f^*\eta$$

for all forms $\omega, \eta$ on a manifold $M$ and smooth maps $f: N \to M$, is sufficient to completely characterise the exterior product. This guess relies on a few sub-guesses, more in the next message...

Amar Hadzihasanovic (May 06 2021 at 13:15):

First a fact: once we have defined the exterior product of forms on $\mathbb{R}^n$, we are able to define it on all manifolds. But these are generated under exterior products and sums by 0-forms (smooth functions $\mathbb{R}^n \to \mathbb{R}$) and by the 1-forms $dx_1,\ldots,dx_n$, so in fact we just need to show that the equations characterise exterior products of these.

Amar Hadzihasanovic (May 06 2021 at 13:21):

The lucky thing is that the wedge product $dx_i \land dx_j$ for $i \neq j$ is of the form $\pi_1^*\omega \land \pi_2^* \eta$! In fact iterating pullbacks of projections we can bring the definition of each $dx_{i_1} \land \ldots \land dx_{i_k}$ down to defining $dx_1 \land dx_2$ as a 2-form on $\mathbb{R}^2$, which is of the form $\pi_1^* (dx) \land \pi_2^*(dx)$.

Amar Hadzihasanovic (May 06 2021 at 13:25):

Now, the first guess:
Defining the integral of $dx_1 \land dx_2$ on 2-chains of the form $c \otimes d$ in $\mathbb{R}^2$ is sufficient to define it on each 2-chain in $\mathbb{R}^2$.
That would follow from the ability to decompose every 2-chain into a combination of 2-chains of the form $c \otimes d$. This seems true if I think of the plane topologically, I don't know if smoothness complicates things!

Amar Hadzihasanovic (May 06 2021 at 13:32):

If the first guess is correct, we would have now a definition $\int_c dx_1 \land dx_2$ on all 2-chains $c$ in $\mathbb{R}^2$.

The second guess is:
From the integrals $\int_c dx_1 \land dx_2$, we obtain more in general integrals $\int_c f(x_1,x_2) dx_1 \land dx_2$ for all smooth functions $f: \mathbb{R}^2 \to \mathbb{R}$
(and similarly for more variables).
This would follow from the ability two write $f$ as the determinant of the Jacobian of some smooth function $F: \mathbb{R}^2 \to \mathbb{R}^2$, so that $\int_c f(x_1,x_2) dx_1 \land dx_2 = \int_{F \circ c} dx_1 \land dx_2$, a kind of “generalised existence of antiderivatives”... Is this true?

Amar Hadzihasanovic (May 06 2021 at 13:36):

If both the first guess and the second guess are correct, we should now be able to integrate all $p$-forms on $\mathbb{R}^n$ on all $p$-chains.

Now the third guess:
A $p$-form is determined uniquely by the data of its integral on all $p$-chains.

I am surprised that I could not find whether this is true, even though it seems an obvious question about whether Stokes is a “definite” duality (before passing to (co)homology)... Maybe I'm just bad at searching!

Fawzi Hreiki (May 06 2021 at 13:37):

Well the final fact is de Rham’s theorem no?

Fawzi Hreiki (May 06 2021 at 13:37):

This works because of the pairing between de Rham cohomology and singular homology which you need Stoke’s theorem to prove is well defined

Amar Hadzihasanovic (May 06 2021 at 13:39):

Fawzi Hreiki said:

Well the final fact is de Rham’s theorem no?

Is the final fact my “third guess”? Maybe! I was confused by the fact that de Rham's theorem refers to cohomology theories, while I'm asking a question about the chains and cochains from which the cohomology theories are defined... Can you say more? :)

Fawzi Hreiki (May 06 2021 at 13:46):

You have a map from de Rham cohomology to singular homology which sends a form $\omega$ to the map $c \mapsto \int_c \omega$. de Rham's theorem tells you that this gives an isomorphism of groups between the de Rham cohomology and the singular homology.

Fawzi Hreiki (May 06 2021 at 13:47):

Stoke's theorem is necessary to prove that this map works (i.e. is independent of the chosen representatives)

Fawzi Hreiki (May 06 2021 at 13:48):

What clarified this greatly for me was the SDG approach to differential forms which literally uses infinitesimal co-chains.

Fawzi Hreiki (May 06 2021 at 13:49):

For the classical POV, chapter 7 in Tu's 'Introduction to Manifolds' is good

Fawzi Hreiki (May 06 2021 at 13:49):

For the SDG POV, there's this paper by Anders Kock

Fawzi Hreiki (May 06 2021 at 13:52):

For a more abstract take on the classical POV, Wedhorn's book 'Manifolds, Sheaves, and Cohomology' develops the basic theory of manifolds in the language of locally ringed spaces and has lots of good stuff on cohomology

Fawzi Hreiki (May 06 2021 at 13:52):

He's an algebraic geometer by trade so it makes for an interesting read

Amar Hadzihasanovic (May 06 2021 at 13:53):

Ok, so de Rham's theorem says that the map $\omega \mapsto (c \mapsto \int_c \omega)$ descends to an isomorphism of groups when you take the (co)homology... Do you know if the map itself is injective, though?

Fawzi Hreiki (May 06 2021 at 13:55):

I'm being a bit sloppy here. The map is already between (co)homology groups. It sends $[ \omega ]$ to $[ c ] \mapsto \int_c \omega$ where the square brackets denote isomorphism classes in the respective (co)homologies

Fawzi Hreiki (May 06 2021 at 13:56):

Stoke's theorem implies that this is independent of the chosen representatives and is hence a homomorphism. de Rham's theorem says that this homomorphim is an isomorphism.

Amar Hadzihasanovic (May 06 2021 at 13:58):

Sorry, maybe I'm being really dumb, but I'm confused. The integral $\int_c \omega$ should be defined for all $p$-forms $\omega$ and $p$-chains $c$, right? There's no quotients involved at this level. I understand that Stokes' theorem allows us to descend to the quotients involved in the def. of (co)homology.

Amar Hadzihasanovic (May 06 2021 at 13:59):

But it still seems to make sense to ask whether $\omega \mapsto (c \mapsto \int_c \omega)$ is injective, in itself?

Amar Hadzihasanovic (May 06 2021 at 13:59):

That's what my “third guess” is asking.

Fawzi Hreiki (May 06 2021 at 14:00):

This map is only about cohomology classes of differential forms.

Amar Hadzihasanovic (May 06 2021 at 14:00):

Why? What makes it ill-defined?

Fawzi Hreiki (May 06 2021 at 14:03):

Well I'm not exactly sure why you wouldn't want to pass to (co)homology classes seeing as two equivalent forms (wrt de Rham cohomology) give the same integral

Fawzi Hreiki (May 06 2021 at 14:04):

Check out the paper by Kock which I linked. It's only 8 pages long and fairly understandable.

Fawzi Hreiki (May 06 2021 at 14:08):

And it certainly explains things better that I can.

Amar Hadzihasanovic (May 06 2021 at 14:09):

Well that restricts me to closed forms and integration along cycles. You can see it this way: I'd like to know whether the pairing $(c, \omega) \mapsto \int_c \omega$ is “definite” for arbitrary forms and arbitrary chains, in the sense that $\int_c \omega = 0$ for all chains $c$ implies that $\omega = 0$.

Amar Hadzihasanovic (May 06 2021 at 14:10):

I appreciate the references that you linked, but I don't think I need to justify my curiosity about a specific question :)

Fawzi Hreiki (May 06 2021 at 14:13):

That makes sense. I think the best place to look for this might be Bott-Tu

Reid Barton (May 06 2021 at 14:15):

I think this is true: if $\omega \ne 0$ then we can choose a point at which $\omega$ does not vanish, choose local coordinates at that point and write $\omega$ in the standard basis $dx_{i_1} \wedge \cdots \wedge dx_{i_n}$ ($i_1 < \cdots < i_n$, for an $n$-form), and choose one of the terms that has a nonzero coefficient, say positive. Then nearby the original point, that coefficient is still positive. So a suitable small chain near the original point that varies only in the corresponding directions will pair positively with $\omega$.

Reid Barton (May 06 2021 at 14:17):

On the other hand, I don't really believe the first guess (sorry if it was already discussed and I overlooked it). How will you define the integral over a chain in $\mathbb{R}^2$ like $\gamma : [0, 1] \to \mathbb{R}^2$, $\gamma(t) = (t, t)$? It's homologous to a sum of tensor products, namely going from $(0, 0)$ to $(1, 0)$ and going from $(1, 0)$ to $(1, 1)$; but not equal.

Amar Hadzihasanovic (May 06 2021 at 14:29):

Thanks Reid. You are right about the first guess, but I think I was sloppy at stating it. It should be sufficient that every chain is homologous to a sum of tensor products, right? Then I can still compute the integral by passing to a homologous chain.

Amar Hadzihasanovic (May 06 2021 at 14:29):

Your argument re: the third guess seems convincing!

Reid Barton (May 06 2021 at 14:35):

But now if we pass to a homologous chain then we only get a well-defined answer if we start with a closed form, right?

Reid Barton (May 06 2021 at 14:37):

So while there is a pairing defined on the chain level, I'm not sure if you can characterize it in the way you want to without either passing to homology, or adding additional axioms (like the integral of a closed form over a boundary is 0, or something like that).

Amar Hadzihasanovic (May 06 2021 at 14:47):

The first guess is only about the iterated wedges $dx_1 \land \ldots \land dx_n$ which are all closed, so it should be fine, I think...

Reid Barton (May 06 2021 at 14:51):

Oh, I overlooked that you were only really interested in the top-dimensional forms as well, so maybe my example of a 1-dimensional chain in $\mathbb{R}^2$ isn't relevant either

Amar Hadzihasanovic (May 06 2021 at 14:53):

Ah but I see, I guess I have not established how wedges interact with the differential operator (so I may not "know" that those are closed). That's something that I was hoping would come out by duality with how boundaries interact with the tensor of chains, but I'm not sure how.

Amar Hadzihasanovic (May 06 2021 at 14:55):

I'm still not sure what's relevant and what isn't, so it's good to think beyond the top-dimensional case, to be safe... :)

John Baez (May 06 2021 at 18:15):

Amar Hadzihasanovic said:

Now, I believe we have the equation

$$\int_{c \otimes d} \pi_1^*\omega \land \pi_2^*\eta = (\int_c \omega) (\int_d \eta).$$

I think that's true.

John Baez (May 06 2021 at 18:31):

Amar Hadzihasanovic said:

I'd like to know whether the pairing $(c, \omega) \mapsto \int_c \omega$ is “definite” for arbitrary forms and arbitrary chains, in the sense that $\int_c \omega = 0$ for all chains $c$ implies that $\omega = 0$.

Yes this is true; Barton sketched how you prove it.

Amar Hadzihasanovic (May 08 2021 at 12:17):

I've made some progress on the original question; I think I now have a nice way of seeing the wedge product of differential forms as “derived” by duality from the tensor of chain complexes.

The starting point is that I think proofs of de Rham's theorem actually show something more than the statement of de Rham's theorem: namely, that closed differential forms on a smooth manifold (so in particular on $\mathbb{R}^n$) are the same as $\mathbb{R}$-valued singular cochains cocycles on the manifold. The singular cochain complex has a structure of real dg-algebra with the cup product $\smile$, which really is obtained by a kind of convolution using the tensor product of chain complexes. And on cocycles, this coincides with the wedge product of forms!

So the idea is the following: define the complex of closed differential forms on $\mathbb{R}^n$ with the wedge product to be the complex $Z^\bullet(\mathbb{R}^n)$ of $\mathbb{R}$-valued singular cocycles together with its dg-algebra structure given by the cup product.

Now, take the ring $C^\infty(\mathbb{R}^n)$ of smooth functions $\mathbb{R}^n \to \mathbb{R}$; as a cochain complex concentrated in degree 0, this also forms a real dg-algebra with multiplication.
Moreover by identifying $C^\infty(\mathbb{R}^n)$ with a subspace of the $\mathbb{R}$-valued 0-cochains with $f \mapsto (x \mapsto f(x))$, we get a linear map $\tau: C^\infty(\mathbb{R}^n) \to Z^1(\mathbb{R}^n)$ simply by restricting the differential.

Now this may be improper but I think that $\tau$ works as a kind of twisting cochain from $C^\infty(\mathbb{R}^n)$ to $Z^\bullet(\mathbb{R}^n)$, when the first is equipped with the “right co-action” of itself $f \mapsto 1 \otimes f$, and the second is seen as a module of itself...

So my new guess is that the de Rham complex $\Omega^\bullet(\mathbb{R}^n)$, as a real dg-algebra with the wedge product $\land$, is the tensor product of $(C^\infty(\mathbb{R}^n), \cdot)$ and of $(Z^\bullet(\mathbb{R}^n), \smile)$ twisted by $\tau$.

[Edit: this can't be right as stated, see later messages; but the de Rham complex may be a quotient of this twisted tensor product]

Translated, this would say that in each degree $i$, $\Omega^i(\mathbb{R}^n)$ is isomorphic to the tensor product $C^\infty(\mathbb{R^n}) \otimes_\mathbb{R} Z^i(\mathbb{R}^n)$ of real vector spaces, that the wedge product is defined by $(f \otimes \omega) \land (g \otimes \eta) = fg \otimes (\omega \smile \eta)$, and that the differential is defined by $d(f \otimes \omega) = 1 \otimes (df \smile \omega)$.

(I think I've seen @Tim Hosgood mention twisting cochains as a focus of his work so perhaps he can tell me if what I'm saying makes sense? I'm very ignorant about these things and only googled them in the last couple of days.)

Amar Hadzihasanovic (May 08 2021 at 12:24):

Now both $(C^\infty(-), \cdot)$ and $(Z^\bullet(-), \smile)$ with this definition can be extended to functors from the category $\mathrm{Cart}$ of cartesian spaces and smooth maps to the category of real dg-algebras, which would also induce a functor $(\Omega^\bullet(-), \land)$. And then presumably I can extend this along colimits to “smooth spaces” as sheaves on $\mathrm{Cart}$, which would in particular give me a sensible definition of $(\Omega^\bullet(-), \land)$ on smooth manifolds. (I haven't checked any of this, it just seems plausible).

Amar Hadzihasanovic (May 08 2021 at 12:32):

So if the guess is correct, that would give me a definition of the de Rham complex and the wedge product, and moreover it would also tell me “tautologically” how to integrate closed differential forms on all chains: just evaluate the cocycle...

Now the question is: can I derive “axiomatically” how to integrate non-closed forms from this data?
Namely require that e.g. “Stokes”, “change of variable by smooth maps”, and “linearity” hold for integrals of generic forms (with this definition of the de Rham complex), and get a unique possible definition?

Amar Hadzihasanovic (May 08 2021 at 12:35):

Something I can definitely do is integrate generic forms on cycles: by linearity it suffices to integrate those of the form $f \otimes \omega$; every cycle in $\mathbb{R}^n$ is a boundary, so

$$\int_c f \otimes \omega = \int_{\partial C} f \otimes \omega = \int_C d(f \otimes \omega) = \int_C df \smile \omega$$

and now $df \smile \omega$ is closed, so I know its integral on $C$.

Amar Hadzihasanovic (May 08 2021 at 12:36):

(Oh I guess I'm also imposing that $\int_c 1 \otimes \omega := \int_c \omega$.)

Amar Hadzihasanovic (May 08 2021 at 12:40):

So the next question is whether this determines how to integrate on arbitrary chains in cartesian spaces. Can every integral on a chain in a cartesian space be turned into an integral on a cycle (possibly by passing to a larger space)?

John Baez (May 08 2021 at 15:56):

Amar Hadzihasanovic said:

So the idea is the following: define the complex of closed differential forms on $\mathbb{R}^n$ with the wedge product to be the complex $Z^\bullet(\mathbb{R}^n)$ of $\mathbb{R}$-valued singular cocycles together with its dg-algebra structure given by the cup product.

Not every $\mathbb{R}$-valued singular cocycle comes from a closed differential form, does it?

John Baez (May 08 2021 at 15:57):

At least not if you mean the usual thing by "differential form", which requires smoothness.

John Baez (May 08 2021 at 15:59):

For example, every singular 0-chain on $\mathbb{R}^n$ is a finite $\mathbb{Z}$-linear combination of points in $\mathbb{R}^n.$ So there's a unique $\mathbb{R}$-valued singular 0-cochain that maps your favorite point $x \in \mathbb{R}^n$ to 1 and all other points to zero.

John Baez (May 08 2021 at 16:00):

But this singular 0-cochain doesn't come from a 0-form, i.e. a smooth function.

John Baez (May 08 2021 at 16:01):

So if you're defining differential forms in the manner you suggest, you're getting some sort of generalization of the usual concept of differential form. Right? Or am I confused?

Amar Hadzihasanovic (May 08 2021 at 17:06):

Um, so I did write wrongly that there's a correspondence between forms and generic cochains; of course there's many more cochains (e.g. all functions are 0-cochains, not just the smooth ones). I got confused with the statement of some theorems I read that seemed to state that a section of the map forms $\mapsto$ cochains exist. On the other hand, I don't need this, I only need it for cocycles (and closed forms). But you're saying that it's still not true.

Amar Hadzihasanovic (May 08 2021 at 17:07):

On the other hand I think that the only 0-cocycles on $\mathbb{R}^n$ are the constant functions, so your example is not a cocycle. Am I wrong?

Amar Hadzihasanovic (May 08 2021 at 17:10):

It should be the case that if $f$ is a 0-cocycle, on any curve $\gamma$ I have $0 = df(\gamma) = f(\partial \gamma) = f(\gamma(1)) - f(\gamma(0))$.

Amar Hadzihasanovic (May 08 2021 at 17:11):

Because $\mathbb{R}^n$ is path-connected, $f$ is constant.

Amar Hadzihasanovic (May 08 2021 at 17:18):

By the way I realised that my "new guess" can't be right as stated because in the tensor product I could get different presentations of the same form (let's call them "pre-forms")...

But the de Rham complex could still be a quotient of that tensor; presumably if I am able to define integration as asked by the "question", it would be the quotient where I identify "pre-forms" whose integrals are equal on all chains.

Amar Hadzihasanovic (May 08 2021 at 17:19):

(Ah, I just saw that John edited his message so it is correcting my initial wrong statement!)

John Baez (May 08 2021 at 17:21):

Amar Hadzihasanovic said:

On the other hand I think that the only 0-cocycles on $\mathbb{R}^n$ are the constant functions, so your example is not a cocycle. Am I wrong?

I meant "cochain", thanks. I've fixed that. I wasn't trying to talk about "cocycles" at all.

Let me check your claim now:

Since a singular 0-cocycle $f$ must have $df = 0$, we have $f(\partial c) = (df)(c) = 0$ for every singular 1-chain $c$. We can take $c$ to be a 1-chain that's a singular 1-simplex starting at a point $x$ and ending at a point $y$, so $\partial c = y - x$. So, we must have $f(x) = f(y)$ if there's a path from $x$ to $y$. So yes, the only singular 1-cocycles on $\mathbb{R}^n$ (or any path-connected space) are constant functions.

John Baez (May 08 2021 at 17:21):

Oh, heh - you said this already! I tend to respond to what I read before reading on... :blushing:

Amar Hadzihasanovic (May 08 2021 at 17:23):

I know it happens, thanks for taking the time to think about it at all...

Amar Hadzihasanovic (May 08 2021 at 17:25):

So yes the question is whether every cocycle comes from a closed differential form. I convinced myself that it should be true but I didn't check my sources carefully...

John Baez (May 08 2021 at 18:16):

Umm, I don't think every cocycle comes from a closed differential form. I believe every singular n-cochain on $\mathbb{R}^n$ is a cocycle. Does that sound right?

If so, I think I can describe one that doesn't come from a differential form. Pick a point $p \in \mathbb{R}^n$ and define a singular n-cochain $f$ as follows. To define it, we just need to say what it does to singular n-simplices, and then extend by linearity. So, let's say that $f(s) = 1$ if the singular n-simplex s contains the point x, and $f(s) = 0$ otherwise.

John Baez (May 08 2021 at 18:18):

The intuitive idea is that $f$ is

$\delta_p dx_1 \wedge \cdots \wedge dx_n$

where $\delta_p$ is the Dirac delta at the point $p$. So, this is not a smooth n-form, but it's a limit of smooth n-forms in a certain topology.

John Baez (May 08 2021 at 18:21):

My proposed counterexamples to your claims have a certain repetitive quality. :upside_down: The reason I can invent them is because I've studied currents.

Amar Hadzihasanovic (May 08 2021 at 19:08):

Oh yes, you are absolutely right. On the other hand this fails for a bit of a stupid reason, namely, as you said, that every singular $n$-cochain on $\mathbb{R}^n$ is a cocycle because there's no non-degenerate $(n+1)$-simplices on which to integrate its differential.

So let me try to amend my definition: instead of taking $Z^\bullet(\mathbb{R}^n)$, take the truncation to degree $\leq n$ of $Z^\bullet(\mathbb{R}^\infty)$. This way you don't get cocycles that are cocycles for trivial reasons.

John Baez (May 08 2021 at 20:08):

I believe that still won't work.

John Baez (May 08 2021 at 20:10):

I believe it just takes more work to describe the counterexamples. I believe that there will again be cocycles that aren't given by closed smooth differential forms, which are limits of closed smooth differential forms.

John Baez (May 08 2021 at 20:11):

The topology here is such that singular cochains $f_i$ converge to a singular cochain $f$ iff $f_i(c) \to f(c)$ for every singular chain $c$.

Amar Hadzihasanovic (May 08 2021 at 20:56):

Thanks John, yes, that sounds very likely.
I suppose that having an injection of closed differential forms into cocycles (that we seem to have), which is closed under the cup product of cocycles, still gives a way of defining the exterior product of closed differential forms.

Amar Hadzihasanovic (May 08 2021 at 21:03):

Any clue on the other question, that is: how far does integration of forms on cycles in $\mathbb{R}^n$ take us? Can I integrate a form on an arbitrary chain only knowing

• how to integrate on cycles, and
• that integration satisfies linearity, Stokes, and “change of variables” (i.e. $\int_c f^*\omega = \int_{f_* c} \omega$ for all smooth maps $f$ between cartesian spaces)?

(By now I'm expecting a “no”...)

Dmitri Pavlov (May 08 2021 at 21:09):

The precise statement here is that the smooth additive singular cochain complex of a smooth manifold M is isomorphic to the de Rham complex of M. Smooth means both that singular simplices are smooth maps and cochains depend smoothly on singular simplices. This is Theorem 15 in Félix and Lavendhomme's On de Rham's theorem in synthetic differential geometry. They do it cubically, but one can also establish the simplicial version analogously.

John Baez (May 08 2021 at 22:30):

Nice, @Dmitri Pavlov!

John Baez (May 08 2021 at 22:39):

Amar Hadzihasanovic said:

Any clue on the other question, that is: how far does integration of forms on cycles in $\mathbb{R}^n$ take us? Can I integrate a form on an arbitrary chain only knowing

• how to integrate on cycles, and
• that integration satisfies linearity, Stokes, and “change of variables” (i.e. $\int_c f^*\omega = \int_{f_* c} \omega$ for all smooth maps $f$ between cartesian spaces)?

Say I know how integrate a specific 1-form $\omega$ over all 1-cycles in $\mathbb{R}$. All these 1-cycles are boundaries and $\omega$ is necessarily closed so we always get zero. Knowing that won't help me integrate $\omega$ over $[0,1]$ (made into a 1-chain in the obvious way).

Amar Hadzihasanovic (May 09 2021 at 06:40):

@Dmitri Pavlov Thanks, that sounds good!

Amar Hadzihasanovic (May 09 2021 at 06:46):

@John Baez I don't think it's that simple: that's why I was postulating change of variable across different cartesian spaces. For example, take an arbitrary 1-form $\omega$ on $\mathbb{R}^2$, consider the map $f: \mathbb{R} \to \mathbb{R}^2$, $t \mapsto (\cos 2\pi t, \sin 2\pi t)$. By assumption I can compute $\int_{f_*[0,1]} \omega$ because it's an integral on a cycle, and by the “change of variables” axiom it is equal to $\int_{[0,1]} f^* \omega$. So I can compute the integral on $[0,1]$ of the pullback of any 1-form on $\mathbb{R}^2$ through $f$.

Amar Hadzihasanovic (May 09 2021 at 06:49):

In general, a sufficient condition for “integrating on cycles is sufficient” would be:
For all $k$-simplices $c$ and $k$-forms $\omega$ on $\mathbb{R}^n$, there exists a smooth map $f: \mathbb{R}^n \to \mathbb{R}^m$ and a $k$-form $\omega'$ on $\mathbb{R}^m$ such that

• $f \circ c$ is a cycle, and
• $\omega = f^* \omega'$.

Amar Hadzihasanovic (May 09 2021 at 07:45):

Well of course that can't be true as it is because e.g. in the example above $f^*\omega$ would be of the form $g(t)dt$ where $g$ is periodic of period 1, and in general I think it would force a certain kind of “periodicity” on the forms that we can integrate... But it may still be a first step?

Amar Hadzihasanovic (May 09 2021 at 08:40):

I think smoothness may be a bit of a red herring here... Perhaps I can try to simplify the problem by looking at continuous maps.

Say that a $k$-dimensional “pre-form” on $\mathbb{R}^n$ is now an element of the product $C^0(\mathbb{R}^n) \otimes_{\mathbb{R}} Z^n(\mathbb{R}^n)$, that is, it is a linear combination with real coefficients of things of the form $f \otimes \omega$ where $f$ is a continuous function on $\mathbb{R}^n$ and $\omega$ is a singular $k$-cocycle. Let $d(f \otimes \omega) := 1 \otimes (df \smile \omega)$ where $df$, as before, is the differential of $f$ seen as a 0-cocycle, and for a continuous map $g: \mathbb{R}^m \to \mathbb{R}^n$ let $g^*(f \otimes \omega) := (f \circ g) \otimes \omega(g \circ -)$.

For a $k$-chain $c$ in $\mathbb{R}^n$, define $\int_c 1 \otimes \omega := \omega(c)$. Impose that

• $\int_{-} {-}$ is linear in its two arguments,
• the Stokes duality holds, that is $\int_{\partial C} f \otimes \omega = \int_C d(f \otimes \omega)$, and
• change of variables holds for all continuous maps $g: \mathbb{R}^n \to \mathbb{R}^m$, that is $\int_{g \circ c} f \otimes \omega = \int_{c} g^*(f \otimes \omega)$.

The questions are:

• Does this determine $\int_c f \otimes \omega$ uniquely for all $c, f, \omega$?
• If so, does it agree with integration of differential forms when $\omega$ comes from a closed differential form on $\mathbb{R}^n$ and $f$ is smooth, under the interpretation $f \otimes \omega \mapsto f\omega$?

Amar Hadzihasanovic (May 09 2021 at 10:18):

I have a sketch of how one could prove that the answer to the first question is yes.

First of all, let's work with cubes instead of simplices. Using change of variables we can reduce the problem to integration on the embedding of the $k$-cube as $[0,1]^k$ into $\mathbb{R}^n$.

Let $w: \mathbb{R} \to \mathbb{R}$ be the continuous “triangle wave” function defined by, $2m + t \mapsto t$ and $2m + 1 + t \mapsto 1-t$ for all $m \in \mathbb{Z}$ and $t \in [0,1]$. Then let $w_{k,n}$ be $\prod_{i=1}^k w \times \prod_{i=1}^{n-k} \mathrm{id}: \mathbb{R}^n \to \mathbb{R}^n$.

Because $w_{k,n}$ is equal to the identity on $[0,1]^k$, we have that $\int_{[0,1]^k} f \otimes \omega$ = $\int_{[0,1]^k} (f\circ w_{k,n}) \otimes w_{k,n}^*\omega)$. Let's call the new integrand $f' \otimes \omega'$.

The idea is that outside of the support $[0,1]^k$, you now have “reflected repetitions” of $f$ and $\omega$... Using change of variables wrt reflection and translation automorphisms of $\mathbb{R}^n$, together with linearity of the integral, we should be able to show that

$$\int_{[0,1]^k} (f' \otimes \omega') = \frac{1}{2^k} \int_{[0,2]^k} (f' \otimes \omega').$$

Now the idea is to take a map $\mathbb{R}^n \to \mathbb{R}^{n+k}$ that wraps $[0,2]^k$ into a $k$-torus $(S^1)^k$ and embeds it into $\mathbb{R}^{2k}$, and does nothing on the remaining variables.

The guess is that we can extend $f' \otimes \omega'$ continuously from $(S^1)^k \times \mathbb{R}^{n-k}$ to a pre-form $g \otimes \eta$ on $\mathbb{R}^{n+k}$ that pulls back to it. Then integrating $g \otimes \eta$ on the $k$-torus, which is a cycle, gives us the integral of $f' \otimes \omega'$.

Amar Hadzihasanovic (May 09 2021 at 10:42):

(I am convincing myself that this should work, mainly because of how flexible continuous maps are... Please crush my hopes soon, if necessary)

Amar Hadzihasanovic (May 09 2021 at 11:05):

In fact I may be able to put the smoothness back in if I wanted, e.g. by replacing $w$ with a smooth wave function that's a diffeomorphism restricted to $[0,1]$.

Amar Hadzihasanovic (May 09 2021 at 13:15):

Oh, I just realised that there is a much simpler argument that can be substituted for the last part, to avoid the whole “embed a $k$-torus in a larger space” thing!

The $w_{k,n}$ have a nice property: they are idempotent, because they are equal to the identity on $[0,1]^k \times \mathbb{R}^{n-k}$ and their image is contained in this subspace.

Amar Hadzihasanovic (May 09 2021 at 13:22):

Now assuming that I have proven that

$$\int_{[0,1]^k} f \otimes \omega = \frac{1}{2^k} \int_{[0,2]^k} w^*_{k,n}(f \otimes \omega)$$

as sketched above, I can just change variables again along the same function, and by idempotency this will do nothing on the integrand; but it will turn the integration domain into a cycle!

[Edit: Actually the equation doesn't work in this way; in fact the RHS will be 0 because “$\omega$ flips sign” on $[1,2]$ and the halves will cancel out... Need a way to circumvent this.]

That is, letting $c$ be the embedding of $[0,2]^k$ into $\mathbb{R}^n$, we have

$$\int_c w^*_{k,n}(f \otimes \omega) = \int_c w^*_{k,n}(w^*_{k,n}(f \otimes \omega)) = \int_{w_{k,n}\circ c} w^*_{k,n}(f \otimes \omega)$$

and $w_{k,n} \circ c$ is a cycle!