Let $H^k(M,\mathbb R)$ be the De Rham cohomology of a manifold $M$.

There is a canonical map $H^k(M;\mathbb Z) \to H^k(M;\mathbb R)$ from the integral cohomology to the cohomology with coefficients in $\mathbb R$, which is isomorphic to the De Rham cohomology. As a previous question already revealed, the images of this map are precisely the classes of differential $k$-forms $[\omega]$ that yield integers when integrated over a $k$-cycle $\sigma$,

$$ \int_{\sigma} \omega \in \mathbb{Z} \quad\text{ whenever } d\sigma = 0$$

Let us call them "integral forms".

Motivated by the cup product on cohomology, my question/request is the following:

Give a direct proof that the wedge product $[\omega\wedge\eta]\in H^{k+l}(M,\mathbb R)$ of two integral forms $\omega\in \Omega^k(M)$ and $\eta\in \Omega^l(M)$ is again an integral form.

This should be true because the cup product is mapped to the wedge product, but the point of the exercise is to prove this statement directly, without constructing the singular cohomology $H^k(M,\mathbb Z)$ or homology first.

Maybe I also have to make sure that the condition of being an integral form is something that can be "checked effectively" without singular homology; this might be subject to a new question.

Greg Graviton
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  • I don't know what the wedge product of two integral forms is, would you mind explaining this in more detail. – Matt Calhoun Apr 01 '11 at 18:23
  • It's simply the [wedge product](http://mathworld.wolfram.com/WedgeProduct.html) of differential forms. – Greg Graviton Apr 01 '11 at 20:47
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    A basic idea would be to decompose a $k+l$ cycle as a $k$-dimensional family of $l$-cycles with a measure 0 set of degeneracies. For example a torus can be decomposed as a 1 dimensional family of 1-manifolds, some of which are circles and some of which are disjoint unions of two circles, with 4 degenerate points. (The standard Morse theoretic picture.) I don't know exactly how to make this precise. – Cheerful Parsnip Apr 06 '11 at 14:04

3 Answers3


$\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}$I'm not sure if this is an answer to the question, since it does refer to $H_{\ast}(M, \ZZ)$, but I think it sheds some interesting light on why the problem is hard.

We begin with a flawed proof attempt. Write $\delta$ for the diagonal map $M \to M \times M$, and $\pi_1$ and $\pi_2$ for the projections from $M \times M$ onto its first and second factor. Let $\rho$ be an integer $k+\ell$-cycle and let $\alpha$ and $\beta$ be a $k$-form and an $\ell$-form on $M$. Then $\int_{\rho} \alpha \wedge \beta = \int_{\delta(\sigma)} \pi_1^{\ast}(\alpha) \wedge \pi_2^{\ast}(\beta)$. Suppose $\rho$ were homologous in $M \times M$ to $\sum \sigma_i \times \tau_i$, for various cycles $\sigma_i$ and $\tau_i$ in $H_{\ast}(M)$, with $\dim \sigma_i + \dim \tau_i=k+\ell$. Then we would have $$\int_{\rho} \alpha \wedge \beta = \sum \int_{\sigma_i \times \tau_i} \pi_1^{\ast}(\alpha) \wedge \pi_2^{\ast}(\beta) =\sum_{(\dim \sigma_i, \dim \tau_i) = (k, \ell)} \int_{\sigma_i} \alpha \int_{\tau_i} \beta.$$ This would prove the result. (The integrals over terms where $\dim \sigma_i \neq k$ would drop out. If $\dim \sigma_i<k$, then $\pi_1^{\ast}(\alpha)|_{\sigma_i \times \tau_i}=0$ so $\int_{\sigma_i \times \tau_i} \pi_1^{\ast}(\alpha) \wedge \pi_2^{\ast}(\beta)=0$, and likewise if $\dim \tau_i < \ell$.)

Unfortunately, this need not be true. Saying that $\delta(\rho)$ should always be homologous to such a class is saying that $\delta_{\ast} H_m(M)$ should lie in the image of $\bigoplus_{i+j=m} H_i(M, \ZZ) \otimes H_j(M, \ZZ) \to H_{m}(M \times M, \ZZ)$. But the map $\bigoplus_{i+j=m} H_i(M, \ZZ) \otimes H_j(M, \ZZ) \to H_{m}(M \times M, \ZZ)$ need not be surjective -- the cokernel is $\bigoplus_{i+j=m-1} \mathrm{Tor}_1(H_i(M, \ZZ), H_j(M, \ZZ))$ by the Kunneth theorem. There is no reason that $\delta_{\ast}$ needs to land in the image of $\bigoplus H_i \otimes H_j$; I give an example where it doesn't in this Mathoverflow thread,

So we have to work harder. The good news is that the full statement of Kunneth's theorem includes the statement that $$0 \to \bigoplus_{i+j=m} H_i(M, \ZZ) \otimes H_j(M, \ZZ) \to H_{m}(M \times M, \ZZ) \to \bigoplus_{i+j=m-1} \mathrm{Tor}_1(H_i(M, \ZZ), H_j(M, \ZZ)) \to 0$$ is (noncanonically) split. Choose such a splitting $$\iota: \bigoplus_{i+j=m-1} \mathrm{Tor}_1(H_i(M, \ZZ), H_j(M, \ZZ)) \to H_m(M, \ZZ).$$ Then we can write $\delta(\rho) = \sum \sigma_i \times \tau_i + \iota(\sum \phi_j)$ for some $\phi_j$ in Tor groups. But the good news is that Tor groups are torsion, so an integer multiple of $\phi_j$ is homologous to zero. This means that $\int_{\iota(\phi_j)} \pi_1^{\ast}(\alpha) \wedge \pi_2^{\ast}(\beta)$ must be zero. So we still have $\int_{\rho} \alpha \wedge \beta = \sum \int_{\sigma_i \times \tau_i} \pi_1^{\ast}(\alpha) \wedge \pi_2^{\ast}(\beta)$ and now the proof concludes as before. $\square$.

What I find interesting about this argument is two things:

(1) We never had to build cup product in $H^{\ast}(M, \ZZ)$, or prove its compatability with de Rham cohomology, we just tried to shove cycles around naively.

(2) Proving Kunneth with the full statement that the sequence is split is not hard. But I have only ever seen it done in a very algebraic way, not by explicitly writing a cycle on $X \times Y$ as $\sum \sigma_i \times \tau_j + \phi$ for some torsion $\phi$. Moreover, there is no canonical choice of splitting. This may explain why no one seems to be able to prove this in a natural way.

David E Speyer
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    Thanks David! I think this is actually as best an answer as we can hope for. The problem is that the condition for being an integral form uses homology with coefficients $\mathbb{Z}$ in an essential way -- it's not enough to know homology, say, over $\mathbb{C}$ to know what an integral form is. – Greg Graviton Apr 02 '16 at 14:16
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    On the issue of non-canonicity, it might be possible to avoid it by arguing that the mapping $\bigoplus_{i+j=k} H_i(M;\mathbb{Z})\otimes H_j(M;\mathbb{Z}) \to H_k(M\times M;\mathbb{Z}) $ is "surjective up to torsion" more directly, i.e. to throw away the torsion contributions before the need to exhibit a splitting. Geometrically, I would also be happy to restrict $M$ to be an orientable manifold. – Greg Graviton Apr 02 '16 at 14:19
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    $\def\RR{\mathbb{R}}\def\PP{\mathbb{P}}$I agree that you can directly argue that the map becomes surjective after taking torsion free quotients. But, vaguely speaking, I feel like geometric arguments about moving cycles around are more like trying to give a splitting, which might explain why we can't do it. Taking $M$ orientable doesn't work; as I just wrote over on MO, the diagonal image of $H_3(\RR \PP^3)$ in $H_3(\RR \PP^3 \times \RR \PP^3)$ is not in the image of $\bigoplus H_i \otimes H_j$, and $\RR \PP^3$ is orientable. – David E Speyer Apr 02 '16 at 16:13
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    Oh, so there exist orientable manifolds where some submanifolds are not orientable, I hadn't thought of that. – Greg Graviton Apr 03 '16 at 08:37

My cohomology is a bit rusty, so I'm sorry for any mistakes or for being sketchy. It may also be that my suggestion is too close to the singular cohomology solution you want to avoid, although I'll be using Cech cohomology instead.

Let's first make a covering $\mathcal{U}=(U_j)_{j\in I}$ of $M$ where all $U_j$, as well as $U_J=\bigcap_{j\in J} U_j$ for $J\subset I$, are contractible (or empty). The basic idea is to use the double complex $$ \begin{array}{ccccccccc} &&0&\rightarrow&C^0(\mathbb{R})&\rightarrow&C^1(\mathbb{R}) &\rightarrow&\cdots\\ &&\downarrow&&\downarrow&&\downarrow&&\downarrow\\ 0&\rightarrow&\Omega^0&\rightarrow&C^0(\Omega^0)&\rightarrow&C^1(\Omega^0) &\rightarrow&\cdots\\ &&\downarrow&&\downarrow&&\downarrow&&\downarrow\\ 0&\rightarrow&\Omega^1&\rightarrow&C^0(\Omega^1)&\rightarrow&C^1(\Omega^1) &\rightarrow&\cdots\\ &&\downarrow&&\downarrow&&\downarrow&&\downarrow\\ \end{array} $$ where $C^q(\Omega^p)=C^q(\mathcal{U},\Omega^p)$ consists of $(\omega_J)_{J\in I_p}$ where $I_p$ is the set of size $p+1$ subsets of $I$ and $\omega_J\in\Omega^p(U_J)$: i.e. the setup for defining the Cech cohomology. We then have maps $\delta:C^q(\Omega^p)\rightarrow C^{q+1}(\Omega^p)$, and $d:C^q(\Omega^p)\rightarrow C^q(\Omega^q)$.

For convenience, I'll pretend $C^q(\Omega^{-1})=C^q(\mathbb{R})$ and $C^{-1}(\Omega^p)=\Omega^p$ (i.e. global $p$-form which I should probably have denoted $\Gamma(\Omega^p)$ instead of just $\Omega^p$).

Note that $H^p(C^q(\Omega^\bullet))=0$ for $p\ge 0$ since closed forms are exact on all the contractible sets $U_J$.

The proof that $H^k_\text{dR}(M)=H^k(\Omega^\bullet)$ is equal to $H^k(M;\mathbb{R})=H^k(C^\bullet(\mathbb{R}))$ is then done by diagram chasing. Start with $\omega\in\Omega^k$ with $d\omega=0$. This maps to $(\omega^{(0)}_J)_{J\in I_0}\in C^0(\Omega^k)$ by restriction of $\omega$ to each $U_J$, $J\in I_0$. Since $d\omega^{(0)}_J=0$, we have $\omega^{(0)}_J=d\upsilon^{(0)}_J$ for some $\upsilon^{(0)}_J\in\Omega^{k-1}(U_J)$. This $(\upsilon^{(0)}_J)$ is then mapped to $\omega^{(1)}\in C^1(\omega^{k-1})$. And so we go, alternatingly moving up and right, until we get an element of $C^k(\mathbb{R})$.

Similar diagram chasing is used to show that this mapping is one-to-one module the images $d\Omega^{k-1}$ and $\delta C^{k-1}(\mathbb{R})$. (Assuming I haven't mixed things up here.)

A $k$-form $\omega$ is integral in $H^k_\text{dR}(M)\equiv H^k(M;\mathbb{R})$ if it can be mapped by the above procedure to an element in $C^k(\mathbb{Z})$.

So, here's the trick. Let $\omega$ and $\nu$ be closed $p$- and $q$-forms resp. corresponding to integral cohomology classes. We can then replicate the entire diagram chasing for $\omega$, but apply it to $\omega\wedge\nu$ where through each step we leave $\nu$ unchanged, finally mapping $\omega$ to an element $\omega^{(p)}\in C^p(\mathbb{Z})$. In this way, we chase $\omega\wedge\nu$ to an element in $C^p(\Omega^q)$ which looks like the restriction of $\nu$ to $U_J$ sets, multiplied by the integer values of $\omega^{(p)}_J$.

Afterwards, we continue to do the same with $\nu$: the $\omega^{(p)}=(\omega^{(p)}_J)_{J\in I_p}$ part now consist of just constants by which the forms are multiplied.

As I warned, my cohomoloogy is somewhat rusty, so I may quite likely have missed a number of technical issues. However, I think this basic approach should work. And it should be a good excercise in diagram chasing.

Mike F
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Einar Rødland
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I mostly use cohomology as a blackbox so I may be overlooking technical difficulties but here's how I'd try to prove your statement.

First I would note that being integral is a local property: the class $[\omega]$ of a closed form is integral iff its restrictions $[\omega|_{U_i}]$ to a covering $M = \bigcup U_i$ is integral. One way is obvious: if $\sigma \subset U_i$, then $\int_\sigma \omega|_{U_i} = \int_\sigma \omega$. For the other, we can chose a (finite) triangulation $\sigma = \bigcup \sigma_\alpha$ such that each $\sigma_\alpha$ is included in one of the $U_i$ so that we can write $\int_\sigma \omega = \sum_\alpha \int_{\sigma_{\alpha}} \omega|_{U_{i(\alpha)}}$.

The point is that you can check integrality by using the triangulation you like.

Then I would write the cup product as $H^i_{dR}(X) \otimes H^j_{dR}(X) \to H^{i+j}_{dR}(X\times X) \to H^{i+j}_{dR}(X)$ the external product followed by restriction to the diagonal. And I'd check that each map sends integral integral classes to integral classes using triangulations.

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    I'm not sure what you mean, but if it's right it must not be what is written. If your $U_i$ are contractible, they will have no cohomology (beyond degree 0, of course). – Max Mar 29 '11 at 23:11
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    Why are the summands $\int_{\sigma_\alpha} \omega |_{U_{i(\alpha)}}$ integers? You are saying that each of the simplexes $\sigma_\alpha$ can be chosen to be closed? – Greg Graviton Mar 30 '11 at 06:20
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    How being integral can be a local property if cohomology itself is a global thing? :) – Alexander Shamov Aug 08 '12 at 21:47