Cup and wedge product in singular and de Rham cohomology

Question

De Rham's theorem asserts that the map $I: H_{dR}^p(M) \to H_{sing}^p(M, \mathbb{R})$ defined as $$I(\omega)= [\sigma^p] \mapsto\int_{\sigma^p}\omega $$ is an isomorphism ($\sigma^p \in [\sigma^p] $ is a smooth representative).

On $H_{sing}^\ast(M, \mathbb{R})$ it is defined a cup product $\cup : H_{sing}^p(M, \mathbb{R})\times H_{sing}^q(M, \mathbb{R})\to H_{sing}^{p+q}(M, \mathbb{R})$ $$ \omega^q\cup \eta^p (\sigma^{p+q}) = \omega(\sigma|_{[e_0,\dots,e_q]})\cdot \eta(\sigma|_{[e_{q},\dots,e_{q+p}]}),$$

while on $H_{dR}^\ast(M)$ we have the wedge product: $(\omega^q, \eta^p) \mapsto \omega\wedge \eta$.

Are these two product the same up to isomorphism? In other words, I would like to prove that $I(\omega\wedge \eta) = I(\omega)\cup I(\eta)$.

This seems possible since the two products satisfy the same commutativity relations and they have the same behaviour wrt the codifferential $d$.

So I tried to prove -without success- that $$ \int_{\sigma^{q+p}}\omega^q\wedge \eta^p = \int_{\sigma|_{[e_0,\dots,e_q]}}\omega^q \cdot \int_{\sigma|_{[e_{q},\dots,e_{q+p}]}}\eta^p $$ but the standard simplex is not a product space so I cannot apply -as my intuition suggests- a Fubini type argument.

Any suggestion?

Answer 1

The deRham map gives an arrow $\theta:H_{dR}(X) \to H(X)^*$, and the algebra structure on the right is induced from the coalgebra structue on $H(X)$ given by the Alexander-Whitney coproduct: $\Delta(\sigma) = \sum \sigma_i \otimes \sigma^i$ where $\sigma_i$ is the front $i$-face and $\sigma^i$ is the back $i$-face. Denote by $\langle -,-\rangle : H_{dR}(X)\otimes H(X)^*\to \mathbb R$ pairing that gives rise to $\theta$.

In terms of the pairing, you are correct this is the same as stating that, for any two forms $\omega,\eta$ of degee $p,q$ and any singular chain $\sigma$ of degree $p+q$, we have that $$\tag{1}\langle \omega\wedge \eta,\sigma\rangle = \langle \omega,\sigma_p\rangle\langle \eta,\sigma^p\rangle$$ which is the same as $\langle \mu,-\rangle =\mu_\mathbb R\langle - \otimes -,\Delta\rangle$. You also note that there is a problem of writing simplices as products. There is a way of fixing this problem, which is (as Spivak does and in another context, Serre does), for example, done by working with singular cubes instead of singular chains.

Concretely, compute $H_*(X)$ using maps $\sigma : [0,1]^*\to X$. The differential is another, but the coproduct is simply given by restricting to coordinates, as in the case of simplices (though more terms appear). Concretely, if $\sigma : [0,1]^p\to M$ is a singular cube, the coproduct of $\sigma$ is given by a sum

$$\sum_{A\sqcup B = [p]}\sigma_A^0\otimes \sigma_B^1 $$ where $(-)_S^\varepsilon$ means you set the $S$-coordinates of the subset $S$ of $[p] = \{1,\ldots,p\}$ to $\varepsilon$ and keep the coordinates of the complement. See page 17 in Serre's thesis. In this case, if $\sigma$ is a singular cube and $\omega = fdx_I$ is a form of the same degree $p$, it is straightforward to check that

$$\langle \omega,\sigma\rangle = \int_{[0,1]^p} f(\sigma) J_\sigma dx_1\cdots dx_p$$

and then $(1)$ is an immediate consequence of Fubini's theorem, as you wanted. There is a natural isomorphism between $H(X)$ computed in two different ways, either from cubical or simplicial singular chains, coming from a map from the cubical complex of $X$ to the singular complex of $X$ that needs to respect the coproduct (this is the nontrivial part of you question, I think, but you can find this map say in Serre's thesis), and this completes the proof that the deRham isomoprhism is a map of algebras.

Answer 2

I just wanted to point out that $$\int_{\sigma^{q+p}}\omega^q\wedge \eta^p \neq \int_{\sigma|_{[e_0,\dots,e_q]}}\omega^q \cdot \int_{\sigma|_{[e_{q},\dots,e_{q+p}]}}\eta^p$$ in general, since simplices are not closed. The formula $$\omega^q\cup \eta^p (\sigma^{p+q}) = \omega(\sigma|_{[e_0,\dots,e_q]})\cdot \eta(\sigma|_{[e_{q},\dots,e_{q+p}]})$$ is valid only for singular simplex $\sigma:\Delta^{p+q}\to M$.