Reference Request: Equivariant cup product in singular cohomology

Solution 1:

I think I have about 70% of a complete answer now. The main idea basically comes from chapter 1 in this paper by Greenblatt.

Let me change our setting:

Consider a finite dimensional manifold $M$ together with a (finite) CW-decomposition. Let $$p \colon \tilde{M} \to M$$ be a regular cover with an abelian group of deck transformations (so that we don't have to worry about left and right actions). Denote $$\pi:= \text{Deck}(\tilde{M}).$$

Let $C_{\bullet}(\tilde{M})$ be the cellular chain complex (note that $\tilde{M}$ has CW-decomposition because $M$ has one). Let $G$ be a $\pi$-module.

Now we can define the equivariant cochain complex $$\hom_{\mathbb{Z}[\pi]}\left(C_{\bullet}(\tilde{M}),G\right).$$

Note the following: The group ring $\mathbb{Z}[\pi \times \pi]$ is isomorphic to $\mathbb{Z}[\pi] \otimes \mathbb{Z}[\pi]$. Thus we can define the equivariant cochain complex $$\hom_{\mathbb{Z}[\pi \times \pi]}\left(C_{\bullet}(\tilde{M}) \otimes C_{\bullet}(\tilde{M}), G \otimes G\right).$$

This allows us to define a cochain map $$ \bigoplus_{k=p+q}\hom_{\mathbb{Z}[\pi]}\left(C_p(\tilde{M}),G\right) \otimes \hom_{\mathbb{Z}[\pi]}\left(C_q(\tilde{M}),G\right) \to \hom_{\mathbb{Z}[\pi \times \pi]}\left( \bigoplus_{k=p+q}C_p(\tilde{M}) \otimes C_q(\tilde{M}), G \otimes G\right),$$ where $f_1 \otimes f_2$ gets sent to, say $\Phi(f_1 \otimes f_2)$ which is a cochain that is defined by $$\Phi(f_1 \otimes f_2)(\sigma_1 \otimes \sigma_2):=f_1 \sigma_1 \otimes f_2 \sigma_2.$$ One should note that the cochain $\Phi(f_1 \otimes f_2)$ is equivariant:

For $\lambda=\lambda_1 \otimes \lambda_2 \in \mathbb{Z}(\pi \times \pi)$ we get $$\lambda \cdot \Phi(f_1 \otimes f_2)(\sigma_1 \otimes \sigma_2)= \lambda \cdot (f_1 \sigma_1 \otimes f_2 \sigma_2)=f_1(\lambda_1 \cdot \sigma_1) \otimes f_2(\lambda_2 \cdot \sigma_2)=\Phi(f_1 \otimes f_2)(\lambda \cdot (\sigma_1 \otimes \sigma_2)).$$

Another direct computation shows that $\Phi$ is a cochain map, i.e. $$\delta \Phi= \Phi \delta.$$

At this point we wish to replace the chain complex $(C_{\bullet}(\tilde{M}) \otimes C_{\bullet}(\tilde{M}))_k$ with $C_k(\tilde{M} \times \tilde{M})$. This can be done by means of an Eilenberg-Zilber morphism, which can be defined in a nice way in the setting of a cellular chain complex, namely: $$\text{EZ} \colon C_k(\tilde{M} \times \tilde{M}) \to (C_{\bullet}(\tilde{M}) \otimes C_{\bullet}(\tilde{M}))_k, \; \sigma_k=\sigma_p \times \sigma_q \mapsto \sigma_p \otimes \sigma_q.$$

For the sake of completeness, observe that the obvious map $$\Psi \colon \hom_{\mathbb{Z}[\pi]}\left(C_p(\tilde{M}),G\right) \otimes \hom_{\mathbb{Z}[\pi]}\left(C_q(\tilde{M}),G\right) \to \bigoplus_{k=p+q}\hom_{\mathbb{Z}[\pi]}\left(C_p(\tilde{M}),G\right) \otimes \hom_{\mathbb{Z}[\pi]}\left(C_q(\tilde{M}),G\right)$$ induces a map in homology, simply because $$\delta \Psi(f_1\otimes f_2)(\sigma_1 \otimes \sigma_2)=(\delta f_1)\otimes f_2(\sigma_1 \otimes \sigma_2) + (-1)^{\vert \sigma_1 \vert}f_1 \otimes (\delta f_2)(\sigma_1 \otimes \sigma_2).$$

This allows us to define what we will call an equivariant cross product by composing the three maps $\text{EZ},\Phi$ and $\Psi$ as follows: $$f_1 \times f_2:=\text{Ez}^* \circ \Phi \circ \Psi \left(f_1 \otimes f_2\right),$$ where $\text{EZ}^*$ is the equivariant dual of $\text{EZ}$, i.e.

$$\text{EZ}^* \colon \hom_{\mathbb{Z}[\pi \times \pi]}\left((C_{\bullet}(\tilde{M})\otimes C_{\bullet}(\tilde{M}))_k,G\otimes G\right) \to \hom_{\mathbb{Z}[\pi \times \pi]}\left(C_k(\tilde{M} \times \tilde{M}),G \otimes G \right),$$ such that $$\text{EZ}^*(f_1\otimes f_2)(\lambda \cdot \tilde{\sigma} )= f_1 \otimes f_2 (\lambda \cdot \text{EZ}(\tilde{\sigma}))=\lambda \cdot \text{EZ}^*(f_1 \otimes f_2)(\tilde{\sigma}).$$

Usually at this point one would take the diagonal embedding $$\Delta \colon \tilde{M} \to \tilde{M} \times \tilde{M}$$ and compose the cross product with the induced map $\Delta^*$ to define a cochain level cup product, BUT the diagonal embedding is not a cellular map, i.e. it does not send cells to cells (in general). Therefore, $\Delta$ does not induce a chain map on the cellular complex.

By means of the Cellular Approximation Theorem (see Hatcher page 349. Theorem 4.8) we can replace the diagonal embedding $\Delta$ with cellular map, say $$\Delta_{\text{cell}} \colon \tilde{M} \to \tilde{M} \times \tilde{M}$$ that is homotopic to $\Delta$.

This is as far as I got. I suspect that using $\Delta_{\text{Cell}}$ one can define an equivariant dual $\Delta_{\text{Cell}}$ in a similar fashion to $\text{EZ}$ and $\text{EZ}^*$, but one needs to be more careful with the coefficients and so on (maybe using free products $G*G$ and $\mathbb{Z}[\pi]* \mathbb{Z}[\pi]$ would work)...

Edit:

Our map $\Phi$ is basically the natural chain map from Lemma 1.8 in Greenblatts Paper (see reference above). There he assumes that the chain complex needs to be finitely generated, which I don't think is the case for the cellular complex of a cover $\tilde{M}$ as above. I think (but I need to check) he only uses this assumption to obtain that $\Phi$ is an isomorphism, which we don't need.

If this is the case however, then maybe we could work with the singular chain complex of $\tilde{M}$ and any choice of a corresponding Eilenberg-Zilber morphism from the get go. This would have the advantage of sidestepping the cellular approximation argument and would eventually lead to a more concrete equivariant cup product.