I've been looking around for a standard treatment of what I think sould be called "equivariant cup product in singular cohomology", but couldn't find anything promissing.

I did played around with what I expect this to be, so maybe you can tell me whether

(a) this is known and written out somewhere (idealy in some axiomatic way), or

(b) this is the right way to define the cup product in this setting.

So let's get started:

Consider an oriented finite dimensional smooth Riemannian manifold $(M,g)$ together with a regular covering map $$\pi \colon \tilde{M} \to M,$$ where $\tilde{M}$ is equipped with the pullback metric $\tilde{g}:=\pi^*g$.

Let us assume that the group of deck transformations $\Gamma$ is isomorphic to $\mathbb{Z}^k$, for some $k$ positive integer. Then we can define the cochain complex $$\left(S^{\bullet}(\tilde{M};\mathbb{Z}) \otimes_{\mathbb{Z}[\Gamma]}N, \delta \otimes \text{id}\right),$$ where $\mathbb{Z}[\Gamma]$ is the group ring w.r.t. $\Gamma$ and $N$ is some module over this group ring (note that we don't have to specify left and right actions, for $\mathbb{Z}[\Gamma]$ is commutative). $S^{\bullet}$ obviously denotes singular cohomology.

My idea now was to define a "cochain cup product" on this chain complex as follows:

$$\cup \colon \left(S^p(\tilde{M};\mathbb{Z})\otimes_{\mathbb{Z}[\Gamma]}N\right) \otimes_{\mathbb{Z}[\Gamma]}\left(S^q(\tilde{M};\mathbb{Z})\otimes_{\mathbb{Z}[\Gamma]}N\right) \longrightarrow S^{p+q}(\tilde{M};\mathbb{Z})\otimes_{\mathbb{Z}[\Gamma]}\left(N\otimes_{\mathbb{Z}[\Gamma]}N\right),$$ where $$(\varphi \otimes n_1)\cup (\psi \otimes n_2):=(\varphi \cup \psi) \otimes (n_1\otimes n_2)$$ and the cup on the RHS is the usual cup product in singular cohomlogy.

Note that lots of the assumption I made are superfluous or could be phrased more generally, but this is more or less the setting I'm working with (maybe it turns out that in this setting some complications become trivial and so on...).

And $\mathbb{Z}_2$ coefficients are sufficient for what I need (maybe this also makes things easier as $\mathbb{Z}_2$ is a field).

Every helpful comment is greatly appreciated, thank you!


It seems that my guess was incorrect. In Eilenberg and Steenrod's "Foundations of Algebraic Topology" on page 209 there is a definition of equivariant homology AND cohomology. Applying this general definition to our special case gives us the following equivariant cochain complex


Also, they claim in the book that equivariant homology/cohomology is a homology theory (where everything is equivariant, roughly speaking). So maybe understanding the corresponding Eilenberg-Zilber morphism and the induced map coming from the diagonal embedding (or more generally any diagonal approximation) migth lead to an answer...


I found some more papers by Eilenberg and by Steenrod. Eilenberg gives a definition of an equivariant cup product in equivariant singular cohomology (see "Homology of Spaces with Operators"), but he uses an extra assumption that ensures that the cup product of two equivariant cochains is an equivariant cochain again. This assumption however, does not hold in my setting (if I understand correctly).

The paper by Steenrod "Homology with Local Coefficients" gives a definition using a CW-decomposition and cellular homology, which seems too far away from what I possibly want.


1 Answer 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)...


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.


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