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Let $(A_1^\bullet,\partial_1)$ and $(A_2^\bullet,\partial_2)$ be complexes of $\mathbb{C}$-vector spaces.

We suppose that $(A_1^\bullet,\partial_1)$ and $(A_2^\bullet,\partial_2)$ are equipped with an action of a discrete group $G$, i.e., $G$ acts on $A_1^\bullet$ (resp. $A_2^\bullet$) by preserving the grading and commuting with $\partial_1$ (resp. $\partial_2$).

Then $A_1^\bullet$ and $A_2^\bullet$ are $\mathbb{C}[G]$-module. We suppose further that $A_1$ is a free $\mathbb{C}[G]$-module.

Let $(A^\bullet,\partial)$ be the product complex, i.e, $A^k= \bigoplus_{p+q=k}A_1^p\otimes_\mathbb{C} A_2^q$ and $\partial = \partial_1+(-1)^q\partial_2$.

Let $F^p = \bigoplus_{p'\geqslant p}A_1^{p'}\otimes_\mathbb{C} A_2^q$ be the classical filtration.

Let $(E_r)$ be the spectral sequence associated to $((A^\bullet)^G,\partial,F^\bullet)$.

The question : does $(E_r)$ converge at $r=2$ ?

PS

I have known that the answer is positive if $G$ is a finite group.

A comprehensible example for the above construction is : let $X$ and $Y$ be topological manifolds, let $G=\pi_1(X)$, we suppose that $G$ acts on $Y$, let $\widetilde{X}$ be the universal cover of $X$, let $A^\bullet_1$ be the (simplical, Cech or de Rham) complex of $\widetilde{X}$, $A^\bullet_2$ be the complex of $Y$, then $(A^\bullet)^G$ is the complex of $\widetilde X\times_GY$.

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  • $\begingroup$ Is this tensor product over $\mathbb C$ or over $\mathbb C[G]$? (I assume you must mean $\otimes$ instead of $\oplus$ in the definition of the product complex). $\endgroup$ Nov 23 '14 at 21:31
  • $\begingroup$ It is tensor product over $\mathbb{C}$, and yes, i wrote $\oplus$ by mistake, thanks a lot. $\endgroup$ Nov 23 '14 at 21:34
  • $\begingroup$ The answer is not positive in general, not even for $G$ finite. Take $G=\{1\}$. You can also construct more involved examples, of course. $\endgroup$ Feb 23 '16 at 11:36
  • $\begingroup$ @FernandoMuro I don't understand your comment (or maybe I don't understand the question correctly). $G=\{1\}$ gives the algebraic Künneth formula at $E_2$, which holds without Tor terms because $\mathbb C$ is a field. Note however that the tensor product of singular cochain complexes is not the singular cochain complex of the product, so the example at the end is misleading. $\endgroup$ Feb 23 '16 at 12:09
  • $\begingroup$ @SebastianGoette You're right, I somehow missed the track of the fact that we have a tensor product. $\endgroup$ Feb 23 '16 at 12:12
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In the following I'm going to assume that $G$ is an infinite group. Then the answer to your question is yes, but probably not for the reason that you expect.

Let's consider your filtration on $(A^\bullet)^G$. By definition we have $$ (F^p)^G = \bigoplus_{p' \geq p} (A_1^p \otimes A_2^q)^G $$ because taking invariants commutes with direct sums. The $E_0$-term of this spectral sequence is, in degree $(p,q)$, the quotient $$ ((F^p)^G/(F^{p+1})^G)^{p+q} = (A_1^p \otimes A_2^{q})^G. $$ The $d_0$-differential is induced by the boundary map on $A_2^\bullet$.

By assumption, for each $p$ the group $A_1^p$ is a free $\Bbb C[G]$-module. Therefore, $$ (A_1^p \otimes A_2^q)^G = \bigoplus (\Bbb C[G] \otimes A_2^q)^G. $$

Given a $\Bbb C[G]$-module $M$, any element $x \in \Bbb C[G] \otimes M$ can be written uniquely as a sum $\sum_{g \in G} g \otimes a_g$ for some unique $a_g \in M$, all but finitely many of which are nonzero. Because the group $G$ is infinite, the only way that such an element can be invariant under the $G$-action is if it is zero.

Therefore, the spectral sequence is zero at the $E_0$-term, and so it collapses pretty quickly.

(Based on your description, however, it sounds like you might not actually want free modules: the cochains on $\widetilde X$ are usually not free, but instead are a product of coinduced modules $\Bbb C^G \cong \prod_{g \in G} \Bbb C$. This is isomorphic to $\Bbb C[G]$ if $G$ is finite. You have to be a little bit careful then, because there may not be an equivalence $C^*(\widetilde X \times Y) \simeq C^*(\widetilde X) \otimes C^*(Y)$ without either assuming $Y$ is equivalent to a finite complex or that you're taking a completed tensor product.)

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    $\begingroup$ So, the sensible thing to ask for would be a Leray-Serre spectral sequence in homology, which would involve coinvariants instead of invariants? $\endgroup$ Feb 23 '16 at 17:45
  • $\begingroup$ @SebastianGoette Yes, that seems most likely. $\endgroup$ Feb 23 '16 at 18:54

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