# Spectral sequence of a bicomplex equipped with a group action

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 $\widetilde X\times_GY$.

• 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). Nov 23 '14 at 21:31
• It is tensor product over $\mathbb{C}$, and yes, i wrote $\oplus$ by mistake, thanks a lot. Nov 23 '14 at 21:34
• The answer is not positive in general, not even for $G$ finite. Take $G=\{1\}$. You can also construct more involved examples, of course. Feb 23 '16 at 11:36
• @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. Feb 23 '16 at 12:09
• @SebastianGoette You're right, I somehow missed the track of the fact that we have a tensor product. Feb 23 '16 at 12:12

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.)

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