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 $\widetilde X\times_GY$.

notpositive in general, not even for $G$ finite. Take $G=\{1\}$. You can also construct more involved examples, of course. $\endgroup$algebraicKünneth 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. $\endgroup$