# Tensor product of spectral sequences?

I'm wondering about a cross product for spectral sequences. I've got an idea, and I wonder if it is written up anywhere, or if it even holds water.

Let's start with three spectral sequences, $E, F$ and $G$. Assume that $G_1^{*,*} \cong E_1^{*,*}\otimes F_1^{*,*}$ as chain complexes. Then the ordinary Künneth theorem gives us a map

$\times_2: E_2^{*,*} \otimes F_2^{*,*} \to G_2^{*,*}.$

Now $E_2^{*,*} \otimes F_2^{*,*}$ has a differential -- the standard one for the tensor product of chain complexes, and I guess I have to hope that $\times_2$ is a chain map. Given this, we apply Künneth again, and get

$\times_3: E_3^{*,*} \otimes F_3^{*,*} \to H^{*,*}( E_2^{*,*} \otimes F_2^{*,*}) \to G_3^{*,*}.$

Repeating the process leads to cross products $\times_r :E_r^{*,*} \otimes F_r^{*,*} \to G_r^{*,*}$ and presumably converging to the appropriate cross product at the end.

There is no reqson for $\times_2$ to be a chain map. Pick for example a spectral sequence such that the differential in $E_1$ is zero, such that the one on $d_2$ is not, and pick a any product $\times_1$ on $E_1$ such that it is not a chain map on $E_2=E_1$.
You can get such structures, though. For example, suppose that $X$, $Y$ qnd $Z$ are filtered complexes and that $m:X\otimes Y\to Z$ is a chain map compatible with the filtrations. Then if $E_X$, $E_Y$ and $E_Z$ are the spectral sequences gotten from $X$, $Y$ and $Z$, respectively, $m$ gives you maps $m_r:E^X_r\otimes E^Y_r\to E^Z_r$, all compatible with the differentials.
This is not exactly an answer to your question, instead i have some advice on how to think about spectral sequences. I think one of the main reasons you have the negative above answer is because it is not coming from geometry, or at least the structure you are asking about is not appear to be coming from geometry. What really helped me was going through Mosher and Tangora's section on spectral sequences. Massey's Exact Couples clarify this difference a bit. The structure in the spectral sequences come from the A's in the exact couples, not the E's. Recall that an exact couple is given by 3 maps $i:A \to A$, $j:A \to E$, and $k: E \to A$ such that the diagram we get is exact at each point. The differentials in the spectral sequence is $d=jk:E \to E$. This is just the first differential, the others are given by using the "derived" $j$ and $k$, that is $d^{r-1}=j^{r-1}k^{r-1}$. I am sure you already know this stuff, just wanted to leave it here though as a pointer for others.