Tensor product of spectral sequences?

Question

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.

Answer 1

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.

You'll find more details, for example, in McLeary's User guide.