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.



Tensor product of spectral sequences?
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*.


