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 K"unneth 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 K"unneth 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.

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.