I want to determine the ring structure of $H^\ast(\mathbb{R}P^3 \times \mathbb{R}P^3; \mathbb{Z}/p)$ for prime $p$.
If $p$ is an odd prime, then each $H^k(\mathbb{R}P^3; \mathbb{Z}/p)$ is zero (by looking at the cellular chain complex). So I'm really looking at the case $p=2$. Then I have $H^k(\mathbb{R}P^3;\mathbb{Z}/2)=\mathbb{Z}/2$ when $k=0,1,2,3$ and zero otherwise.
I know the Kunneth Formula for Cohomology: $H^k(X \times Y) \cong \bigoplus\limits_{p+q=n} H^p(X) \otimes H^q(Y) \bigoplus\limits_{p+q=n+1} \mbox{Tor}(H^p(X),H^q(Y))$.
Since $\mathbb{Z}/2 \otimes \mathbb{Z}/2 \cong \mathbb{Z}/2$, Tor$(\mathbb{Z}/2, \mathbb{Z}/2)=\mathbb{Z}/2$, and Tor$(0, \mathbb{Z}/2)=0$,
I get $H^0(\mathbb{R}P^3 \times \mathbb{R}P^3; \mathbb{Z}/2)\cong \mathbb{Z}/2$,
$H^1(\mathbb{R}P^3 \times \mathbb{R}P^3; \mathbb{Z}/2)\cong (\mathbb{Z}/2)^5$ where two copies come from tensor product of cohomology groups and 3 copies come from Tor, $H^2(\mathbb{R}P^3 \times \mathbb{R}P^3; \mathbb{Z}/2)\cong (\mathbb{Z}/2)^7$ where 4 copies come from the product of cohomology groups an 3 copies come from Tor. I can keep doing this until the dimension of the largest cell, 9, but my main question is what can I do next to actually determine the ring structure?
In addition to that, this method seems pretty tedious. Is there a better route I can take toward calculating the cohomology groups of a direct product?
Any help is appreciated!
Best Answer
Note: I want to add this as a remark
Since you are working over field coefficient, one can apply Kunneth Formula without any Tor term. Hence we have:
$H^{*}(\mathbb{RP}^{3}\times\mathbb{RP}^{3};\mathbb{Z}/p)\cong H^{*}(\mathbb{RP}^{3};\mathbb{Z}/p)\otimes H^{*}(\mathbb{RP}^{3};\mathbb{Z}/p)$.
Inorder to calculate the cohomology $H^{*}(\mathbb{RP}^{3};\mathbb{Z}/p)$ use the restriction map $q:\mathbb{Z}\rightarrow\mathbb{Z}/p$.