# Cohomology ring on $\Sigma_g\times\Sigma_h$

I want to determine the cohomology ring structure of the space $$X=\Sigma_g\times\Sigma_h$$, where $$\Sigma_g$$, $$\Sigma_h$$ denote the orientable closed surface of genus $$g$$ and $$h$$, respectively. I already know the cup product of the torus $$T^2$$: if $$a$$ and $$b$$ are generators of $$H^1(T^2)\cong\mathbb{Z}^2$$, we have that $$a\cup b=-b\cup a=z$$, where $$z$$ is a generator of $$H^2(T^2)$$ and $$a^2=b^2=0$$. All other $$\cup: H^k\times H^l \to H^{k+l}$$ are trivial. We may write this as $$H^*(T^2)=\frac{\mathbb{Z}[a,b]}{(a^2,b^2,ab+ba)},\deg(a)=\deg(b)=1$$. Since $$\Sigma_g$$ is a connected sum of $$g$$ tori, we get \begin{align} H^*(\Sigma_g)=\frac{\mathbb{Z}[a_1,b_1,\dots, a_g,b_g]}{J}, \end{align} where J is the ideal generated by $$a_i^2,b_i^2,a_ia_j,b_ib_j,a_ia_j,a_ib_i+b_ia_i$$ with $$1\leq i,j\leq g$$, $$i \neq j.$$ Let us say $$H^*(\Sigma_h)$$ is obtained in the same way with writing $$c_i,d_i$$ instead of $$a_i,b_i$$ and calling the the respective ideal $$K$$ instead of $$J$$. Now I want to use Künneth formula, which provides us with a ring isomorphism \begin{align} H^*(\Sigma_g)\otimes H^*(\Sigma_h)\cong H^*(\Sigma_g\times\Sigma_h). \end{align} I do not understand yet how to explicitly evaluate this tensor product though, is it just \begin{align} H^*(\Sigma_g\times\Sigma_h)=\frac{\mathbb{Z}[a_1,b_1,\dots, a_g,b_g,c_1,d_1,\dots,c_h,d_h]}{J,K}, \end{align} or does this tensor product act in a more complicated way on these rings? I have only dealt with the tensor product in very easy special cases like finitely presented abelian groups so far, so I do not really have an intuition for what it means for rings. Is this generally the right way to go about this problem?

## 1 Answer

You also need to kill the relators $$a_ic_j+c_ja_i, b_ic_j+c_jb_i, a_id_j+d_ja_i,b_id_j+d_jb_i$$, for $$1\leq i\leq g$$, $$1\leq j\leq h$$ .

• Thanks for the answer, in what way is this ideal I have to devide out related to $J$ and $K$? May 30, 2020 at 20:29
• It is $J+K$ plus commutator relations between the generators of the two factors. It is the same as how you get the cohomology ring of the torus from 2 copies of a circle $\langle a|a^2\rangle$ and $\langle b|b^2\rangle$ result in $\langle a,b|a^2,b^2,ab+ba\rangle$
– tkf
May 30, 2020 at 21:00
• But how do I know what the relations between the generators of the two factors are, because in the case of the torus I can explicitly calculate ab=-ba (or see it geometrically using Poincaré duality and intersection forms), but here I really should try to get this information from the tensor product somehow, right? May 30, 2020 at 21:22
• The additional relations are all implied by the fact that cohomology rings are graded commutative. That is $ab=ba$ whenever one of $a$ or $b$ has even degree. If they are both of odd degree then $ab=-ba$.
– tkf
May 30, 2020 at 21:36