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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?

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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 $1\leq i\leq g$, $1\leq j\leq h$ .

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  • $\begingroup$ Thanks for the answer, in what way is this ideal I have to devide out related to $J$ and $K$? $\endgroup$ May 30, 2020 at 20:29
  • $\begingroup$ 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 $\langle a,b|a^2,b^2,ab+ba\rangle$ $\endgroup$
    – tkf
    May 30, 2020 at 21:00
  • $\begingroup$ 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? $\endgroup$ May 30, 2020 at 21:22
  • $\begingroup$ 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 $ab=-ba$. $\endgroup$
    – tkf
    May 30, 2020 at 21:36

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