Cohomology ring of a product of two spheres ($H^{\ast }(S^n\times S^n;\mathbb{Z})$)

For positive $n$, let $x,y$ be the pullbacks of the orientation classes from the two factors. Then $x^2=y^2=0$, $xy$ generates degree $2n$, and $yx=(-1{)}^{n}xy$. When $n$ is even, every graded ring automorphism restricts on $H^n$ to a signed permutation matrix; when $n$ is odd the ring admits every matrix in $G{L}_2(\mathbb{Z})$.

For positive unequal dimensions $p,q$, let $u,v$ be the pullbacks of the integral orientation classes of the factors. The Künneth theorem gives the free abelian basis $1,u,v,uv$, with degrees $0,p,q,p+q$. The products satisfy $u^2=v^2=0$ and $vu=(-1{)}^{pq}uv$, by naturality and graded commutativity of the cup product. In particular, for $p=2,q=4$, the cohomology ring is $\mathbb{Z}[u,v]/(u^2,v^2)$, with degrees two and four.

For even $n\ge 2$, attach an $(n+1)$-cell to $S^n\times S^n$ along the diagonal. The cellular chain complex has boundary $(1,1)$ for the new cell. The degree-$n$ cohomology generator restricts to $a-b$, and the top generator restricts to $ab$. Naturality of the cup product therefore gives $u^2=-2v$, since $(a-b{)}^2=-2ab$. Thus the resulting cohomology ring is $\mathbb{Z}[u,v]/(u^2+2v,uv,v^2)$, with $|u|=n$ and $|v|=2n$. This distinguishes the attachment from a wedge of spheres, whose positive-degree products vanish.