Cohomology ring ($H^{\ast }(X;R)$)

The cohomology ring is the graded ring whose multiplication is the cup product. It records intersection information that the graded cohomology groups alone do not detect.

The property that every degree-one element has square zero is invariant under a graded ring isomorphism. Over ${\mathbb{F}}_2$ it distinguishes the cohomology ring of a torus from the mod-two cohomology ring of an inversion mapping torus of dimension at least two.

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.

Two finite CW complexes can have isomorphic integral cohomology rings but different cohomology rings modulo $p$. For example, attaching a three-cell of degree $p$ to the projective line in ${\mathbb{CP}}^2$ preserves the nonzero square of the degree-two class modulo $p$, whereas the corresponding Moore space wedged with $S^4$ has zero square. The spaces therefore cannot be homotopy equivalent.

Choose degree-one classes ${\alpha }_i,{\beta }_i$ dual to a symplectic basis of one-cycles and let $\omega$ generate $H^2({\Sigma }_g;\mathbb{Z})$. The only nonzero products of positive-degree basis elements are