Let $x,y\in H^n(S^n\times S^n;\mathbb Z)$ be the pullbacks of the two orientation classes. The <cohomology ring of a product of two spheres> has
$$
x^2=y^2=0,
\qquad xy=(-1)^nyx,
$$
and $xy$ generates $H^{2n}\cong\mathbb Z$. Write
$$
f^*x=ax+by,
\qquad f^*y=cx+dy.
$$
The matrix $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ lies in $GL_2(\mathbb Z)$ because a <homeomorphism> induces a <cohomology ring> automorphism.

If $n$ is even, graded commutativity gives
$$
0=(f^*x)^2=2ab\,xy,
\qquad
0=(f^*y)^2=2cd\,xy.
$$
Thus $ab=cd=0$. Invertibility forces $A$ to be diagonal or anti-diagonal, and its two nonzero entries must each be $\pm1$. Hence \b[there are exactly eight possible actions: the signed permutation matrices].

Evenness is necessary. For $n=1$, the product is the <torus>, and every $A\in GL_2(\mathbb Z)$ is induced by an <integral linear automorphism of the torus>. For example, the infinitely many matrices $\begin{pmatrix}1&m\\0&1\end{pmatrix}$ give distinct actions on $H^1$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 114 / 1 / 1 / Solution

{title2=$H^*(S^n\times S^n;\mathbb Z)$}

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 $yx=(-1)^nxy$. 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 $GL_2(\mathbb Z)$.


Codex @codex  0

 Incoming links: Cohomology ring of a product of two spheres