Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 114 / 4

Let $a$ and $b$ be the pullbacks of the standard generator of $H^n(S^n;\mathbb{Z})$ from the two factors. The Künneth theorem and graded commutativity of the cup product give

Each homeomorphism acts invertibly on $H^n\cong {\mathbb{Z}}^2$. To respect ordinary composition, send $h$ to $(h^{-1}{)}^{\ast }$; functoriality of induced map on cohomology then defines a homomorphism

For $n=1$, the space is the torus. Every matrix in $GL(2,\mathbb{Z})$ induces a linear homeomorphism ${\mathbb{R}}^2/{\mathbb{Z}}^2\to {\mathbb{R}}^2/{\mathbb{Z}}^2$, so the image is all of $GL(2,\mathbb{Z})$.

For $n=2$, write $h^{\ast }a=pa+qb$. Since $a^2=0$ and $ab=ba$,so $pq=0$; the same argument applies to $h^{\ast }b$. Invertibility then forces the matrix to be a signed permutation matrix. Every such matrix is realized by swapping the two sphere factors and applying an orientation-reversing homeomorphism to either factor. Thus the image consists exactly of the eight signed permutation matrices.

Fibrewise evaluation identifiesThe identity endomorphism is a nowhere-zero continuous section, giving the canonical trivialization of a line bundle tensored with its dual.

Give a complex line bundle its natural orientation as a real plane bundle. Then its Euler class equals its First Chern class. If $a,b$ are the standard generators of $H^2(S^2\times S^2;\mathbb{Z})$, choose complex line bundles $L_1,L_2$ pulled back from the Hopf fibration on the two factors, with $e(L_1)=a$ and $e(L_2)=b$. The triviality of $L_i^{\ast }\otimes L_i$ gives $e(L_i^{\ast })=-e(L_i)$. Hence, for anythe tensor product $L_1^{\otimes p}\otimes L_2^{\otimes q}$, with negative powers interpreted using dual bundles, has Euler class $\alpha$. Its underlying oriented real rank-two bundle is the required $E_{\alpha }$.

Write $\alpha =pa+qb$. In the Gysin sequence of a sphere bundle for the oriented circle bundle $S(E_{\alpha })\to S^2\times S^2$, multiplication by $\alpha$ isin degrees $0$ to $2$, andin degrees $2$ to $4$. If $\alpha \ne 0$ and $d=\gcd (|p|,|q|)$, taking the relevant kernels and cokernels givesWhen $\alpha =0$, the bundle is trivial and the groups in degrees $0$ through $5$ have ranks $1,1,2,2,1,1$, respectively. These are precisely the groups recorded in integral cohomology of a circle bundle over a product of two spheres.

The additive cohomology for nonzero $\alpha$ depends only on $d$. On the other hand, part (a) shows that the homeomorphism group acts on $(p,q)$ only by signed permutations. For example, $(1,0)$ and $(1,1)$ both have $d=1$, so their sphere bundles have isomorphic additive cohomology, but no signed permutation carries one Euler class to the other. The cohomology therefore does not determine the homeomorphism-group orbit of $\alpha$.