Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 114 / 4 / Solution

The Lefschetz fixed-point theorem says that a continuous self-map $f$ of a compact triangulable space has a fixed point whenever its Lefschetz numberis nonzero. In the smooth nondegenerate case, the Lefschetz-Hopf fixed-point theorem expresses this number as the sum of the local fixed-point indices.

Identify $H_1({\mathbb{R}}^2/{\mathbb{Z}}^2;\mathbb{Z})$ with ${\mathbb{Z}}^2$ by sending $v\in {\mathbb{Z}}^2$ to the class of the loop $t\mapsto tv\mathrm{mod}{\mathbb{Z}}^2$. The image of this loop under $f$ lifts from $0$ to the path $t\mapsto \tilde{f}(tv)$, whose endpoint is $\tilde{f}(v)=A(v)$. Its homology class is therefore $A(v)$. Thus $A=f_{\ast }$ under this identification, and in particular $A$ is a homomorphism represented by an integer matrix.

The expansion inequality implies, after taking a derivative, thatfor every tangent vector. The supplied linear-algebra fact shows that every eigenvalue of $D{\tilde{f}}_x$ has modulus at least $\mu >1$, so $1$ is not an eigenvalue. Every fixed point of $f$ is consequently nondegenerate and contributes local index $+1$ or $-1$.

On the torus, the induced maps on $H_0,H_1,H_2$ have traces $1,\mathrm{tr}A,\det A$. HenceThe original expansion inequality applied to lattice vectors gives $|Av|\ge \mu |v|$ for $v\in {\mathbb{Z}}^2$, and homogeneity and density extend it to all $v\in {\mathbb{R}}^2$. If ${\lambda }_1,{\lambda }_2$ are the possibly complex eigenvalues of $A$, then $|{\lambda }_j|\ge \mu$, soSince every fixed point contributes an index of absolute value one, the Lefschetz-Hopf formula gives

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