 Solution

Write the hyperbolic toral automorphism as $T_A([v])=[Av]$ on ${\mathbb{T}}^2={\mathbb{R}}^2/{\mathbb{Z}}^2$, where $A\in \mathrm{GL}(2,\mathbb{Z})$. Its determinant is $\pm 1$. The eigenvalues cannot be a nonreal conjugate pair: their product would have absolute value one, forcing each to have absolute value one, contrary to hyperbolicity. The same product condition and hyperbolicity imply two distinct real eigenvalues, labelled so thatWe will proveThis is the entropy of a hyperbolic toral automorphism; the absolute value is important when the expanding eigenvalue is negative.

Choose real eigenvectors $v_s,v_u$ and the adapted norm $\Vert a{v}_s+b{v}_u{\Vert }_{\ast }=\max (|a|,|b|)$. Its quotient metric on the torus isIt is a compatible metric, so part (a) allows its use. Let $\rho ={\min }_{0\ne \ell \in {\mathbb{Z}}^2}\Vert \ell {\Vert }_{\ast }>0$ and $D=\Vert A{\Vert }_{\ast }=|{\lambda }_u|$. Fix $ϵ>0$ small enough that $2ϵ<\rho$ and $(D+1)ϵ<\rho$.

If $[v]$ lies in the Bowen ball $B_n(0,ϵ)$, each $T_A^k[v]$ has a unique lift $z_k$ with $\Vert z_k{\Vert }_{\ast }<ϵ$. Uniqueness follows because two such lifts differ by a lattice vector of norm less than $2ϵ<\rho$. Moreover, $z_{k+1}-A{z}_k$ is a lattice vector, andIt must be zero. Thus $z_k=A^k{z}_0$: there is no hidden lattice jump between successive lifted iterates.

Write $z_0=a{v}_s+b{v}_u$. The simultaneous bounds $\Vert A^k{z}_0{\Vert }_{\ast }<ϵ$ for $0\le k<n$ are exactlyThe stable coordinate is largest at time zero and the unstable coordinate at time $n-1$. Conversely, every vector in this rectangle satisfies all the Bowen ball bounds. The quotient is injective on this rectangle by $2ϵ<\rho$. If $J=|\det (v_s,v_u)|$, its normalized Haar measure is thereforeThe quotient metric is translation invariant and $T_A$ is linear, so $d_n(x,y)=d_n(0,y-x)$; every Bowen ball has this same measure, independently of its centre.

Take an $ϵ$-separated set of largest cardinality $s_n(d,ϵ)$. It is maximal under inclusion, so its radius-$ϵ$ Bowen balls cover the torus: a point outside all these balls could be added. The covering bound givesIts radius-$ϵ/2$ Bowen balls are disjoint, by the triangle inequality for $d_n$. Their measures sum to at most one, givingTaking logarithms, dividing by $n$, and letting $n\to \infty$ yields $h_d(T_A,ϵ)=\log |{\lambda }_u|$ for every sufficiently small $ϵ$. The limit as $ϵ\downarrow 0$ proves the answer. Contraction in the stable direction contributes no exponential growth; only expansion in the unstable direction does.