Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 19 / 5 / Solution

Sunada's theorem states that if a finite group $T$ acts isometrically on a closed Riemannian manifold $X$, and two Gassmann equivalent subgroups $H_1,H_2$ act freely, then $H_1\backslash X$ and $H_2\backslash X$ are isospectral for the function Laplace-Beltrami operator. Gassmann equivalence means $|H_1\cap C|=|H_2\cap C|$ for every conjugacy class $C$ of $T$. For each Laplace-Beltrami operator eigenspace $E_{\lambda }$, its invariant subspace has dimensionThe character of a representation is a class function, so the two multiplicities agree. A necessary condition for nonisometry is that the subgroups are not conjugate in $T$: conjugation by an ambient deck transformation would induce an isometry. This is not sufficient when the common covering has additional geometric symmetries.

Use the oriented hyperbolic triangle groupand the epimorphism $\rho :\Gamma \to T$ given by $a\mapsto s$, $b\mapsto d$. Its base orbifold is the doubled hyperbolic triangle with angles $\pi /3,\pi /3,\pi /6$, having signature $(0;3,3,6)$. Let $K=\ker \rho$ and $K_i={\rho }^{-1}(H_i)$.

The groups $K_i$ are torsion free. Indeed, torsion elements in an oriented triangle group are conjugate to nontrivial powers of its three vertex rotations. Such powers map either to elements of order three, which cannot lie in a subgroup of order $96/12=8$, or to the order-two cube $z=(sd{)}^3$. The latter is central and absent from both $H_i$, so it is absent from every conjugate as well. Thus the degree-twelve orbifold coversare smooth, closed, connected orientable surfaces. The common cover $X=K\backslash {\mathbb{H}}^2$ is smooth too; $H_i$ act freely on it. Orbifold Euler characteristic givesFor example, this formula follows by triangulating the base and weighting a vertex of order $r$ by $1/r$ before multiplying by the cover degree. Equivalently, each doubled triangle has hyperbolic area $\pi /3$, so $N_i$ has area $4\pi$ and Gauss-Bonnet theorem gives the same genus.

There is one extra issue in obtaining the same hyperbolic limit on a single marked surface: the two limiting hyperbolic covers must themselves be isometric. Nonconjugacy and the displayed element orders alone do not prove this. An explicit realization of the supplied example resolves the issue. The group seed is recorded in the original genus-two construction; the following coset actions and reflection identities give a direct verification. Label the twelve sheets so that the two right-coset actions of $s,d$ areThese are concrete permutation data for an order-$96$ realization with the required two subgroups. The permutationsatisfies, by direct substitution,Reflection of the doubled triangle reverses both vertex rotations. Apply this reflection on every fundamental domain and relabel sheets by $P$. The two identities ensure that identified edges go to identified edges, so this descends to a hyperbolic isometry $F_0:N_1\to N_2$. Fix $N=N_1$, let $g$ be its hyperbolic metric, and use $F_0$ to mark $N_2$ by the same smooth surface. Thus both pulled-back limiting Riemannian metrics are exactly $g$. The explicit reversed-action check is what makes the common limit justified, rather than treating it as a consequence of Sunada's theorem.

Choose smooth orbifold Riemannian metrics $h_k$ on the base converging to its hyperbolic metric $h_0$, with no nonidentity local isometry on the regular part. The density assertion in Sunada's local isometry lemma applies in orbifold charts as well: use equivariant perturbations at the finitely many cone points and ordinary local perturbations elsewhere. Smoothness here is orbifold smoothness, so the lifted Riemannian metrics on $X$ and the two smooth covers are ordinary smooth Riemannian metrics. Such choices can be made in successively smaller smooth neighbourhoods of $h_0$; cone orders and the topological coverings stay fixed.

For area separation of convergent Sunada families, adjust each $h_k$ by a constant positive factor so thatExplicitly, multiply the initially chosen Riemannian metric by the ratio of the desired area to its actual area. In dimension two this ratio is exactly the area multiplier. The ratios tend to one, so smooth convergence persists, and constant scaling preserves the absence of local isometries. Lift these adjusted Riemannian metrics to $N_i$ and pull them back by the fixed markings to obtain $g_{ik}$, for $i=1,2$. The three required properties follow below.