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

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.

For each fixed $k$, the two quotient Riemannian metrics are not isometric. If an isometry $F:N_1\to N_2$ existed, choose a small open set avoiding the finite preimages of cone points and their inverse images under $F$. Composing $F$ with the local inverse of the first base projection and the second base projection gives a local isometry of the base Riemannian metric $h_k$. Local rigidity forces this map to be the identity. Hence $p_{2}F=p_1$ on a dense open set, and by continuity everywhere.

The alleged isometry would therefore be an isomorphism of the two orbifold covers over the same base. Covering-space classification makes $K_1,K_2$ conjugate in $\Gamma$, and applying $\rho$ makes $H_1,H_2$ conjugate in $T$, contrary to the supplied hypothesis. Thus $g_{1k},g_{2k}$ are nonisometric.

For $k\ne \ell$, the surface areas are different:Area is preserved by isometries, so no surface from the $k$th pair is isometric to a surface from the $\ell$th pair. Together these arguments prove the full pairwise nonisometry assertion.

For each $k$, the lifted Riemannian metric on $X$ is $T$-invariant and the two subgroups still act freely. Sunada's theorem therefore givesTo rule out any other equalities, the leading diagonal heat kernel expansion from Question 3 gives, on a closed surface,Thus the function spectrum determines area: multiply the heat trace by $4\pi t$ and take the limit. Since the chosen areas are distinct for distinct $k$, no cross-pair spectra can agree. This proves exactly the prescribed spectral coincidences.

The base Riemannian metrics satisfy $h_k\to h_0$ in the smooth topology, and the fixed smooth covering maps pull this convergence back to both $N_i$. Although the base has cone points, each lifted chart is an ordinary smooth chart, so convergence holds there as well. The fixed isometry $F_0$ used to mark the second limit givesThe area normalizations multiply by constants tending to one and therefore do not change these limits. There are precisely two sequences, indexed by $i=1,2$; the appearance of $i=1,2,3,\dots$ in the printed last condition is a redundant indexing error, not a requirement for additional families. The common limit is isometric even though every finite pair is nonisometric, because the hyperbolic limit recovers the reflection symmetry broken by the locally rigid perturbations.