Constructing multi-cusped hyperbolic manifolds that are isospectral and not isometric

The eigenfunctions associated to the discrete Laplace spectrum of $M$ are the set of eigenfunctions of the Laplacian that are invariant under $\Gamma$ and which are $L^{2}$-integrable over some (and hence any) fundamental domain for $\Gamma$. Any such function is also invariant under $\Gamma^{\prime}$, and since the fundamental domain of $\Gamma^{\prime}$ is a finite union of fundamental domains of $\Gamma$, the function will also be $L^{2}$ integrable over a fundamental domain for $\Gamma^{\prime}$. It follows that $M^{\prime}$ has an infinite discrete Laplace spectrum if $M$ does.∎

Recall that $HgK$ is the union of the cosets $Hgk$ as $k$ varies over the elements of $K$. As right cosets of $H$ in $G$, two cosets $Hgk_{1}$ and $Hgk_{2}$ intersect if and only if they are equal. Observe that $Hgk_{1}=Hgk_{2}$ if and only if there is an element $h\in H$ such that $gk_{1}=hgk_{2}$, or equivalently, if and only if $k_{1}k_{2}^{-1}\in g^{-1}Hg$ (and thus is an element of $K\cap g^{-1}Hg$). This shows that $Hgk_{1}=Hgk_{2}$ if and only if $(K\cap g^{-1}Hg)k_{1}=(K\cap g^{-1}Hg)k_{2}$. We have therefore shown that the map $f$ given by $f(Hgk)=(K\cap g^{-1}Hg)k$ is a bijection between the cosets of $H$ in $HgK$ and of $K\cap g^{-1}Hg$ in $K$. We can now conjugate by $g$ to obtain a bijection between the cosets of $H$ in $HgK$ and the cosets of $(gKg^{-1}\cap H)$ in $gKg^{-1}$. ∎

That any element of $G\gamma P_{x}$ lies in $\Gamma_{x,y}$ is clear. Suppose therefore that $\delta\in\Gamma_{x,y}$ and that $\delta x=gy=g(\gamma x)$. Then $(g\gamma)^{-1}\delta x=x$, hence $\gamma^{-1}g^{-1}\delta\in P_{x}$ and there exists $p\in P_{x}$ such that $\gamma^{-1}g^{-1}\delta=p$. This implies that $\delta=g\gamma p\in G\gamma P_{x}$ and completes the proof of the lemma. ∎

That $GP\subseteq\Gamma$ is clear as both $G$ and $P$ are subgroups of $\Gamma$. Now let $\gamma\in\Gamma$. Since $\Gamma c=Gc$ there exists an element $g\in G$ such that $\gamma c=gc$. It follows that $(g^{-1}\gamma)c=c$, hence $g^{-1}\gamma\in P$ and there exists $p\in P$ such that $g^{-1}\gamma=p$. This implies that $\gamma=gp$, concluding the proof. ∎

Write $\Gamma$ as a disjoint union of cosets $G\gamma_{i}$:

Since $\Gamma$ acts transitively on $[c]$, every element of $[c]$ is in the $G$ orbit of $\gamma_{i}d_{1}$ for some $i$. For each $j\in\{1,\dots,m\}$, fix $\delta_{j}\in\Gamma$ such that $\delta_{j}d_{1}=d_{j}$. By Lemma 4.2, $\Gamma_{d_{1},d_{j}}=G\delta_{j}\mathrm{Stab}_{\Gamma}(d_{1})$. Lemma 4.1 shows that $\Gamma_{d_{1},d_{j}}$ is the union of $n$ cosets of $G$, where $n$ is the index of $\delta_{j}\mathrm{Stab}_{\Gamma}(d_{1})\delta_{j}^{-1}\cap G$ in $\delta_{j}\mathrm{Stab}_{\Gamma}(d_{1})\delta_{j}^{-1}$. As $\delta_{j}\mathrm{Stab}_{\Gamma}(d_{1})\delta_{j}^{-1}=\mathrm{Stab}_{\Gamma}(\delta_{j}d_{1})=\mathrm{Stab}_{\Gamma}(d_{j})$, we see that $n=[\mathrm{Stab}_{\Gamma}(d_{j}):\mathrm{Stab}_{\Gamma}(d_{j})\cap G]$.

Putting all of this together, we see that $\Gamma$ is the disjoint union of $\Gamma_{d_{1},d_{j}}$ as $j$ varies over $\{1,\dots,m\}$. Since each of these is the disjoint union of $[\mathrm{Stab}_{\Gamma}(d_{j}):\mathrm{Stab}_{\Gamma}(d_{j})\cap G]$ cosets of $G$, we conclude that

which completes our proof.∎

We first prove that if every cusp of $M$ remains a cusp of $N$ then $[\Gamma:G]=[\mathrm{Stab}_{\Gamma}(d):\mathrm{Stab}_{G}(d)]$ for all cusps $[d]$ of $G$. Fix a cusp $[d]$ of $G$ and define $P=\mathrm{Stab}_{\Gamma}(d)$. We must show that $[P:P\cap G]=[\Gamma:G]$. To that end, suppose that $p_{1},p_{2}\in P$. Then

We have therefore exhibited a bijection between the cosets of $G$ in $GP=\Gamma$ (the equality follows from Lemma 4.3) and the cosets of $(P\cap G)$ in $P$, hence $[\Gamma:G]=[P:P\cap G]$.

As the reverse direction is an immediate consequence of Theorem 4.4, our proof is complete. ∎

In light of Theorem 4.4 it suffices to prove that if $[d_{i}],[d_{j}]$ are cusps of $G$ contained in the cusp $[c]$ of $\Gamma$ then $[\mathrm{Stab}_{\Gamma}(d_{i}):\mathrm{Stab}_{\Gamma}(d_{i})\cap G]=[\mathrm{Stab}_{\Gamma}(d_{j}):\mathrm{Stab}_{\Gamma}(d_{j})\cap G]$. To that end, let $\gamma\in\Gamma$ be such that $\gamma d_{i}=d_{j}$. Then

hence, as $G=\gamma G\gamma^{-1}$, we have

which completes the proof. ∎

Let $c$ represent a fixed cusp of $\Gamma$ and $P=\mathrm{Stab}_{\Gamma}(c)$. We begin our proof by noting that Theorem 4.4 shows that $[\Gamma:G]=[P:P\cap G]$, hence we may select a collection of coset representatives for $P\cap G$ in $P$ which is also a collection of coset representatives for $G$ in $\Gamma$. Let $\{\delta_{1},\dots,\delta_{n}\}\subset P$ be such a collection.

An arbitrary term of $E_{M,c}(w,s)$ is of the form $y(\sigma^{-1}\gamma w)^{s}$ where $\gamma\in\Gamma$ represents a non-identity coset $P\gamma$ of $P$ in $\Gamma$ and $\sigma$ is the orientation preserving isometry of hyperbolic space taking the point at infinity to the cusp point $c$. Here we use the coordinates $z=(x,y)\in{\bf H}^{d}={\bf R}^{d-1}\times{\bf R}^{+}$ for the upper half-space. Using our decomposition of $\Gamma$ into cosets of $G$ we see that there exists $\delta_{j}$ and $g\in G$ such that $\gamma=\delta_{j}g$. Because $\delta_{j}\in P$, the coset $P\gamma=P\delta_{j}g$ is equal to the coset $Pg$ as cosets of $P\backslash\Gamma$. In particular this implies that we may choose representatives for the cosets $P\backslash\Gamma$ to all lie in $G$. Note that for all $g_{1},g_{2}\in G$ we have

It follows that

∎

We begin by constructing a finite cover $\widetilde{M}$ of $M$ which has an infinite discrete spectrum. The manifold $M_{q}$ will arise as a finite cover of $\widetilde{M}$ and will therefore have an infinite discrete spectrum by virtue of Lemma 3.1. To that end, let $H$ be an index $11$ subgroup of $\operatorname{PSL}(2,11)$. Such a subgroup is well-known to exist, and the cover of $M$ associated to the pullback subgroup of $H$ by $\rho^{\prime}$ is a degree $11$ cover of $M$. Denote this cover by $\widetilde{M}$. We claim that $\widetilde{M}$ has one cusp. Let $P$ be the subgroup of $\pi_{1}(M)$ stabilizing the cusp of $M$. As was commented above, $\rho^{\prime}(P)$ must be a cyclic subgroup of $\operatorname{PSL}(2,11)$ of order $11$. Since $H$ has index $11$ in $\operatorname{PSL}(2,11)$ and $|\operatorname{PSL}(2,11)|=660=2^{2}\cdot 3\cdot 5\cdot 11$ it must be the case that $\rho^{\prime}(P)\cap H$ is trivial. It follows that $[P:P\cap\pi_{1}(\widetilde{M})]=11=[\pi_{1}(M):\pi_{1}(\widetilde{M})]$, hence $\widetilde{M}$ has one cusp by Corollary 4.5. It now follows from [4, Theorem 2.4] that $\widetilde{M}$ has an infinite discrete spectrum. We note that [4, Theorem 2.4] has two hypotheses: that $\widetilde{M}$ be non-arithmetic and that $\widetilde{M}$ not be the minimal element in its commensurability class. That $\widetilde{M}$ is non-arithmetic is clear, since it is a finite cover of $M$, which is non-arithmetic. It is equally clear that $\widetilde{M}$ is not the minimal element of its commensurability class, since such an element cannot be a finite cover of another hyperbolic $3$-manifold.

We claim that $\pi_{1}(\widetilde{M})$ also admits a $p$-rep to $\operatorname{PSL}(2,7)$. In particular, we will show the homomorphism to $\operatorname{PSL}(2,7)$ obtained by composing the inclusion map $\pi_{1}(\widetilde{M})\hookrightarrow\pi_{1}(M)$ with $\rho:\pi_{1}(M)\rightarrow\operatorname{PSL}(2,7)$ is a $p$-rep. To see this, note that because $\gcd(11,|\operatorname{PSL}(2,7)|)=1$, the map $g\mapsto g^{11}$ is a bijection from $\operatorname{PSL}(2,7)$ to itself, hence our claim follows from the fact that for every $\gamma\in\pi_{1}(M)$ the element $\gamma^{11}$ lies in $\pi_{1}(\widetilde{M})$.

Given a proper, non-zero ideal $I$ of $\mathcal{O}_{k}$ we have a composite homomorphism

called the level $I$ congruence homomorphism. It follows from the Strong Approximation Theorem of Nori [14] and Weisfeiler [23] that for all but finitely many prime ideals $\mathfrak{p}$ of $\mathcal{O}_{k}$ the level $\mathfrak{p}$ congruence homomorphism $\phi_{\mathfrak{p}}$ is surjective.

By Dirichlet’s Theorem on Primes in Arithmetic Progressions we may choose a prime $p$ satisfying $p\equiv 5\pmod{168}$ which does not divide the discriminant of $k$. Let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_{k}$ lying above $p$ which has inertia degree $f$ satisfying $\gcd(f,3)=1$. Note that the existence of such a prime ideal $\mathfrak{p}$ follows from the well-known equality in algebraic number theory

where $p\mathcal{O}_{k}=\mathfrak{p}_{1}\cdots\mathfrak{p}_{g}$, $e(\mathfrak{p}_{i}/p)$ denotes the ramification degree of $\mathfrak{p}_{i}$ over $p$ and $f(\mathfrak{p}_{i}/p)$ denotes the inertia degree of $\mathfrak{p}_{i}$ over $p$. In particular our assertion follows from the hypothesis that $[k:\mathbf{Q}]$ not be divisible by $3$ and the fact that all of the ramification degrees $e(\mathfrak{p}_{i}/p)$ are equal to one (since $p$ doesn’t divide the discriminant of $k$ and thus does not ramify in $k$).

We observed above that it follows from the Strong Approximation Theorem that for all but finitely many primes the associated congruence homomorphism is surjective. In light of our use of Dirichlet’s Theorem on Primes in Arithmetic Progressions in the previous paragraph we may assume that $\mathfrak{p}$ was selected so that $\phi_{\mathfrak{p}}$ is surjective. Let $M_{q}$ be the cover of $\widetilde{M}$ associated to the kernel of $\phi_{\mathfrak{p}}$. The cover $M_{q}$ of $\widetilde{M}$ is normal of degree

which proves (ii) upon setting $q=p^{f}$.

Assertion (iii) follows from assertion (ii) and Corollary 4.6 since the image under $\phi_{\mathfrak{p}}$ of a cusp stabilizer $P_{i}$ will be an abelian subgroup of $\operatorname{PSL}(2,p^{f})$ and thus will have order at most $\frac{p^{f}(p^{f}-1)}{2}$ by the classification of subgroups of $\operatorname{PSL}(2,q)$ (see [2]).

We now prove assertion (i). We will abuse notation and denote by $\rho$ the $p$-rep from $\pi_{1}(\widetilde{M})$ onto $\operatorname{PSL}(2,7)$. Because this $p$-rep was obtained by composing the inclusion of $\pi_{1}(\widetilde{M})$ into $\pi_{1}(M)$ with the $p$-rep from $\pi_{1}(M)$ onto $\operatorname{PSL}(2,7)$ (which was also denoted $\rho$), it suffices to prove assertion (i) with $\widetilde{M}$ in place of $M$. Let $N=\frac{p^{3f}-p^{f}}{2}=[\pi_{1}(\widetilde{M}):\pi_{1}(M_{q})]$. As $\rho_{q}(\pi_{1}(M_{q}))$ contains $\rho_{q}(\gamma^{N})=\rho(\gamma^{N})=\rho(\gamma)^{N}$ for all $\gamma\in\pi_{1}(\widetilde{M})$ and $\rho:\pi_{1}(M)\rightarrow\operatorname{PSL}(2,7)$ is surjective, the surjectivity of $\rho_{q}$ follows from the fact (easily verifiable in SAGE [16]) that $\operatorname{PSL}(2,7)$ is generated by the $N$th powers of its elements whenever $p\equiv 5\pmod{168}$ and $\gcd(f,3)=1$.

Let $P_{0}$ be the subgroup of $\pi_{1}(M_{q})$ which fixes some cusp of $M_{q}$ and $P$ be the subgroup of $\pi_{1}(\widetilde{M})$ fixing the corresponding cusp of $\widetilde{M}$. Because $\rho:\pi_{1}(\widetilde{M})\rightarrow\operatorname{PSL}(2,7)$ is a $p$-rep, $\rho(P)$ consists entirely of parabolic elements and therefore is a subgroup of $\operatorname{PSL}(2,7)$ of order $7$. Note that $[P:P_{0}]=d$ for some divisor $d$ of $N$. We will show that $N$, and thus $d$, is not divisible by $7$. Because $p$ was chosen so that $p\equiv 5\pmod{168}$, we also have $p\equiv 5\pmod{7}$ (since $168=2^{3}\cdot 3\cdot 7$). It is now an easy exercise in elementary number theory to show that $N=\frac{p^{3f}-p^{f}}{2}$ is not divisible by $7$ whenever $\gcd(f,3)=1$. Having shown that $\gcd(d,7)=1$, we observe that if $\gamma\in P$ has non-trivial image in $\operatorname{PSL}(2,7)$ then $\gamma^{d}\in P_{0}$ and thus $\rho_{q}(\gamma^{d})=\rho(\gamma)^{d}$ is non-trivial in $\operatorname{PSL}(2,7)$. Since $\rho_{q}(P_{0})$ is a subgroup of $\rho(P)$ and thus also consists entirely of parabolic elements, this proves assertion (i). ∎

Our proof will largely follow the proof of the analogous result of Garoufalidis and Reid [4, Theorem 3.1].

We begin by proving that the manifolds $M_{1}$ and $M_{2}$ are non-isometric. Let $\Gamma_{1},\Gamma_{2}$ be such that $M_{1}={\bf H}^{3}/\Gamma_{1}$ and $M_{2}={\bf H}^{3}/\Gamma_{2}$. If $M_{1}$ and $M_{2}$ are isometric then there exists $g\in\mathrm{Isom}({\bf H}^{3})$ such that $g\Gamma_{1}g^{-1}=\Gamma_{2}$. Such an element $g$ necessarily lies in the commensurator $\mathrm{Comm}(\Gamma)$ of $\Gamma$, and since $\Gamma=\mathrm{Comm}(\Gamma)$ we see that $g\in\Gamma$. By hypothesis there exists a surjective homomorphism $\rho:\Gamma\rightarrow G$. Projecting onto $G$ we see that $\rho(g)H_{1}\rho(g)^{-1}=H_{2}$, which contradicts our hypothesis that $H_{1}$ and $H_{2}$ be non-conjugate.

To prove that $M_{1}$ and $M_{2}$ are isospectral we must show that their scattering determinants have the same poles with multiplicities and that they have the same discrete spectrum. Since $M_{1}$ and $M_{2}$ have the same covering degree over $M_{0}$, that their scattering determinants have the same poles with multiplicities follows immediately from Theorem 5.1, which in fact shows that all of their Eisenstein series coincide. That $M_{1}$ and $M_{2}$ have the same discrete spectrum follows from Lemma 7.2 with $k=\mathbf{C}$, $V=L^{2}_{disc}(M_{0})$ and $\Delta$ the Laplacian.

That $M_{1}$ and $M_{2}$ have the same complex length spectra follows from the proof given by Sunada [18, Section 4].

That $M_{1}$ and $M_{2}$ have infinite discrete spectra follows from [4, Theorem 2.4].

∎