A Gassmann-Sunada triple is a triple $(U,V,W)$ of groups, with $V, W$ subgroups of $U$, such that $U$ and $V$ meet every conjugacy class in $U$ in the same number of elements, and such that $V$ and $W$ are not conjugate. Such triples have been widely used for constructing isospectral manifolds. However, usually when I encounter these triples in the manifold context, I read something like "... and if one now can find a suitable manifold such that $U$ acts in the right way, then ...."

My first question is:

(a) Starting from a triple $(U,V,W)$ as above, how does one naturally construct isospectral manifolds, which are not isometric ?

(So I want to see the manifolds arise only from the data $(U,V,W)$, and I also want a guarantee that the manifolds are not isometric.)

Next, I wonder if someone can give me a reference on the second question:

(b) Construct isospectral manifolds $\mathcal{M}_U$ and $\mathcal{M}_V$ from $(U,V,W)$ in the "natural way" (see the answers on (a)). Now let $T$ be a group properly containing $U$, and assume it "induces the same Gassmann-Sunada triple" (for instance, $T$ also acts on the same sets $X$ and $Y$ which give the permutation groups $(U,U/V)$ and $(U,U/W)$). My guess is that this "new" triple gives the same manifolds as those coming from $(U,V,W)$. Can anyone fill in the details here ?

Bye ...

  • $\begingroup$ I don't have a helpful answer offhand, just a suggestion- it may be useful to look up and work through some examples of isospectral, non-isomorphic hyperbolic surfaces. $\endgroup$
    – Neal
    May 11, 2017 at 13:19
  • 1
    $\begingroup$ Correction: "$U$ and $V$ meet every conjugacy class..." should be "$V$ and $W$ meet every conjugacy class..." $\endgroup$
    – KConrad
    May 11, 2017 at 15:05
  • $\begingroup$ Here's a construction of isospectral, non-isometric Heisenberg manifolds: projecteuclid.org/download/pdf_1/euclid.mmj/1029003354 $\endgroup$
    – Neal
    May 11, 2017 at 17:00
  • $\begingroup$ Regarding your question (b): The set up I'm used to (notation as in Ben Linowitz's answer) uses a surjection $\pi_1(M) \to G$. Thus, it wouldn't make sense to change $G$ to a larger $G'$. $\endgroup$ May 11, 2017 at 17:17
  • $\begingroup$ @ DavidSpeyer: I was thinking, for example, at the case $U = \mathrm{PSL}_3(q)$, acting naturally on the projective plane $\mathbb{P}^2(q)$, with $V$ the stabilizer of a point ($x$), and $W$ the stabilizer of a line ($Y$). Then $(U,V,W)$ is Gassmann-Sunada. Now take a group $\widetilde{U}$ containing $U$ inside the automorphism group of $\mathbb{P}^2(q)$. My guess was that $(\widetilde{U},\widetilde{U}_x,\widetilde{U}_Y)$ is a Gassmann-Sunada triple which gives the same isospectral manifolds ? $\endgroup$
    – THC
    May 12, 2017 at 12:04

3 Answers 3


Given a finite group $G$, define two subgroups $H_1, H_2$ of $G$ to be almost conjugate if $H_1$ and $H_2$ are not conjugate but have the property that every conjugacy class in $G$ intersects $H_1$ the same number of times it intersects $H_2$. Thus $(G, H_1, H_2)$ is a Gassmann-Sunada triple. Geometrically, the interest in these triples is due to the following theorem of Sunada:

Theorem (Sunada): Let $M$ be a closed Riemannian manifold, $(G, H_1, H_2)$ be a Gassmann-Sunada triple and $$f: \pi_1(M)\rightarrow G$$ be a surjective homomorphism. The manifolds associated to $f^{-1}(H_1)$ and $f^{-1}(H_2)$ are isopsectral.

Note that while the groups produced by Sunada's theorem are not conjugate inside of $\pi_1(M)$, they could still be conjugate in some larger group. So care must be taken to ensure that one is actually producing non-isometric manifolds.

To show the process that one follows when employing Sunada's theorem, let's say that you are interested in constructing large families of pairwise isospectral, non-isometric hyperbolic surfaces. This was done, for instance, by Brooks, Gornet and Gustafson in their paper

Brooks, Robert; Gornet, Ruth; Gustafson, William H., Mutually isospectral Riemann surfaces, Adv. Math. 138, No.2, 306-322 (1998). ZBL0997.53031.

In order to get started we need to find a finite group $G$ which contains a large number of almost conjugate subgroups. The Heisenberg group over a finite field is an example of such a group, and is the group selected by Brooks, Gornet and Gustafson in their paper. The next step is to prove the existence of a surjective homomorphism from $\pi_1(S)$ (where $S$ is a hyperbolic surface) to our Heisenberg group. There is no real trick to producing these homomorphisms. For instance, whereas Brooks, Gornet and Gustafson produce the homomorphisms by hand, McReynolds, in his paper

McReynolds, D.B., Isospectral locally symmetric manifolds, Indiana Univ. Math. J. 63, No. 2, 533-549 (2014). ZBL1314.58021.https://arxiv.org/abs/math/0606540

makes use of the Strong Approximation Theorem (in the more general setting of locally symmetric spaces). Our surjective homomorphism in hand, Sunada's theorem now gives us isospectral covers of $S$. The final step is to make an argument about why we can choose $S$ in such a way that these covers (or at least many of them) are pairwise non-isometric. In the Brooks, Gornet and Gustafson paper this argument is made on page 320 and utilizes Margulis' characterization of the commensurators of non-arithmetic Fuchsian groups.


In my opinion, given a Gassmann-Sunada triple $(U,V,W)$, there is no any "natural" way to associate a pair of non-isometric isospectral manifolds. In general, one looks for a manifold $M$ such that $U$ can be embbeded (or something similar) in the isometry group of $M$.

Now, I will give you a way to associate a pair of isospectral manifolds to any triple $(U,V,W)$, but I wouldn't call it "natural". It is just a very particular (and simple in my opinion) way to do it.

We assume that $U$ is a finite group. Then, $U$ can be embbeded in some symmetric group $S_n$ of $n$ letters (is there a natural way?). Of course, $S_n$ is a subgroup of $SO(n)$. Then, $SO(n)/V$ and $SO(n)/W$ are isospectral manifolds!

As Ben mentioned in his answer, these manifolds are going to be non-isometric if $V$ and $W$ are not conjugate in $SO(n)$, which I think it doesn't necessarily hold by assuming that $V$ and $W$ are not conjugate in $U$.

I hope this helps you to your first question (a). I'm sorry, but I couldn't understand your question (b), particularly when you wrote "induces the same Gassmann-Sunada triple".


The essential computation isn't that hard. Let $G$, $H_1$, $H_2$, $M$ and $f:\pi_1(M) \to G$ be as in Ben Linowitz's answer. Let $X \to M$ be the Deck cover with Galois group $G$ corresponding to $f$ and let $Y_i = X/H_i$.

Fix a real number $\lambda$. Let $V$ be the $\lambda$-eigenspace of $\nabla^2$ on $C^{\infty}(X)$; we note that $V$ is finite dimensional. Write $\rho: G \to GL(V)$ for the action.

A function on $Y_i$ is a $\lambda$-eigenfunction of $\nabla^2$ if and only if it pulls back to a $\lambda$-eigenfunction on $X$, and a function on $X$ is pulled back from $Y_i$ if and only if it is $H_i$-invariant. So the multiplicity of $\lambda$ as an eigenvalue on $Y_i$ is the dimension of the space of invariants $V^{H_i}$.

Note that $\pi_i := \tfrac{1}{H_i} \sum_{h \in H_i} \rho(h_i)$ is an idempotent with image $V^{H_i}$. So $$\dim V^{H_i} = \mathrm{Tr}(\pi_i) = \frac{1}{|H_i|} \sum_{h \in H_i} \mathrm{Tr} \ \rho(h).$$ Using conjugacy invariance of trace, and the obvious fact that $|H_1| = |H_2|$ concludes the result.

Note that we could replace $C^{\infty}(X)$ with sections of some other vector bundle $\mathcal{E}$ on $X$, and $\nabla^2$ with some other differential operator $D$, so long as the $G$-action extended to an action on $\mathcal{E}$ and this action commutes with $D$.


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