TITLE: From Gassmann-Sunada triples to isospectral manifolds
QUESTION [10 upvotes]: 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 ...

REPLY [5 votes]: 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".

REPLY [5 votes]: 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 $D$.