## Laplacian spectra and linearly equivalent G-sets September 19, 2007

Posted by Ben Webster in differential geometry, representation theory.

So, to build on my previous post about zeta functions and group actions, let me talk about a second application of linearly equivalent actions, this time in differential geometry.

So, pick your favorite compact Riemannian manifold $X$, and consider the Laplacian $\Delta_X:L^2(X)\to L^2(X)$. This is the negative of a symmetric elliptic differential operator, and thus has discrete spectrum, which each eigenvalue having finite multiplicity. Let $0>\lambda_1\geq \lambda_2 \geq \cdots$ be these eigenvalues (with multiplicity) in decreasing order. These eigenvalues are always negative (except for the constant function, an eigenvector for 0), since the Laplacian of a function is always negative at local maxima, and positive at local minima.

One way of packaging these eigenvalues is in the heat trace $Z_X(t)=\sum_{i=1}^\infty e^{\lambda_it}=tr(\eta(t),L^2(X))$, which is the trace of the heat kernel $\eta(t)$ on $X$. You can think of this as a bit like the zeta function of the manifold.

To me, the best way of thinking about these is as stationary solutions to the heat equation. Eigenvectors are ways of distributing heat such that over time, we just get an even decay to 0 (i.e., the distribution of heat over time is of the form $e^{\lambda t}f(x)$).

Now, if one takes a Galois cover $Y$ of $X$ (in this context, “Galois” means that the group of deck transformations acts transitively on the pre-image of a point in $X$), then we have a finite group $G$ acting freely on $Y$ with quotient $X$. Thus, if we take the Hilbert space $L^2(Y)$, then $G$ acts on here and this action commutes with the Laplacian. Thus, each eigenspace for the Laplacian is invariant under this $G$-action, and is the sum of its intersections with isotypic components for that action of $G$.

That is, if $\chi$ ranges over the irreducible representationsof $G$, then we get a bunch of commuting projections to the isotypic components $\pi_\chi$, satisfying $\sum_{\chi}\pi_\chi=1$. Since $\sum_{\chi}\pi_\chi\eta(t)=\eta(t)$, we get an identity for $Z_Y(t)$ in terms of the components $Z_\chi(t)= tr(\pi_\chi\eta(t), L^2(Y))$.

Of course, $L^2(Y/H)=L^2(Y)^H$. Since each eigenspace intersected with an isotypic component is a direct sum of the same representation a number of times, applying $H$ invariants simply multiplies its contribution by $\frac{\dim V^H_\chi}{\dim V_\chi}$. Applying this to each isotypic component, we get the formula

$Z_{Y/H}(t)=\sum_{\chi}\frac{\dim V^H_\chi}{\dim V_\chi}Z_\chi(t).$

So, how do these multiplicities $\dim V_\chi^H$ depend on the subgroup $H$? Well, Frobenius reciprocity tells us that $\dim V^H=\dim\mathrm{Hom}(V,\mathbb{C}[G/H])$, so these are precisely the multiplicities appearing in the permutation representation.

Thus, just as before, we get the

Theorem. (Sunada, 1986) The quotients of $Y$ by two subgroups $H,H'$ will be isospectral (i.e. have the same Laplacian spectrum) if $\mathbb{C}[G/H]\cong \mathbb{C}[G/H']$.

Now, this may seem like a rather trivial observation, but it came after people had been thinking about isospectral manifolds for over 20 years. I think it’s a great example of a very simple observation in one field seeming very interesting in another.

1. David Speyer - September 20, 2007

Possible typo. In the statement of the theorem, G/H should be Y/H, yes?

2. Greg Kuperberg - September 20, 2007

No, it is stated correctly. If it said Y/H, then the theorem would be a tautology.

Another way to say it is: If Z and Z’ are two regular finite covering spaces of a space X, and if their deck translation groups have the same group algebra, then Z and Z’ are isospectral.

(“regular” is the perhaps unfortunate conventional term for Galois covering spaces.)

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It is on my mind that isospectral manifolds seem similar to isospectral graphs. Certainly the finite covering trick is a way to make isospectral graphs, because the argument here didn’t anywhere use that X is a manifold. Or, more concretely, if you draw a graph in X, its two lifts in Z and Z’ are isospectral. But is there a way to go in the other direction, to produce isospectral manifolds from isospectral graphs?

Or, the real question: Consider the computational problem of determining whether two manifolds are isometric. (E.g., say that they are triangulated PL manifolds with rational edge lengths.) Is it as hard as the graph isomorphism problem? Or conceivably harder since you can retriangulate? (I would be a bit surprised if the latter were true.)

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