Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 20 / 3

On a closed compact Riemannian manifold, the heat kernel for the positive Laplace-Beltrami operator is the smooth kernel $K_M(t,x,y)$ of $e^{-t{\Delta }_M}$ for $t>0$. It solves $({\partial }_t+{\Delta }_x)K_M=0$ and tends to the delta kernel as $t\downarrow 0$, so thatFor an orthonormal eigenbasis with eigenvalues repeated according to multiplicity,The exponential factors and elliptic regularity ensure smooth convergence for positive time. If a boundary is present, the same construction uses a specified boundary condition preserved by the covering; the closed-manifold case is the one needed below.

Let $p:N\to M=N/U$ be the finite normal locally isometric covering map. The heat equation for $f\circ p$ on $N$ is the lift of the one for $f$ on $M$. Integrate the lifted solution over a fundamental domain for the deck transformation group $U$, splitting the integral over $N$ into its translates. This givesThere is no averaging factor in this kernel formula: each translate accounts for one lift of the integration variable. To verify it directly, the sum solves the lifted heat equation and its integral against $f$ converges to $f(px)$; uniqueness of the heat solution identifies the kernel. Replacing either lift $x$ or $y$ only permutes the sum, since isometries preserve the heat kernel.

Integrating the diagonal and using the covering degree gives the heat trace formulaSuppose $U$ is contained in a finite group $T$ of isometries of $N$. For $s\in T$, change variables $x=sy$ and use invariance of both volume and the heat kernel. ThenThus $F_t$ is a class function on $T$, and for its conjugacy classes $\mathcal{C}$,

Consequently, if $U_1,U_2\le T$ are Gassmann equivalent and act freely, their orders are equal and the displayed class-by-class sums give $Z_{N/U_1}(t)=Z_{N/U_2}(t)$ for every $t>0$. Equal heat traces determine equal eigenvalues with multiplicities: the least eigenvalue at which the multiplicities differed would give a nonzero leading exponential in their trace difference as $t\to \infty$, a contradiction. Hence the two smooth quotient manifolds are isospectral. This proves the Sunada theorem, with the free-action hypothesis ensuring that the quotients are manifolds rather than orbifolds. The theorem alone makes no nonisometry claim.