= Solution
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 $t\downarrow0$, so that
$$
u(t,x)=\int_M K_M(t,x,y)f(y)\,dV_M(y),\qquad u(0,x)=f(x).
$$
For an <orthonormal eigenbasis> with <eigenvalues> repeated according to multiplicity,
$$
K_M(t,x,y)=\sum_{j=0}^\infty e^{-t\lambda_j}\phi_j(x)\overline{\phi_j(y)},\qquad
Z_M(t)=\operatorname{Tr}(e^{-t\Delta_M})=\int_MK_M(t,x,x)\,dV_M(x).
$$
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 gives
$$
\boxed{K_M(t,px,py)=\sum_{u\in U}K_N(t,x,uy)}.
$$
There 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> formula
$$
\boxed{Z_M(t)=\frac1{|U|}\sum_{u\in U}F_t(u)},\qquad
F_t(u)=\int_NK_N(t,x,ux)\,dV_N(x).
$$
Suppose $U$ 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>. Then
$$
F_t(sus^{-1})=\int_NK_N(t,sy,suy)\,dV_N(y)=F_t(u).
$$
Thus $F_t$ is a <class function> on $T$, and for its <conjugacy classes> $\mathcal C$,
$$
Z_{N/U}(t)=\frac1{|U|}\sum_{\mathcal C\subset T}|U\cap\mathcal C|\,F_t(c_{\mathcal C}).
$$
Consequently, if $U_1,U_2\leq T$ 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 \b[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.
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