 Solution

Work on a closed compact smooth Riemannian manifold $M$ without boundary, as in the usual discrete spectral setting here. Use the paper's $\Delta ={\mathrm{tr}}_g{\mathrm{Hess}}_g$, and put $P=-\Delta$. The Riemannian heat kernel $H(t,x,y)$ is the smooth kernel, for $t>0$, representing the solution ofThus $({\partial }_t-{\Delta }_x)H=0$ and its distributional initial value in $x$ is the Dirac delta distribution at $y$. The heat trace isThe heat invariants are the coefficients $a_r(M)$ in the short-time heat trace expansion specified below.

To derive its spectral form, take an orthonormal eigenbasis $P{\varphi }_j={\lambda }_j{\varphi }_j$, with $0={\lambda }_0\le {\lambda }_1\le \cdots$, repeating eigenvalues according to multiplicity. For the initial datum ${\varphi }_j$, the function $e^{-t{\lambda }_j}{\varphi }_j$ solves the heat equation. Uniqueness identifies it with the heat-kernel solution. Expanding arbitrary initial data in this basis therefore gives the spectral expansion of the Riemannian heat kernel:Standard elliptic estimates and eigenvalue growth imply smooth convergence for $t$ bounded away from zero: polynomial bounds on derivatives of eigenfunctions are dominated by the exponential factors. On the diagonal all summands are nonnegative, so monotone convergence also justifies integration without needing an interchange of conditionally convergent terms. Since each eigenfunction has unit $L^2$ norm,Equivalently, if the paper's Laplacian eigenvalues are written ${\nu }_j=-{\lambda }_j$, the formula is ${\sum }_j{e}^{t{\nu }_j}$; a growing exponential is not the heat semigroup.

Here are the needed heat kernel expansion facts, stated without proof. On a closed $d$-dimensional smooth Riemannian manifold,with the diagonal remainder uniform in $x$ after any finite truncation. The coefficients are smooth local geometric quantities; $u_1=R/6$. Compactness therefore allows integration term by term:These integrated coefficients are the heat invariants. The uniform leading remainder in particular gives $Z_M(t)=(4\pi t{)}^{-d/2}(\mathrm{Vol}(M)+O(t))$.

Isospectral manifolds have identical heat traces by the derived spectral formula. If their dimensions were $d<e$, multiplying the common trace by $t^{d/2}$ would give a finite positive limit on the $d$-dimensional manifold but divergence on the $e$-dimensional one. Thus the dimensions are equal, and equality of the leading coefficient gives equal Riemannian volume. Explicitly,The absence of a boundary fixes this particular integer-power expansion. With a boundary one must first specify boundary conditions and include the appropriate half-integer boundary terms; the leading dimension-and-volume conclusion remains valid in the usual compact smooth setting.