= Solution
Work on a <closed> <compact> smooth <Riemannian manifold> $M$ without <boundary>, as in the usual discrete spectral setting here. Use the paper's $\Delta=\operatorname{tr}_g\operatorname{Hess}_g$, and put $P=-\Delta$. The <Riemannian heat kernel> $H(t,x,y)$ is the smooth kernel, for $t>0$, representing the solution of
$$
\partial_tu=\Delta u,\qquad u(0,\cdot)=f,\qquad
u(t,x)=\int_M H(t,x,y)f(y)\,dV(y).
$$
Thus $(\partial_t-\Delta_x)H=0$ and its distributional initial value in $x$ is the <Dirac delta distribution> at $y$. The <heat trace> is
$$
Z_M(t)=\int_M H(t,x,x)\,dV(x)=\operatorname{Tr}(e^{-tP}).
$$
The <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> $P\phi_j=\lambda_j\phi_j$, with $0=\lambda_0\le\lambda_1\le\cdots$, repeating <eigenvalues> according to <multiplicity>. For the initial datum $\phi_j$, the function $e^{-t\lambda_j}\phi_j$ 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>:
$$
H(t,x,y)=\sum_{j=0}^\infty e^{-t\lambda_j}\phi_j(x)\overline{\phi_j(y)}.
$$
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,
$$
\boxed{Z_M(t)=\sum_{j=0}^\infty e^{-t\lambda_j}.}
$$
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>,
$$
H(t,x,x)\sim(4\pi t)^{-d/2}\sum_{r=0}^\infty u_r(x)t^r,\qquad u_0(x)=1,
$$
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:
$$
Z_M(t)\sim(4\pi t)^{-d/2}\sum_{r=0}^\infty a_r(M)t^r,
\qquad a_r(M)=\int_Mu_r(x)\,dV(x),\qquad a_0(M)=\operatorname{Vol}(M)>0.
$$
These integrated coefficients are the <heat invariants>. The uniform leading remainder in particular gives $Z_M(t)=(4\pi t)^{-d/2}(\operatorname{Vol}(M)+O(t))$.
<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,
$$
\boxed{d=2\lim_{t\downarrow0}\frac{\log Z_M(t)}{\log(1/t)},\qquad
\operatorname{Vol}(M)=\lim_{t\downarrow0}(4\pi t)^{d/2}Z_M(t).}
$$
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.
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