 Solution

Use the positive Laplace-Beltrami operator on a compact Riemannian manifold without boundary. Its heat kernel is the smooth integral kernel $K(t,x,y)$, $t>0$, of the solution operators for the heat equation:The last condition holds, in particular, in $L^2$ for smooth initial functions. The heat trace isThe heat invariants are the coefficients in the small-time heat kernel expansion of this trace:Equivalently, the local diagonal coefficients $u_j(x,x)$ integrate to $a_j$; $R_g$ is the scalar curvature. Some normalizations absorb $(4\pi {)}^{-d/2}$ into the coefficients. These definitions here use the displayed normalization consistently.

To prove uniqueness, let $K_1,K_2$ satisfy the heat kernel conditions and let $w$ be the difference of their solutions with the same smooth initial function. It solves $w_t=-\Delta w$ and tends to zero in $L^2$ as $t\downarrow 0$. The heat equation energy identity, using integration by parts on the closed manifold, gives for $t>0$Integrating from $s>0$ to $t$ gives $\Vert w(t){\Vert }_{L^2}\le \Vert w(s){\Vert }_{L^2}$; sending $s\downarrow 0$ proves $w(t)=0$. Thus for every smooth test function $f$,For fixed positive $t$ and $x$, smoothness makes this difference a smooth function of $y$. Taking its conjugate as test function shows its squared norm is zero. The heat kernel is unique.

The compact elliptic spectral theorem for the self-adjoint elliptic Laplace-Beltrami operator gives a complete orthonormal basis of smooth eigenfunctions ${\varphi }_j$, with discrete eigenvalues $0\le {\lambda }_0\le {\lambda }_1\le \cdots$ counted with multiplicity. Solving the heat equation for initial data ${\varphi }_j$ gives $e^{-t{\lambda }_j}{\varphi }_j$, so the unique heat kernel has its eigenfunction expansion and its heat trace isThe series converges for every $t>0$. This proves that the spectrum with multiplicities determines the heat trace.

Conversely, its large-time behaviour recovers the distinct eigenvalues and their multiplicities successively. The multiplicity of zero is $m_0={\lim }_{t\to \infty }{Z}_M(t)$. After subtracting the already determined terms, letbe the positive remaining sum, and let $\mu$ be its smallest remaining eigenvalue. ThenIndeed, after factoring $e^{-t\mu }$, the next distinct eigenvalue is separated from $\mu$ by a positive gap. The remaining tail tends to zero: for fixed $t_0>0$ its absolute sum is bounded by a constant times $e^{-({\mu }_{\mathrm{next}}-\mu )(t-t_0)}$, using finiteness of $Z_M(t_0)$. Induction recovers every spectral term. Thus the heat trace and the spectrum with multiplicities determine one another.

Finally the leading coefficient among the heat invariants is positive, soThe common heat trace of isospectral manifolds therefore forces them to have the same dimension. Once the dimension is recovered, its leading coefficient also recovers their common volume.