 Solution

Let $({\varphi }_j)$ be a smooth orthonormal eigenbasis of the positive Laplace-Beltrami operator, with eigenvalues $({\lambda }_j)$. The orthonormal eigenbasis definesThe large-$j$ lower bound of part (c), valid also after the finitely many zero modes of a disconnected compact manifold, gives ${\lambda }_j\ge a{j}^{2/n}$ for sufficiently large $j$. Henceby comparison with ${\int }_0^{\infty }{e}^{-ta{x}^{2/n}}dx$. The operator is positive and diagonal with these summable coefficients, so it is a trace-class operator and its operator trace is their sum.

Its proposed heat kernel isWe verify smoothness and the kernel action. The elliptic regularity estimate established in Question 4(d) gives, by induction,The Sobolev embedding theorem therefore gives, for every derivative order $a$, constants $C_a$, $r_a$ with $\Vert {\varphi }_j{\Vert }_{C^a}\le C_a(1+{\lambda }_j{)}^{r_a}$. Every differentiated term of the kernel series is bounded by $C_{a,b}{e}^{-t{\lambda }_j}(1+{\lambda }_j{)}^{r_a+r_b}$. An exponential absorbs any fixed power: for $t\ge \tau >0$ this is at most $C_{a,b,\tau }{e}^{-\tau {\lambda }_j/2}$, a summable bound. Time derivatives merely introduce more powers of ${\lambda }_j$. Thus the series converges uniformly with all spatial and time derivatives on $[\tau ,\infty )\times M\times M$. In particular $K_t$ is smooth for every $t>0$.

Integrating its uniformly convergent series against $f\in L^2(M)$ is legitimate since the compact manifold has finite volume and $L^2(M)\subseteq L^1(M)$. It gives exactly the spectral expression above, soFinally $K_t(x,x)={\sum }_j{e}^{-t{\lambda }_j}|{\varphi }_j(x){|}^2$ is nonnegative. Integrating term by term, either by uniform convergence or by Tonelli theorem, and using $\Vert {\varphi }_j{\Vert }_2=1$ proves the heat kernel trace formula