## 1.2 The General Heat Equation

We can further generalize our previous problem and replace $\Delta$ by any elliptic self-adjoint operator. We will soon specialize to the case of the Laplace–Beltrami operator, but this section is completely general.

Our original motivation was to understand the index of operators, so let us start from there. Suppose we are given a compact Riemannian manifold $M$ and $E, F$ Hermitian vector bundles on $M$. We are also given an elliptic differential operator $D\colon \Gamma (E) \to \Gamma (F)$ over $M$ with formal adjoint $D^*\colon \Gamma (F) \to \Gamma (E)$ (in the case of the Laplace–Beltrami operator, we have $D = D^* = \mathrm{d}+ \mathrm{d}^*$ and $E = F = \Omega ^*$).

Similar to the case of the Laplace–Beltrami operator, we define

$\begin{aligned} \Delta _E & = D^* D \colon \Gamma (E) \to \Gamma (E)\\ \Delta _F & = D D^*\colon \Gamma (F) \to \Gamma (F). \end{aligned}$Then $\ker \Delta _E = \ker D$ and $\ker \Delta _F = \ker D^*$. So we have

$\begin{aligned} \operatorname{index}D & = \dim \ker D - \dim \operatorname{coker}D \\ & = \dim \ker D - \dim \ker D^*\\ & = \dim \ker \Delta _E - \dim \ker \Delta _F. \end{aligned}$This is the form of the index that will be of interest to us.

By analogy with the classical heat equation, we consider the equation

$\left(\frac{\partial }{\partial t} + \Delta _E\right) f = 0$on $E \times (0, \infty ) \to M \times (0, \infty )$, with $t \in (0, \infty )$.

Hodge theory gives us an easy way to solve this, at least formally. We can decompose

$L^2(E) = \bigoplus _\lambda \Gamma _\lambda (E),$where $\Gamma _\lambda (E)$ is the $\lambda$-eigenspace of $\Delta _E$. The spectral theorem tells us the eigenvalues are discrete and tend to infinity. Let $\{ \psi _\lambda \}$ be an orthonormal eigenbasis. Then we can write a general solution as

$f(x, t) = \sum _\lambda c_\lambda e^{-\lambda t} \psi _\lambda (x)$for some constants $c_\lambda$.

If we want to solve this with initial condition $f_0$, i.e. we require $f(-, t) \to f_0$ as $t \to 0$ in $L^2$, then we must pick
^{1}

Thus, we can write the heat kernel as

$H_t(x, y; \Delta _E) = \sum _\lambda e^{\lambda t} \psi _\lambda (x) \psi _\lambda (y)^T.$Then we have

$f(x, t) = \int _M H_t(x, y; \Delta _E) f_0(y)\; \mathrm{d}y.$We remark that for each fixed $t$, the heat kernel $H_t(x, y)$ is a section of the exterior product $E \boxtimes E^* \to M \times M$.

To ensure the sum converges, we need to make sure the eigenvalues grow sufficiently quickly. This is effectively Weyl's law, but we for our purposes, we can use a neat trick to obtain a weaker bound easily.

If $\Delta$ is any self-adjoint elliptic differential operator, then there is a constant $C$ and an exponent $\varepsilon$ such that for large $\Lambda$, the number of eigenvalues of magnitude $\leq \Lambda$ is at most $C \Lambda ^\varepsilon$.

In particular, if $\psi$ is an eigenfunction of eigenvalue at most $\Lambda$, then we can bound

$\| \psi \| _{C^0} \leq C(1 + \Lambda ) \| \psi \| _{L^2}.$Let $\{ \psi _\lambda \}$ be an orthonormal eigenbasis. Then for any constants $a_\lambda$ and fixed $x \in M$, we have

$\left|\sum _{|\lambda | \leq \Lambda } c_\lambda \psi _\lambda (x) \right| \leq C(1 + \Lambda ) \left(\sum _{\lambda \leq \Lambda } |c_\lambda |^2\right)^{1/2}.$We now pick $c_\lambda$ to be the numbers $\bar{\psi }_\lambda (x)$, recalling that we have fixed $x$. Then

$\sum _{|\lambda | \leq \Lambda } \bar{\psi }_\lambda (x) \psi _\lambda (x) \leq C(1 + \Lambda ) \left(\sum _{\lambda \leq \Lambda } \bar{\psi }_\lambda (x) \psi _\lambda (x)\right)^{1/2}.$Equivalently, we have

$\sum _{|\lambda | \leq \Lambda } \bar{\psi }_\lambda (x) \psi _\lambda (x) \leq C^2(1 + \Lambda )^2.$Integrating over all of $M$, the left-hand side is just the number of eigenvalues of magnitude $\leq \Lambda$. So we are done.

Elliptic regularity lets us bound the higher Sobolev norms of $H_t$ as well, and so we know $H_t$ is in fact smooth by Sobolev embedding.

It is fruitful to consider the “time evolution” operator $e^{-t \Delta _E}$ that sends $f_0$ to $f(-, t)$ defined above. This acts on the $\lambda$ eigenspace by multiplication by $e^{-\lambda t}$. Thus, the trace is given by

$h_t(\Delta _E) \equiv \operatorname{tr}e^{-t \Delta _E} = \sum _\lambda e^{-\lambda t} \dim \Gamma _\lambda (E).$The key observation is that $D$ gives an isomorphism between the $\lambda$ eigenspace of $\Delta _E$ and the $\lambda$ eigenspace of $\Delta _F$ as long as $\lambda > 0$, and the $0$ eigenspaces are exactly the kernels of $\Delta _E$ and $\Delta _F$. So we have an expression

$\operatorname{index}\Delta _E = h_t(\Delta _E) - h_t(\Delta _F)$for any $t > 0$.

This expression is a very global one, because $h_t(\Delta _E)$ depends on the eigenfunctions of $\Delta _E$. However, the fact that $e^{-\lambda \Delta _E}$ is given by convolution with $H_t(x, y; \Delta _E)$ gives us an alternative expression for the trace, namely

$h_t(\Delta _E) = \int _M H_t(x, x; \Delta _E)\; \mathrm{d}x.$This is still non-local, but we will later find that the asymptotic behaviour of $H_t(x, x; \Delta _E)$ as $t \to 0$ is governed by local invariants. Since the index is independent of $t$, we can take the limit $t \to 0$ and get a local expression for the index.

Take the flat torus $T^d = \mathbb {R}^d / \mathbb {Z}^d$, and pick $D = \mathrm{d}\colon \Omega ^0(M) \to \Omega ^1(M)$, so that $\Delta$ is the classical Laplacian

$\Delta = -\sum \frac{\partial ^2}{\partial x_i^2}.$The eigenvectors are given by $e^{2\pi ix\cdot \xi }$ for $\xi \in \mathbb {Z}^d$ with eigenvalue $4\pi ^2 |\xi |^2$. So the heat kernel is

$H_t(x, y) = \sum _{\xi \in \mathbb {Z}^d} e^{-4\pi t |\xi |^2} e^{2\pi i(x - y)\cdot \xi } = \prod _{i = 1}^d \sum _{k = -\infty }^\infty e^{-4 \pi t k^2 + 2 \pi i (x_i - y_i) k}.$Number theorists may wish to express this in terms of the Jacobi theta function:

$H_t(x, y) = \prod _{i = 1}^d \vartheta (x_i - y_i; 4it).$We can explicitly evaluate the trace to be

$h_t = \int _{T^d} H_t(x, x) \; \mathrm{d}x = \left(\sum _{k = -\infty }^\infty e^{-4\pi t k^2}\right)^d$In the limit $t \to 0$, we can replace the sum with the integral to get

$h_t \sim \left(\int _{-\infty }^\infty e^{-4\pi t k^2}\; \mathrm{d}k\right)^d = (4t)^{-d/2}.$