6.2 Kernel of Operator

We now turn to the heat equation. As usual, we define

$$\Delta _1 = D^* D,\quad \Delta _2 = D D^*,$$

and consider $e^{-t \Delta _i}$. Our objective is to understand

$$K(t) = \iint K_{1, t}(y, u; y, u) - K_{2, t}(y, u; y, u) \; \mathrm{d}y\; \mathrm{d}u,$$

where $K_{i, t}(y, u; z, v)$ is the kernel of $e^{-t \Delta _i}$. We do not make any claims about how this relates to an index (or not), since we are working on a non-compact manifold and life is tough.

We first consider $\Delta _1$, which is explicitly

$$\Delta _1 = -\frac{\partial ^2}{\partial u^2} + A^2.$$

We again perform separation of variables. We write our potential solution as $\sum f_\lambda (u, t) \psi _\lambda (x)$. Our boundary conditions are then

$$Pf = 0,\quad (1 - P) Df = 0.$$

Note that the first equation constrains $f_\lambda$ for $\lambda \geq 0$, and the second constrains $f_\lambda$ for $\lambda < 0$. So we have a single constraint for each $f_\lambda$.

To find the fundamental solution for $e^{- t \Delta _1}$, we have to find a fundamental solution for the operator

$$\frac{\partial }{\partial t} - \frac{\partial ^2}{\partial u^2} + \lambda ^2.$$

Up to the $\lambda$, this is just the classical heat equation.

For the boundary condition when $\lambda \geq 0$ we can simply write down the fundamental solution to be

$$K_{\lambda , t}(u, v) = \frac{e^{-\lambda ^2 t}}{\sqrt{4\pi t}} \left(\exp \left(-\frac{(u - v)^2}{4t}\right) - \exp \left(-\frac{(u + v)^2}{4t}\right)\right).$$

This is easily seen to vanish when $u = 0$ or $v = 0$.

When $\lambda < 0$, we need to do something more complicated. People good at solving differential equations will find the kernel to be

$$K_{\lambda , t}(u, v) = \frac{e^{-\lambda ^2 t}}{\sqrt{4\pi t}} \left(\exp \left(-\frac{(u - v)^2}{4t}\right) + \exp \left(-\frac{(u + v)^2}{4t}\right)\right) - |\lambda | e^{|\lambda |(u + v)} \operatorname{erfc}\left(\frac{u + v}{2\sqrt{t}} + |\lambda | \sqrt{t}\right),$$

where $\operatorname{erfc}$ is the complementary error function

$$\operatorname{erfc}(x) = \frac{2}{\sqrt{\pi }} \int _x^\infty e^{- \xi ^2}\; \mathrm{d}\xi .$$

Equipped with these, we can then write the heat kernel of $\Delta _1$ as

$$K_{1, t}(y, u; z, v) = \sum _\lambda K_{\lambda , t}(u, v) \bar{\psi }_\lambda (y) \psi _\lambda (z).$$

The formula for $K_{2, t}$ is basically the same, except the boundary conditions are swapped. So we find the difference to be

$$\begin{aligned} & \hphantom {{}={}}K_{1, t}(y, u; y, u) - K_{2, t}(y, u; y, u)\\ & = \sum _\lambda \operatorname{sign}\lambda \left(-\frac{e^{-\lambda ^2t - u^2/t}}{\sqrt{\pi t}} + |\lambda | e^{2|\lambda | u} \operatorname{erfc}\left(\frac{u}{\sqrt{t}} + |\lambda | \sqrt{t}\right)\right) |\psi _\lambda (y)|^2\\ & = \sum _\lambda \operatorname{sign}\lambda \, \frac{\partial }{\partial u}\left(\frac{1}{2} e^{2|\lambda |u} \operatorname{erfc}\left(\frac{u}{\sqrt{t}} + |\lambda |\sqrt{t}\right)\right) |\psi _\lambda (y)|^2, \end{aligned}$$

where we set $\operatorname{sign}0 = +1$.

Integrating first over $u$ then over $y$, we get

$$K(t) = -\sum _\lambda \frac{\operatorname{sign}\lambda }{2} \operatorname{erfc}(|\lambda |\sqrt{t}).$$

Recall that in the compact case, this expression is identically equal to the index of the operator $D$. In this case, we have

$$K(t) \to -\frac{1}{2} \dim \ker A \text{ as } t \to \infty .$$

Writing $h = \dim \ker A$, we see that in fact $K(t) + \frac{h}{2} \to 0$ exponentially as $t \to \infty$.

To understand this better, we define the $\eta$ function

$$\eta (s) = \sum _{\lambda \not= 0} \operatorname{sign}\lambda \, |\lambda |^s,$$

where of course we sum over eigenvalues of $A$ with multiplicity. One should think of this as some sort of Dirichlet $L$-function for the spectrum. To relate this to $K(t)$, we first observe that

$$K'(t) = \frac{1}{\sqrt{4\pi t}} \sum _\lambda \lambda e^{-\lambda ^2 t}.$$

So after integration by parts once, we find that, at least formally,

$$\int _0^\infty \left(K(t) + \frac{1}{2}h\right) t^{s - 1} \; \mathrm{d}t = - \frac{\Gamma (s + 1/2)}{2s \sqrt{\pi }} \sum _{\lambda \not= 0} \frac{\operatorname{sign}\lambda }{|\lambda |^{2s}} = - \frac{\Gamma (s + 1/2)}{2s\sqrt{\pi }} \eta (2s).$$

This is in fact literally true when $\mathrm{Re}(s)$ is big enough, since Lemma 1.1 tells us how quickly the eigenvalues grow, and $\operatorname{erfc}(x) < e^{-x^2}$ for all $x$.

If we assume that $K(t)$ has an asymptotic expansion

$$K(t) \sim \sum _{k = 0}^{\infty } a_k t^{k - d/2},$$

which we will show in the next section by devious means, then we can explicitly do the integral to get

$$\eta (2s) = -\frac{2 s \sqrt{\pi }}{ \Gamma (s + 1/2)} \left(\frac{h}{2s} + \sum _{k = 0}^N \frac{a_k}{k - d/2 + s} + \Theta _N(s)\right),$$

where $\Theta _N$ is holomorphic for $\mathrm{Re}(s) > -\frac{1}{2}(N + 1 - d/2)$. Taking $N$ to be large enough, we know that $\eta$ admits a meromorphic continuation to the whole plane, and the value at $0$ is

$$\eta (0) = - (2a_{d/2} + h).$$

This is the final result we seek.