The Heat Kernel — Kernel of Operator

## 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.