The Heat Kernel — Kernel of Operator

6.2 Kernel of Operator

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

Δ1=DD,Δ2=DD, \Delta _1 = D^* D,\quad \Delta _2 = D D^*,

and consider etΔie^{-t \Delta _ i}. Our objective is to understand

K(t)=K1,t(y,u;y,u)K2,t(y,u;y,u)  dy  du, K(t) = \iint K_{1, t}(y, u; y, u) - K_{2, t}(y, u; y, u) \; \mathrm{d}y\; \mathrm{d}u,

where Ki,t(y,u;z,v)K_{i, t}(y, u; z, v) is the kernel of etΔie^{-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 Δ1\Delta _1, which is explicitly

Δ1=2u2+A2. \Delta _1 = -\frac{\partial ^2}{\partial u^2} + A^2.

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

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

Note that the first equation constrains fλf_\lambda for λ0\lambda \geq 0, and the second constrains fλf_\lambda for λ<0\lambda < 0. So we have a single constraint for each fλf_\lambda .

To find the fundamental solution for etΔ1e^{- t \Delta _1}, we have to find a fundamental solution for the operator

t2u2+λ2. \frac{\partial }{\partial t} - \frac{\partial ^2}{\partial u^2} + \lambda ^2.

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

For the boundary condition when λ0\lambda \geq 0 we can simply write down the fundamental solution to be

Kλ,t(u,v)=eλ2t4πt(exp((uv)24t)exp((u+v)24t)). 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=0u = 0 or v=0v = 0.

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

Kλ,t(u,v)=eλ2t4πt(exp((uv)24t)+exp((u+v)24t))λeλ(u+v)erfc(u+v2t+λt), 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 erfc\operatorname{erfc} is the complementary error function

erfc(x)=2πxeξ2  dξ. \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 Δ1\Delta _1 as

K1,t(y,u;z,v)=λKλ,t(u,v)ψˉλ(y)ψλ(z). K_{1, t}(y, u; z, v) = \sum _\lambda K_{\lambda , t}(u, v) \bar{\psi }_\lambda (y) \psi _\lambda (z).

The formula for K2,tK_{2, t} is basically the same, except the boundary conditions are swapped. So we find the difference to be

=K1,t(y,u;y,u)K2,t(y,u;y,u)=λsignλ(eλ2tu2/tπt+λe2λuerfc(ut+λt))ψλ(y)2=λsignλu(12e2λuerfc(ut+λt))ψλ(y)2, \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 sign0=+1\operatorname{sign}0 = +1.

Integrating first over uu then over yy, we get

K(t)=λsignλ2erfc(λt). 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 DD. In this case, we have

K(t)12dimkerA as t. K(t) \to -\frac{1}{2} \dim \ker A \text{ as } t \to \infty .

Writing h=dimkerAh = \dim \ker A, we see that in fact K(t)+h20K(t) + \frac{h}{2} \to 0 exponentially as tt \to \infty .

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

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

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

K(t)=14πtλλeλ2t. 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,

0(K(t)+12h)ts1  dt=Γ(s+1/2)2sπλ̸=0signλλ2s=Γ(s+1/2)2sπη(2s). \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 Re(s)\mathrm{Re}(s) is big enough, since Lemma 1.1 tells us how quickly the eigenvalues grow, and erfc(x)<ex2\operatorname{erfc}(x) < e^{-x^2} for all xx.

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

K(t)k=0aktkd/2, 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

η(2s)=2sπΓ(s+1/2)(h2s+k=0Nakkd/2+s+ΘN(s)), \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 ΘN\Theta _ N is holomorphic for Re(s)>12(N+1d/2)\mathrm{Re}(s) > -\frac{1}{2}(N + 1 - d/2). Taking NN to be large enough, we know that η\eta admits a meromorphic continuation to the whole plane, and the value at 00 is

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

This is the final result we seek.