[Nakahara GTP] 12.7 The heat kernel and generalized $\zeta$-functions

 This article is one of Manifold, Differential Geometry, Fibre Bundle.

As we mentioned in section 12.2, there are several methods of proving the AS index theorem. The heat kernel is relatively accessible to physicists and it also has many applications to other problems in physics. The generalized $\zeta$-function is related to the heat kernel and also has relevance in physics.

12.7.1 The heat kernel and index theorem 

Let $E$ be a complex vector bundle over an $m$-dimensional compact manifold $M$. Let $\Delta:\Gamma(M,E)\rightarrow\Gamma(M,E)$ be an elliptic operator with eigenvectors $|n\rangle$ such that $$\begin{align}\Delta|n\rangle=\lambda_n|n\rangle.\end{align}$$ We denote the set of eigenvalues are non-negative. Suppose there are $n_0$ modes $|0,i\rangle$, $1\le i\le n_0$ with vanishing eigenvalue. In other words, $$\begin{align}\dim\ker \Delta=n_0.\end{align}$$ These modes are called the zero modes. Define the heat kernel $h(t)$ by $$\begin{align}h(t)\equiv e^{-t\Delta}.\end{align}$$ It is convenient to represent $h(t)$ in the coordinate basis as $$\begin{align}h(x,y;t)\equiv \langle x|h(t)|y\rangle =\langle x|\sum_n e^{-t\Delta}|n\rangle\langle n|y\rangle\nonumber\\ =\sum_n e^{-t\lambda_n}\langle x|n\rangle\langle n|y\rangle.\end{align}$$ Multiple eigenstates should be counted as many times as they appear. We assume $\langle x|n\rangle$ is orthonormal: $\int\langle n|x\rangle\langle x|m\rangle dx=\delta_{mn}$. The convergence of (3) for $t>0$ is guaranteed since $\Delta$ is non-negative. Taking the limit $t\rightarrow \infty$, we have $$\begin{align}\lim_{t\rightarrow\infty}h(x,y;t)=\int^{n_0}_{i=1} \langle x|0,i\rangle\langle 0,i|y\rangle\end{align}$$ where the summation is over the zero modes $|0,i\rangle$ only. Thus, $h=e^{-t\Delta}$ tends to be the projection operator onto the space of zero modes as $$\begin{align}e^{-t\Delta}\stackrel{t\rightarrow \infty}{\rightarrow}\sum^{n_0}_{i=1}|0,i\rangle\langle 0,i|.\end{align}$$ Define $$\begin{align}\tilde{h}(t)\equiv \int h(x,x;t)\, dx=\sum_n e^{-i\lambda_n}.\end{align}$$ (trace of $h$) Then it follows from (5) that $$\begin{align}n_0=\lim_{t\rightarrow \infty}\tilde{h}(t).\end{align}$$

It is easy to verify that $h$ satisfies the heat equation, $$\begin{align}\left( \frac{\partial}{\partial t}+\Delta_x\right) h(x,y;t)=0.\end{align}$$ If $\Delta$ is the conventional Laplacian, (9) reduces to the ordinary heat equation. The initial condition is $$\begin{align}h(x,y;0)=\sum_n \langle x|n\rangle\langle n|y\rangle=\delta(x-y)\end{align}$$ where the last equality follows from the completeness of the eigenvectors.

Exercise 12.4. Let $u(x,t)$ be a solution of (9) such that $u(x,0)=u(x)$. Show that $$\begin{align}u(x,t)=\int h(x,y;t)u(y)\, dy.\end{align}$$ (Hint: First verify that (11) satisfies that initial condition, next that it is a solution of the heat equation.)

It is known that the solution of (9) has an asymptotic expansion for $t\rightarrow \epsilon$ given by $$\begin{align}h(x,x;e\epsilon)=\sum_i a_i(x)\epsilon^i.\end{align}$$ Similarly, $h(t)$ has an expansion $$\begin{align}\tilde{h}(\epsilon)\equiv \sum_i a_i\epsilon^i\end{align}$$ where $a_i=\int a_i(x)dx$.

Let $E$ and $F$ be complex vector bundles over $M$ and $D:\Gamma(M,E)\rightarrow \Gamma(M,F)$ be an elliptic operator. We dfine two Laplacians $$\begin{align}\Delta_E\equiv D^\dagger D:\Gamma(M,E)\rightarrow \Gamma(M,E)\nonumber,\\ \Delta_F\equiv DD^\dagger:\Gamma(M,F)\rightarrow \Gamma(M,F).\end{align}$$ It is important to note that they have the same non-vanishing eigenvalues including the degeneracy. To see this, let $\Delta_E|\lambda\rangle=\lambda|\lambda\rangle$. Then there is a vector $D|\lambda\rangle\in \Gamma(M,F)$ such that $$\Delta_F(D|\lambda\rangle)=DD^\dagger D|\lambda\rangle=D\Delta_E|\lambda\rangle=\lambda (D|\lambda\rangle).$$ Note that $D|\lambda\rangle\ne 0$ since $\ker \Delta_E=\ker D$. Conversely, if $|\mu)\in \Gamma(M,F)$ satisfies $\Delta_F|\mu)=\mu |\mu)$, then $D^\dagger|\mu)\in \Gamma(M,E)$ is an eigenvector of $\Delta_E$ with the same eigenvalue $\mu$. Thus, we have gound the symmetry $$\begin{align}\mbox{Spec'}\Delta_E=\mbox{Spec'}\Delta_F\end{align}$$ where the prime denotes the zero eigenmodes are omitted. (spectrum of kernel?)

Define two heat kernels $h_E$ and $h_F$ by $$\begin{align}h_E(x,y,t)=\sum e^{-\lambda_n}\langle x|n\rangle \langle n|y\rangle, \nonumber\\ h_F(x,y,t)=\sum e^{-\mu_n}\langle x|m)( m|y\rangle.\end{align}$$ We have $$\begin{align}\lim_{t\rightarrow\infty}\tilde{h}_E(t)=\dim\ker\Delta_E=\dim\ker D,\nonumber\\ \lim_{t\rightarrow\infty}\tilde{h}_E(t)=\dim\ker\Delta_E=\dim\ker D.\end{align}$$ What is more interesting is the index of $D$. Since $\ker D=\ker \Delta_E$ and $\ker D^\dagger=\ker \Delta_F$, we have $$\begin{align}\mbox{ind} D=\dim\ker D-\dim\ker D^\dagger=\dim\ker \Delta_E-\dim\ker \Delta_F\nonumber\\ =\lim_{t\rightarrow \infty} [\tilde{h}_E(t)-\tilde{h}_F(t)]=\tilde{h}_E(t)-\tilde{h}_F(t).\end{align}$$ The final equality follows since the $t$-dependent part of $\tilde{h}_E(t)-\tilde{h}_F(t)$ cancel out by the symmetry (15). We expand $\tilde{h}_E(t)$ and $\tilde{h}_F(t)$ as $$\begin{align}\tilde{h}_E(t)=\sum a^E_it^i\quad \tilde{h}_F(t)=\sum a^F_it^i.\end{align}$$ Picking up $t$-independent terms, we have $$\begin{align}\mbox{ind} D=a^E_0-a^F_0=\int [a_0^E(x)-a^F_0(x)]\, dx\end{align}$$ where $a^{E,F}_0(x)$ are defined in (12). 

In general, $a^{E,F}_0(x)$ are local invariants written in terms of curvature two-forms. In section 13.2, we use the heat kernel to prove index theorem $$\mbox{ind} D=\nu_+-\nu_-=\int_M\mbox{ch}(\mathcal{F})|_{vol}$$ for the twisted spin complex over a manifold with $\hat{A}(TM)=1$.

Exercise 12.5. Let $D,D^\dagger,\Delta_E$ and $\Delta_E$ be as before. Show that $$\begin{align}I(s)\equiv \mbox{tr}\left[ \frac{s}{\Delta_E+s}-\frac{s}{\Delta_F+s}\right]\quad \mbox{Re} s>0\end{align}$$ is independent of $s$. Show also that $I(s)=\mbox{ind}D$.

12.7.2 Spectral $\zeta$-functions

Let $E$ and $F$ be vector bundle over $M$. Define a new function $$\begin{align}\zeta_E(x,y;s)\equiv \sum'\langle x|n\rangle \langle n|y\rangle \lambda^{-s}_n\quad \mbox{Re} s>0\end{align}$$ where $\Delta_E|n\rangle=\lambda_n|n\rangle$ and the prime denotes the omission of the zero modes ($\lambda_n=0$). A function $\zeta_F(x,y;s)$ may similarly be defined for $\Delta_F$. The functions $h_E$ and $\zeta_E$ are related by Mellin transformation. To see this, we recall the definition of the $\Gamma$-function, $$\begin{align}\Gamma(s)\zeta(x,y;s)=\sum_n'\int^\infty_0t^{s-1}e^{-\lambda_nt}\langle x|n\rangle\langle n|y\rangle dt\nonumber\\ =\int^\infty_0 t^{s-1}\left[ h(x,y;t)-\sum_i\langle x|0,i\rangle\langle 0,i|y\rangle \right] dt.\end{align}$$ We also note that $$\begin{align}\zeta_\Delta(s)\equiv \int_M\zeta(x,x;s)\, dx=\sum_n'\lambda_n^{-s}\end{align}$$ is the spectral $\zeta$-function definition.

Exercise 12.6. Verify that $$\begin{align}\Delta^{-s}f(x)=\int \zeta(x,y;s)f(y)\, dy\end{align}$$ where the general power of an operator may be defined in the sense of an eigenvalue, namely we put $\Delta^{-s}|n\rangle=\lambda^{-s}_n|n\rangle$. $\mbox{Re}s$ in assumed to be sufficiently large so that (24) is well defined. (Hint: Use that completeness of the eigenvectors.)

Example 12.7. Let $M=S^1=\{e^{i\theta}\}$ and $E=F=\mbox{a trivial line bundle over }S^1$ (a cylinder). Take an operator $\Delta\equiv -\partial^2/\partial\theta^2$. From the eigenvalue equation, $$-\frac{\partial^2e^{in\theta}}{\partial\theta^2}=n^2e^{in\theta}\quad n\in\mathbb{Z}$$ we find that $$\lambda_n=n^2\quad \langle\theta|n\rangle=(2\pi)^{-1/2}e^{in\theta}.$$ The heat kernel is $$\begin{align}h(\theta_1,\theta_2;t)=\sum e^{-n^2t}\langle \theta_1|n\rangle\langle n|\theta_2\rangle\nonumber\\ =\frac{1}{2\pi}\left( 1+\sum' e^{-n^2t}e^{in(\theta_1-\theta_2)}\right)\end{align}$$ while $$\begin{align}\zeta(\theta_1,\theta_2;s)=\sum'n^{-2s}\langle\theta_1|n\rangle\langle n|\theta_2\rangle\nonumber\\ =\frac{1}{2\pi}\sum' n^{-2s}e^{in(\theta_1-\theta_2)}.\end{align}$$ We easily verify that $\tilde{h}(t)=1+\sum'e^{-n^2t}$ satisfies $$1+2\int^\infty_1e^{-x^2t}dx<\tilde{h}(t)<1+2\int^\infty_0e^{-x^2t}dx.$$ We then find from these inequalities that $$\int^{+\infty}_{-\infty}e^{-x^2t}dx-1<\tilde{h}(t)<\int^{+\infty}_{-\infty}e^{-x^2t}dx+1$$ or by putting the value $$\int e^{-x^2t}dx=\sqrt{\pi}t^{-1/2}$$ we find $$\sqrt{\pi}t^{-1/2}-1<\tilde{h}(t)<\sqrt{\pi}t^{-1/2}+1.$$ This shows that $$\begin{align}\lim_{t\rightarrow 0^+}\tilde{h}(t)\sim \sqrt{\pi}t^{-1/2}.\end{align}$$ In general, the asymptotic series starts with $t^{-\dim M/2}$.