[Nakahara GTP] 12.1 Elliptic operators and Fredholm operators

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

12.1.1 Elliptic operators 

Let $E$ and $F$ be complex vector bundles over a manifold $M$. A differential operator $D$ is a linear map $$\begin{align}D:\Gamma(M,E)\rightarrow \Gamma(M,F).\end{align}$$ Take a chart $U$ of $M$ over which $E$ and $F$ are trivial. We denote the local coordinates of $U$ as $x^\mu$. We introduce the following multi-index notation, $$M\equiv (\mu_1,\mu_2,\cdots,\mu_m)\quad \mu_j\in\mathbb{Z},\ \mu_j\ge 0\\ |M|\equiv \mu_1+\mu_2+\cdots+\mu_m\\ D_M=\frac{\partial^{|M|}}{\partial x^M}\equiv \frac{\partial^{\mu_1+\cdots+\mu_m}}{\partial(x^1)^{\mu_1}\cdots\partial(x^m)^{\mu_m}}.$$ If $\dim E=k$ and $\dim F=k'$, the most general form of $D$ is $$\begin{align}[Ds(x)]^\alpha=\sum_{|M|\le N,\ 1\le a\le k}A^{M\alpha}_a(x)D_Ms^a(x)\quad 1\le a\le k'\end{align}$$ where $s(x)$ is a section of $E$. Note that $x$ denotes a point whose coordinates are $x^\mu$. This slight abuse simplifies the notation. $A^M\equiv (A^M)^\alpha_a$ is a $k\times k'$ matrix which may depend on the position $x$. The positice integer $N$ in (2) is called the order of $D$. We are interseted in the case in which $N=1$ (the Dirac operator) and $N=2$ (the Laplacian). For example, if $F$ is a spin bundle over $M$, the Dirac operator $D\equiv i\gamma^\mu\partial_\mu+m:\ \Gamma(M,E)\rightarrow \Gamma(M,E)$ acts on a section $\psi(x)$ of $E$ as $$[D\psi(x)]^\alpha=i(\gamma^\mu)^\alpha_\beta\partial_\mu\psi^\beta(x)+m\psi^\alpha(x).$$

The symbol of $D$ is a $k\times k'$ matrix $$\begin{align}\sigma(D,\xi)\equiv \sum_{|M|=N}A^{M\alpha}_a(x)\xi_M\end{align}$$ where $\xi$ is a real $m$-tuple $\xi=(\xi_1,\cdots,\xi_m)$. The symbol is also defined independently of the coordinates as follows. Let $\pi:E\rightarrow m$ be a vector bundle and let $p\in M$, $\xi\in T^*_pM$ and $s\in \pi^{-1}_E(p)$. Take a section $\tilde{s}\in\Gamma(M,E)$ such that $\tilde{s}(p)=s$ and a function $f\in \mathcal{F}(M)$ such that $f(p)=0$ and $df(p)=\xi\in T^*_pM$. Then the symbol may be defined by $$\begin{align}\sigma(D,\xi)s=\frac{1}{N!}D(f^N\tilde{s})|_p.\end{align}$$ The factor $f^N$ automatically picks up the $N$th-order term due to the condition $f(p)=0$. Equation (4) yields the same symbol as (3). (Differential at a point and given order)

If the matrix $\sigma(D,\xi)$ is invertible for each $x\in M$ and each $\xi\in \mathbb{R}^m-\{0\}$, the operator $D$ is said to be elliptic. Clearly this definition makes sense only when $k=k'$ (square matrix). If should be noted that the symbol for a composite operator $D=D_1\circ D_2$ is a composite of the symbols, namely $\sigma(D,\xi)=\sigma(D_1,\xi)\, \sigma(D_2,\xi)$. This shows that composites of elliptic operators are also elliptic. In general, powers and roots of elliptic operators are elliptic.

Example 12.1. Let $x^\mu$ be the natural coordinates in $\mathbb{R}^m$. If $E$ and $F$ are real line bundles over $\mathbb{R}^m$, the Laplacian $\Delta:\Gamma(\mathbb{R}^m,E)\rightarrow \Gamma(\mathbb{R}^m,F)$ is defined by $$\begin{align}\Delta\equiv \frac{\partial^2}{\partial(x^1)^2}+\cdots+\frac{\partial^2}{\partial(x^m)^2}.\end{align}$$

$A^M$ is $m\times 1$ matrix. According to (3), the symbol is $$\sigma(\Delta,\xi)=\sum_\mu(\xi_\mu)^2.$$ This is in agreement with the result obtained from (4), $$\sigma(\Delta,\xi)s=\frac{1}{2}\Delta(f^2\tilde{s})|_p=\frac{1}{2}\sum\frac{\partial^2}{\partial(x^\mu)^2}(f^2\tilde{s})|_p\\ =\frac{1}{2}\left. \left( f^2\Delta\tilde{s}+2f\Delta f\tilde{s}+2f\sum \frac{\partial f}{\partial x^\mu}\frac{\partial\tilde{s}}{\partial x^\mu}+2\sum\frac{\partial f}{\partial x^\mu}\frac{\partial f}{\partial x^\mu}\tilde{s}\right)\right|_p\\ =\sum(\xi_\mu)^2s.$$ (Use $f(p)=0$, $df=\xi$) This symbol is clearly invertible for $\xi\ne 0$, and hence $\Delta$ is elliptic.

12.1.2 Fredholm operators 

Let $D:\Gamma(M,E)\rightarrow \Gamma(M,F)$ be an elliptic operator. The kernel of $D$ is the set of null eigenvectors $$\begin{align}\ker D\equiv \{s\in \Gamma(M,E)|Ds=0\}.\end{align}$$ Suppose $E$ ans $F$ are endowed with fibre metrics, which will be denoted $\langle\ ,\ \rangle_E$ and $\langle\ ,\ \rangle_F$, respectively. The adjoint $D^\dagger:\Gamma(M,F)\rightarrow \Gamma(M,E)$ of $D$ is defined by $$\begin{align}\langle s',Ds\rangle_F\equiv \langle D^\dagger s',s\rangle_E\end{align}$$ where $s\in \Gamma(M,E)$ and $s'\in \Gamma(M,F)$. We defined the cokernel of $D$ by $$\begin{align}\mbox{coker}D\equiv \Gamma(M,F)/\mbox{im} D.\end{align}$$

Among elliptic operators we are interested in a class of operators whose kernels and cokernels are finite dimensional. An elliptic operator $D$ which satisfies this condition is called a Fredholm operator. The analytical index $$\begin{align}\mbox{ind}D\equiv \dim \ker D-\dim \mbox{coker}D\end{align}$$ is well defined for a Fredholm operator. Henceforth, we will be concerned only with Fredholm operators. It is known from the general theory of operators that elliptic operators on a compact manifold are Fredholm operators. Theorem 12.1 shows that $\mbox{ind}D$ is also expressed as $$\begin{align}\mbox{ind}D=\dim\ker D-\dim\ker D^\dagger.\end{align}$$

Theorem 12.1. Let $D:\Gamma(M,E)\rightarrow \Gamma(M,F)$ be a Fredholm operator. Then $$\begin{align}\mbox{coker}D\simeq \ker D^\dagger\equiv \{s\in\Gamma(M,F)|D^\dagger s=0\}.\end{align}$$

Proof. Let $[s]\in\mbox{coker} D$ be given by $$[s]=\{s'\in \Gamma(M,F)|s'=s+Du,\ u\in\Gamma(M,E)\}.$$ We show that there is a surjection $\ker D^\dagger\rightarrow \mbox{coker} D$, namely ans $[s]\in\mbox{coker}D$ has a representative $s_0\in\ker D^\dagger$. Define $s_0$ by $$\begin{align}s_0\equiv s-D\frac{1}{D^\dagger D}D^\dagger s.\end{align}$$ We find $s_0\in\ker D^\dagger$ since $D^\dagger s_0=D^\dagger s-D^\dagger D(D^\dagger D)^{-1}D^\dagger s=D^\dagger s-D^\dagger s=0$. Next, let $s_0,s_0'\in\ker D^\dagger$ and $s_0\ne s_0'$. We show that $[s_0]\ne [s_0']$ in $\Gamma(M,F)/\mbox{im}D$. If $[s_0]=[s_0']$, there is an element $u\in\Gamma(M,E)$ such that $s_0-s_0'=Du$. Then $0=\langle u,D^\dagger(s_0-s_0')\rangle_E=\langle u,D^\dagger Du\rangle_E=\langle Du,Du\rangle_F\ge 0$, hence $Du=0$, which contradicts our assumption $s_0\ne s_0'$. Thus, the map $s_0\mapsto [s]$ is a bijection and we have established that $\mbox{coker}D\simeq \ker D^\dagger$.

12.1.3 Elliptic complexes

Consider a sequence of Fredholm operators, $$\begin{align}\cdots\rightarrow \Gamma(M,E_{i-1})\stackrel{D_{i-1}}{\rightarrow}\Gamma(M,E_i)\stackrel{D_i}{\rightarrow}\Gamma(M,E_{i+1})\stackrel{D_{i+1}}{\rightarrow}\cdots\end{align}$$ where $\{E_i\}$ is a sequence of vector bundles over a compact manifold $M$. The sequence $(E_i,D_i)$ is called an elliptic complex if $D_i$ is nilponent (that is $D_i\circ D_{i-1}=0$) for any $i$. The reader may refer to $\Gamma(M,E_i)=\Omega_i(M)$ and $D_i=d$ (exterior derivative) for example. The adjoint of $D_i:\Gamma(M,E_i)\rightarrow \Gamma(M,E_{i+1})$ is denoted by $$D^\dagger_i:\Gamma(M,E_{i+1})\rightarrow\Gamma(M,E_i).$$ The Laplacian $\Delta_i:\Gamma(M,E_i)\rightarrow \Gamma(M,E_i)$ is $$\begin{align}\Delta_i\equiv D_{i-1}D^\dagger_{i-1}+D^\dagger_iD_i.\end{align}$$ The Hodge decomposition also applies to the present case, $$\begin{align}s_i=D_{i-1}s_{i-1}+D^\dagger_is_{i+1}+h_i\end{align}$$ where $s_{i\pm1}\in\Gamma(M,E_{i\pm1})$ and $h_i$ is in the kernel of $\Delta_i$, $\Delta_ih_i=0$.

Analogously to the de Rham cohomology groups, we define $$\begin{align}H^i(E,D)\equiv \ker D_i/\mbox{im}D_{i-1}.\end{align}$$ As in the case of de Rham theory, it can be shown that $H^i(E,D)$ is isomorphic to the kernel of $\Delta_i$. Accordingly, we have $$\begin{align}\dim H^i(E,D)=\dim \mbox{Harm}^i(E,D)\end{align}$$ where $\mbox{Harm}^i(E,D)$ is a vector space spanned by $\{h_i\}$. The index of this elliptic complex is defined by $$\begin{align}\mbox{ind}D\equiv \sum^m_{i=0}(-1)^i\dim H^i(E,D)=\sum^m_{i=0}(-1)^i\dim \ker \Delta_i.\end{align}$$ The index thus defined generalizes the Euler characteristic, see example 12.2.

How is this related to (10)? Consider the complex $\Gamma(M,E)\stackrel{D}{\rightarrow}\Gamma(M,F)$. We may formally add zero on both sides, $$\begin{align}0\stackrel{i}{\hookrightarrow} \Gamma(M,E)\stackrel{D}{\rightarrow}\Gamma(M,F)\stackrel{\varphi}{\rightarrow} 0\end{align}$$ where $i$ is the inclusion. The index according to (19) is $$\dim\ker D-\{\dim\Gamma(M,F)-\dim\mbox{im}D\}=\dim\ker D-\dim\mbox{coker}D$$ where we have noted that $\dim\mbox{im}i=0$, $\ker\varphi=\Gamma(M,F)$ and $\mbox{coker}D=\ker\varphi/\mbox{im}D$. Thus, (19) yields the same index as (10).

It is often convenient to work with a two-term elliptic complex which has the same index as the original elliptic complex $(E,D)$. This rolloing up is carried out by defining $$\begin{align}E_+\equiv \oplus_r E_{2r},\ E_-\equiv \oplus_r E_{2r+1}\end{align}$$ which are called the even bundle and the odd bundle, respectively. Correspondingly we consider the operators $$\begin{align}A\equiv \oplus_r (D_{2r}+D^\dagger_{2r-1}),\ A^\dagger\equiv \oplus_r (D_{2r+1}+D^\dagger_{2r}).\end{align}$$ We readily verify that $A:\Gamma(M,E_+)\rightarrow \Gamma(M,E_-)$ and $A^\dagger:\Gamma(M,E_-)\rightarrow \Gamma(M,E_+)$. From $A$ and $A^\dagger$, we construct the two Laplaticans $$\begin{align}\Delta_+\equiv A^\dagger A=\oplus_{r,s}(D_{2r+1}+D^\dagger_{2r})(D_{2s}+D^\dagger_{2s-1})\nonumber \\ =\oplus_r(D_{2r-1}D^\dagger_{2r-1}+D^\dagger_{2r}D_{2r})=\oplus_r\Delta_{2r} \\ \Delta_-\equiv AA^\dagger = \oplus_r\Delta_{2r+1}.\end{align}$$ Then we have $$\begin{align}\mbox{ind}(E_\pm,A)=\dim\ker \Delta_+-\dim\ker\Delta_-\nonumber\\ =\sum(-1)^r\dim\ker \Delta_r=\mbox{ind}(E,D).\end{align}$$

Example 12.2. Let us consider the de Rham complex $\Omega(M)$ over a compact manifold $M$ without a boundary, $$\begin{align}0\stackrel{i}{\rightarrow}\Omega^0(M)\stackrel{d}{\rightarrow}\Omega^1(M)\stackrel{d}{\rightarrow}\cdots \stackrel{d}{\rightarrow}\Omega^m(M)\stackrel{d}{\rightarrow}0\end{align}$$ where $m=\dim M$ and $d$ stands for $d_r:\Omega^r(M)\rightarrow \Omega^{r+1}(M)$. $H^r(E,D)$ defined by (25) agrees with the de Rham cohomology group $H_r(M,\mathbb{R})$. The index is identified with the Euler characteristic, $$\begin{align}\mbox{ind}(\Omega^*(M),d)=\sum^m_{r=0}(-1)^r\dim H^r(M;\mathbb{R})=\chi(M).\end{align}$$ We found in chapter 7 that $b^r\equiv \dim H^r(M,\mathbb{R})$ agrees with the number of linearly independent harmonic $r$-forms: $\dim H^r(M,\mathbb{R})=\dim\mbox{Harm}^r(M)=\dim\ker\Delta_r$, where $\Delta_r$ is the Laplacian $$\begin{align}\Delta_r=(d+d^\dagger)^2=d_{r-1}d^\dagger_{r-1}+d^\dagger_r d_r\end{align}$$ $d^\dagger_r:\Omega^{r+1}(M)\rightarrow \Omega^r(M)$ being the adjoint of $d_r$. Now we find that $$\begin{align}\chi(M)=\sum^m_{r=0}(-1)^r\dim\ker\Delta_r.\end{align}$$ This relation is very interesting since the LHS is a purely topological quantity which can be computed by triangulating $M$, for example, while the RHS is given by the solution of an analytic equation $\Delta_ru=0$. We noted in example 11.6 that $\chi(M)$ is given by integrating the Euler class over $M:\chi(M)=\int_Me(TM)$. Now (28) reads $$\begin{align}\sum^m_{r=1}(-1)^r\dim\ker \Delta_r=\int_Me(TM).\end{align}$$ This is a typical form of the index theorem. The RHS is an analytic index while the LHS is a topological index given by the integral of certain characteristic classes. In section 12.3, we derive (29) from the Atiyah-Singer index theorem.

The two-term complex is given by $$\begin{align}\Omega^+(M)\equiv \oplus_r\Omega^{2r}(M)\quad \Omega^-(M)\equiv \oplus_r\Omega^{2r+1}(M).\end{align}$$ The corresponding operators are $$\begin{align}A\equiv \oplus_r(d_{2r}+d^\dagger_{2r-1})\quad A^\dagger\equiv \oplus_r(d_{2r-1}+d^\dagger_{2r}).\end{align}$$ It is left as an exercise to show that $$\begin{align}\mbox{ind}(\Omega^\pm(M),A)=\dim\ker A_+-\dim\ker A_-=\chi(M).\end{align}$$