# M721: Index Theory

Ruminations of a Graduate Class

## Symbols

This post contains various definitions of the symbol of a differential operator. We will state a local version, then a global version and then we will finally view the symbol in its most abstract form: a section of a bundle over the total space of a cotangent bundle.

Review of Local Definitions

Let’s start by recalling that a differential operator of order ${m}$ on the manifold is ${X={\mathbb R}^n}$ is defined by:

$\displaystyle D=\sum_{|\alpha|\leq m}A_\alpha(\hspace{1ex})D^\alpha:\:C^\infty({\mathbb R}^n,{\mathbb R}^N)\rightarrow C^\infty({\mathbb R}^n,{\mathbb R}^N)\ \ \ \ \ (1)$

where ${A_\alpha(\hspace{1ex}): X={\mathbb R}^n\rightarrow M_{M\times N}({\mathbb R})}$ is smooth and if ${\alpha=(\alpha_1,\ldots,\alpha_n)}$, then ${\displaystyle D^\alpha=\frac{\partial^{\alpha_1}}{\partial x_1^{\alpha_1}}\frac{\partial^{\alpha_2}}{\partial x_2^{\alpha_2}}\cdots\frac{\partial^{\alpha_n}}{\partial x_n^{\alpha_n}}}$

Let ${x\in X={\mathbb R}^n}$ and ${\xi\in{\mathbb R}^n=T^*X}$.  The Symbol of ${D}$, denoted by ${\sigma_D}$ is then

$\displaystyle \sigma_D(x,\xi)=\sum_{|\alpha|\leq m}A_\alpha(x)\xi_1^{\alpha_1}\xi_2^{\alpha_2}\cdots\xi_n^{\alpha_n}\quad\in M_{M\times N}({\mathbb R})=\text{Hom}({\mathbb R}^N,{\mathbb R}^M)$

A differential operator ${D:\:C^\infty({\mathbb R}^n,{\mathbb R}^N)\rightarrow C^\infty({\mathbb R}^n,{\mathbb R}^N)}$ is said to be elliptic if for all ${x\in X}$ and every ${\xi\neq 0}$ we have that ${\sigma_D(x,\xi)}$ is invertible.

Global definition of the Symbol

Consider a globally defined differential operator

$\displaystyle D:\: C^\infty(E)\rightarrow C^\infty(F)$

for ${x_0\in X}$ and ${\xi\in T_{x_0}^*X}$ we want to define a linear map
$\displaystyle \sigma_D(x_0,\xi):E_{x_0}\rightarrow F_{x_0}$
in a coordinate free way.

With this in mind let ${e\in E_{x_0}}$ and choose:
1.    ${f:X\rightarrow{\mathbb R}}$ such that ${\mbox{d} f_{x_0}=\xi}$
2.    ${s\in C^\infty(E)}$ such that ${s(x_0)=e}$

Then we define

$\displaystyle \sigma_D(x_o,\xi)(e)=\frac{1}{m!}D(f^ms)(x_0)\ \ \ \ \ (2)$

Notice that even though this is a coordinate free definition of the symbol, it is still unclear how it changes in ${x}$ and ${\xi}$. We will later see that ${\sigma_D}$ is actually smooth on ${(x,\xi)}$. Before this, we should prove that this definition is in fact independent on the choice of ${f}$ and ${s}$.

${\sigma_D}$ does not depend on ${f}$

Claim 1 If ${g: X\rightarrow {\mathbb R}}$ is a smooth function such that ${\mbox{d} g_{x_0}=\xi}$ and ${g(x_0)=0}$, then
$\displaystyle D((f^m-g^m)s)(x_0)=0$

Proof: For any differential operator ${D}$, any section ${s}$ and any function ${\varphi:X\rightarrow{\mathbb R}}$,

$\displaystyle [D,\varphi](s)=D(\varphi s)-\varphi D(s)$

Setting ${\varphi=f^m-g^m}$ we have

$\displaystyle [D,f^m-g^m](s)(x_0)=D((f^m-g^m)s)(x_0)\ \ \ \ \ (3)$

Induction on the order of ${D}$ and (3) will give us the result:

Let ${D\in \text{DO}_0(E,F)}$, then by definition

$\displaystyle D((f^m-g^m)s)=(f^m-g^m)D(s)$

and so

$\displaystyle D((f^m-g^m)s)(x_0)=0$

Now assume the claim is true for every differential operator of order less than ${m}$ and suppose ${D\in \text{DO}_m(E,F)}$. By definition, ${[D,f^m-g^m]\in \text{DO}_{m-1}(E,F)}$
Thus, by induction

$\displaystyle [D,f^m-g^m](s)(x_0)=0$

and notice that (3) gives us

$\displaystyle [D,f^m-g^m](s)(x_0)=D((f^m-g^m)s)(x_0)$

so that

$\displaystyle D((f^m-g^m)s)(x_0)=0$

$\Box$

${\sigma_D}$ does not depend on ${s}$

Claim 2 Let ${s_1,s_2\in C^\infty(E)}$ be such that ${e=s_1(x_0)=s_2(x_0)}$, then
$\displaystyle D(f^m(s_1-s_2))(x_0)=0$

Proof: It is easier if we use the easy direction of Peetre’s Theorem so that we can use the fact that ${D}$ is local, that is

$\displaystyle \text{supp}(Ds)\subseteq \text{supp}(s)$

equivalently

$\displaystyle X\setminus\text{supp}(s)\subseteq X\setminus\text{supp}(Ds)$

equivalently

$\displaystyle s(x_0)=0\Rightarrow Ds(x_0)=0$

So, since ${f^m(s_1-s_2)(x_0)=0}$, we have ${D(f^m(s_1-s_2))(x_0)=0}$ as sought. $\Box$

Let us finish the section with a short remark:

${\sigma_D(x_0,\xi)}$ is homogeneous of degree ${m}$ in ${\xi}$. That is, for every ${\rho>0}$,
$\displaystyle \sigma_D(x_0,\rho\xi)=\rho^m\sigma_D(x_0,\xi)$

Proof: Simply take ${\rho f}$ instead of ${f}$ in the definition for ${\sigma_D}$.$\Box$

Local=Global

Lemma 1 For ${D=\sum_{|\alpha|\leq m}A_\alpha(\hspace{1ex})D^\alpha:\:C^\infty({\mathbb R}^n,{\mathbb R}^N)\rightarrow C^\infty({\mathbb R}^n,{\mathbb R}^N)}$ a differential operator of order ${m}$, the two definitions of symbol coincide under the identification ${{\mathbb R}^n\cong T_{x_0}^*{\mathbb R}^n}$ given by ${\xi\rightarrow\sum_{i=1}^n\xi_i\mbox{d} x_i}$

Proof: Let ${x_0,\xi\in{\mathbb R}^n}$. The function ${f(x)=\langle x-x_0,\xi\rangle}$ satisfies the conditions stated in the coordinate free definition of ${\sigma_D}$.  Let ${s}$ be the constant section ${e}$, that is, ${s(x)=e}$ for every ${x\in X}$.

$\displaystyle \sigma_D(x_0,\xi)=\frac{1}{m!}D(f^me)(x_0)$

where by (1)

$\displaystyle \frac{1}{m!}D(f^me)(x_0)=\frac{1}{m!}\sum_{|\alpha|\leq m}A_\alpha(x_0)D^\alpha(f^me)(x_0)=\frac{1}{m!}\sum_{|\alpha|\leq m}A_\alpha(x_0)\frac{\partial^{\alpha_1}}{\partial x_1^{\alpha_1}}\frac{\partial^{\alpha_2}}{\partial x_2^{\alpha_2}}\cdots\frac{\partial^{\alpha_n}}{\partial x_n^{\alpha_n}}(f^me)(x_0)$

Notice that here

$\displaystyle D^\alpha(f^me)=\left(\begin{array}{c}D^\alpha(f^me_1)\\D^\alpha(f^me_2)\\ \vdots\\ D^\alpha(f^me_N)\end{array}\right)= \left(\begin{array}{c}D^\alpha(f^m)e_1\\D^\alpha(f^m)e_2\\ \vdots\\ D^\alpha(f^m)e_N\end{array}\right)= D^\alpha(f^m)e$

since ${e}$ is a constant section.

Also notice that
1.    ${D^\beta(f^m)(x_0)=0}$ for every ${|\beta|\leq m-1}$: This is because there is always a factor of ${f}$ in the expression for ${D^\beta(f^m)}$ whenever ${|\beta|\leq m-1}$.
2.    ${D^\alpha(f^m)(x_0)=m!\,\xi_1^{\alpha_1}\xi_2^{\alpha_2}\cdots\xi_n^{\alpha_n}}$: This is a simple calculation.

Consolidating all the information we conclude

$\displaystyle \sigma_D(x_0,\xi)=\frac{1}{m!}D(f^me)(x_0)=\frac{1}{m!}\sum_{|\alpha|= m}A_\alpha(x_0)D^\alpha(f^me)(x_0)=\sum_{|\alpha|= m}A_\alpha(x_0)\xi_1^{\alpha_1}\xi_2^{\alpha_2}\cdots\xi_n^{\alpha_n}$

$\Box$

Symbol as a section

By consolidating definitions (*) and (1) of ${\sigma_D}$ we get ${\sigma_D\in C^{\infty}(\pi^*(\text{Hom}_{\mathbb R}(E,F)))}$. Here ${\pi}$ is the bundle map ${\pi:T^*X\rightarrow X}$ and we are just looking at the diagram

To be explicit, if ${\omega\in T^*X}$, then ${\omega=(x_0,\xi)}$ with ${\xi\in T_{x_0}^*X}$. So

$\displaystyle \sigma_D(\omega)=\sigma_D(x_0,\xi): E_{x_0}\rightarrow F_{x_0}$

that is, ${\sigma_D(x_0,\xi)\in \text{Hom}_{\mathbb R}( E_{x_0}, F_{x_0})}$ and we are using the identification ${\text{Hom}_{\mathbb R}( E_{x_0}, F_{x_0})\cong\pi^*(\text{Hom}_{\mathbb R}( E_{\omega}, F_{\omega}))}$.

Smoothness follows from the smoothness of the local definition and the fact that both definitions coincide locally.

Finally, let

$\displaystyle \text{sym}_m(E,F)=\{\sigma\in C^{\infty}(\pi^*(\text{Hom}_{\mathbb R}(E,F))) \mid \text{ for all } \rho>0,\: \omega\in T^*X\:,\: \sigma(\rho\omega)=\rho^m\sigma(\omega)\}$

then we have

Proposition 2 There is an exact sequence

$\displaystyle 0\rightarrow \text{DO}_{m-1}(E,F)\rightarrow \text{DO}_{m}(E,F)\rightarrow\text{Symbol}_m(E,F)$

Notice that this proposition (re)captures the fact that the symbol of an operator only sees’ the top’ degree of the operator.

Fundamental Theorem of Elliptic Operators

Now that we have a global definition of the symbol of a differential operator, we can state what it means for a differential operator to be elliptic. Namely, ${D}$ is elliptic if for every ${\omega\in T^*X\setminus\{X\}}$ (i.e ${\omega}$ is in the complement of the zero section of the cotangent bundle), the map ${\sigma_D(\omega)}$ is invertible.  The most important result involving elliptic operators is the following theorem:

Theorem 3 Fundamental Theorem of Elliptic Operators
If ${D:\: C^\infty(E)\rightarrow C^\infty(F)}$ is an elliptic differential operator over a compact manifold ${X}$, then both ${\text{ker}D}$ and ${\text{coker}D}$ are finite dimensional vector spaces.