# M721: Index Theory

## 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.

Written by jpinzon84

September 11, 2012 at 10:42 pm

Posted in Uncategorized

## Two beautiful theorems about C(X)

with one comment

Let ${X}$ be a topological space. Let ${C(X)}$ be the set of continuous functions from ${X}$ to ${\mathbb{R}}$. ${C(X)}$ can also be thought of as set of smooth sections of the trivial bundle ${ X \times \mathbb{R} \longrightarrow X}$. Anyway, we get a contravariant functor

$\displaystyle C: Top \longrightarrow Ring$

where ${Top}$ is the category of ${topological}$ ${spaces}$. ${Ring}$ is the category of ${rings}$ with ${1}$ and every ${ring}$ ${homomorphism}$ sends ${1}$ to ${1}$. The two beautiful ${theorems}$ to be discussed here are the following. (Hewitt)For ${X}$, ${Y}$ compact Hausdorff there is a bijection

$\displaystyle Top(X, Y) \longrightarrow Ring(C(Y), C(X))$

(Swan)If ${X}$ is compact Hausdorff then taking sections gives a bijection from

$\displaystyle \lbrace \text{ isomorphism class of vector bundles over } X \rbrace \longrightarrow \lbrace \text{finitely generated projective C(X)-modules} \rbrace$

These two beautiful theoremshave some remarkable consequences If ${X}$, ${Y}$ are compact and Hausdorff then,

$\displaystyle X \cong Y \Longleftrightarrow C(X) \cong C(Y)$

${Theorem}$ \textup{2} leads to the following result in ${K}$${Theory}$

$\displaystyle K^{0}(X) \cong K_{0}(C(X))$

${Theorem}$ ${1}$ is a consequence of the following ${Lemma}$ Let ${X}$ be compact, Hausdorff topological space

1. For ${x_{0} \in X}$,$\displaystyle M(x_{0}) = \lbrace f \in C(X): f(x_{0}) = 0 \rbrace$

is a maximal ideal,

2. If ${M \lhd C(X)}$ is a maximal ideal, then ${\exists !}$ ${x_{0} \in X}$ such that ${M = M(x_{0})}$,
3. ${MaxSpec(C(X)) \cong X}$ where ${MaxSpec(R)}$ is the set of all maximal ideals of a ring ${R}$ equipped with ${Zariski}$ topology. The isomorphism takes ${M(x_{0})}$ to ${x_{0}}$

Proof:

1. Clearly ${M(x_{0})}$ is maximal as ${C(X) / M(x_{0}) \simeq \mathbb{R}}$ which is a field.
2. Notice, if ${f \notin M(x_{0})}$, then ${f(x_{0}) \neq 0}$ If ${I}$ is an ${ideal}$ such that ${I \nsubseteq M(x_{0})}$ for all ${x_{0}}$ in ${X}$ then for every ${x \in X}$, ${\exists}$ ${f_{x} \in I}$ such that ${f_{x}(x) \neq 0}$. Each ${f_{x}}$ there exists ${U_{x} \ni x}$ such that ${f_{x}(t) \neq 0}$ ${\forall t \in U_{x}}$. Since ${X}$ is compact ${\lbrace U_{x_{0}}, \ldots U_{x_{n}} \rbrace}$ cover ${X}$. Using ${bump}$ ${functions}$ ${b_{i}}$ which do not vanish on ${U_{x_{i}}}$ respectively, define$\displaystyle f = |f_{x_{0}}|b_{0} + \ldots +|f_{x_{n}}|b_{n}$

. Observe, ${f(x) \neq 0}$ ${\forall x \in X}$. Define ${g(x) = \frac{1}{f(x)}}$. Clearly ${g(x) \in C(X)}$ and ${f.g = 1}$. Thus ${I = C(X)}$. Thus the only maximal ideals of ${C(X)}$ is of the form ${M(x_{0})}$ for some ${x_{0} \in X}$.

3. For any ideal ${I}$ of a ring ${R}$ define,$\displaystyle V(I) = \lbrace M \text{ maximal in } R : I < M \rbrace$

${V(I)}$ is the basis for all ${closed}$ sets in the space ${MaxSpec(R)}$ under ${Zariski}$ topology. The map$\displaystyle F : X \longrightarrow MaxSpec(X)$

which sends$\displaystyle x \longmapsto M(x_{0})$

is already a bijection. All we need to show is$\displaystyle C \text{ closed} \Leftrightarrow F(C) \text{ closed}$

${(\Rightarrow):}$
IF ${C}$ closed then define ${I_{C} = \lbrace f \in C(X): f(C) = 0 \rbrace}$ and ${F(C) = V(I_{C}) = \bigcup_{x \in C} M(x)}$ \vspace{5pt}
${(\Leftarrow):}$
IF ${C}$ be a basic closed set in ${MaxSpec(C(X))}$, ie, ${C= V(I)}$ for some ${I \in C(X)}$ then, define$\displaystyle D = \lbrace x \in X: f(x) = 0 \forall f \in I \rbrace$

. Then ${D}$ is clearly a closed set and clearly ${D = F^{-1}(C)}$.

$\Box$ Proof: (of Theorem 1) In fact the ${C}$ gives the map between the ${Hom}$ sets of the respective category.
One-one
Let ${f: X \longrightarrow Y}$. Then

$\displaystyle C(f)(g) =f^{*}g = g \circ f$

, where ${g \in C(Y)}$. If ${C(f) = C(f')}$ then

${\Rightarrow g \circ f = g \circ f' \forall g \in C(Y)}$
${\Rightarrow f(x) = g(x)}$ by using bump functions near each point
Onto
Given a map ${F:C(Y) \longrightarrow C(X)}$, we induce a map

$\displaystyle \overline{F} : Spec(C(X))\longrightarrow Spec(C(Y))$

By ${Lemma}$ \textup{5} we get a map

$\displaystyle \overline{F}: X \longrightarrow Y$

It is clear that ${\overline{F}^{*} = F}$. $\Box$ Proof: ( sketch of proof of theorem 2)
Notice that Let ${G}$ be the map
G: ${\lbrace}$ isomorphism class of vector bundles over ${X}$ ${\rbrace \longrightarrow \lbrace}$ finitely generated C(X)-modules ${\rbrace}$
where given a vector bundle ${\xi}$
${G(\xi)}$ = ${\lbrace}$ smooth sections of ${\xi \rbrace}$.
Since ${X}$ is compact, any vector bundle ${\xi}$ is a ${subbundle}$ of a trivial bundle of finite dimension, ie ${X \times \mathbb{R}^{n}}$. Hence ${G(\xi)}$ is a sub-module of ${\bigoplus_{1}^{n} C(X)}$ due to the following isomorphism.
${ \tau: \bigoplus_{1}^{n} C(X) \cong \lbrace }$ smooth sections on the trivial bundle ${X \times \mathbb{R}^{n} \rbrace }$
Thus ${G(\xi)}$ is a finitely generated module. Moreover every bundle ${\xi}$ of finite dimension over a compact space has a complement, say ${\xi^{-1}}$, hence ${G(\xi) \oplus G(\xi^{-1})=\bigoplus_{1}^{n} C(X)}$. Hence its projective. Given a finitely generated projective module over ${C(X)}$, say ${M}$, find ${n}$ and a ${C(X)}$ module ${N}$, such that

$\displaystyle M \oplus N = \bigoplus_{1}^{n} C(X)$

Then define ${G^{-1}(M) = \tau^{-1}(M)}$. This is a ${vector}$ ${bundle}$ over ${X}$. The proof is non-trivial and is a ${theorem}$ of ${Swan}$. $\Box$

Written by prasit0605

September 5, 2012 at 1:33 am

Posted in Uncategorized