M721: Index Theory

Ruminations of a Graduate Class


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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


{\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)


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


\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


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}.

Then (2) reads

\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}


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

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