M721: Index Theory

Ruminations of a Graduate Class

Archive for September 2012

Symbols

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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

\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}.

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}

\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.

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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

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