# M721: Index Theory

## A global definition of a differential operator

Background

${E\rightarrow X}$ and ${F\rightarrow X}$ are smooth vector bundles over a smooth manifold ${X}$. We have seen how to define differential operators locally, that is, over ${\mathbb{R}^n}$. How should we define global differential operators?

Definitions

Let us write ${Op(E,F) = Hom_\mathbb{R}(C^\infty E, C^\infty F)}$. We define a Leibniz bracket ${Op(E,F)\times C^\infty(X)\rightarrow Op(E,F)}$ by ${(D,f)\mapsto [D,f] = Df - fD,}$ which acts on a section ${s\in C^\infty E}$ by ${[D,f]s = D(fs) - fDs}$. That is,

$\displaystyle DO_m(E,F) = \{ D\in Op(E,F)\ |\ \forall f\in C^\infty(X)\ [D,f]\in DO_{m-1}(E,F)\}.$

This is similar to the Lie bracket of two vector fields, ${[X,Y]f = X(Yf) - Y(Xf)}$.

We set ${DO_{-1}(E,F) = \{0\}}$ and inductively define differential operators of order at most ${m}$ by requiring that ${D\in DO_m(E,F)}$ provided ${[D,f]\in DO_{m-1}(E,F)}$ for each ${f\in C^\infty(X)}$.

Recall that for a section $s \in C^\infty E$, its support $supp (s)$ is the closure of $\{ x \in X \mid s(x) \not = 0 \}$.

We would like to compare differential operators to local operators, those operators which only use local information in the following sense: An operator ${T\in Op(E,F)}$ is local if, for each ${s\in C^\infty E}$, ${supp(Ts) \subset supp(s)}$.

Lemma 1 A differential operator is local.

Proof: The only obvious way to prove this lemma is by induction. It is clearly true for ${m = -1}$: if ${s}$ is a section, then ${supp (s) = \emptyset\subset supp (s)}$.

Now suppose that this is true for ${m-1}$. Let ${D\in DO_m(E,F)}$, ${s\in C^\infty E}$ be arbitrarily chosen. Let ${U}$ be any open set for which ${supp (s)\subset U}$. By Urysohn’s lemma, there is some smooth function ${f}$ with ${f|_{supp (s)} = 1}$ and ${supp (s) \subset supp (f)\subset U}$. In particular, ${fs = s}$. (For brevity’s sake, call such an ${f}$ a “support function.”) Then since ${[D,f]s = D(fs) - f(Ds)}$, we observe that

$\displaystyle Ds = D(fs) = [Df]s + f(Ds).$

The support of the sum of two sections is the union of the supports of the sections since the sum is zero exactly when both sections are zero. Therefore,

$\displaystyle supp (Ds) = supp [D,f]s \cup supp (f(Ds)).$

The support of the product of two sections is the intersection of the supports of the sections, since the product is zero exactly when at least one of the sections is zero. Thus ${supp (f(Ds)) = supp(f) \cap supp(Ds)\subset supp(f)}$.

Additionally, ${supp[D,f]s \subset supp(s)}$ by the inductive assumption. So

$\displaystyle supp (Ds) \subset supp(s) \cup supp(f) = supp (f) \subset U.$

This containment holds for arbitrary ${U}$ containing ${supp(s)}$. Therefore, the support of ${Ds}$ is contained in the intersection of the closures of these open sets:

$\displaystyle supp(Ds) \subset \cap \bar{U} = supp (s).$

$\Box$

One might naively wonder if this proves anything substantial: Are there ever any operators that aren’t local? In fact, yes; here’s an example. Let ${X = \mathbb{S}^1}$, ${E = F = \mathbb{S}^1\times\mathbb{R}}$, so that the sections of ${E}$ and ${F}$ both constitute smooth functions on ${\mathbb{S}^1}$. Let ${\phi}$ be a function with support on the northern hemisphere of ${\mathbb{S}^1}$, i.e., ${supp(\phi) = \{e^{i\theta}\ |\ \theta\in[0,\pi]\}}$. Define an operator ${A_\phi\in Op(E,F)}$ by

$\displaystyle A_\phi f(\omega) = \int_\mathbb{S}^1 f(\theta)\phi(\omega-\theta)d\theta.$

We see that ${A_\phi}$ cannot be local. Consider, say, a function ${f}$ with support only on the southern hemisphere of ${\mathbb{S}^1}$, so that ${supp(f) \cap supp(\phi) = \{1,-1\}}$. But by examining the integral ${A}$, it’s clear that ${supp(Af)}$ is not a subset of ${supp (f)}$.

This gives intuition to the word “local:” a local operator ${D}$ acting on a section ${s}$ of ${E}$ determines ${Ds(x)}$ based only on the behavior of ${s}$ in a neighborhood of ${x}$, rather than on any global information. This corollary justifies the intuition: If on a neighborhood ${U}$ in ${X}$ two sections of ${E}$ agree, i.e., ${(s_1 - s_2)|_U = 0}$, then, restricted to ${U}$, ${Ds_1 = Ds_2}$.

Proof: Put ${s = s_1 - s_2}$. Then ${supp(Ds_1 - Ds_2) = supp (Ds) \subset supp (s) = supp(s_1 - s_2)}$ and ${supp (s) \cap U = \emptyset}$. Therefore, ${supp(Ds_1 - Ds_2)\cap U = \emptyset}$, so on ${U}$ we have that ${Ds_1 = Ds_2}$. $\Box$

Since we’ve seen that differential operators are local, the natural next question is whether there are any non-differential local operators. This is answered by a theorem of Peetre, proved in the 1960s:

Theorem 2 All local operators are differential operators.

The next question is whether in local coordinates differential operators can be represented in the form

$\displaystyle \sum_{|\alpha|\leq m} A_\alpha(x)D^\alpha.$