Global AnalysisDifferential Operators

# 1 Differential Operators

Fix a manifold $M$, and Hermitian vector bundles $E_0, E_1 \to M$ with inner products.

Definition

A differential operator $L: \Gamma (M, E_0) \to \Gamma (M, E_1)$ of order $k$ is a $\mathbb {C}$-linear map that is local, i.e. the value of $Lu$ near a point $p \in M$ depends only on the values of $u$ near $p$, and in a coordinate chart, it is of the form

$Lf = \sum _{|\alpha | \leq k} A^\alpha (x) \mathrm{D}_\alpha f$

for some $A^\alpha \in \Gamma (\operatorname{Hom}(E_0, E_1))$.

Note that under our definition, any differential operator of order $k$ is also a differential operator of order $k + 1$.

Just as we can define tangent vectors as derivations, we have the following coordinate-free definition of differential operators (which we will not use), with a similar proof:

Fact

Let $L: \Gamma (M, E_0) \to \Gamma (M, E_1)$ be linear. Then

1. $L$ is a differential operator of order $0$ iff $[L, f] = 0$ for all $f \in C^\infty (M)$.

2. $L$ is a differential operator of order $k$ iff $[L, f]$ is a differential operator of order $k - 1$ for all $f \in C^\infty (M)$.

Integration by parts implies that we have

Lemma

For any differential operator $L: \Gamma (M, E_0) \to \Gamma (M, E_1)$, there is a formal adjoint $L^*: \Gamma (M, E_1) \to \Gamma (M, E_0)$ such that for any $f_i \in \Gamma (M, E_i)$, we have

$(Lf_0, f_1)_{L^2} = (f_0, L^* f_1)_{L^2}.$

Definition

Let $L$ be a differential operator of order $k$. The (principal) symbol of $L$ is the family of operators $\operatorname{sym}_k(L)(x, \xi )\in \operatorname{Hom}((E_0)_x, (E_1)_x)$ for $(x, \xi ) \in T^*M$ given locally by

$\operatorname{sym}_k (L)(x, \xi ) = \sum _{|\alpha | = k} A^\alpha (x) \xi _\alpha .$

Formally, if $\pi : T^*M \to M$ is the projection, then $\operatorname{sym}_k(L) \in \Gamma (\operatorname{Hom}(\pi ^* E_0, \pi _* E_1))$.

In a coordinate-free manner, if $s \in \Gamma (M, E_0)$ and $f \in C^\infty (M)$ with $f(x) = 0$, then

$\operatorname{sym}_k (L)(x, (\mathrm{d}f)_x)(s(x)) = \frac{1}{k!} L(f^k s)(x).$

We say $L$ is elliptic at $x \in M$ if $\operatorname{sym}_k(x, \xi )$ is invertible for all $\xi \in T^*_x M \setminus \{ 0\}$, and $L$ is elliptic if it is elliptic everywhere.

While the coordinate-free definition seems rather artificial, it is actually useful when we want to do computations later on.

It will be convenient to note that the adjoint of an elliptic operator is elliptic. More generally,

Lemma

For any operators $L, L'$, we have

\begin{aligned} \operatorname{sym}_k(L^*)(x, \xi ) & = \pm (\operatorname{sym}_k(L)(x, \xi ))^*\\ \operatorname{sym}_k(L \circ L')(x, \xi ) & = \operatorname{sym}_k(L)(x, \xi ) \circ \operatorname{sym}_k(L')(x, \xi ).\end{aligned}
Hence the composition and adjoints of elliptic operators is elliptic.

Example

Consider the exterior derivative $\mathrm{d}: \Omega ^p(M) \to \Omega ^{p + 1}(M)$. Using the coordinate-free definition, we compute the symbol as

$\operatorname{sym}_1(\mathrm{d})(x, (\mathrm{d}f)_x)(\omega _x) = (\mathrm{d}(f\omega ))_x = (\mathrm{d}f \wedge \omega )_x$

whenever $f(x) = 0$. So the symbol of $\mathrm{d}$ is

$(\xi , \omega ) \mapsto \xi \wedge \omega .$
Note that for $p > 0$, this is not invertible. Instead, what we have is an elliptic complex.
Definition

An elliptic complex is a sequence of vector bundles $E_0, \ldots , E_m$ with first-order differential operators

$L_i: \Gamma (M, E_{i - 1}) \to \Gamma (M, E_i)$

such that $L_{i + 1} \circ L_i = 0$ and for any non-zero $\xi \in T^*_x M$, the sequence

is exact outside of the zero section of $T^*M$.

Exercise

The de Rham complex and Dolbeault complex are elliptic complexes.

Ultimately, we will prove Hodge decomposition for elliptic complexes, which subsumes the Hodge decomposition of Riemannian and Kähler manifolds.

To get from an elliptic complex to an elliptic operator, we use the following linear algebraic result:

Lemma

Let $\{ V_i\}$ be finite-dimensional vector spaces, and

be an exact sequence. Let $V = \bigoplus V_i$. Then $f + f^*: V \to V$ is an isomorphism.

Proof
It suffices to show $f + f^*$ is injective. Suppose $(f + f^*)x = 0$. Then

$(f + f^*)^2 x = (ff^* + f^* f)x = 0.$

So we get

$0 = \langle ff^*x, x\rangle + \langle f^*fx\rangle = \langle f^* x, f^* x\rangle + \langle fx, fx\rangle .$

So $fx = f^*x = 0$. By exactness, $x = fy$ for some $y$, and then

$0 = \langle f^* fy, y\rangle = \langle fy, fy\rangle = \langle x, x \rangle .$

So $x = 0$.

Proof

Corollary

If $(E_*, L_*)$ is an elliptic complex, define $E = \bigoplus E_i$. Then

$D = L + L^*: \Gamma (M, E) \to \Gamma (M, E)$

is elliptic.