# 1 Differential Operators

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

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

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:

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

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

$L$ 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

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

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.

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

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

$\begin{aligned} \operatorname{sym}_k(L^*)(x, \xi ) & = \pm (\operatorname{sym}_k(L)(x, \xi ))^*\\ \operatorname{sym}_k(L \circ L')(x, \xi ) & = \operatorname{sym}_k(L)(x, \xi ) \circ \operatorname{sym}_k(L')(x, \xi ).\end{aligned}$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

$(\xi , \omega ) \mapsto \xi \wedge \omega .$An *elliptic complex* is a sequence of vector bundles $E_0, \ldots , E_m$ with first-order differential operators

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 $T^*M$.

The de Rham complex and Dolbeault complex are elliptic complexes.

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

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.

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 $x = 0$.

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.