Global Analysis — Differential Operators

1 Differential Operators

Fix a manifold MM, and Hermitian vector bundles E0,E1ME_0, E_1 \to M with inner products.

Definition

A differential operator L:Γ(M,E0)Γ(M,E1)L: \Gamma (M, E_0) \to \Gamma (M, E_1) of order kk is a C\mathbb {C}-linear map that is local, i.e. the value of LuLu near a point pMp \in M depends only on the values of uu near pp, and in a coordinate chart, it is of the form

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

for some AαΓ(Hom(E0,E1))A^\alpha \in \Gamma (\operatorname{Hom}(E_0, E_1)).

Note that under our definition, any differential operator of order kk is also a differential operator of order k+1k + 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:Γ(M,E0)Γ(M,E1)L: \Gamma (M, E_0) \to \Gamma (M, E_1) be linear. Then

  1. LL is a differential operator of order 00 iff [L,f]=0[L, f] = 0 for all fC(M)f \in C^\infty (M).

  2. LL is a differential operator of order kk iff [L,f][L, f] is a differential operator of order k1k - 1 for all fC(M)f \in C^\infty (M).

Integration by parts implies that we have

Lemma

For any differential operator L:Γ(M,E0)Γ(M,E1)L: \Gamma (M, E_0) \to \Gamma (M, E_1), there is a formal adjoint L:Γ(M,E1)Γ(M,E0)L^*: \Gamma (M, E_1) \to \Gamma (M, E_0) such that for any fiΓ(M,Ei)f_ i \in \Gamma (M, E_ i), we have

(Lf0,f1)L2=(f0,Lf1)L2. (Lf_0, f_1)_{L^2} = (f_0, L^* f_1)_{L^2}.\fakeqed

Definition

Let LL be a differential operator of order kk. The (principal) symbol of LL is the family of operators symk(L)(x,ξ)Hom((E0)x,(E1)x)\operatorname{sym}_ k(L)(x, \xi )\in \operatorname{Hom}((E_0)_ x, (E_1)_ x) for (x,ξ)TM(x, \xi ) \in T^*M given locally by

symk(L)(x,ξ)=α=kAα(x)ξα. \operatorname{sym}_ k (L)(x, \xi ) = \sum _{|\alpha | = k} A^\alpha (x) \xi _\alpha .

Formally, if π:TMM\pi : T^*M \to M is the projection, then symk(L)Γ(Hom(πE0,πE1))\operatorname{sym}_ k(L) \in \Gamma (\operatorname{Hom}(\pi ^* E_0, \pi _* E_1)).

In a coordinate-free manner, if sΓ(M,E0)s \in \Gamma (M, E_0) and fC(M)f \in C^\infty (M) with f(x)=0f(x) = 0, then

symk(L)(x,(df)x)(s(x))=1k!L(fks)(x). \operatorname{sym}_ k (L)(x, (\mathrm{d}f)_ x)(s(x)) = \frac{1}{k!} L(f^ k s)(x).

We say LL is elliptic at xMx \in M if symk(x,ξ)\operatorname{sym}_ k(x, \xi ) is invertible for all ξTxM{0}\xi \in T^*_ x M \setminus \{ 0\} , and LL 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,LL, L', we have

symk(L)(x,ξ)=±(symk(L)(x,ξ))symk(LL)(x,ξ)=symk(L)(x,ξ)symk(L)(x,ξ).\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 ).\fakeqed \end{aligned}

Hence the composition and adjoints of elliptic operators is elliptic.

Example

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

sym1(d)(x,(df)x)(ωx)=(d(fω))x=(dfω)x \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)=0f(x) = 0. So the symbol of d\mathrm{d} is

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

An elliptic complex is a sequence of vector bundles E0,,EmE_0, \ldots , E_ m with first-order differential operators

Li:Γ(M,Ei1)Γ(M,Ei) L_ i: \Gamma (M, E_{i - 1}) \to \Gamma (M, E_ i)

such that Li+1Li=0L_{i + 1} \circ L_ i = 0 and for any non-zero ξTxM\xi \in T^*_ x M, the sequence

\begin{useimager} 
    \[
      \begin{tikzcd}[column sep=large]
        0 \ar[r] & \pi_* E_0 \ar[r, "\symb_1(L_1)"] & \pi^* E_0 \ar[r, "\symb_1(L_2)"] & \cdots \ar[r] & \pi^* E_m \ar[r] & 0
      \end{tikzcd}
    \]
  \end{useimager}

is exact outside of the zero section of TMT^*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 {Vi}\{ V_ i\} be finite-dimensional vector spaces, and

\begin{useimager} 
    \[
      \begin{tikzcd}
        0 \ar[r] & V_0 \ar[r, "f_1"] & V_1 \ar[r, "f_2"] & \cdots \ar[r] & V_m \to 0
      \end{tikzcd}
    \]
  \end{useimager}

be an exact sequence. Let V=ViV = \bigoplus V_ i. Then f+f:VVf + f^*: V \to V is an isomorphism.

Proof
It suffices to show f+ff + f^* is injective. Suppose (f+f)x=0(f + f^*)x = 0. Then (f+f)2x=(ff+ff)x=0. (f + f^*)^2 x = (ff^* + f^* f)x = 0. So we get 0=ffx,x+ffx=fx,fx+fx,fx. 0 = \langle ff^*x, x\rangle + \langle f^*fx\rangle = \langle f^* x, f^* x\rangle + \langle fx, fx\rangle . So fx=fx=0fx = f^*x = 0. By exactness, x=fyx = fy for some yy, and then 0=ffy,y=fy,fy=x,x. 0 = \langle f^* fy, y\rangle = \langle fy, fy\rangle = \langle x, x \rangle . So x=0x = 0.

Corollary

If (E,L)(E_*, L_*) is an elliptic complex, define E=EiE = \bigoplus E_ i. Then

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

is elliptic.