Global Analysis: Global elliptic regularity

3.2 Global elliptic regularity

Let $M$ be a compact manifold, $E_0, E_1$ complex vector bundles, and $L$ an elliptic differential operator from $E_0$ to $E_1$ of order $k$ . By patching local results together, we conclude that

Theorem (Global elliptic regularity)

There is a constant $A$ such that for all $s \geq 0$ and $u \in H^{s + k}(M; E_0)$ , we have
$\| u\| _{s + k} \leq A(\| u\| _s + \| Lu\| _s).$
If $Lu \in H^s$ , then $u \in H^{s + k}$ . In particular, if $Lu = 0$ , then $u \in C^\infty (M)$ .

This implies a lot of very nice properties about elliptic operators. First observe the following result:

Proposition

Let $U, V, W$ be Hilbert spaces and $L: U \to V$ bounded, $K: U \to W$ compact. If there is an $A$ such that

\| u\| _U \leq C (\| L u\| _V + \| K u\| _W),

then $\ker L$ is finite-dimensional and $\operatorname{im}L$ is closed.

Proof

□

We show that the unit ball of

\ker L

is compact. If

(u_n)

is a sequence in the unit ball of

\ker L

, then

\| u_n - u_m\| _U \leq A \| Ku_n - Ku_m\| .

Since $K$ is compact, there is a subsequence $u_{n_i}$ such that $K u_{n_i}$ is Cauchy. So $u_{n_i}$ is Cauchy. So we are done.

To show $\operatorname{im}L$ is closed, by restricting to the complement of the kernel, we may assume $L$ is injective. We will show that there is a $c$ such that

\| u\| _U \leq c \| L u\| _V.

If not, pick a sequence $u_n$ with $\| u_n\| _U = 1$ but $\| L u\| _V \to 0$ . By compactness, we may assume that $Ku_n$ is Cauchy. Then we see that $u_n$ must also be Cauchy, and the limit $u$ must satisfy $\| u\| _U = 1$ and $Lu = 0$ , a contradiction.

Proof

□

Corollary

$L: H^{s + k} \to H^s$ has finite-dimensional kernel and closed image.

The compactness of the manifold

M

is crucial for the inclusion

H^{s + k} \to H^s

to be compact.

We want to show that in fact $L$ is Fredholm, i.e. both the kernel and the cokernel are compact. Here we simply use duality. We need to show that

K = \{ v \in H^{-s} : \langle Lu, v\rangle = 0\text{ for all }u \in H^{s + k}\}

is finite-dimensional. But this is exactly the kernel of $L^*$ . So we deduce

Corollary

$L$ is Fredholm, and in fact

H^s = \ker L^* + \operatorname{im}(L: H^{s + k} \to H^s).

The first factor is independent of $s$ (since all elements are $C^\infty$ ), and taking the limit $s \to \infty$ , we get

\Gamma (M, E_1) = \ker L^* \oplus \operatorname{im}L.

A bit of care is actually needed to take the limit

s \to \infty

, but we leave that for the reader.