Global AnalysisHodge Theory

# 4 Hodge Theory

We now arrive at the main theorem we were working towards.

Theorem (Hodge Decomposition Theorem)

Let $(E_*, L)$ be an elliptic complex with $D = L + L^*$ and $\Delta = D^2$ as before. Then

$\Gamma (M, E) = \ker D \oplus \operatorname{im}D.$

Moreover,

1. $\ker D$ is finite-dimensional.

2. $\ker D = \ker \Delta = \ker L \cap \ker L^*$

3. $\operatorname{im}\Delta = \operatorname{im}D = \operatorname{im}LL^* \oplus \operatorname{im}L^*L = \operatorname{im}L \oplus \operatorname{im}L^*$

4. $\ker L = \operatorname{im}L \oplus \ker \Delta$, and $\ker \Delta \to H(E_*)$ is an isomorphism.

Proof
(1) follows from regularity. (2) follows from the identities

$\langle u, \Delta u\rangle = \langle D u, D u\rangle = \langle L u, L u\rangle + \langle L^* u + L^* u\rangle ,\quad u \in \Gamma (M, E_i).$

For (3), we can also decompose $\Gamma (M, E) = \ker \Delta \oplus \operatorname{im}\Delta$. Since $\operatorname{im}\Delta \subseteq \operatorname{im}D$, they must be equal. Moreover,

$\operatorname{im}D \subseteq \operatorname{im}(LL^*) \oplus \operatorname{im}(L^*L) \subseteq \operatorname{im}L \oplus \operatorname{im}L^*,$

but clearly $\operatorname{im}L \oplus \operatorname{im}L^* \perp \ker \Delta$ since

$\langle L u + L^* v, w\rangle = \langle u + v, D w\rangle .$

So we must have equality throughout, and $\operatorname{im}L$ is clearly orthogonal to $\operatorname{im}L^*$. For (4), it is clear that $\ker L \supseteq \operatorname{im}L \oplus \ker \Delta$, and since $\ker L \perp \operatorname{im}L^*$, that must be an equality.

Proof

Theorem (Spectral theorem)

Let $M$ be a closed Riemannian manifold and $E \to M$ a Hermitian vector bundle. Let $D: \Gamma (M, E) \to \Gamma (M, E)$ be formally self-adjoint of order $k \geq 1$. Then we have an orthogonal decomposition

$L^2(M, E) = \bigoplus _{\lambda \in \mathbb {R}} \ker (D - \lambda ).$

Moreover, each $\ker (D - \lambda )$ is finite-dimensional, and for any $\Lambda$, there are only finitely many eigenvalues of magnitude $< \Lambda$.

The idea is to apply the spectral theorem for compact self-adjoint operators to the inverse of $D$. Of course, $D$ need not be invertible. So we do the following:

Proof
Consider the operator $L = 1 + D^2: \Gamma (M, E) \to \Gamma (M, E)$. It is then clear that $L = L^*$ is elliptic and injective. So $L: H^{2k} \to L^2$ is invertible (since the complement of the image is $\ker L^*$), with inverse $S: L^2 \to H^{2k}$. Since $L$ induces a bijection between the smooth sections, so does $S$. Let $T$ be the composition $L^2 \overset {S}{\to } H^{2k} \hookrightarrow L^2$. Then this is compact and self-adjoint (can check this for smooth sections, and use that its “inverse” $L$ is formally self adjoint).

By the spectral theorem of compact self–adjoint operators (and positivity of $T$),

$L^2(M; E) = \bigoplus _{\mu > 0} \ker (T - \mu ).$

Moreover, each factor is finite-dimensional, and $0$ is the only accumulation point of the spectrum.

We will show that $\ker (T - \mu )$ decomposes as a sum of eigenspaces for $D$. We first establish that

$\ker (T - \mu ) = \operatorname{im}(1 - \mu L)^\perp = \ker (1 - \mu L).$

Since $L$ is self-adjoint, the second equality follows by elliptic regularity. The first equality follows from the computation

$\langle x, (1 - \mu L)u\rangle = \langle x, u\rangle - \mu \langle x, Lu\rangle = \langle Tx, Lu\rangle - \mu \langle x, Lu\rangle = \langle (T- \mu )x, Lu\rangle ,$

plus the density of $\Gamma (M, E)$ and surjectivity of $L: \Gamma (M, E) \to \Gamma (M, E)$.

Now since $D$ commutes with $L$, we know $D$ acts as a self-adjoint operator on the finite-dimensional vector space $\ker (T - \mu ) = \ker (1 - \mu L)$. Moreover, restricted to this subspace, we have

$D^2 = \frac{1}{\mu } - 1.$

So by linear algebra, $\ker (T - \mu )$ decomposes into eigenspaces of $D$ of eigenvalues $\pm \sqrt{\frac{1}{\mu } - 1}$, and the theorem follows.

Proof