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,

$\ker D$ is finite-dimensional.
$\ker D = \ker \Delta = \ker L \cap \ker L^*$
$\operatorname{im}\Delta = \operatorname{im}D = \operatorname{im}LL^* \oplus \operatorname{im}L^*L = \operatorname{im}L \oplus \operatorname{im}L^*$
$\ker L = \operatorname{im}L \oplus \ker \Delta$ , and $\ker \Delta \to H(E_*)$ is an isomorphism.

Proof

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(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

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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

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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

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