# 4 Hodge Theory

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

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.

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 $< \Lambda$.

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

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

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