Image of the Hodge-de Rham Laplacian

Question

I am reading about index of the Hodge-de Rham operator d$+$d$^*$, for a compact Riemannian manifold $M$, and its equality with the Euler characteristic of $M$. In the proof, which is an easy consequence of Hodge decomposition, there is usually stated the "obvious" fact that $$ \text{Image}((\text{d} + \text{d}^*)^2) = \text{Image}((\text{d}^*\text{d} + \text{d}^*\text{d}) = \text{image}(\text{d}) + \text{image}(\text{d}^*). $$ The inclusion of the LHS in the RHS is obvious. However, I cannot see why the opposite inclusion is true.

Answer 1

Assume that you know the Hodge decomposition in the form

$$ \Omega^k(M) = \Delta(\Omega^k(M)) \oplus H^k(M) $$

where $H^k(M)$ denotes the subspace of the harmonic $k$-forms on $M$ and this decomposition is orthogonal with respect to the inner product on $k$-forms.

Let $\eta = d\omega + d^{*}\mu \in \Omega^k(M)$ and decompose it as $\eta = \Delta(\eta_1) + \eta_2$ with $\Delta(\eta_2) = 0$. We want to show that $\eta_2 = 0$. On one hand, we have

$$ \left< \eta_2, \eta_2 \right> = \left< \eta, \eta_2 \right> - \left< \Delta(\eta_1), \eta_2 \right> = \left< \eta, \eta_2 \right> $$

because the decomposition is orthogonal while on the other hand we have

$$ \left< \eta, \eta_2 \right> = \left< d\omega + d^{*} \mu, \eta_2 \right> = \left< \omega, d^{*} \eta_2\right> + \left< \mu, d \eta_2 \right> = 0$$ because $\Delta(\eta_2) = 0$ iff $d \eta_2 = 0$ and $d^{*} (\eta_2) = 0$.