Hodge decomposition seems to say harmonic and exact is zero (and possibly (harmonic and co-exact) and (exact and co-exact))

Question

What I understand. Please correct if wrong:

Part of Hodge Decomposition Theorem says that for a compact oriented Riemannian (smooth) $m$−manifold $(M,g)$ (I think M need not be connected, but you may assume connected if need be or you want) and for a smooth $k$-form $\omega$, i.e. $\omega \in \Omega^k(M)$

A. $\omega$ decomposes into exact, co-exact and harmonic: $\omega =$ (respectively) $\omega_d \oplus \omega_\delta \oplus \omega_\Delta $$\in \Omega^k(M) = B^kM \bigoplus \mathscr H^k(M,g) \bigoplus image(\delta_{k+1})$

B. $\omega$ is zero if $\omega$ is at least 2 of the ff: harmonic, exact, co-exact. Specifically:

• B.1. 'harmonic and exact is zero'

• B.2. 'harmonic and co-exact is zero'

• B.3. 'exact and co-exact is zero'

C. I think (B.1) is equivalent to the injectivity of the map $\phi: \mathscr H^k(M,g) := \ker(\Delta_k) \to$$ H^k_{dR}M := \frac{Z^kM}{B^kM}$, which maps a harmonic form to its de rham cohomology class, $\phi(\omega)=[\omega]:=\omega + B^kM$ (or '$\omega\bmod B^kM$').

Proof: $\phi$ is the restriction of $\Phi:Z^kM \to H^k_{dR}M$ (from closed $k$-forms) to harmonic $k$-forms. Then $\ker(\phi) = \ker(\Phi) \cap Domain(\phi) = B^kM \cap \mathscr H^k(M,g) = $$image(d_{k-1}) \cap \ker(\Delta_k)$. QED

• C.1. Note on notation: I personally would like to have $\phi$ and $\Phi$ to have subscripts $k$ since their domains and ranges depend on $k$, but I'm omitting the subscript $k$ because I'm following notation in the following powerpoint.

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Questions about this powerpoint by Ryan Vaughn:

1. What exactly is going on in slide 46? This is part of proving that $\phi$ is injective.

• 1.1. What I think: Here, Ryan Vaughn seems to conclude, for the decomposition of harmonic $\omega = \omega_d \oplus \omega_\delta \oplus \omega_\Delta$, that $\omega_{\Delta}$ is zero because $\omega_{\Delta}$ is exact and because of (B.1). However, I really don't think decomposition is necessary because the other components of $\omega$ will obviously be zero.

• 1.2. I assume that the decomposition itself (I mean without the 'orthonormal' part) doesn't rely on (B.1) (or (B.2)-(B.3)), otherwise I think this would be circular. I'm guessing that (B.1) (or (B.2)-(B.3)) is used in the 'ortho(normal)' part of the decomposition theorem. Thus, it makes sense to talk about $\omega = \omega_d \oplus \omega_\delta \oplus \omega_\Delta$ even if we don't know (B.1) (or (B.2)-(B.3)).

2. How does Ryan Vaughn prove (B.1), if Ryan Vaughn did? Otherwise, how does one prove (B.1) (of course without assuming the full Hodge Decomposition Theorem; soooo of course proving (B.1) is part of proving Hodge Decomposition Theorem)?

What I tried so far:

2.1 - I think the following is somehow relevant, if true:

a smooth $k$-form is harmonic if and only if closed and co-closed, i.e. $\ker (\Delta_k) = $$ \ker(d_k) \cap \ker(\delta_k) = Z^kM \cap \ker(\delta_k)$

2.2 - If true: It seems 'harmonic and exact' is equivalent to 'co-closed and exact'. Please explain how to show 'co-closed and exact is zero'.

Answer 1

On slide $43$ he writes $[\omega] = [\omega_d] + [\omega_{\delta}] + [\omega_{\Delta}]$. Then on slide $44$, he replaces $\omega_{\delta}$ by $0$ because a form which is both $d$-closed and $d^*$-exact on a closed manifold is zero. To see this, note that if $\alpha = d^*\beta$ and $d\alpha = 0$ then

$$\|\alpha\|^2 = \langle\alpha,\alpha\rangle = \langle\alpha, d^*\beta\rangle = \langle d\alpha, \beta\rangle = \langle 0, \beta\rangle = 0$$

so $\alpha = 0$. On slide $45$, he replaces $\omega_d$ by $0$ because a form which is both $d^*$-closed and $d$-exact on a closed manifold is zero (which follows from a similar argument). Therefore $\phi(\omega) = [\omega_{\Delta}]$.