# Hodge decomposition seems to say co-exact component is zero

Hodge Decomposition Theorem says smooth $$k$$-forms on on compact oriented Riemannian (smooth) $$m$$−manifold $$(M,g)$$ (I think M need not be connected, but assume connected if need be or you want) decompose into exact, co-exact and harmonic.

1. If a smooth $$k$$-form $$\omega = \omega_d \oplus \omega_\delta \oplus \omega_\Delta$$ is closed, then is its co-exact component $$\omega_\delta$$ zero?

2. What exactly is going on in slides 28-41?

• 2.1. It seems to be proving (1), but actually it appears we have the stronger assumption that $$\omega$$ is harmonic. In this case, for a smooth $$k$$-form $$\omega$$, I think $$\omega$$ is harmonic if and only if the exact and co-exact components, resp $$\omega_d$$ and $$\omega_\delta$$, are both zero. However, I think for a smooth closed $$k$$-form $$\omega$$, we have only that $$\omega_\delta = 0$$ (and $$\omega_d]$$, all we have is $$[\omega_d]=[0]$$, but I believe this doesn't even require $$\omega$$ to be closed).

Update: Based on this, I think: (1) is true, and I think the name for this fact is called 'short Hodge decomposition (theorem)'. As for (2), I think author forgot to explicitly exclude the assumption of harmonic from the claim, which is meant assume closed (instead of harmonic).

• What is $[\omega]$? Jan 1 at 0:56
• @ArcticChar de rham cohomology class, based on slide 27? (thanks, though it was just a round bracket remark)
– BCLC
Jan 1 at 0:58
• Well, $\omega$ is harmonic if and only if $\omega_d = \omega_\delta=0$. This is in the statement of Hodge decomposition. I don't know why they try to prove it again... Jan 1 at 1:24
• @ArcticChar That's why I'm thinking what is being proved is for the case of $\omega$ is closed rather than that $\omega$ is harmonic. notice the other claim is merely that $[\omega_d]=[0]$ instead of that $\omega_d=0$
– BCLC
Jan 1 at 8:23
• @ArcticChar merry christmas, happy new year, and happy holidays!
– BCLC
Jan 1 at 8:34