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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.

About this powerpoint,

  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).

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    $\begingroup$ What is $[\omega]$? $\endgroup$ Jan 1 at 0:56
  • $\begingroup$ @ArcticChar de rham cohomology class, based on slide 27? (thanks, though it was just a round bracket remark) $\endgroup$
    – BCLC
    Jan 1 at 0:58
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    $\begingroup$ 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... $\endgroup$ Jan 1 at 1:24
  • $\begingroup$ @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 $\omega_d=0$ $\endgroup$
    – BCLC
    Jan 1 at 8:23
  • $\begingroup$ @ArcticChar merry christmas, happy new year, and happy holidays! $\endgroup$
    – BCLC
    Jan 1 at 8:34
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(copying 'Update' into an answer): 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).

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