# Solvability of Poisson's equation for $p$-forms directly from Hodge decomposition

If $M$ is a closed manifold, it's well known that $\Delta\phi=\alpha$, where $\alpha$ is a $p$-form, has a solution if and only if $\alpha$ is orthogonal to the space of harmonic $p$-forms. This is proved on page 224 of Warner, but the proof depends heavily on his approach to Hodge theory. There are other methods that avoid his main technical lemma 6.6. So, can one prove this fact using only the "standard" facts of the theory?

For functions it is easy, for one has $$\alpha=\delta \psi+h.$$ But $(\alpha,h)=0$, so $\alpha=\delta\psi$. Then decompose $\psi$ to get $\alpha=\delta d\zeta$ for some function $\zeta$. But on functions, $\Delta=\delta d$. But for $p$-forms in general one has $$\alpha=d\xi+\delta \psi+h,$$ since $\Omega^{p-1}$ is nontrivial. Using the decompositions again, one has $$\alpha=d\delta \rho+\delta d\sigma,$$ where $\rho,\sigma$ are $p$-forms. I don't see how $\rho$ and $\sigma$ are related, but it looks close.

## 1 Answer

If by "standard" facts of the theory, you mean the fact that every smooth $p$-form $\alpha$ can be decomposed as $\alpha=d\xi+\delta\psi+h$ with $h$ harmonic, here's a way to get from there to where you want to go.

The key is to observe that when we decompose $\alpha$ as above, we can require in addition that $\delta\xi=0$ and $d\psi=0$. To see why, use the decomposition to write $\xi = d\beta+\delta\gamma+h_1$, and note that $d\xi = d\delta\gamma$, so we can replace $\xi$ by $\delta\gamma$. A similar argument applies to $\psi$.

Now suppose $\alpha$ is a $p$-form that is orthogonal to the harmonic forms. As you noted, we can write $\alpha = d\delta\rho + \delta d\sigma$. In addition, we can assume $d\rho=0$ and $\delta\sigma=0$. Thus $$\Delta(\rho+\sigma) = d\delta\rho + \delta d\rho + d\delta\sigma + \delta d\sigma = d\delta\rho + \delta d\sigma = \alpha.$$ It has to be noted, however, that Warner's "technical lemma 6.6" is exactly the heart of the matter. This is the estimate that's required to show that $\Delta$ has closed range, which in turn is the key ingredient in proving the decomposition theorem. There's no way to prove the Hodge theorem without some form of that estimate.

• Yes, that's exactly the trick I was looking for. – Ryan Unger May 29 '17 at 17:53
• Clarification: Lemma 6.6 is not "standard" insofar as the exact statement is not needed for the proof of the main theorem. One can use a variational approach (Jost), which requires estimates on $\delta$, not $\Delta$, or the pseudodifferential approach, which is in my mind completely different, conceptually. – Ryan Unger May 29 '17 at 17:55
• One might be able to hammer out a proof using the other techniques, but I think what you did here is much better. – Ryan Unger May 29 '17 at 17:57
• @0celouvskyopoulo7: Just want to point out that Jost's variational approach doesn't avoid Warner's "technical lemma." Jost's Lemma 2.2.2 in the second edition of Riemannian Geometry and Geometric Analysis (Lemma 3.4.2 in the sixth edition) says that a sequence of forms that are bounded in his $H^{1,2}$ norm has a convergent subsequence in the $L^2$ norm. In fact, his lemma immediately implies Warner's -- if you look at Jost's definition of the $H^{1,2}$ norm, you'll see that a sequence that satisfies the hypotheses of Warner's Theorem 6.6 automatically satisfies those of Jost's lemma. – Jack Lee May 31 '17 at 21:17
• Warner has a bound on the Laplacians of the forms, not the first derivatives. So it doesn't immediately follow from Rellich (modulo brain fart). I'm not saying it's hard, but I do think that what Jost does is to shift the analysis away from the second order problem to make it a first order problem. Of course the whole business boils down to Rellich in the end, which is what I meant by my third comment above. – Ryan Unger May 31 '17 at 21:58