Let $M$ be a smooth $d$-dimensional oriented Riemannian manifold, and let $p \in M$. Let $0 \le k \le d$ be fixed.

Does there exist an open neighbourhood $U$ of $p$, which admit a non-zero harmonic $k$-form? i.e $\omega \in \Omega^k(U)$ satisfying $d\omega=\delta \omega=0$?

Since a form $\omega$ is harmonic if and only if $\star \omega$ is harmoinc ($\star$ is the Hodge dual operator), the answers for a given $k,d-k$ are the same.

For $k=d$, one can take $\omega$ to be the Riemannian volume form.

For $k=1$, we can take $\omega=df$ where $f$ is a harmonic function. Then $\delta \omega=\delta df=0$. Locally, there are always harmonic functions- we can solve the Dirichlet problem on a small ball with boundary, that is finding a harmonic function which is zero on the boundary.

This solves the cases $k=0,1,d-1,d$.

So, we are left with the cases $2 \le k \le d-2$.


1 Answer 1


Yes, locally there is always an infinite-dimensional space of harmonic $k$-forms (as long as the dimension $d$ is at least 2). One way to see this is a variational approach, since harmonic forms are critical points of an energy functional.

Assume that $U \subset M$ is a small ball around $p$ and that $\alpha$ is some smooth $(k-1)$-form defined in a neighborhood of $\partial U$. Let $\beta$ be a $(k-1)$-form minimizing $\int_U \langle d\beta, d\beta \rangle$ with boundary values $\beta = \alpha$ on $\partial U$, and let $\omega = d\beta$. Then $d\omega = 0$, so for harmonicity we only need to show that $\delta \omega = 0$, which turns out to be exactly the Euler-Lagrange equation for this functional: If $\gamma$ is a $(k-1)$-form vanishing on $\partial U$, we get that $$ 0 = \frac{d}{dt}\int_U \langle \omega + td\gamma, \omega + t d\gamma) \rangle = 2 \int_U \langle d\gamma,\omega \rangle = 2 \int_U \langle \gamma, \delta \omega \rangle, $$ which then implies $\delta \omega = 0$. So $\omega$ is a harmonic $k$-form with boundary values $\omega = d\alpha$ on $\partial U$. Since $\alpha$ was an arbitrary smooth $(k-1)$-form in a neighborhood of $\partial U$, the space of boundary values, and thus the space of locally harmonic forms is infinite-dimensional.

Obviously, a lot of details have to be filled in, in particular the smoothness of the minimizer. These are treated in textbooks on elliptic differential equations.

  • $\begingroup$ Thanks, this is interesting. How do you know a minimizer $\beta$ exists? Do you take a minimizing sequence, then extract a weakly convergent subsequence, and take its limit? (I guess the limit would be a minimizer due to lower-semicontinuity of the Dirichlet functional w.r.t weak convergence, and it will satsify the boundary condition in the trace sense, right? only now enters elliptic regularity to establish the smoothness of the minimizer?). If you have any reference for the existence part it would be great. $\endgroup$ May 22, 2018 at 9:52
  • $\begingroup$ @AsafShachar: Exactly, that would be the idea, and first do it in some appropriate Sobolev space, at the end show that it is actually represented by a smooth function. I don't have a reference for exactly this, but there are similar methods in the proof of the Hodge theorem that every cohomology class is represented by a unique harmonic form (for smooth compact manifolds), where one minimizes the same integral over a cohomology class. I saw this in Jost, Riemannian Geometry and Geometric Analysis, though he also skips some details. $\endgroup$ May 22, 2018 at 15:29
  • $\begingroup$ @AsafShachar For a complete treatment (and certainly where Jost took his proof from), see Morrey's Multiple Integrals in the Calculus of Variations, Chapter 7. $\endgroup$
    – Ryan Unger
    May 24, 2018 at 1:36

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