Compactly supported harmonic forms

Question

If $(\mathcal{M},g)$ is a compact Riemannian manifold (without boundary), then it is well-known that a $k$-form $\alpha$ is harmonic, i.e. $\Delta\alpha=0$, where $\Delta$ is the Laplace-Beltrami operator if and only if $\mathrm{d}\alpha=0$ and $\delta\alpha=0$.

Now, if $\mathcal{M}$ is non-compact (without boundary), then this is in general not true. However, is it true if $\alpha$ is compactly supported? In the compact case, the statement is proven using the Hodge decomposition, however, in the non-compact, I was not able to find a similar result.

Answer 1

Note that $\Delta = d\delta + \delta d$ and $\delta$ is the formal adjoint to $d$. The direction $d\alpha = \delta \alpha =0\Rightarrow \Delta \alpha$ is obvious and $\alpha$ does not have to be of compact support. For the other direction, if $\Delta \alpha = 0$, then

\begin{align} 0=\int_M \langle \Delta \alpha, \alpha\rangle d\mu_M &= \int \langle d\delta \alpha + \delta d\alpha, \alpha\rangle d\mu_M \\ &= \int_M (\| d\alpha\|^2 + \|\delta \alpha\|^2) d\mu_M. \end{align}

This implies $d\alpha = \delta\alpha = 0$. In the proof we used that $\alpha$ is of compact support, so that the integration is well-defined.

Answer 2

It's true, as @ArcticChar proved, that if $\alpha$ is harmonic and compactly supported then $d\alpha = \delta\alpha = 0$.

But this is not very useful on a noncompact manifold, because if $\alpha$ is harmonic and compactly supported on a noncompact manifold, then it vanishes on a nonempty open set, so it's identically zero by unique continuation.

A more useful result is that if $M$ is complete and noncompact and $\alpha$ is a harmonic form in the Sobolev space $H^1(M)$ ($L^2$ forms whose weak first derivatives are in $L^2$), then $d\alpha = \delta \alpha = 0$. The same proof applies, using the fact that smooth compactly supported forms are dense in $H^1(M)$ [see Aubin, Some Nonlinear Problems in Riemannian Geometry, Theorem 2.6].