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.
Best Answer
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].