TITLE: Kernel of the Laplacian on a compact manifold
QUESTION [9 upvotes]: Is there a way to characterise the kernel of the Laplace-Beltrami operator on a compact manifold without boundary? 
Or is it just "the set of functions $u$ such that $-\Delta u = 0$?"

REPLY [5 votes]: If I am not mistaken, all such functions are constant.  I argue as follows:  the Laplacian on a Riemannian manifold $M$ may be defined by the equation
$\Delta u = \nabla^2 u = \nabla \cdot \nabla u, \tag{1}$
provided both $\nabla$ and the divergence of vector fields $X$, $\nabla \cdot X$, are defined with respect to the manifold's Riemannian metric $\langle \cdot, \cdot \rangle$, just as they are if the metric is Euclidean.  This being the case, we have the identity
$\nabla \cdot (u \nabla u) = \langle \nabla u, \nabla u \rangle + u \nabla^2 u, \tag{2}$
exactly as for Euclidean metrics.  We integrate (2) over $M$, apply the divergence theorem to the left hand side, and obtain
$\int_M \langle \nabla u, \nabla u \rangle dV = -\int_M u \nabla^2 u dV, \tag{3}$
(here $dV$ is the volume element on $M$) which, if $\nabla^2 u = 0$ for some (sufficiently smooth) function $u$, yields
$\int_M \langle \nabla u, \nabla u \rangle dV = 0. \tag{4}$
Since $\langle \nabla u, \nabla u \rangle$ is non-negative and smooth, (4) in turn implies that
$\langle \nabla u, \nabla u \rangle = 0, \tag{5}$
or
$\nabla u = 0 \tag{6}$
everywhere; and (6) can only apply everywhere if $u$ is constant.
The reason that (3) follows from (2) for boundaryless $M$ is that the divergence theorem says that
$\int_N \nabla \cdot X dV = \int_{\partial N} \langle X, \vec n \rangle dS \tag{7}$
for any manifold with boundary $N$, where $dS$ is the volume element on $\partial N$ and $\vec n$ is the outward pointing normal on $\partial N$.  If we pick a point $p \in M$ and a small open ball $B$ around $p$, we may take $N = M \setminus B$; then (7) applies to $N$ and $\partial N$, and if we let $B$ become smaller and smaller then
$\int_{\partial N} \langle X, \vec n \rangle dS \rightarrow 0, \tag{8}$
but
$\int _N \nabla \cdot X dV \rightarrow \int_M \nabla \cdot X dV; \tag{9}$
together (8) and (9) show that
$\int_M \nabla \cdot X dV = 0 \tag{10}$
for all sufficiently smooth vector fields $X$ on $M$; taking $X = u \nabla u$ we see that (3) follows from (2), completing the proof that $u$ must be constant.  QED.
Remarks:  Of course the demonstrationof (10) presented above is not competely rigorous, lacking as it does certain details mostly having to do with establishing the limits affirmed in (8) and (9); but these should be relatively easy for the astute reader to flesh out; the basic idea is, I believe, sound.  Finally, we note that Michael cast his answer in the language of differential forms, taking $\Delta = d \delta + \delta d$.  On functions $u$, $\delta u = 0$ so $(d \delta + \delta d) u = \delta du$, recalling that $\delta = \ast d \ast$ up to a sign, where $\ast$ is the Hodge endomorphism defined on differential forms, as in http://en.m.wikipedia.org/wiki/Hodge_dual; but $\delta d$ and $\nabla^2 = \nabla\cdot \nabla$ agree on functions, as explained in http://en.m.wikipedia.org/wiki/Laplace%E2%80%93Beltrami_operator.  So there´s really no discrepancy between Michael´s definition of $\Delta = \nabla^2$ and mine.  End of Remarks.