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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 $-\Delta u = 0$?"

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As an alternative to the proofs below, one could also show that all harmonic functions are constant on a compact manifold using the maximum principle. – Phillip Andreae Aug 13 '14 at 1:51
    
I thought that this was supposed to be a functional analysis question. We are not supposed to assume smoothness, which is needed for the maximum principle($C^2$). Instead we need to say there is a minimizer of $\int_X |\nabla u |^2 $ (which happens to be bounded below, lowersemicontinuous which is all we need to show this) over $H_1(X)$ functions, and that minimizers of this are going to be mild solutions of $-\Delta u=0$ and each of these functions will have an equivalence class that is constant. – Hari Rau-Murthy Dec 16 '15 at 8:16
up vote 6 down vote accepted

These solutions are called harmonic functions, and there's a lot of results about them. For compact manifolds, like you stated, the only possible solutions are constant functions.

To see this, you must recall that $\Delta = dd^* + d^*d$, where $d$ is the exterior derivative, and then you can show (the argument uses integration) that $\Delta u = 0$ is equivalent to $du = 0$ and $d^*u = 0$. The former implies that $u$ is constant.

Remark: Here is the argument, which is quite simple: suppose $\Delta u = 0$. Then,

$$0 = \int_M \langle d^*d u, u \rangle~\mathrm{dvol} = \int_M \langle d u, du \rangle~\mathrm{dvol} = \int_M \|du\|^2~\mathrm{dvol},$$ since $d^*u = 0$. Here, $\mathrm{dvol}$ is the volume element of $M$. Now of course, for the last integral to be zero, $du$ has to be zero, so $u$ is constant.

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Thank you, a succinct answer – DoNotIncludeSynonyms Aug 14 '14 at 9:56
    
about the comment of the OP on using the maximum principle, this manifold has no boundary. Of course the boundary term that comes in proving the maximum principle is not there. You are making good use of the fact that this manifold has no boundary in saying that the only harmonic functions here are constant. – Hari Rau-Murthy Dec 16 '15 at 5:45

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.

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Thanks, nice explanations – DoNotIncludeSynonyms Aug 14 '14 at 9:57

I might be wrong, but I think only constant functions are in the kernel. Consider discretizing the manifold with a very fine mesh and approximating the Laplace-Beltrami operator with a finite volume or finite difference scheme - then the resulting discrete operator will be a weighted graph laplacian, which is positive semi-definite, and has only one zero eigenvalue per connected component, corresponding to the constant indicator function on that component of the graph.

To make this argument complete, one would have to show

  1. that any compact manifold without boundary can be discretized arbitrarily well, and
  2. that the discretized version of the operator converges to the true operator on manifolds just as it does on flat space.

I'm pretty sure that being 1. is a standard result. For 2. I don't see why it shouldn't work, since the operator is local so you can do everything on charts.

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Thank you a lot. – DoNotIncludeSynonyms Aug 14 '14 at 9:58

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