I'm trying to work out a problem in John Lee's Introduction to Smooth Manifold:

A harmonic function on a Riemannian manifold is a function $u$ satisfies

$-\Delta u = -\frac{1}{\sqrt{\mathrm{det} g}} \frac{\partial}{\partial x^i}(g^{ij}\sqrt{\mathrm{det}g} \frac{\partial u}{\partial x^j})=0$

(a) If $M$ is a compact, connected Riemannian manifold without boundary, then the only harmonic functions on $M$ are constants.

(b) If $M$ is a compact, connected Riemannian manifold with boundary, and $u,v$ are harmonic functions that agree on the boundary, then $u \equiv v$.

In Why must harmonic functions on compact Riemannian manifolds be constant? Ivo Terek gives a solution to (a). My idea is to use the mean value property of Laplacian equation, but when we pullback $u$ to a chart of $M$, $u$ does not satisfy Laplacian equation unless $g_{ij}=\delta_{ij}$. Does this have something to do with elliptic equation? I've heard of this, but I haven't taken advanced PDE course. How to deal with problem (b) if we don't use the theory of PDE? Thanks in advance.

  • 2
    $\begingroup$ Try using Green's identity. The argument is very similar to the one you linked. Perhaps consider $u-v$. $\endgroup$
    – Thorgott
    Aug 8, 2022 at 14:34
  • 1
    $\begingroup$ @Thorgott Thank you. So we apply Green's identity: $\int_M f\Delta g = \int_{\partial M} f\langle\nabla g,N\rangle-\int_M \langle\nabla f,\nabla g\rangle$, let $f=g=u-v$, then we are done. And by the way, is $-\Delta u$ an elliptic operator when pullback to a chart, so we can apply theory of PDE? $\endgroup$
    – user725757
    Aug 9, 2022 at 1:37

1 Answer 1


I'll provide two more (less direct) reasons in addition to the Green's identity mentioned by Thorgott. One uses maximal principle. Since $M$ is compact without boundary, and a harmonic function $f$ is continuous, $f$ achieves its maximum in (the interior of) $M$. By maximal principle, it must be a constant function.

Another deeper point of view comes from Hodge theory. One can generalize harmonic functions to harmonic $p$-forms, where functions are $0$-forms. The space of harmonic $p$-forms is isomorphic to the $p$th de Rham cohomology $H^p(M)$. Thus the space of harmonic function on compact $M$ without boundary is isomorphic to $H^0(M)$, which is $1$-dimensional if $M$ is connected. Constant functions are harmonic, and thus they span this one-dimensional space.

  • $\begingroup$ Thank you for providing these backgrounds. $\endgroup$
    – user725757
    Aug 10, 2022 at 5:49

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