On a compact Riemannian manifold $(M,g)$, a function $f$ is called harmonic if $\Delta_{g} f = 0$, and it is known that the only harmonic function on a compact riemannian manifold is constant function. I wonder how one would prove this.

So locally $\Delta_{g}$ is just a second order elliptic operator, thus has the weak maximum principle, which says that the max must appear on the boundary. However I do not think this is enough to prove that harmonic functions are constants. Instead, we need the strong maximum principle which I think holds when the operator is strongly elliptic. Is $\Delta_{g}$ strongly elliptic?

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    $\begingroup$ It is not true bounded harmonic functions are constant. For instance Anderson and Schoen (Ann. Math. 1985) construct bounded harmonic functions on any complete Riemannian manifold whose sectional curvatures are bounded between two negative numbers $\endgroup$ – Quarto Bendir Apr 22 '20 at 23:07

Ivo Terek's answer gives the most elegant answer when $M$ is oriented. (If un-oriented, pass to the orientation double cover.)

I'll answer directly your question about the strong maximum principle and ellipticity. Relative to a coordinate system, one has $$\Delta f=g^{ij}\frac{\partial^2f}{\partial x^i\partial x^j}-\frac{1}{2}g^{ij}g^{k\ell}\Big(\frac{\partial g_{j\ell}}{\partial x^i}+\frac{\partial g_{i\ell}}{\partial x^j}-\frac{\partial g_{ij}}{\partial x^\ell}\Big)\frac{\partial f}{\partial x^k}.$$ Treating this as a function on a bounded open subset of Euclidean space, then, uniform ellipticity would mean that there are positive constants $\lambda$ and $\Lambda$ with $\lambda I\leq[g^{ij}]\leq\Lambda I$, as $n\times n$ matrices. This is not necessarily the case, imagine for instance polar coordinates on the sphere, so that $g=d\theta^2+\sin^2\theta\,d\varphi^2$ and so, in this case, the upper bound $\Lambda$ does not exist. One could also imagine examples where a positive lower bound $\lambda$ does not exist.

However, the restriction of this function to any precompact subregion of the coordinate chart ensures the existence (and finiteness and positivity) of the constants, by compactness and continuity of the matrix-valued function $[g^{ij}]$.

So all you need to accept is the following statement: there is a coordinate cover $\{(U_\alpha,\varphi_\alpha)\}$ of the manifold and for each $\alpha$ an open precompact set $K_\alpha\subset U_\alpha$ such that $\{K_\alpha\}$ remains an open cover of the manifold. Relative to any coordinate chart $(K_\alpha,\varphi_\alpha)$, the operator $f\mapsto \Delta f$ is uniformly elliptic, and so one can replay the proof of the Hopf maximum principle and recover the usual strong maximum principle.

This leads to a result slightly stronger than in Ivo Terek's answer, since there the manifold must be compact for the integration to work. Here one gets the strong maximum principle, where it only has to be assumed that a maximum (or minimum) is achieved.

Also of interest is S.-T. Yau's Liouville theorem (Indiana Univ. Math. J. 1976) which says that if $(M,g)$ is a complete Riemannian manifold and $u\geq 0$ is harmonic with $\int_M u^p\,d\mu_g<\infty$ for some $p>2$, then $du=0$.


The product rule for the Laplacian of two functions is $$\triangle(fh) = f(\triangle h) + h(\triangle f) + 2\langle \nabla f,\nabla h\rangle.$$Stokes' theorem says that the integral of a divergence (hence of a Laplacian) over a compact manifold without boundary vanishes. So we may integrate the above to get $$0 = \int_M \triangle(fh)\,{\rm d}M = \int_Mf(\triangle h)\,{\rm d}M + \int_Mh(\triangle f)\,{\rm d}M + 2\int_M\langle\nabla f,\nabla h\rangle\,{\rm d}M$$If $\triangle f = 0$ and we take $h = f/2$ we obtain $$\int_M \|\nabla f\|^2\,{\rm d}M = 0 \implies \nabla f = 0 \implies f\mbox{ is constant}.$$(It goes without saying that we assume $M$ connected)

  • $\begingroup$ Marvelous answer. But just wondering, where are we actually using the compactness? Say for example I look at an open subset $U$ of a compact riemannian manifold and a harmonic $f$ function defined only on $U$. Why wouldn't by the same logic that $f$ must be constant? $\endgroup$ – koch Sep 11 '20 at 0:11
  • $\begingroup$ We're using compactness to apply Stokes' theorem freely. In non-compact manifolds you have to restrict yourself to functions with compact support to make sure the integrals are well-defined and finite, and all that fluff.. $\endgroup$ – Ivo Terek Sep 11 '20 at 0:15

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