# Problem on Harmonic Functions on Riemannian Manifolds

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.

• Try using Green's identity. The argument is very similar to the one you linked. Perhaps consider $u-v$. Aug 8, 2022 at 14:34
• @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? Aug 9, 2022 at 1:37

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.