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Consider a hyperbolic manifold $M=H^n/\Gamma$ with finite volume. Suppose that there exists a harmonic function $u$ defined on $M$. Then is $u$ a constant? If $M$ is compact, yes. So I want to know the non-compact case. Are there counterexamples?

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    $\begingroup$ Did you check Compactifications os Symmetric Spaces by Guivarc'h, Ji and Taylor ? If I remember correctly, their main theorem state that the Martin boundary is the Satake-Furstenberg boundary. $\endgroup$
    – M. Dus
    Commented Dec 14, 2019 at 6:37

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The answer is no, a result of Greene-Wu is that any connected, non-compact Riemannian manifold has a proper embedding by harmonic functions in some Euclidean space [see R. Greene and H.Wu: GREENE Embedding of open riemannian manifolds by harmonic functions Annales de l’institut Fourier, tome 25, no 1 (1975), p. 215-235, http://www.numdam.org/article/AIF_1975__25_1_215_0.pdf ].

However, any bounded harmonic function on a complete Riemannian manifold with finite volume is constant. In the book of Pigola and Setti [proposition 3.5 in Global divergence theorems in nonlinear PDEs and geometry. Ensaios Matematicos], you will more general result. But in the case of hyperbolic manifold $M$ with finite volume, I suspect that any harmonic function $h$ such that $h(x)=\mathcal{O}\left(e^{\tau\ \mathrm{dist}(o,x)}\right)$ (for some $\tau<\dim M-1$) is constant.

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    $\begingroup$ do you mean "finite volume" rather than "constant volume"? $\endgroup$
    – YCor
    Commented Dec 14, 2019 at 18:11
  • $\begingroup$ Yes I changed the text and modified the last sentence. $\endgroup$
    – ubik
    Commented Dec 15, 2019 at 5:20

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