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Aug 30, 2015 at 12:38 comment added Nikita Kalinin to Otis: for each $n\geq 6$ there exists a regular triangulation of the hyperbolic plane by the hyperbolic equilateral triangles with angles $2\pi/n$. The vertices and edges of this triangulation constitute a graph $G$. An integer-valued function on the vertices of $G$ is discrete harmonic if the value of $f$ at every vertex $v$ is equal to the average of its values at the neighbors of $v$. Then we can define polynomial growth and ask what is the dimension of the integer-valued discrete harmonic functions on $G$ of a polynomial growth.
Aug 29, 2015 at 22:17 comment added Otis Chodosh No problem! I am not sure what you mean by your comment, maybe you will get a better answer if you define your terms precisely. In particular, what is a "integer discrete harmonic function" and what is the "regular triangulation" of the harmonic plane?
Aug 29, 2015 at 9:40 comment added Nikita Kalinin to Otis: Thank you! In fact I was interested in the integer discrete harmonic functions on the regular triangulations of the hyperbolic plane. But it seems that the relation to the usual harmonic functions is not direct.
Aug 29, 2015 at 9:38 history edited Nikita Kalinin CC BY-SA 3.0
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Aug 28, 2015 at 19:38 comment added Otis Chodosh Do you mean $|f(x)| \leq c (1+d(x,p))^n$? In that case, there is an easy way to see that for $n=2$, $H_n$ is infinite dimensional. This is because for any continuous function on the circle, we can solve the (Euclidean) Dirichlet problem on the unit ball $B_1(0)\subset \mathbb{R}^2$. This yields a bounded harmonic function. However, begin harmonic is conformally invariant in two dimensions, so this function is also harmonic for the ball model of hyperbolic space. Moreover, it is clearly bounded.
Aug 28, 2015 at 19:17 history asked Nikita Kalinin CC BY-SA 3.0