Timeline for Dimension of the set of the polynomial growth harmonic function on the hyperbolic plane
Current License: CC BY-SA 3.0
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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 |