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For $r$, $s$, small positive integers, do the complex numbers on the unit disc (without the hyperbolic metric) corresponding to the vertices of the hyperbolic tiling with Schläfli symbol $\{r,s\}$ satisfy the Blaschke condition?

The Blaschke condition for a sequence $(a_n)$ of complex numbers in the open unit disc is $\sum_n (1-|a_n|)<\infty$, see Wikipedia: Blaschke product.

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    $\begingroup$ What IS the Blaschke condition? $\endgroup$ – Igor Rivin Feb 13 '17 at 0:39
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I cannot make sense of the other answer and I think the sum is infinite so let me add mine.

If I understood the question correctly the tiling with Schläfli symbol $\{p, q\}$ is just the tiling of the hyperbolic plane by triangles with angles $\pi/2, \pi/p, \pi/q$. The Fuchsian group $\Gamma \subset \mathrm{PU}(1,1)$ generated by reflections in the faces of one such triangle acts cofinitely on the vertices of the tiling and so it suffices to see whether for a given $z$ with $|z| < 1$ we have $\sum_{\gamma\in\Gamma} (1 - |\gamma z|) < +\infty$ or not. This series seems to have an infinite sum: to simplify assume that $z = 0$ (the finiteness or infiniteness does not depend on the basepoint) and let $N(r)$ be the number of points of the orbit $\Gamma 0$ inside the ball $\{|z| < r\}$. This is the same as those $\gamma 0$ which are at distance at most $d_r = \log((1+r)/(1-r))$ from 0 in the hyperbolic metric. By a result of Margulis this implies that $$ N(r) \sim_{r \to 1} c_\Gamma e^{d_r} $$ for some $c_\Gamma > 0$ (which equals the covolume of $\Gamma$ in $\mathbb H^2$, that is $\pi/2 - \pi/p - \pi/q$). So $$ N(r) \sim_{r \to 1} 2c_\Gamma(1-r)^{-1}. $$ Integrating by parts shows that $$ \sum_{\gamma : |\gamma 0| \le r} (1 - |\gamma 0|) = (1 - r)N(r) + \int_0^r N(r) dr $$ and since the first term has a finite limit, as the integral diverges when $r \to 1$ so does the sum.

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I do not like the statement of the question, because you are identifying the hyperbolic plane with the Poincaré disk model (apparently -- nothing is said about the model, so it could be actually the Beltrami-Klein model too). In my experience, the Minkowski hiperboloid model is much more "natural" and much better for working with hyperbolic geometry, and the Poincaré disk model is great only for visualizations and some specific arguments.

Anyway, some intuition why the sum in question should be infinite. I find this intuition convincing, but I guess it would take some work to turn this into a real proof.

Consider the upper halfplane model instead, and only the points with $x \in [0,1]$ and $y \leq 1$. We will show that the sum of $y$ coordinates of all the grid points in this sector is infinite. This will also show that your sum in Poincaré disk model is infinite too, because the Poincaré disk model behaves the same as the halfplane model in the limit (like this).

Isometries of the halfplane model include horizontal translations $(x,y) \mapsto (x+a,y)$ and scaling $(x,y) \mapsto (ax,ay)$. Take $a<1$. Since regular hyperbolic tilings have constant density, there are as many grid points in the rectangle $[0,1] \times [a^{k+1},a^k]$ as in $[0,a] \times [a^{k+2}, a^{k+1}]$ (by scaling isometry). Therefore, there will be $1/a$ as many points in the rectangle $[0,1] \times [a^{k+2}, a^{k+1}]$ (by translation isometry). Hence, the number of points in this rectangle is asymptotically $\Theta(1/a^k)$. The sum of $y$ coordinates of all points in the rectangle $[0,1] \times [a^{k+1}, a^k]$ will be $\Theta(a^k)$ ($y$ coordinate) times $\Theta(1/a^k)$, which is $\Theta(1)$. Since our region is the sum of $[0,1] \times [a^{k+1}, a^k]$ for all $k \geq 0$, the sum in question is $\sum_{k=0}^\infty \Theta(1) = \infty$.

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If you were to map {r,s} as a conformal projection into the unit disk, the vertices would belong to some cyclotomic number not greater than {rs}.

For both disk models, {r,s} takes the form of something like $\Pi (1+a/k^n)$, where $k>1$, and this converges.

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