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Let $P$ be a fixed pentagon in the hyperbolic plane $\mathbb H^2$ with all the angles equal to $\pi / 3$. Let $w_1, w_2, \dots, w_5$ be the sides of $P$ going in the counterclockwise order. We are looking for five holomorphic functions $f_1, f_2, \dots f_5 \colon \mathbb H^2 \to \mathbb C$ such that $\sum_{j=1}^5 f_j(z) \equiv 1$ and if $z \in w_j$ for some $j$ then the following numbers are real

$$f_j(z),\quad \tau f_{j+1} (z),\quad \overline \tau f_{j-1} (z),\quad f_{j+1}(z) + f_{j-2}(z), \quad f_{j-1}(z) + f_{j+2}(z) \in \mathbb R,$$ where $\tau = \exp(2\pi i / 3)$. Is it true that the space of solutions $(f_1, f_2, f_3, f_4, f_5)$ has finite dimension? Is it true that there is at most one solution?

image of pentagon in the hyperbolic plane

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  • $\begingroup$ I am confused with your notation. You say $z\in\{ w_j\}$ and the following statements depend on $j$. Can we remove the curly braces? $\endgroup$
    – Alexandre Eremenko
    Feb 8, 2019 at 21:20
  • $\begingroup$ @AlexandreEremenko , thanks, I've corrected the question. I meant that the given condition holds for any $j$. The curly brackets were added by j.c. , I hope he doesn't mind I have removed them. $\endgroup$
    – Misha
    Feb 9, 2019 at 21:54
  • $\begingroup$ The function $g=f_1^3$ satisfies $g(Tz)=\overline{g(z)}$, where $T$ is a symmetry in $w_1,w_2$ or $w_5$. Thus the group generated by these three symmetries must be discrete, else $f_1\equiv 0$ which quickly yields a contradiction. This looks a rare event (say, if $w_2$ and $w_5$ intersect at a point belonging to $\mathbb{H}^2$ and the angle between them is not a rational multiple of $\pi$, the composition of symmetries w.r.t. them already generates a non-discrete group.) $\endgroup$
    – Fedor Petrov
    Feb 14, 2019 at 18:46
  • $\begingroup$ @FedorPetrov , 1) Those symmetries belong to the symmetry group of the corresponding tessellation of $\mathbb H^2$ wich seems to be discrete. <br><br> 2) Let us suppose that lines $w_1, w_2, w_5$ form a triangle $T$. If $T$ lies "below" $w_1$ then it has angles $(2\pi/3, 2\pi/3, \dots)$, so its area is negative. If $T$ lies "above" $w_1$ then it has angles $(\pi/3, \pi/3, \dots)$, so its area is at most $\pi/3$ and the pentagon $P$ of area $3\pi-5\pi/3 = \frac43\pi$ can not fit inside $T$. So I think the sad fact is that $w_2$, $w_5$ can not intersect each other. $\endgroup$
    – Misha
    Feb 15, 2019 at 10:43
  • $\begingroup$ @Misha is it discrete for any equiangular pentagon? Or for generic? Why? Well, they do not intersect of course indeed, but I guess the composition of two symmetries is in general an element of infinite order even when it is not a rotation. $\endgroup$
    – Fedor Petrov
    Feb 15, 2019 at 10:46

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