Constructing right-angled regular polygons in the hyperbolic plane

679 Views Asked by At

I want to prove the existence of right-angled regular polygons in the hyperbolic plane.

A polygon is regular if all its sidelengths and internal angles are equal. For which n exist right-angled regular n-gons in the hyperbolic plane?. What is their sidelength (as a function of n)?

Hint: You may assume that for each n all regular polygons are congruent.

My answer to the first question is that $n \geq 5$.

I tried to prove this claim in the following way:

Consider the plane $E:= \; \{(0,0,1)^T+ \lambda_1 (1,0,1)^T+\lambda_2 (0,1,1)^T \mid \lambda_1,\lambda_2 \in R\}$.

Draw into E a regular euclidean n-gon K with center $N:=(0,0,1)^T$ such that each vertex has euclidean distance $r \in \ ]0,1[$ to the center. In particular draw the n-gon K in such a manner that the points

$v_1=(r,0,1)^T$

$v_i=L^{i-1} \cdot (r,0,1)^T$, i $\in \{2,...,n-1\}$

where L=$\begin{bmatrix} cos(\delta) & sin(\delta) & 0 \\ - sin(\delta) & cos(\delta) &0 \\ 0 & 0 & 1 \end{bmatrix}$

and $\delta= \frac{(n-2) \cdot \pi}{n}$

are the vertices of K.

Denote by $s_1,...,s_n$ the sides of K. In particular let

$s_1$ be the side connecting $v_1$ and $v_2$

and

$s_2$ be the side connecting $v_2$ and $v_3$. Let $\langle\, \cdot, \cdot\rangle$ denote the Lorentz inner product

Denote by $U_i:=\; \{x \in R^{2,1} \mid \langle\,n_i,x\rangle=0\}$ the plane passing through $s_i$.

Since K is contained inside the unit circle of the plane E the planes $U_i$ will intersect the upper sheet of the hyperboloid such that we get a regular hyperbolic polygon.

It follows from the construction that $U_1=span\{v_1,v_2\}$ and $U_2=span\{v_2,v_3\}$

We may use the Gram-Schidt-Algorithm to determine unit-length space- like vectors $n_1$ and $n_2$ such that

$\langle\, n_1, v_1\rangle=0$

$\langle\, n_1, v_2\rangle=0$

$\langle\, n_2, v_2\rangle=0$

$\langle\, n_2, v_3\rangle=0$

which are the normals of the planes $U_1$ and $U_2$ respectively.

The formulas obtained from the Gram-Schmidt-Algorithm show that $n_1$ and $n_2$ only depend on r and also that the functions

$r \mapsto n_1(r)$

$r \mapsto n_2(r)$

are continous. Since the interior angle of the hyperbolic n-gon can be measured via

$cos(\alpha(r))= -\langle\, n_1(r), n_2(r)\rangle$

the function $r \mapsto \alpha(r)$ is continous.

Since the angle function is bounded by 0 and $\frac{(n-2) \cdot \pi}{n}$ it follows from the continuity that for $n \geq 5$ we find $r \in \ ]0,1[$ such that $\alpha(r)=\frac{\pi}{2}$.

Hence we find right-angled regular polygons in the hyperbolic plane for $n \geq 5$.

For the second question I attempted to do the following:

Construct a right-angled regular n-gon via the above method. Let a be its side length. Consider two adjacent vertices $u_1$ and $u_2$.

Let $d(\cdot, \cdot)$ denote the metric in the hyperboloid model.

Consider the hyperbolic circles

$C_1:= \; \{x \in R^{2,1} \mid d(N,x)=b\}$.

$C_2:= \; \{x \in R^{2,1} \mid d(N,x)=c\}$

where $C_1$ contains all the vertices of the n-gon and $C_2$ lies inside the n-gon and touches each side once. Then the radius of $C_1$ bisects each interior angle of the n-gon. Since the interior angles were $\frac{\pi}{2}$ we get two smaller angles $\frac{\pi}{4}$ . Moreover the radius of $C_2$ is the perpendicular bisector of the side $s_1$ that connects $u_1$ and $u_2$. Denote the point of intersection by M. Consider the lines $l_1$ and $l_2$ defined by N and $u_1$ and N and M respectively. Since $u_1$ is a vertex of the n-gon the line $l_1$ and the radius of $C_1$ coincide and so d(N,$u_1$)=b. Likewise, since M lies on the radius of $C_2$ we have d(N,M)=c. Also, since M is the midpoint of $s_1$ we have $d(u_1,M)=d(u_2,M)=\frac{a}{2}$.

Hence we may consider the hyperbolic triangle with vertices N, $u_1$, M. The side lengths are

a’=$\frac{a}{2}$

b

c

The interior angles are

$\alpha$

$\beta=\frac{\pi}{2}$

$\gamma=\frac{\pi}{4}$.

Using the hyperbolic side cosine theorem we get:

$cosh(a’)=\frac{cos(\alpha)+cos(\beta) \cdot cos(\gamma)}{sin(\beta) \cdot sin(\gamma)}$ $=\sqrt(2) \cdot cos(\alpha)$

and hence:

$a=2 \cdot arcosh(\sqrt(2) \cdot cos(\alpha))$ (1)

My questions are:

1) Is there an easier way to determine a formula for the angle function $r \mapsto \alpha(r)$ than using Gram-Schmidt?

2) How can I determine the angle $\alpha$ to solve equation (1) ? I know that it should be something dependent on n but how can I find an explicit formula?