Constructing right-angled regular polygons in the hyperbolic plane

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?

• Wouldn't it be easier just to determine the measurements of a $45^\circ$-$45^\circ$-$(360^\circ/n)$ triangle? – Blue Nov 27 '18 at 23:02
• Didn't read whole question, but I'd decompose the regular $n$-gon into $2n$ right triangles. Which is half the triangle Blue suggested in the comment before. Using either of these you can use the hyperbolic law of cosines to obtain any length given the three angles (and the curvature), but with a right angle in the mix it becomes even easier. – MvG Nov 28 '18 at 22:31