Note: Hyperbolic Pentagon Moduli

There are no right angled quadrilaterals in the hyperbolic plane, but there are right angled pentagons. Existence of at least one is easy: we can construct a regular hyperbolic pentagon (with all sides the same) by constructing an appropriate triangle, and repeating.

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But just like the square isn’t the only right angled quadrilateral in Euclidean geometry, neither is this the only right angled pentagon in the hyperbolic plane. The goal of this note is to completely classify right angled hyperbolic pentagons. In particular, we prove

Thus the moduli space of right angled pentagons is parameterized by $$\mathcal{P}\cong a,b>0\mid \sinh (a)\sinh (b)>1$$

The Proof

The main idea is simple to state. Draw sides of length $a,b$ perpendicular to each other, then draw common perpendiculars at the endpoints. These either intersect, or they don’t. If they do, no pentagon. If they do not, then there’s a unique shortest common perpendicular. This fifth side forms a right angled pentagon with the others.

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To work out the details, we need to find a way to tell when exactly the two sides we extend orthogonally from the segments of length $a,b$ remain disjoint. And to be really get quantitative, when they are ultraparallel we should like to know the length of their common perpendicular. The idea is to bring in the trigonometry of hyperbolic triangles, by drawing the geodesic through the endpoints of segments $a$ and $b$. This subdivides each of the right angles there into pairs, $\alpha ,\overline{alpha}$ opposite $a$ and $\beta ,\overline{\beta }$ opposite $b$. Call the length of this segment $X$.

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The Law of Cosines

The top of this shape is a right triangle, so we have access to many trigonometric relations between $a,b,\overline{\alpha },\overline{\beta }$ and $X$. The other portion is still a mystery: maybe its a triangle (finite or ideal) or maybe its infinite in area (if the geodesics are ultraparallel). There’s a nice trick here - in the triangle case, one is naturally tempted to employ the law of cosines: but in fact the hyperbolic law of cosines[1^] works in both cases. First, for the triangle - we can compute the angle $\chi$ opposite $X$ via

[1^]: There are actually two hyperbolic laws of cosines. The other one is a direct analog of the Euclidean case; and the one we utilize here has no analog in Euclidean geometry as it allows one to calculate a side length in terms of only information about angles!

$$\cos \chi =\cos \alpha \cos \beta +\sin \alpha \sin \beta \cosh X$$ But in the ultraparallel case, the length $d$ of the common perpendicular satisfies the same relation: $$\cosh d=\cos \alpha \cos \beta +\sin \alpha \sin \beta \cosh X$$

Thus, given any leg of length $X$ with angles $\alpha ,\beta$ at the endpoints we can compute the common right hand side of this formula, and fully understand the situation:

• If it is less than or equal to[2^] $1$, then it the resulting shape is a triangle, and its new angle is the $\arccos$ of the quantity.
• If it is greater than $1$, then it forms a quadrilateral whose new side is the $\mathrm{arccosh}$ of the quantity.

[2^]: When it equals precisely 1, the triangle has an idea vertex opposite $X$ so the angle is zero.

Relating to the Triangle

We want an expression in terms of $a$ and $b$, but right now the law of cosines is in terms of $\overline{\alpha },\overline{\beta }$ and $X$. First, as $\alpha +\overline{\alpha }=\pi /2$ and $\beta +\overline{\beta }=\pi /2$ the trigonometric functions of either of these determine the other. This observation, together with the trigonometry of right triangles implies

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$$\cos \alpha =\sin \overline{\alpha }=\frac{\sinh a}{\sinh X}\cos \beta =\sin \overline{\beta }=\frac{\sinh b}{\sinh X}$$

$$\sin \alpha =\cos \overline{\alpha }=\frac{\tanh b}{\tanh X}\sin \beta =\cos \overline{\beta }=\frac{\tanh a}{\tanh X}$$

And, of course the pythagorean identity for the sides $$\cosh X=\cosh a\cosh b$$

The Calculation

Starting with the mystery quantity determined by the law of cosines, we substitute terms computed from the triangle above:

$$\begin{array}{rlr}?& =\cos \alpha \cos \beta +\sin \alpha \sin \beta \cosh X& =\frac{\sinh a}{\sinh X}\frac{\sinh b}{\sinh X}+\frac{\tanh b}{\tanh X}\frac{\tanh a}{\tanh X}\cosh X\end{array}$$

Expanding the definitions of $\tanh a$, $\tanh b$ in the second term and using the pythagorean identity for $\cosh X$:

$$\begin{array}{rlrl}\frac{\tanh a}{\tanh X}\frac{\tanh b}{\tanh X}\cosh X& =\frac{1}{{\tanh }^{2}X}\left(\frac{\sinh a}{\cosh a}\frac{\sinh b}{\cosh b}\right)\cosh X& =\frac{1}{{\tanh }^{2}X}\left(\frac{\sinh a}{\cosh a}\frac{\sinh b}{\cosh b}\right)\cosh a\cosh b& =\frac{\sinh a\sinh b}{{\tanh }^{2}X}\end{array}$$

Plugging this back in:

$$\begin{array}{rlrlr}?& =\frac{\sinh a\sinh b}{{\sinh }^{2}X}+\frac{\sinh a\sinh b}{{\tanh }^{2}X}& =\sinh a\sinh b(\frac{1}{{\sinh }^{2}X}+\frac{1}{{\tanh }^{2}X})& =\sinh a\sinh b({\mathrm{csch}}^{2}X+{\coth }^{2}X)& =\sinh a\sinh b\end{array}$$

where the final line uses the hyperbolic trigonometric identity for cosecant and cotangent. After all that cancellation we see that our mystery quantity is just $\sinh a\sinh b$, and when this quantity is $\le 1$ the edges intersect (perhaps ideally) and there is no possible hyperbolic pentagon constructible from the configuration. But when $\sinh a\sinh b>1$, the sides are ultraparallel and there is a unique geodesic intersecting each orthogonally. That is, there is a unique hyperbolic pentagon with all right angles, proving our theorem.

Description of All Sides

With a precise characterization of all right angled pentagons in hand, we turn to the quantitative problem of determining their side lengths. Label the sides $a,b,\dots ,e$ clockwise around the pentagon as below:

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First, the law of cosines determines $d$ in terms $a$ and $b$. This identity turns out to be incredibly useful well beyond the study of pentagons, so we box it off:

Applying this rule to other pairs of sides, we see that $\cosh a=\sinh c\sinh d$, and as $d$ and $a$ are known, this gives $c$. Likewise, $\sinh b\sinh c=\cosh e$ and since we already know $b$ and $c$ this fixes $e$. Unpacking these give explicit descriptions in terms of $a,b$:

Appendix: The Regular Pentagon

Here we give the specific measurements for the regular pentagon, where $a=b=c=d=e$. Call this common side length $x$. Then by the trigonometry of pentagons we see $$\sinh x\sinh x=\cosh x$$ And as ${\sinh }^{2}x={\cosh }^{2}x-1$,the quantity $X=\cosh x$ satisfies the quadratic polynomial $X^2-1=X$, or

$$X^2=X+1$$

The defining equation for the golden ratio, which has $\varphi =\frac{1+\sqrt{5}}{2}$ as its unique positive root.

An Independent Argument

Here’s an alternative elementary argument that is independent of all the work we did above. We return to the subdivision by triangles alluded to at the beginning of the post, but subdivide each once further to get right triangles:

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The ten right triangles above have an angle of $\pi /4$ (half of the right angle at the pentagon’s vertex) and an angle of $\pi /5$ in the center. Such a triangle exists as the angle sum is less than $\pi$: $$\frac{\pi }{2}+\frac{\pi }{4}+\frac{\pi }{5}=\frac{19\pi }{20}<\pi$$

We know all three angles of this triangle, which in hyperbolic space ( unlike Euclidean geometry) determines it up to isometry. Its area is the angle deficit: $$A_{\mathrm{Triang}}=\pi -\frac{\pi }{2}-\frac{\pi }{4}-\frac{\pi }{5}=\frac{\pi }{20},⟹,A_{\mathrm{Pentagon}}=\frac{\pi }{4}$$

And its side lengths follow from trigonometry: if we call $v$ the distance to the vertex, $r$ the distance to an edge’s center and $e$ the length of an edge,

$$\cosh v=\cot \frac{\pi }{4}\cot \frac{\pi }{5}=\cot \frac{\pi }{5}=\sqrt{1+\frac{2}{\sqrt{5}}}$$

$$\cosh r=\frac{\cos \frac{\pi }{4}}{\cos \frac{\pi }{5}}=\sqrt{1+\frac{1}{\sqrt{5}}}$$

$$\cosh \frac{e}{2}=\frac{\cos \frac{\pi }{5}}{\cos \frac{\pi }{4}}=\frac{1+\sqrt{5}}{2\sqrt{2}}$$

Or approximately $v\approx 0.842$, $r\approx 0.626$, and recovering what we found above, $e\approx 1.061$.

A corollary of this is that there is a unique regular pentagon (even though there are many right angled pentagons where a pair of sides $a,b$ are equal)