Finding Fibonacci in the Hyperbolic Plane

Summary

We use a combinatorial approximation of the hyperbolic plane to investigate properties of hyperbolic geometry such as exponential growth of perimeter and area of disks, and the linear isoperimetric inequality. This calculations give a surprising link to Fibonacci numbers.

MSC:

• 51M09

This paper is a result of a mathematical mistake I made seven years ago. As a graduate student, a math professor of mine asked me to compute the perimeters and areas of some hyperbolic polygons as an introduction to hyperbolic geometry. I sat down, drew some pictures, found a pattern, and gave him an answer. We were both happy, and went on with our study of hyperbolic spaces.

Last year, while I was teaching hyperbolic geometry for the first time, I remembered that problem. I had fun solving it and it had a nice and neat solution, so I put it on the homework set for my students. They were much more careful than either my former self or my former professor, and they discovered what neither of us had: there was a mistake in my solution from seven years ago! The correct solution ended up being more of a challenge to prove than I had planned, but it was also a much more surprising and beautiful result.

The problem

Consider the family of complexes X_n made by gluing together equilateral triangles along their edges so that there are n triangles at every vertex. If n is 3, 4, or 5, we get a finite complex; in fact, we get the three Platonic solids with triangular faces (see Figure 1). If n = 6, then we get a copy of the Euclidean plane (tiled by equilateral triangles).

Fig. 1 The complexes (a) $X_3,$ (b) $X_4,$ (c) X₅, (d) part of X₆. Note that for $n\ge 6,$, X_n is infinite. All models made from Geometiles®.

If n > 6, then we get a strange and infinite surface that cannot be drawn to scale in the Euclidean plane, though we can make it out of physical triangles. The result is a floppy and frilly object that refuses to lie flat, like a leaf of kale (see Figure 2). We can also draw it in the plane, if we accept a drawing that is not accurate with respect to distances or angles (see Figure 3).

Fig. 2 Part of X₇, also made from Geometiles ®.

Fig. 3 Part of X₇. (Not drawn to scale!) The yellow triangles lie in D(1(1) $$\left[\begin{array}{c}{B}_{r+1}\\ Y_{r+1}\end{array}\right]={\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]}^r\left[\begin{array}{c}7\\ 0\end{array}\right]=7·{\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]}^r\left[\begin{array}{c}1\\ 0\end{array}\right].$$(1) ), the red in D(2), and the purple in D(3). The vertex v is in the middle of the yellow region.

Let us scale our triangles so that each edge has length 1. Pick any vertex, v. We can define a combinatorial disk in X_n of radius r, denoted $D_n(r)$, iteratively as follows: Pick any vertex v. Let $D_n(0)=\left\{v\right\}$, and let $D_n(r+1)$ be the union of $D_n(r)$ with all of the triangles that share a vertex with $D_n(r)$.

Let $P_n(r)$ denote the perimeter of the combinatorial disk in X_n. If we choose $n=6$, we can see that $D_6(r)$ is a hexagon with edge length r, so $P_6(r)=6r$.

When n = 7, it is a more complicated story. Figure 3¹ shows $D_7(1),D_7(2),D_7(3)$. Counting the perimeter of each combinatorial disk, you get the following sequence:

Table

r	1	2	3	4	5	6
$P_7(r)$	7	21	56	91	385	938

Do you see the pattern? It is not immediately obvious! But if we divide every entry by 7, we get the numbers 1, 3, 8, 21, 55, 134, and now the pattern emerges: this list consists of every other Fibonacci number.

Theorem 1.

Let F_n denote the nth Fibonacci number. Then $P_7(r)=7F_{2r}$.

The goal of the next section is to prove this theorem.

Remark.

To simplify the notation, for the rest of the paper we will write D(r) to denote $D_7(r)$ and P(r) to denote $P_7(r)$.

2 Fibonacci appears

We will prove Theorem 1 by induction on the radius r. We can establish the base case by checking D(1(1) $$\left[\begin{array}{c}{B}_{r+1}\\ Y_{r+1}\end{array}\right]={\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]}^r\left[\begin{array}{c}7\\ 0\end{array}\right]=7·{\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]}^r\left[\begin{array}{c}1\\ 0\end{array}\right].$$(1) ). For the inductive step, we will count the number of vertices in $\partial D(r)$ instead of the number of edges.

Consider Figure 4. Every vertex in $\partial D(r)$ is adjacent to 7 triangles in total. Either 2 or 3 of those triangles lie in D(r). We will partition the vertices in $\partial D(r)$ into two sets; the blue vertices are those that are adjacent to 2 triangles in D(r), and the yellow vertices are those that are adjacent to 3 triangles in D(r). Let B_r be the number of blue vertices in $\partial D(r)$ and let Y_r be the number of yellow vertices.

Fig. 4 Part of X₇. Vertices are colored blue and yellow as in the proof. The red and green regions each contain one vertex in $\partial D(2)$ and the vertices in $\partial D(3)$ that belong or partially belong to it.

Two vertices are adjacent if they are separated by an edge. Every vertex in $\partial D(r+1)$ is adjacent to exactly 1 or 2 vertices in $\partial D(r)$. We will say that a vertex v in $\partial D(r+1)$ belongs to a vertex $v\prime \in \partial D(r)$ if $v\prime$ is the only vertex in D(r) adjacent to v. We will say v partially belongs to $v\prime$ if $v\prime$ is one of the two vertices in D(r) adjacent to v.

Then the total number of vertices in $D(r+1)$ is the sum of $$\#\left\{\text{vertices that belong to }v\right\}+\frac{1}{2}\#\left\{\text{vertices that partially belong to }v\right\},$$taken over every vertex in $\partial D(r)$.

For every blue vertex b in $\partial D(r)$, there are 2 blue vertices in $\partial D(r+1)$ that belong to b and 2 yellow vertices that partially belong to b. For every yellow vertex y in $\partial D(r),$ $\partial D(r+1)$ contains 1 blue vertex that belongs to y and 2 yellow vertices that partially belong to y. In summary, we have that $P(r+1)=B_{r+1}+Y_{r+1}$, where: $$B_{r+1}=2B_r+1Y_r\text{ and }{Y}_{r+1}=1B_r+1Y_r.$$

Written in matrix form, that is: $$\left[\begin{array}{c}{B}_{r+1}\\ Y_{r+1}\end{array}\right]=\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]\left[\begin{array}{c}{B}_r\\ Y_r\end{array}\right].$$

In particular, since $B_1=7$ and $Y_1=0$, we get (1) $$\left[\begin{array}{c}{B}_{r+1}\\ Y_{r+1}\end{array}\right]={\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]}^r\left[\begin{array}{c}7\\ 0\end{array}\right]=7·{\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]}^r\left[\begin{array}{c}1\\ 0\end{array}\right].$$(1)

On the other hand, the Fibonacci numbers also satisfy a linear recurrence relation, namely: $$\left[\begin{array}{c}{F}_{r+1}\\ F_r\end{array}\right]={\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]}^r\left[\begin{array}{c}1\\ 0\end{array}\right].$$

Since we are only interested in even order Fibonacci numbers, we have $$\left[\begin{array}{c}{F}_{2r+1}\\ F_{2r}\end{array}\right]={\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]}^{2r}\left[\begin{array}{c}1\\ 0\end{array}\right]={\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]}^r\left[\begin{array}{c}1\\ 0\end{array}\right].$$

This differs from the result of equation (1)(1) $$\left[\begin{array}{c}{B}_{r+1}\\ Y_{r+1}\end{array}\right]={\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]}^r\left[\begin{array}{c}7\\ 0\end{array}\right]=7·{\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]}^r\left[\begin{array}{c}1\\ 0\end{array}\right].$$(1) by a factor of 7, so we get $7F_{2r+1}=B_{r+1}$ and $7F_{2r}=Y_{r+1}$. Therefore: $$7F_{2r+2}=7(F_{2r+1}+F_{2r})=B_{r+1}+Y_{r+1}=P(r+1).$$

Hyperbolicity, exponential growth, and isoperimetry

The title of this paper promised that we would find Fibonacci in the hyperbolic plane, so where is the hyperbolic plane in all this?

One of the standard models of the hyperbolic plane is the Poincaré disk. In this model, the “plane” is the interior of the unit disk, and geodesics are arcs of Euclidean circles or lines that cross the boundary of the disk at right angles.

Many results in hyperbolic geometry are extremely counterintuitive at first. One of the most surprising facts about hyperbolic triangles is that the sum of the interior angles is always strictly less than π (in radians). Furthermore, the hyperbolic area of a hyperbolic triangle is determined entirely by the sum of the interior angles: A triangle $\Delta \mathit{\text{ABC}}$ with interior angles α, β, γ has area $\pi -(\alpha +\beta +\gamma )$. Two more surprising facts, which are consequences of this, are that (1(1) $$\left[\begin{array}{c}{B}_{r+1}\\ Y_{r+1}\end{array}\right]={\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]}^r\left[\begin{array}{c}7\\ 0\end{array}\right]=7·{\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]}^r\left[\begin{array}{c}1\\ 0\end{array}\right].$$(1) ) any two hyperbolic triangles which are similar must also be congruent, and (2) there exists a largest triangle, namely the so-called ideal triangle whose angles are all 0. A good resource if you would like to read more about hyperbolic geometry is the book by Stahl [1].

As we scale an equilateral hyperbolic triangle, the sum of the interior angles changes: the larger the triangle, the smaller the sum. Since the triangle is equilateral, it is also equiangular, so the angle at any one corner changes as well. Therefore, by scaling our triangles to the appropriate size, we can find equilateral triangles with interior angles $2\pi /n$ for any n > 6. In particular, if we pick n = 7, we will get equilateral triangles with interior angles perfectly sized to fit 7 triangles around a common vertex. These triangles have area equal to $\pi -\frac{6\pi }{7}=\frac{\pi }{7}$, by the fact above. The perimeter of the triangle is a little more difficult to calculate. If we continue to place these equilateral triangles, with 7 at each vertex, we get a tiling of the hyperbolic plane, as illustrated in Figure 5.

Fig. 5 A tiling of the hyperbolic plane by equilateral triangles, with 7 triangles at every vertex.

We built X₇ out of Euclidean triangles with side length 1, but we can interpret equation (1)(1) $$\left[\begin{array}{c}{B}_{r+1}\\ Y_{r+1}\end{array}\right]={\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]}^r\left[\begin{array}{c}7\\ 0\end{array}\right]=7·{\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]}^r\left[\begin{array}{c}1\\ 0\end{array}\right].$$(1) as a statement about the number of edges around the boundary of D(r), rather than its perimeter. In fact, we can call the number of edges around the boundary of D(r) the combinatorial perimeter of D(r), and we will denote it P(r). (Note that P(r) will always be a constant multiple of the actual perimeter of D(r)). In this interpretation, it doesn’t matter if we use Euclidean or hyperbolic triangles to build X₇. If we build X₇ out of hyperbolic triangles our combinatorial disks D(r) approximate hyperbolic disks.

This implies (at least asymptotically) another counterintuitive fact about hyperbolic space: The perimeter of a hyperbolic circle grows exponentially with the radius. Compare this to the Euclidean case, where the perimeter of a circle grows linearly with its radius!

Another fun fact about the hyperbolic plane is that polygons (and more general regions) satisfy a linear isoperimetric inequality; that is, if R is a hyperbolic polygon with perimeter $|\partial R|$ and area $\left|R\right|$, then there is a constant c (which does not depend on the choice of polygon) so that $|\partial R|\ge c\left|R\right|$. Again, let us compare this to the Euclidean case. Consider any Euclidean polygon R, and let R_k denote a copy of R scaled by k. Then $\left|R_k\right|$ is proportional to k², while $|\partial R_k|$ is proportional to k. Thus, $\underset{k\to \infty }{\mathrm{\text{lim}}}|\partial R_k|/\left|R_k\right|=0$, so there does not exist a constant c such that $|\partial R_k|/\left|R_k\right|\ge c$ for all k. (And we did not even have to consider non-similar polygons!)

For this paper, we will use an approximation of area that is easier to calculate with, called the combinatorial area and denoted A(r), where A(r) is the number of triangles required to make up our disk D(r). Note that if we make X₇ out of hyperbolic triangles, then $\frac{\pi }{7}A(r)=|D(r)|$. If we make A(r) out of Euclidean triangles, then $\frac{\sqrt{3}}{4}A(r)=|D(r)|$. In general, the combinatorial area A(r) will always be a constant multiple of the actual area, $|D(r)|$, in the same way that the combinatorial perimeter is a constant multiple of the actual perimeter. In particular, our disks will satisfy a linear isoperimetric inequality with respect to combinatorial area and perimeter if and only if they satisfy one (possibly with a different coefficient) with respect to actual area and perimeter.

Can we see this linear isoperimetric inequality in our combinatorial model?

Fibonacci strikes again

Let us find a formula for A(r). By counting triangles in Figure 3, we get the following data:

Table

r	1	2	3	4	5	6
$A(r)-A(r-1)$	7	28	77	315	847	2240

This time the pattern is not as easy to spot. Here it is:

Theorem 2.

Let A(r) denote the combinatorial area of D(r). Then we have $$A(r)=7(4F_{2r-2}+3F_{2r-3}-2).$$

To prove the theorem, we will enumerate triangles in $D(r+1)-D(r)$ using the blue and yellow vertices from before. Every blue vertex in $\partial D(r)$ is adjacent to 5 triangles in $D(r+1)-D(r)$. On the other hand, every yellow vertex in $\partial D(r)$ is adjacent to 4 triangles in $D(r+1)-D(r)$. (See Figure 6.)

Fig. 6 The orange region is $D(2).$ Blue vertices are adjacent to 5 triangles in $D(3)-D(2)$, as indicated in red. Yellow vertices are adjacent to 4, as indicated in green.

It’s tempting to say that the change in area is therefore $5B_r+4Y_r$, but be careful: some of the triangles in $D(r+1)-D(r)$ are adjacent to two vertices in $\partial D(r)$. In fact, every vertex is adjacent to exactly two of the double-counted triangles because they are precisely the triangles in $D(r+1)-D(r)$ that share an edge with a triangle in D(r). But since the number of vertices in $\partial D(r)$ is $P_r=B_r+Y_r$, we get: $$A(r+1)-A(r)=5B_r+4Y_r-P_r=4B_r+3Y_r$$whenever $r\ge 1$.

Since $B_r=7F_{2r-1}$ and $Y_r=7F_{2r-2}$, this gives us $$A(r+1)-A(r)=7\left(4F_{2r-1}+3F_{2r-2}\right).$$

Therefore, for $r\ge 1$, we have: $$A(r)=A(1)+\sum _{i=1}^{r-1}\left(A(i+1)-A(i)\right)$$ $$=7+7\sum _{i=1}^{r-1}(4F_{2i-1}+3F_{2i-2}).$$

This looks nasty, but since these are the Fibonacci numbers there is a lovely identity that simplifies the story. You can verify, either by induction or by using the recursion formula, that ${\sum }_{i=1}^r{F}_{2i}=F_{2r+1}-1$, and ${\sum }_{i=1}^r{F}_{2i-1}=F_{2r}.$

Therefore, for r > 1 we get: $$A(r)=7\left(1+4F_{2r-2}+3(F_{2r-3}-1)\right)$$ $$=7(4F_{2r-2}+3F_{2r-3}-2).$$

This is not quite as nice as the formula for the perimeter, but it does allow us to compare the perimeter and the area. In fact, we can use facts about the Fibonacci sequence to calculate $\underset{r\to \infty }{\mathrm{\text{lim}}}A(r)/P(r)$ exactly. You can show, using the matrix form of the Fibonacci recursion, that $F_r=\frac{{\varphi }^r-{\psi }^r}{2}$, where $\varphi =\frac{1+\sqrt{5}}{2}$ (the Golden Ratio), and $\psi =\frac{1-\sqrt{5}}{2}$. Then you can use this to calculate that $$\underset{r\to \infty }{\mathrm{\text{lim}}}\frac{F_{2r-2}}{F_{2r}}=\frac{1}{{\varphi }^2}\hspace{1em}\text{and}\hspace{1em}\underset{r\to \infty }{\mathrm{\text{lim}}}\frac{F_{2r-3}}{F_{2r}}=\frac{1}{{\varphi }^3}.$$

Therefore, $$\underset{r\to \infty }{\mathrm{\text{lim}}}\frac{A(r)}{P(r)}=4\frac{F_{2r-2}}{F_{2r}}+3\frac{F_{2r-3}}{F_{2r}}-2\frac{-2}{F_{2r}}=\frac{4}{{\varphi }^2}+\frac{3}{{\varphi }^3}=\sqrt{5}.$$

This is certainly different from the Euclidean world!

To show that X₇ approximates the hyperbolic plane, we would like to show that $P(r)\ge \mathit{\text{cA}}(r)$ for some constant c > 0. Equivalently, we want to show that $A(r)/P(r)\le \frac{1}{c}$ for some c > 0. And since we have been haunted by the number 7 throughout this entire paper, it would be nice—for the sake of narrative symmetry if nothing else—if 7 shows up again. Luckily for us, it does.

Theorem 3.

For any radius r, $P(r)\ge \frac{1}{7}A(r)$.

Let us see why. Using our formulas from Theorems 1 and 2, we have $$\frac{A(r)}{P(r)}=\frac{4F_{2r-2}+3F_{2r-3}-2}{F_{2r}}$$ $$=4\frac{F_{2r-2}}{F_{2r}}+3\frac{F_{2r-3}}{F_{2r}}-\frac{2}{F_{2r}}.$$

But since F_r is an increasing sequence, we know that $F_q/F_r\le 1$ whenever $q\le r$. So we get: $$\frac{A(r)}{P(r)}\le 4+3-0=7,$$just as we wanted.

Note

Acknowledgment

The author would like the thank the anonymous reviewer for comments that improved the exposition of this paper.

Disclosure statement

No potential conflict of interest was reported by the author(s).

Additional information

Notes on contributors

MurphyKate Montee

MURPHYKATE MONTEE is an assistant professor of mathematics at Carleton College. Her research lies in geometric group theory, an area of math that uses geometric objects (like the combinatorial hyperbolic plane in this article) to understand groups.

Notes

1 Note that the online version of this article has color diagrams. In Figure 3, the innermost triangles are yellow, the middle triangles are red, and the outer triangles are purple.