# The Farey Graph

27 Jun 2021

The Farey graph, or Farey diagram, is an object that appears in many guises throughout math. For me, it appears several times as a complex related to the outer automorphism group of a free group of rank two, but it has connections to things like continued fractions as well. The purpose of this post is introduce the Farey graph and prove a couple of basic properties of it.

The Farey graph has as vertices all rational numbers expressed as fractions $$\frac{p}{q}$$ in lowest terms with $$q \ge 0$$, together with a vertex for $$\frac{1}{0}$$. Two vertices $$\frac{a}{b}$$ and $$\frac{c}{d}$$ span an edge when the matrix $$\begin{pmatrix} a & c \\ b & d \end{pmatrix}$$ has determinant $$\pm 1$$. Above is a picture; thanks to StackExchange user marchetto for the TikZ code for producing the Farey diagram, to which I mainly added numbers. I’d like to demonstrate that the visual properties of the Farey graph hold: namely, that every edge belongs to exactly two triangles, and that edges do not cross.

## Edges belong to two triangles

In this section we will show that every edge of the Farey graph belongs to two triangles. The key tool will be a lemma I learned from Allen Hatcher’s Topology of Numbers.

Lemma. Suppose $$a$$ and $$b$$ are integers with no common divisor. If one integral solution of $$ay - bx = n$$ is $$(x,y) = (c,d)$$, then the general integral solution is $$(x,y) = (c + ka, d + kb)$$ for integer $$k$$.

Proof. Consider a general solution $$(x,y)$$. Given our known solution $$(c,d)$$, we see that $$a(y-d) - b(x -c) = n - n = 0$$. Since $$a$$ and $$b$$ have no common factors, it follows that $$a$$ divides $$x-c$$ and $$b$$ divides $$y-d$$. In fact, we have $$a(y-d) = abk = b(x-c)$$, from which the result follows. $$\blacksquare$$

A variant on the proof shows that if $$ac - bd = -n$$, then the general integral solution for $$ax - by = n$$ is $$(ka - c,kb - d)$$.

Proposition. Suppose that $$\frac ab$$ and $$\frac cd$$ are in the Farey graph, and that $$ad > bc$$. If $$\frac xy$$ is adjacent to both $$\frac ab$$ and $$\frac bc$$, then $$\frac xy = \frac{a\pm c}{b\pm d}$$.

Proof. Suppose $$\frac ab$$ and $$\frac cd$$ are as in the statement. Now suppose $$\frac xy$$ is adjacent to both $$\frac ab$$ and $$\frac cd$$. Suppose first for definiteness that $$ay - bx = 1$$. By the lemma, we have that $$(x,y) = (ka + c, kb + d)$$ for some integer $$k$$. By assumption, we have $$cy - dx = \pm 1$$, so the lemma shows that $$(x,y) = (\ell c \mp a,\ell d \mp b)$$ for some integer $$\ell.$$ Therefore we have $$a(k \pm 1) = (\ell - 1)c$$ and $$b(k \pm 1) = (\ell-1)d$$, from which it follows that $$(k \pm 1)(\ell - 1)(ad - bc) = 0$$. Therefore either $$k = \mp 1$$ or $$\ell = 1$$, both of which imply $$(x,y) = (c \mp a,d \mp b)$$. If instead we have $$ay-bx = -1$$, then the conclusion is the same. We of course have $$\frac{c \mp a}{d \mp b} = \frac{a \pm c}{b \pm d}$$. $$\blacksquare$$

## The action of $$\operatorname{SL}_2(\mathbb{Z})$$

In this section we analyze the action of $$\operatorname{SL}_2(\mathbb{Z})$$ on the Farey diagram, proving that it is transitive on edges and triangles.

First let us define the action of $$\operatorname{SL}_2(\mathbb{Z})$$ on the Farey diagram. The action of $$\operatorname{SL}_2(\mathbb{Z})$$ on $$\mathbb{Z}^2$$ restricts to an action on the vertex set of the Farey diagram. Indeed, $$a$$ and $$b$$ are relatively prime if and only if there exists integers $$x$$ and $$y$$ such that $$ax + by = 1$$ if and only if $$(a,b)$$ is the first column of a matrix in $$\operatorname{SL}_2(\mathbb{Z})$$ if and only if there is an element of $$\operatorname{SL}_2(\mathbb{Z})$$ taking $$(1,0)$$ to $$(a,b)$$. In terms of fractions, if $$z = \frac{p}{q}$$ is a vertex of the Farey diagram, then the action of $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ sends $$z$$ to

$\frac{az + b}{cz + d} = \frac{ap + bq}{cp + dq}.$

This action preserves adjacency: the reader is encouraged to figure out why. Indeed, the action is transitive on edges: can you show that if $$\frac ab$$ and $$\frac cd$$ are adjacent in the diagram, there is an element of $$\operatorname{SL}_2(\mathbb{Z})$$ sending the edge between $$\frac 01$$ and $$\frac 10$$ to the edge between $$\frac ab$$ and $$\frac cd$$?

Finally, put a cyclic order on the vertex set of the Farey diagram by taking the cyclic order induced by the linear order on $$\mathbb{Q} \cup \{\infty\}$$. That is, given three distinct elements $$p$$, $$q$$ and $$r$$ in $$\mathbb{Q}\cup\infty$$, say $$(p,q,r)$$ is cyclically ordered if $$a < b < c$$ or $$b < c < a$$ or $$c < a < b$$ holds. For example $$(\infty,-2,0)$$ is cyclically ordered.

Proposition. Suppose $$p$$, $$q$$ and $$r$$ are the vertices of a triangle in the Farey diagram and that $$(p,q,r)$$ is cyclically ordered. Then there is an element of $$\operatorname{SL}_2(\mathbb{Z})$$ sending $$(0,1,\infty)$$ to $$(p,q,r)$$.

Proof. Suppose $$p = \frac cd$$ and $$r = \frac ab$$. Because $$p$$ and $$r$$ are adjacent in the Farey diagram, the matrix $$\begin{pmatrix} a & c \\ b & d \end{pmatrix}$$ has determinant $$\pm 1$$.

Assume first that $$ad-bc = 1$$, i.e. that $$p < r$$. Then the claim is that $$q = \frac{a+c}{b+d}$$, which is the image of $$\frac 11$$ under the matrix $$\begin{pmatrix} a & c \\ b & d\end{pmatrix}$$. In other words, the claim is that $$(p,\frac{a-c}{b-d},r)$$ is not circularly ordered, for which we need to show that $$p<\frac{a-c}{b-d}<r$$ is false. Assume first that $$b-d$$ is nonnegative. Then $$ad > bc$$ implies $$ab - ad < ab - bc$$, which implies $$\frac ab < \frac{a-c}{b-d}$$, and the claim follows. Next assume that $$b-d$$ is negative. Then $$ad > bc$$ implies that $$ad - dc > bc - dc$$, which implies $$\frac{a-c}{b-d} < \frac cd$$, and the claim follows (Note that we assume that $$d$$ is nonnegative.)

Next assume that $$ad - bc = -1$$, i.e. that $$r < p$$. Then the claim is that $$q = \frac{a-c}{b-d}$$, which is the image of $$\frac 11$$ under the matrix $$\begin{pmatrix} a & - c \\ b & -d\end{pmatrix}$$ in $$\operatorname{SL}_2(\mathbb{Z})$$. In other words, the claim is that $$(p,\frac{a+c}{b+d},r)$$ is not circularly ordered, for which we need to show that it neither $$\frac{a+c}{b+d} < r < p$$ nor $$r < p < \frac{a+c}{b+d}$$ holds. In fact, $$ad < bc$$ implies $$ab + bc > ab + ad$$ holds, which implies $$\frac{a+c}{b+d} > \frac ab$$. But, $$ad < bc$$ implies that $$ad + dc < bc + dc$$, which implies that $$\frac{a+c}{b+d} < \frac cd$$, and the claim follows. $$\blacksquare$$

In fact, the proof above shows that the matrix in $$\operatorname{SL}_2(\mathbb{Z})$$ is unique up to multiplication of all the entries by $$-1$$. This corresponds to the matrix $$\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$$, which generates the center of $$\operatorname{SL}_2(\mathbb{Z})$$.

Corollary. The stabilizer of a triangle is isomorphic to $$C_3 \times C_2 \cong C_6$$.

An explicit generator for the cyclic group of order $$6$$ stabilizing the triangle spanned by $$(0,1,\infty)$$ is $$\begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix}$$.

The proposition also yields an element of $$\operatorname{SL}_2(\mathbb{Z})$$ interchanging $$0$$ and $$\infty$$: this element is $$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$. This element is unique up to multiplication by $$-1$$ and has order $$4$$, so we have proven the following.

Corollary. The stabilizer of an edge is isomorphic to $$C_4$$.

## Edges do not cross

We finish this post by proving that edges of the Farey diagram do not cross, in the sense that if $$p$$ and $$q$$ are adjacent and $$r$$ and $$s$$ are adjacent and $$p$$, $$q$$, $$r$$ and $$s$$ are all distinct, then if $$(p,r,q)$$ is cyclically ordered, then $$(p,s,q)$$ is also cyclically ordered. By results in the previous section, we may assume $$p = 0$$ and $$q = \infty$$. That is, if $$r = \frac ab$$ is positive and $$\frac cd$$ is adjacent to $$\frac ab$$, then $$\frac cd$$ is also positive. By swapping the roles of $$r$$ and $$s$$ if necessary, we may assume that $$ad - bc = 1$$. Since we assume that $$a$$, $$b$$, and $$d$$ are positive integers, then we must have $$c \ge 0$$ in order to satisfy $$ad -bc = 1$$.