Orphans in Forests of Linear Fractional Transformations

This paper studies the set of orphans in certain Calkin-Wilf trees generated by linear fractional transformations.

In [2], Calkin and Wilf introduced a rooted infinite binary tree where every vertex is labeled by a positive rational number according to the following rules: (1) the root is labeled 1/1, (2) the left child of a vertex a/b is labeled a/(a + b), and (3) the right child of a vertex a/b is labeled (a + b)/b. As noted by several authors [4,6], replacing a/b in (2) and (3) by the variable z shows that the vertex labels of the Calkin-Wilf tree are generated by applying one of two linear fractional transformations. For any vertex labelled z in the Calkin-Wilf tree, the left child of z is z z+1 and the right child of z is z + 1.
If the root of the tree is allowed to be a complex number in the upper half-plane, then it follows that the set of all of the descendants of the root are the complex numbers obtained from acting on the root by elements in the subsemigroup of SL 2 (N 0 ) generated freely [7,8] by Here the group action is given by For any complex number z, let ℜ(z) and ℑ(z) represent the real and imaginary parts of z, respectively, and let D 0 = {z ∈ C : ℜ(z) > 0, ℑ(z) > 0}. Nathanson [8] considers the complex Calkin-Wilf trees associated with complex roots in D 0 using the matrices where u and v are positive integers, to generate descendants. This leads to the creation of an infinite forest of complex numbers associated with each pair (u, v). As a word of caution, it is not immediately obvious that Nathanson's generalization of the Calkin-Wilf tree leads to a forest. Some justification for this fact is required (see [8,Thm. 2] for details). One common property seen in various generalizations of the Calkin-Wilf tree [1,3,4,5] is that every element appearing in a tree always has a finite number of ancestors. The goal of this article is to extend this notion to the above forest of complex numbers associated with the pair (u, v). Note that the restriction to elements in D 0 is crucial here. Without such a restriction, every element would have an infinite number of ancestors.
Given a pair (u, v), if w ∈ D 0 has no ancestors in its (uniquely) associated complex Calkin-Wilf tree, then we say that w is a complex (u, v)-orphan. We begin with a characterization of the set of complex (u, v)-orphans due to Nathanson [8].
Theorem 1 (Nathanson, [8]). Let D u,v be the set of complex (u, v)-orphans. Then Proof. Suppose that z = x + iy is a complex (u, v)-orphan. If ℜ(z) > v, then z is the right child of z − v. This is a contradiction, so ℜ(z) ≤ v. It remains to show that |2uz − 1| ≥ 1.
Let w = L −1 u · z. A straightforward calculation shows that In other words, z is a left child unless w / ∈ D 0 . That is, we must have that x(1 − ux) − uy 2 ≤ 0. It follows that By completing the square for x, So z lies on or outside of the circle centered at 1 2u of radius 1 2u . In particular, z − 1 2u ≥ 1 2u , from which the desired result follows. (See Figure 2 for a graphical representation of D u,v .) Theorem 2. Let 0 < y 0 ≤ 1 2u and z ∈ D 1 be such that ℑ(z) ≥ y 0 . Then Proof. As in Theorem 1, suppose that z = x + iy and let w = L −1 u · z. It follows from (1) that In particular, f ′ u,y (x) > 0 for x < 1 u , which clearly holds in this case since |2uz − 1| ≤ 1 and y > 0. This shows that, for a fixed y value, f u,y (x) is minimized when x is as small as possible. Finding the location of the desired minimum is equivalent to determining the smaller x-value of the two points of intersection of the horizontal line of all complex numbers with imaginary part y and the circle of radius 1 2u around 1 2u . A simple computation shows that this occurs at Note that x u,y is a real number since we have that 0 < y ≤ 1 2u and that f u,y (x u,y ) = ǫ u (y).
To complete the proof, it is therefore enough to show that ǫ u (y) ≥ ǫ u (y 0 ). Differentiating ǫ u (y) with respect to y, we see that Since ǫ ′ u (y) > 0 for 0 < y < 1 2u and ǫ ′ u,v (y) → ∞ as y → 1 2u − , it follows that ǫ u (y) is minimized at y = y 0 .
We now obtain the desired result. Proof. Suppose that there is a z ∈ D 0 this is not the descendant of a (u, v)-orphan. That is, assume that z has infinitely many ancestors z = z 0 , z 1 , z 2 , . . . , all in D 0 , where either z i+1 = L −1 u ·z i or z i+1 = R −1 v ·z i for i ≥ 0. If z i ∈ D 1 for all sufficiently large i, then lim i→∞ ℜ(z i ) = −∞, a contradiction. So there is an infinite subsequence i k , k ≥ 0, so that z i k ∈ D 1 . Using induction, it follows from Theorem 2 that ℑ(z i k ) − ℑ(z i 0 ) ≥ kǫ u (ℑ(z i 0 )). So z i k ∈ D 1 for all sufficiently large k, a contradiction.