By R.L. Graham, P. Erdos
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Suppose |ad − bc| = 1. Then (1, 0) and (0, 1) are in the span of (a, b) and (c, d). Indeed, d(a, b) − b(c, d) = (ad − bc, 0) = ±(1, 0) and a(c, d) − c(a, b) = ±(0, 1). Thus (a, b) and (c, d) generate Z2 . Conversely, suppose (a, b) and (c, d) generate Z2 . Then the matrix equation a c x=b b d has a unique solution x ∈ Z2 for all vectors b ∈ Z2 . For b = (0, 1)T , we have x= a c b d −1 0 1 = 1 ad − bc d −c −b a 0 1 = 1 ad − bc −c a ∈ Z2 . It follows that |ad − bc| divides |a| and |c|. Since (a, b) and (c, d) generate Z2 , there exists i, j ∈ Z such that i(a, b) + j(c, d) = (1, 0).
Example. 6, marching upwards from the vertex (xxxy, xxxyxxy) to the root (x, y). 26 CHAPTER 3. STANDARD FACTORIZATION some Christoffel words w ... 6: Paths in the Christoffel tree from (u, v) to the root (x, y) preserve the cutting points for standard factorizations. Note that we have found a characterization of those Christoffel morphisms that preserve Christoffel words. Namely, f : (x, y) → (w1 , w2 ) is such a morphism if and only if (w1 , w2 ) is a standard factorization of a Christoffel word.
1. u = Pal(v) for some v ∈ {x, y}∗ if and only if xuy is a Christoffel word. 2. u = Pal(v) for some v ∈ {x, y}∗ if and only if u has relatively prime periods p and q and |u| = p + q − 2. Proof of 1. 6, if u = Pal(v) then xuy is a Christoffel word. Conversely, let w = xuy be a Christoffel word and let (w1 , w2 ) be its standard factorization. 4, |w1 |x |w2 |x |w1 |y |w2 |y ∈ SL2 (Z) ∩ N2×2 38 CHAPTER 4. PALINDROMIZATION (writing N2×2 for the set of 2 × 2 matrices with nonnegative integer entries).
Old And New Problems And Results In Combinatorial Number Theory by R.L. Graham, P. Erdos
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