CChristoffel words and Markoff triples
Christophe Reutenauer ∗ UQAM
A Markoff triple is a triple of natural integers a, b, c which satisfies theDiophantine Equation a + b + c = 3 abc. These numbers have been introduced in the work of Markoff [M2], wherehe finished his earlier work [M1] on minima of quadratic forms and approx-imation of real numbers by continued fractions. He studies for this certainbi-infinite sequences and shows that they must be periodic; each of thesesequences is a repetition of the same pattern, which turns out to be oneof the words introduced some years before by Christoffel [C1]; Markoff wasapparently not aware of this work of Christoffel. These words were calledChristoffel words much later in [B] and may be constructed geometrically,see [MH], [OZ], [B], [BL], and have also many different interpretations, inparticular in the free group on two generators (see [BLRS] for the theory ofChristoffel words).In this Note, we describe a mapping which associates to each Christof-fel word a Markoff triple and show that this mapping is a bijection. Theconstruction and the result are a variant a theorem of Harvey Cohn [C], seealso [CF] chapter 7 and [P].As them, we use the
Fricke identities . We use also the tree constructionof the Christoffel words, see [BL] and [BdL], and moreover the arithmeticrecursive construction of the solutions of the previous Diophantine equation,see [M2], [F], [CF], [W], [P]. ∗ Christophe Reutenauer; Universit´e du Qu´ebec `a Montr´eal; D´epartement demath´ematiques; Case postale 8888, succursale Centre-Ville, Montr´eal (Qu´ebec) Canada,H3C 3P8 (mailing address); e-mail: [email protected]. Supported by NSERC. a r X i v : . [ m a t h . N T ] S e p Results
We consider lattice paths , which are consecutive elementary steps in the x, y -plane; each elementary step is a segment [( a, b ) , ( a +1 , b )] or [( a, b ) , ( a, b +1)],with a, b ∈ Z .Let p, q be relatively prime integers. Consider the segment from (0 , p, q ) and the lattice path from (0 ,
0) to ( p, q ) located below this segmentand such that the polygon delimited by the segment and the paths has nointerior integer point.The
Christoffel word of slope q/p is the word in the free monoid { x, y } ∗ coding the above path, where x (resp. y ) codes an horizontal (resp. vertical)elementary step. See the figure, where is represented the path with ( p, q ) =(7 ,
3) corresponding to the Christoffel word of slope 3/7. (0, 0) (7, 3)
Figure 1: the Christoffel word xxxyxxyxxy of slope 3/7Note that the definition includes the particular cases ( p, q ) = (1 ,
0) and(0 , x and y . All other Christoffelwords will be called proper .Each proper Christoffel word w has a unique standard factorization w = w w ; it is obtained by cutting the path corresponding to w at the integerpoint closest to the segment. In the figure, the standard factorization is givenby w = xxxyxxy , w = xxy . The words w and w are then Christoffelwords.Define the homomorphism µ from the free monoid { x, y } ∗ into SL ( Z ),defined by µx = (cid:18) (cid:19) and µy = (cid:18) (cid:19) .A Markoff triple is a multiset { a, b, c } of positive integers which satisfiesthe equation a + b + c = 3 abc . The triple is proper if a, b, c are distinct. Itis classical that the only inproper Markoff triples are { , , } and { , , } ,see [CF] Chapter 2. 2 heorem 2.1 A multiset { a, b, c } is a proper Markoff triple if and only if itis equal to { / T r ( µw ) , / T r ( µw ) , / T r ( µw ) } for some unique Christof-fel word w with standard factorization w = w w . It is known that for each Christoffel word w , one has 1 / T r ( µw ) = µ ( w ) , , see [R] Lemme 3.2 or [BLRS] Lemma 8.7. Proof
1. Each couple ( w , w ) forming the standard factorization of aChristoffel word is obtained by applying iteratively the rules ( u, v ) → ( u, uv )or ( u, v ) → ( uv, v ) starting from the couple ( x, y ), see [BL] Proposition2 or [BdL] Lemma 7.3. As a consequence, one obtains inductively that w w w − w − is conjugate to xyx − y − in the free group F generated by x and y .Now, let w be some Christoffel word with standard factorization w = w w . Let a = 1 / T r ( w ) , b = 1 / T r ( w ) , c = 1 / T r ( w ). We use the Frickeidentity T r ( A ) + T r ( B ) + T r ( AB ) = T r ( ABA − B − ) + 2 + T r ( A ) T r ( B ) T r ( AB )for any A, B ∈ SL ( Z ). We take A = µ ( w ), B = µ ( w ) and therefore AB = µ ( w ). Thus 9 a + 9 b + 9 c = 27 abc (and we are done), provided T r ( ABA − B − ) = −
2. Since w w w − w − is conjugate to xyx − y − , it suf-fices to show that T r ( µxµyµx − µy − ) = −
2. Now the matrix µxµyµx − µy − is equal to (cid:18) − − − (cid:19) , which shows what we want.It remains to show that a, b, c are distinct. In the special case where w = xy , this is seen by inspection. Furthermore, µ ( w ) = µ ( w ) µ ( w ) + µ ( w ) µ ( w ) , which implies that µ ( w ) > µ ( w ) , µ ( w ) , since thematrices have positive coefficients. Thus, by the remark before the proof, c > a, b . Now, since we assume w (cid:54) = xy and by the tree construction recalledat the beginning of the proof, w is a prefix of w or w is a suffix of w .Then a < b or a > b repectively.2. Let { a, b, c } be a proper Markoff triple and assume that a < b < c .Then { a, b, ab − c } is a Markoff triple and 3 ab − c < b by [M2], [F], [CF],[W].Suppose first that this triple is proper. By induction, the multiset { a, b, ab − c } is equal to { / T r ( µw ) , / T r ( µw ) , / T r ( µw ) } for someChristoffel word w with standard factorization w = w w . By the first partof the proof, b = 1 / T r ( µw ), since both numbers are the maximum of theirmultiset. Then either (i) a = 1 / T r ( µw ) and 3 ab − c = 1 / T r ( µw ), or (ii) a = 1 / T r ( µw ) and 3 ab − c = 1 / T r ( µw ).3n case (i), we have c = 1 / T r ( µ ( w w )); indeed, for A, B in SL ( Z ), T r ( A B )+ T r ( B ) = T r ( A ) T r ( AB ), hence T r ( µ ( w w )) = T r ( µw ) T r ( µw ) − T r ( µw ) = 3 a. b − ab − c ) = 3 c .In case (ii), we have c = T r ( µ ( w w )); indeed, T r ( AB ) + T r ( A ) = T r ( AB ) T r ( B ), hence T r ( µ ( w w )) = T r ( µw ) T r ( µw ) − T r ( µw ) = 3 b. a − ab − c ) = 3 c .Thus in case (i), { a, b, c } corresponds to the Christoffel word w w (withstandard factorization w .w w ) and in case (ii) to the Christoffel word w w (with standard factorization w w .w ).If { a, b, ab − c } is not proper, then, since a (cid:54) = b , we have { a, b, ab − c } = { , , } and a = 1 and b = 2 or conversely. Thus 3 ab − c = 1 and thisimplies c = 3 . . − { a, b, c } corresponds to the Christoffel word xy . 3. Concerning unicity, suppose that the proper Markoff triple { a, b, c } with a < b < c may be obtained from the two Christoffel words u and v with standard factorization u = u u and v = v v . We may assume thatthey are both distinct from xy . Then by the tree construction of Christoffelwords, u is a prefix of u or u is a suffix of u , and similarly for v , v .Hence, we are by symmetry reduced to two cases: (i) u is a prefix of u and v is a prefix of v , or (ii) u is a prefix of u and v is a suffix of v .In Case (i), u (resp. v ) has the standard factorization u = u u (cid:48) (resp. v = v v (cid:48) ). Moreover a = 1 / T r ( µu ) , b = 1 / T r ( µu ) , c = 1 / T r ( µu )as we saw in the first part. Then the Markoff triple corresponding tothe Christoffel word u u (cid:48) is equal to { a, ab − c, b } . Indeed, T r ( µu (cid:48) ) = T r ( µu ) T r ( µ ( u u (cid:48) )) − T r ( µ ( u u (cid:48) )) = 3 a. b − c , because u u (cid:48) = u and u u (cid:48) = u . Similarly, the Markoff triple corresponding to the Christoffelword v v (cid:48) is also equal to { a, ab − c, b } . Thus, by induction, the Christoffelwords u u (cid:48) and v v (cid:48) are equal, together with their standard factorization,hence u = v .In Case (ii), by the same calculation, the Markoff triple correspondingto the Christoffel word u u (cid:48) is equal to { a, ab − c, b } , where the standardfactorization of u is u u (cid:48) . Symmetrically, v has the standard factorization v = v (cid:48) v and the Markoff triple corresponding to the Christoffel word v (cid:48) v is also equal to { a, ab − c, b } . Thus, by induction, the Christoffel words u u (cid:48) and v (cid:48) v are equal, together with their standard factorization. Hence u = v (cid:48) and u (cid:48) = v . Thus u = u u = u u (cid:48) and v = v v = v (cid:48) v = u u (cid:48) .Thus we are reduced to the following particular case, which shows that Case(ii) cannot happen.4. Let w = uv be the standard factorization of the proper Christoffelword w . Then the Markoff triple corresponding to the Christoffel words u v uv are distinct.Indeed, disregarding the trivial case w = xy , we may assume that u is aprefix of v or v is a suffix of u . In the first case, each entry of the matrix µu is stricly smaller than the corresponding entry of the matric µv . The sametherefore holds for the matrices µuµuµv and µuµvµv . Thus T r ( µ ( u v )) issmaller than T r ( µ ( uv )), which proves that the Markoff triples are distinct,since so are their greatest elements. The second case is similar.It is known that each Markoff number is of the form 1 / T r ( µw ) forsome Christoffel word w , see [R] Cor.3.1 or [BLRS] Th.8.4. It is howevernot known if the mapping associating to each Christoffel w word the Markoffnumber 1 / T r ( µw ) is injective. This is equivalent to the Markoff numbersinjectivity conjecture , see [F] page 614, [CF], [W].
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