aa r X i v : . [ m a t h . G R ] J a n New sequences of non-free rationalpoints
Ilia Smilga ∗ January 12, 2021
We exhibit some new infinite families of rational values of τ , some of themsquares of rationals, for which the group or even the semigroup generated bythe matrices ( ) and ( τ ) is not free.For each α, β ∈ C , we define the group Γ( α, β ) := (cid:28)(cid:18) α (cid:19) , (cid:18) β (cid:19)(cid:29) , and we define S ( α, β ) to be the semigroup generated by the same two matrices. Theproblem of determining values of the product τ := αβ for which the group Γ( α, β ) (conjugate by ( α ) to Γ(1 , τ ) ) is free has already attracted a considerable amount ofattention [San47, Bre55, CJR58, Ree61, LU69, KS94, Gil08]; the corresponding problemfor the semigroup S ( α, β ) has also been studied [BC78, Sł15]. In particular, a lot ofpeople have focused on the special case where τ is the square of a rational [LU69, Ign86,Bea93, TT96] (which is a natural condition as it corresponds to studying Γ( µ, µ ) withrational µ ), or more generally any rational number [Bam00, KK21].To avoid confusion, we should warn the reader that previous authors have variouslyused as the key parameter either the product αβ (e.g. [KS94, KK21]), its half, usuallydenoted by λ (focusing on Γ(2 , λ ) ) (e.g. [CJR58, Ree61, LU69, Bam00, Gil08, Sł15]), orits square root, usually denoted by µ (e.g. [San47, Bre55, BC78, Bea93, TT96]). Wemake the former choice, hence the following terminology: Definition 1.
We say that a number τ ∈ C is free (resp. semigroup-free ) if the group Γ(1 , τ ) (resp. the semigroup S (1 , τ ) ) is free.An elementary observation (first made by Brenner [Bre55]) is that for τ ≥ , thegroup Γ(1 , τ ) is always Schottky, hence free. So for groups, it suffices to study theinterval ( − , (in fact even (0 , , given the symmetry Γ(1 , τ ) = Γ(1 , − τ ) ). Similarly, ∗ The author is supported by the European Research Council (ERC) under the European Union Horizon2020 research and innovation programme (ERC starting grant DiGGeS, grant agreement No. 715982). τ ≥ , the semigroup S (1 , τ ) is Schottky (see also [BC78, Theorem 2.6]), so it sufficesto study the interval ( − , .Beyond this, very little is known. It is conjectured that all rational τ ∈ ( − , arenon-free, and that all rational τ ∈ ( − , are non-semigroup-free. There are no knowncounterexamples, but only a limited number of values for which these two conjectureshave been proved. Indeed, group relations are easiest to find for τ close to . A spectacu-lar development in this direction is [KK21], proving that all rational numbers τ ∈ ( − , whose numerator is at most , or between and , are non-free. On the other hand,as τ gets closer to , finding relations becomes extremely hard.It is even harder to find accumulation points of non-free rational values. It seems thatthe only ones known so far (besides , which is trivial as n is never free) are those ofthe form n [Bea93] and those of the form nn ± [TT96]. As for accumulation points ofnon-semigroup-free rational values, only is known.Our goal in this paper has been to find accumulation points of non-free and non-semigroup-free rational values (or squares of rationals) that lie as close as possible to theendpoints of the corresponding interval. The result that we obtain can be summarisedas follows. Theorem 2.
The set of τ that are rational and non-semigroup-free (hence in particularnon-free) has − ± √ , − φ ± , − and as accumulation points (where φ = √ is thegolden ratio). The set of τ that are squares of rationals and non-semigroup-free (hencein particular non-free) has as an accumulation point.Proof. This follows from the more precise Theorem 3 below, which gives a concretedescription of the infinite families that accumulate at these points. Note that:• the values given in (iv) for positive (resp. negative) k are in fact the convergentsof the continued fraction of φ (resp. of φ − );• the values given in (v) for positive (resp. negative) k are in fact the convergents ofthe continued fraction of √ (resp. of − √ ).The technique used to prove points (iv) and (v) is similar to that used by Beardon[Bea93] and Tan and Tan [TT96]: a one-parameter family of relations leading to valuesof τ that are convergents of some continued fraction. The main difference is that thesetwo papers focused on the case where τ is the square of a rational, whereas we considerall rational τ . Moreover, with similar methods, one can also construct many other accu-mulation points; we have chosen to present only those that are closest to the endpointsof the interval.For each member of each of the following five families, we will actually exhibit anexplicit relation for the corresponding group or semigroup (see Proposition 7). Theorem 3.
All of the following values of τ are non-free. The values given in (ii) arealso non-semigroup-free. For τ as in (iii) with k > or as in (iv) or (v) with ( − k k > ,its negative − τ is non-semigroup-free.(i) τ = (cid:0) k ± k (cid:1) , for all integer k = 0 . ii) τ = (cid:0) n − n (cid:1) , for all n = σ σ u σk u σk +1 , where σ = ( σ , σ ) is any pair of distinctnumbers among { , , } , and ( u σk ) k ∈ Z is the integer sequence determined by thefollowing recurrence relations: ( u σ = u σ = 1; ∀ k ∈ Z , u σk − − σ ( k mod 2) u σk + u σk +1 = 0 (1) (see discussion below for explicit list of values).(iii) τ = k ± k , for all integer k = 0 .(iv) τ = F k +2 F k for all integer k = 0 , where F is the Fibonacci sequence ( F = F = 1 ).Explicitly, these are , , , , , , . . . (2) and their reciprocals.(v) τ = H k +1 P k for all integer k = 0 , where P k are the Pell numbers and H k are thehalf-companion Pell numbers, given by (cid:16) H k P k (cid:17) := ( ) k ( ) . Explicitly, these are , , , , , , . . . (3) and twice their reciprocals.Discussion of the values given in (ii). One readily checks the following symmetry of thesequences u σk : ∀ σ = σ ∈ { , , } , ∀ k ∈ Z , u ( σ ,σ ) k = u ( σ ,σ )1 − k , (4)so it suffices to consider only three of the six sequences. Moreover the sequences u (1 , and u (1 , are symmetric: ∀ σ ∈ { , } , ∀ k ∈ Z , u (1 ,σ ) k = u (1 ,σ ) − k (5)(indeed observe that in this case u σ − = 2 u σ − u σ = 1 = u σ , then use induction), so itsuffices to consider their nonnegative terms. Explicitly, we have: (cid:16) u (1 , ± k (cid:17) k ≥ = 1 , , , , , , , , . . . ; (6) (cid:16) u (1 , ± k (cid:17) k ≥ = 1 , , , , , , , , . . . ; (7) (cid:16) u (2 , k (cid:17) k ∈ Z = . . . , , , , , , , , , , , , , . . . . (8)These give rise to the following values of n : (cid:16) u (1 , k u (1 , k +1 (cid:17) k ≥ = 3 , , , , , , , . . . ; (9) (cid:16) u (1 , k u (1 , k +1 (cid:17) k ≥ = 2 , , , , , , . . . ; (10) (cid:16) u (2 , k u (2 , k +1 (cid:17) k ∈ Z = . . . , , , , , , , , , , . . . . (11)3n fact, instead of relations, we will exhibit so-called “half-relations”, that we will defineright now. We will then see (Proposition 5) that each half-relation easily allows one towrite down an actual relation. Definition 4.
Let τ ∈ C , and let us fix the notations g := ( ) and h τ := ( τ ) . Wesay that a sequence ( a , . . . , a l ) ∈ Z l is a half-relation for τ if:(i) for odd l , the matrix M = g a h a τ · · · h a l − τ g a l satisfies τ c ( M ) − c ( M ) = 0; (12)(ii) for even l , the matrix M = g a h a τ · · · g a l − h a l τ satisfies c ( M ) − c ( M ) = 0 , (13)where c ij denotes the ( i, j ) -th coefficient, so that ∀ M, M =: (cid:18) c ( M ) c ( M ) c ( M ) c ( M ) (cid:19) . (14)The point of half-relations is that they allow us to construct a special kind of relationsfor the group or semigroup corresponding to ± τ : Proposition 5.
A sequence ( a , . . . , a l ) ∈ Z l is a half-relation for τ if and only if theidentity g a h a τ · · · f a l = h a l τ g a l − · · · f a , where f := ( g if l is odd h τ if l is even, (15) or equivalently g a h a τ · · · g − a h − a τ = Id , (16) holds. Such an identity provides a nontrivial relation:• for the group Γ(1 , τ ) if ∀ i, a i = 0 ;• for the semigroup S (1 , τ ) if ∀ i, a i > ;• for the semigroup S (1 , − τ ) if ∀ i, ( − i a i > .Proof. Let M = g a h a τ · · · denote the left-hand side of (15). Then ( a , . . . , a l ) is ahalf-relation for τ if and only if M is a fixed point of:• the involution (cid:0) a bc d (cid:1) ( a c/ττb d ) , or equivalently M ( τ ) M t ( τ ) − , for odd l ;• the involution (cid:0) a bc d (cid:1) ( d bc a ) , or equivalently M (cid:0) − (cid:1) M − (cid:0) − (cid:1) − , foreven l . 4oth of them are antimorphisms; moreover the former switches g and h τ , whereas thelatter fixes both of them. In both cases, the involution maps both sides of (15) to eachother.Now assume that this is true. Then (15), usually rewritten as (16) for groups, is arelation in Γ(1 , τ ) , which is nontrivial if the coefficients are all nonzero. If they are allpositive, then (15) is in fact a relation in S (1 , τ ) . If they have alternating signs (notethat whenever ( a , . . . , a l ) is a half-relation, so is ( − a , . . . , − a l ) ), then (15) (for even l )or (16) (for odd l ) is a relation in S (1 , − τ ) .Let us also observe that being a half-relation is a polynomial condition: more precisely,the left-hand side of (12) (for odd l ) or (13) (for even l ) is a polynomial in a , . . . , a l and τ . One also easily checks by induction that this polynomial is always divisible by τ ,so we can factor τ out: Definition 6.
For each l ≥ , we define the polynomial P l HR ( a , . . . , a l ; τ ) := ( ( c − τ c )( g a h a τ · · · g a l ) if l is odd; τ ( c − c )( g a h a τ · · · h a l τ ) if l is even (17)(where c ij is the i, j -th coefficient, as in (14)).Thus, by construction, whenever τ ∈ C is a root of some polynomial P l HR ( a , . . . , a l ) for some tuple ( a , . . . , a l ) of nonzero (resp. positive) integers, it is non-free (resp. non-semigroup-free). Explicitly, for small l , these polynomials are as follows: P ( a ; τ ) = a P ( a , a ; τ ) = a a P ( a , a , a ; τ ) = a a a τ + a − a + a P ( a , . . . , a ; τ ) = a a a a τ + a a − a a + a a + a a P ( a , . . . , a ; τ ) = a a a a a τ + ( a a a − a a a + a a a + a a a + a a a ) τ + a − a + a − a + a . The observation that non-free values can be obtained as roots of certain polynomialshas already been made previously. Even more specifically, when l = 2 k + 1 is odd, thepolynomial P k +1HR coincides (up to the substitution τ = − λ and a sign switch for theodd a i ’s) with Bamberg’s polynomial B k [Bam00].We are now ready to exhibit the half-relations that prove all the statements of Theo-rem 3. Note that in point (iii), the subcase a) suffices by itself to prove the Theorem inthe general case; but we have also presented two additional special cases, b) and c), forwhich we can shorten the half-relation by one coefficient. Proposition 7.
Fix some k ∈ Z , assumed to be nonzero except in point (ii), and somepair σ = ( σ , σ ) of distinct coefficients among { , , } .(i) The value τ = (cid:0) k − k (cid:1) has (1 , − , − k, k (4 k + 4)) as a half-relation. ii) The value τ = (cid:0) n − n (cid:1) , where n = σ σ u σk u σk +1 , has (1 , σ k +1 ( u σk ) , σ k ( u σk +1 ) , asa half-relation (where σ k is a notation shortcut for σ ( k mod 2) ).(iii) The value τ = k +1 k has following half-relations:a) All sequences ( k, − , , − , k, x ) , where x is any integer.b) If k = 2 t for some integer t , the sequence (1 , − , , − t, − t + 2 t − .c) If k = t ( t +1)2 − for some integer t , the sequence (1 , − , , − t + 1 , − t − .(iv) The value τ = F k +2 F k has (1 , − , , − , − k F k − F k ) as a half-relation (where F isthe Fibonacci sequence, F = F = 1 ).(v) The value τ = H k +1 P k has (( − k P k − P k , − , , − , , − , , − , ( − k P k − P k , x ) asa half-relation for any integer x (where P k are the Pell numbers and H k are thehalf-companion Pell numbers, given by (cid:16) H k P k (cid:17) := ( ) k ( ) ). A remark about point (ii): in fact, one can show that these are the only values of n for which τ = (cid:0) n − n (cid:1) has a half-relation of length with positive coefficients. Proof.
To prove these statements, we check that all of the listed values make the poly-nomial P HR vanish.• For points (i) and (iii), this is a straightforward computation.• For point (iv), we start by computing P (1 , − , , − , N ; τ ) = N τ − (3 N + 2) τ + ( N + 4) . (18)It remains to verify that for all k = 0 , the values τ = F k +2 F k and N = 2( − k F k − F k make this expression vanish. This is straightforward (if tedious), for example byplugging in the closed formula F k = φ k − ( − φ ) − k φ + φ − (where φ := √ is the golden ratio) and expanding.• For point (v), we start by computing P ( N, − , , − , , − , , − , N, x ; τ ) == x (cid:16) N τ − (6 N + 2 N ) τ + (10 N + 10 N + 1) τ − (4 N + 12 N + 4) τ + 2 N + 3 (cid:17) = x (cid:16) N τ − (2 N + 1) τ + 1 (cid:17)(cid:16) N τ − (4 N + 1) τ + (2 N + 3) (cid:17) . (19)It turns out that for all k = 0 , the values τ = H k +1 P k and N = ( − k P k − P k makethe last factor vanish (so that the whole expression vanishes regardless of the value6f x ). This is straightforward (if tedious) to verify, for example by plugging in theclosed formulas P k = α k − ( − α ) − k α + α − , H k = α k + ( − α ) − k α − α − (20)(where α = 1 + √ ) and expanding.• Finally, for point (ii), we start by computing P , σ k +1 ( u σk ) , σ k ( u σk +1 ) , − σ σ u σk u σk +1 ! == P ( σ k ,σ k +1 ) ( u σk , u σk +1 )= ( P σ ( u σk , u σk +1 ) if k is even P σ ( u σk +1 , u σk ) if k is odd, (21)where, for all σ = ( σ , σ ) , P σ is the polynomial given by P σ ( x, y ) := 1 + σ σ + 6 σ x + 6 σ y − xy. (22)Recalling the definition (1) of the sequences u σ , to prove (by induction on k ) thatthe right-hand side of (21) vanishes for all k ∈ Z , it suffices to check the followingtwo identities. On the one hand, we have ∀ σ = σ ∈ { , , } , P σ (1 ,
1) = 0 (23)(indeed the triplet ( σ σ , σ , σ ) is always some permutation of (2 , , ). On theother hand, we easily check that ∀ σ, ∀ x, y, P σ ( x, y ) = P σ (2 σ y − x, y ) = P σ ( x, σ x − y ) . (24) Proof of Theorem 3.
Theorem 3 now easily follows by Proposition 5: indeed, one readilychecks that the coefficients of almost all the half-relations listed above are nonzero, andhave the required signs. There are only three exceptions:• k = 1 in cases (iv) and (v), which yields respectively τ = 2 and τ = 3 , which arewell-known to be non-free. Also in these cases, one can still write down the relation(15) (or (16)); some cancellations occur, but one still recovers the nontrivial relation g h − τ gh − τ gh − τ = Id (for τ = 2 ) and gh − τ gh − τ gh − τ = Id (for τ = 3 ).• k = − in case (i), so that τ = . Then the half-relation written above leads toa trivial relation, so we need to find something else. To wit, (1 , − , , , is ahalf-relation with nonzero coefficients for τ = .7 cknowledgements I would like to thank Gregory Margulis for drawing my attention to this problem, andPaul Mercat for some helpful discussions.
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