aa r X i v : . [ m a t h . C O ] F e b Noncommutative recursions and the Laurentphenomenon
Matthew C. Russell ∗ October 18, 2018
Abstract
We exhibit a family of sequences of noncommutative variables, recursively definedusing monic palindromic polynomials in Q [ x ], and show that each possesses the Laurentphenomenon. This generalizes a conjecture by Kontsevich. Let K r (the Kontsevich map) be the automorphism of a noncommutative plane definedby K r : ( x, y ) (cid:0) xyx − , (1 + y r ) x − (cid:1) . Maxim Kontsevich conjectured that, for any r , r ∈ N , the iterates . . . K r K r K r K r ( x, y )are all given by noncommutative Laurent polynomials in x and y . This is known as theLaurent phenomenon. The conjecture was proved in special cases for certain values of r and r (see [7], [8], [3], and [4]), sometimes also with the positivity conjecture (that allof the Laurent polynomials have nonnegative integer coefficients), and sometimes replacing1 + y r with any monic palindromic polynomial. Eventually, Berenstein and Retakh [2] gavean elementary proof of the Kontsevich conjecture for general r and r . while Rupel [6]subsequently proved it using the Lee-Schiffler Dyck path model (see [5]) while also settlingthe positivity conjecture.Later, Berenstein and Retakh [1] extended their methods to consider a more general class ofrecurrences given by Y k +1 Y k − = h k ( a k − ,k Y k a k,k +1 ), where h k ∈ Q [ x ] and h k ( x ) = h k − ( x )for all k ∈ Z , Y a Y a = a Y a Y , and a k,k ± are defined recursively by a k +2 ,k +1 = a − k − ,k ∗ Department of Mathematics, Rutgers, The State University of New Jersey, Piscataway, NJ 08854. Email: russell2 [at] math [dot] rutgers [dot] edu a k +1 ,k +2 = a − k,k +1 . Proceeding in a similar fashion to their previous paper, they provethe Laurent phenomenon for these recurrences where h k = 1 + x r k .In this paper, we endeavor to expand the methods of Berenstein and Retakh [1] to higher-order recurrences, and to using monic palindromic polynomials instead of 1 + x r . Following section 4 of Berenstein and Retakh [1], let K ≥
2. Let F K denote the Q -algebragenerated by a ± , a ± , . . . , a ± K , b ± , b ± , . . . , b ± K , and let F K ( Y , . . . , Y K ) denote the algebragenerated by F K and Y , . . . , Y K , subject to the relation Y a Y a · · · Y K a K = b K Y K b K − Y K − · · · b Y . (1)For n ∈ Z , define a n and b n recursively by a n + K = b − n (2) b n + K = a − n . (3)Suppose we have a sequence of monic palindromic polynomials h n ∈ Q [ x ] such that h n = h n − K for all n ∈ Z . Let us write h n ( x ) = P d n i =0 P n,i x i , so P n, = P n,d n = 1 for all n .Define h ↓ n ( x ) = d n X i =1 P n,i x i = h n ( x ) − h ↓↓ n ( x ) = d n X i =1 P n,i x i − = h n ( x ) − x and recursively define Y n ∈ F K ( Y , . . . , Y K ) for n ∈ Z \ { , . . . , K } by Y n + K Y n = h n ( a n Y n +1 a n +1 Y n +2 · · · Y n + K − a n + K − ) . (4)Define Y − n,m = a n − Y n a n Y n +1 · · · a m − Y m a m n ≤ mY + n,m = b n Y n b n − Y n − · · · b m Y m b m − n ≥ m, while also defining Y − n,n − = a n − and Y + n,n +1 = b n . Then, (4) becomes Y n + K Y n = h n (cid:0) Y − n +1 ,n + K − (cid:1) . (5) Proposition 1.
For all n ∈ Z , we also have Y n Y n + K = h n (cid:0) Y + n + K − ,n +1 (cid:1) (6) Y n Y − n +1 ,n + K − = Y + n + K − ,n +1 Y n . (7)2 roof. We proceed by induction on n . (We will only prove these for n ≥
1; the proof for n < n = 1. Now, supposethat (7) holds for some n ≥
1. Conjugating (5) on the left by Y n gives Y n Y n + K = Y n h n (cid:0) Y − n +1 ,n + K − (cid:1) Y − n , but, since Y n Y − n +1 ,n + K − Y − n = Y + n + K − ,n +1 Y n Y − n = Y + n + K − ,n +1 (by the inductive hypothesis), we conclude that Y n Y n + K = h n (cid:0) Y + n + K − ,n +1 (cid:1) , which is (6).Now, we desire to prove (7) for n + 1. This is equivalent to proving Y − n +1 ,n + K − Y n + K = Y n + K Y + n + K − ,n +1 (multiply each side by a − n = b n + K on the left and a n + K = b − n on the rightto recover the original expression). We calculate Y − n +1 ,n + K − Y n + K = Y − n Y + n + K − ,n +1 Y n Y n + K = Y − n Y + n + K − ,n +1 h n (cid:0) Y + n + K − ,n +1 (cid:1) = Y − n h n (cid:0) Y + n + K − ,n +1 (cid:1) Y + n + K − ,n +1 = Y − n Y n Y n + K Y + n + K − ,n +1 = Y n + K Y + n + K − ,n +1 , using (7) (the inductive hypothesis) and (6). Let A n be the subalgebra of F K ( Y , . . . , Y K ) generated by F K and Y n , . . . , Y n +2 K − . We nowstate our main result. Theorem 2.
For all n ∈ Z , A n = A . Proof.
It is enough to show that A n +1 = A n . Without loss of generality, we let n = 0. So,we try to show that Y K ∈ A . By (6) and the definition of h K ( x ), we find Y K = Y − K h K (cid:0) Y +2 K − ,K +1 (cid:1) = Y − K d K − X i =0 P K,i (cid:0) Y +2 K − ,K +1 (cid:1) i + Y − K (cid:0) Y +2 K − ,K +1 (cid:1) d K , (8)as h K is a monic palindromic polynomial. 3e would like to find an expression for Y − K (cid:0) Y +2 K − ,K +1 (cid:1) d K . Using (4), we calculate Y = Y − K d X m =1 P ,m (cid:0) Y − ,K − (cid:1) m ! Y − K = Y − Y − K d X m =1 P ,m (cid:0) Y − ,K − (cid:1) m Y − K (cid:0) Y +2 K − ,K +1 (cid:1) d K = Y (cid:0) Y +2 K − ,K +1 (cid:1) d K − Y − K d X m =1 P ,m (cid:0) Y − ,K − (cid:1) m (cid:0) Y +2 K − ,K +1 (cid:1) d K = Y (cid:0) Y +2 K − ,K +1 (cid:1) d K − Y − K d X m =1 P ,m (cid:0) Y − ,K − (cid:1) m (cid:0) Y +2 K − ,K +1 (cid:1) m (cid:0) Y +2 K − ,K +1 (cid:1) d K − m . (9)Now, we would like a formula for (cid:0) Y − ,K − (cid:1) m (cid:0) Y +2 K − ,K +1 (cid:1) m . Lemma 3.
For m ≥
0, we have (cid:0) Y − ,K − (cid:1) m (cid:0) Y +2 K − ,K +1 (cid:1) m = 1 + m − X s =0 (cid:0) Y − ,K − (cid:1) s K − X j =1 Y − ,j − h ↓ j (cid:0) Y + K + j − ,j +1 (cid:1) Y + K + j − ,K +1 ! (cid:0) Y +2 K − ,K +1 (cid:1) s . Proof.
We proceed by induction on m . The case m = 0 is trivial. For the case m = 1, wecalculate Y − ,K − Y +2 K − ,K +1 . Proposition 4.
For 0 ≤ l , Y − ,l Y + K + l,K +1 = 1 + l X j =1 Y − ,j − h ↓ j (cid:0) Y + K + j − ,j +1 (cid:1) Y + K + j − ,K +1 Proof.
The base case l = 0 simply reduces to a b K = 1, which checks. Now, assuming theinductive hypothesis, Y − ,l +1 Y + K + l +1 ,K +1 = Y − ,l Y l +1 a l +1 b K + l +1 Y K + l +1 Y + K + l,K +1 = Y − ,l Y l +1 Y K + l +1 Y + K + l,K +1 = Y − ,l h l +1 (cid:0) Y + K + l,l +2 (cid:1) Y + K + l,K +1 = Y − ,l h ↓ l +1 (cid:0) Y + K + l,l +2 (cid:1) Y + K + l,K +1 + Y − ,l Y + K + l,K +1 = Y − ,l h ↓ l +1 (cid:0) Y + K + l,l +2 (cid:1) Y + K + l,K +1 + 1 + l X j =1 Y − ,j − h ↓ j (cid:0) Y + K + j − ,j +1 (cid:1) Y + K + j − ,K +1 = 1 + l +1 X j =1 Y − ,j − h ↓ j (cid:0) Y + K + j − ,j +1 (cid:1) Y + K + j − ,K +1 , and we have proved our proposition. 4y this proposition, we see Y − ,K − Y +2 K − ,K +1 = 1 + K − X j =1 Y − ,j − h ↓ j (cid:0) Y + K + j − ,j +1 (cid:1) Y + K + j − ,K +1 , which takes care of our base case. Proceeding to the inductive step, we calculate (cid:0) Y − ,K − (cid:1) m +1 (cid:0) Y +2 K − ,K +1 (cid:1) m +1 = (cid:0) Y − ,K − (cid:1) m Y − ,K − Y +2 K − ,K +1 (cid:0) Y +2 K − ,K +1 (cid:1) m = (cid:0) Y − ,K − (cid:1) m K − X j =1 Y − ,j − h ↓ j (cid:0) Y + K + j − ,j +1 (cid:1) Y + K + j − ,K +1 ! (cid:0) Y +2 K − ,K +1 (cid:1) m = (cid:0) Y − ,K − (cid:1) m (cid:0) Y +2 K − ,K +1 (cid:1) m + (cid:0) Y − ,K − (cid:1) m K − X j =1 Y − ,j − h ↓ j (cid:0) Y + K + j − ,j +1 (cid:1) Y + K + j − ,K +1 ! (cid:0) Y +2 K − ,K +1 (cid:1) m = 1 + m − X s =0 (cid:0) Y − ,K − (cid:1) s K − X j =1 Y − ,j − h ↓ j (cid:0) Y + K + j − ,j +1 (cid:1) Y + K + j − ,K +1 ! (cid:0) Y +2 K − ,K +1 (cid:1) s + (cid:0) Y − ,K − (cid:1) m K − X j =1 Y − ,j − h ↓ j (cid:0) Y + K + j − ,j +1 (cid:1) Y + K + j − ,K +1 ! (cid:0) Y +2 K − ,K +1 (cid:1) m = 1 + m X s =0 (cid:0) Y − ,K − (cid:1) s K − X j =1 Y − ,j − h ↓ j (cid:0) Y + K + j − ,j +1 (cid:1) Y + K + j − ,K +1 ! (cid:0) Y +2 K − ,K +1 (cid:1) s , which completes our proof. Lemma 5.
For 0 ≤ l ≤ K and l + 1 ≤ q ≤ K + 1, we have Y − ,l Y + K + l,q = Y K Y + K − ,q l Y t =1 h t (cid:0) b − q − Y + q − ,t +1 Y + K + t − ,q (cid:1) (10) Proof. If l = 0, then (10) trivially holds, as the product is empty. Otherwise, note that Y − ,l Y + K + l,q = Y − ,l − Y l a l b K + l Y K + l Y + K + l − ,q = Y − ,l − Y l Y K + l Y + K + l − ,q = Y − ,l − h l (cid:0) Y + K + l − ,l +1 (cid:1) Y + K + l − ,q = Y − ,l − Y + K + l − ,q h l (cid:0) b − q − Y + q − ,l +1 Y + K + l − ,q (cid:1) . Y − ,l Y + K + l,q = Y − , Y + K,q l Y t =1 h t (cid:0) b − q − Y + q − ,t +1 Y + K + t − ,q (cid:1) = a b K Y K Y + K − ,q l Y t =1 h t (cid:0) b − q − Y + q − ,t +1 Y + K + t − ,q (cid:1) = Y K Y + K − ,q l Y t =1 h t (cid:0) b − q − Y + q − ,t +1 Y + K + t − ,q (cid:1) . Lemma 6.
For s ≥
0, we have (cid:0) Y − ,K − (cid:1) s Y K = Y K (cid:0) Y + K − , (cid:1) s . Proof.
Using (1) and (2), we note that Y − ,K − Y K = a Y a · · · Y K − a K − Y K = a ( Y a · · · Y K − a K − Y K a K ) a − K = b − K ( b K Y K b K − Y K − · · · b Y ) b = Y K b K − Y K − · · · b Y b = Y K Y + K − , . The general claim follows by induction.
Lemma 7.
For m ≥ (cid:0) Y − ,K − (cid:1) m (cid:0) Y +2 K − ,K +1 (cid:1) m = 1 + Y K m − X s =0 K − X j =1 (cid:0) Y + K − , (cid:1) s Y + K − ,j +1 A ( j, s )where A ( j, s ) = j − Y t =1 h t (cid:0) b − j Y + j,t +1 Y + K + t − ,j +1 (cid:1)! h ↓↓ j (cid:0) Y + K + j − ,j +1 (cid:1) Y + K + j − ,K +1 (cid:0) Y +2 K − ,K +1 (cid:1) s . (11) Proof.
From Lemma 5, Y − ,j − h ↓ j (cid:0) Y + K + j − ,j +1 (cid:1) = Y − ,j − Y + K + j − ,j +1 h ↓↓ j (cid:0) Y + K + j − ,j +1 (cid:1) = Y K Y + K − ,j +1 j − Y t =1 h t (cid:0) b − j Y + j,t +1 Y + K + t − ,j +1 (cid:1)! h ↓↓ j (cid:0) Y + K + j − ,j +1 (cid:1) , (cid:0) Y − ,K − (cid:1) s Y − ,j − h ↓ j (cid:0) Y + K + j − ,j +1 (cid:1) = (cid:0) Y − ,K − (cid:1) s Y K Y K − ,j +1 j − Y t =1 h t (cid:0) b − j Y + j,t +1 Y + K + t − ,j +1 (cid:1)! h ↓↓ j (cid:0) Y + K + j − ,j +1 (cid:1) = Y K (cid:0) Y + K − , (cid:1) s Y K − ,j +1 j − Y t =1 h t (cid:0) b − j Y + j,t +1 Y + K + t − ,j +1 (cid:1)! h ↓↓ j (cid:0) Y + K + j − ,j +1 (cid:1) Regrouping and substituting the above into the expression from Lemma 3 gives the desiredresult.We now have our final expression for (cid:0) Y − ,K − (cid:1) m (cid:0) Y +2 K − ,K +1 (cid:1) m . So, by (9) and Lemma 7, wefind Y − K (cid:0) Y +2 K − ,K +1 (cid:1) d K = Y (cid:0) Y +2 K − ,K +1 (cid:1) d K − Y − K d X m =1 P ,m (cid:0) Y +2 K − ,K +1 (cid:1) d K − m − Y − K d X m =1 P ,m Y K m − X s =0 K − X j =1 (cid:0) Y + K − , (cid:1) s Y + K − ,j +1 A ( j, d K + s − m ) ! = Y (cid:0) Y +2 K − ,K +1 (cid:1) d K − Y − K d X m =1 P ,m (cid:0) Y +2 K − ,K +1 (cid:1) d K − m − d X m =1 m − X s =0 K − X j =1 P ,m (cid:0) Y + K − , (cid:1) s Y + K − ,j +1 A ( j, d K + s − m ) . Now, by (8), Y K = Y − K d K − X i =0 P K,i (cid:0) Y +2 K − ,K +1 (cid:1) i + Y (cid:0) Y +2 K − ,K +1 (cid:1) d K − Y − K d X m =1 P ,m (cid:0) Y +2 K − ,K +1 (cid:1) d K − m − d X m =1 m − X s =0 K − X j =1 P ,m (cid:0) Y + K − , (cid:1) s Y + K − ,j +1 A ( j, d K + s − m ) . Using the facts that h = h K (so d = d K and P ,i = P K,i for all i ) and h is palindromic,we find d X m =1 P ,m (cid:0) Y +2 K − ,K +1 (cid:1) d K − m = d X m =1 P ,d − m (cid:0) Y +2 K − ,K +1 (cid:1) d K − m = d − X i =0 P ,i (cid:0) Y +2 K − ,K +1 (cid:1) i = d K − X i =0 P K,i (cid:0) Y +2 K − ,K +1 (cid:1) i .
7e then get the desired cancellation of terms, and can write Y K = Y (cid:0) Y +2 K − ,K +1 (cid:1) d K − d X m =1 m − X s =0 K − X j =1 P ,m (cid:0) Y + K − , (cid:1) s Y + K − ,j +1 A ( j, d K + s − m ) . (12)We have finally shown that Y K ∈ A , which proves our main theorem.As a corollary, we see that these recursions have the Laurent phenomenon: each Y n is anoncommutative Laurent polynomial in Y , . . . , Y K . The Maple package
NonComChecker was written to accompany this paper. It is freely avail-able at math.rutgers.edu/~russell2/papers/recursions13.html .The main function,
VerifyPaper(H,x) , inputs a H , a list of K polynomials in a variable x , which are taken to be h ( x ) , h ( x ) , . . . , h K ( x ) = h ( x ). It then simplifies Y K using alist of equations that it generates, including (1), (2), (3), (5), and (6), and verifies that theresult equals (12). The author would like to thank Vladimir Retakh and Doron Zeilberger for illuminatingdiscussions.
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