Noncommutative Double Bruhat Cells and Their Factorizations. Arkady Berenstein and Vladimir Retakh. 1 Introduction
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1 IMRN International Mathematics Research Notices 2005, No 8 Noncommutative Double Bruhat Cells and Their Factorizations Arkady Berenstein and Vladimir Retakh 1 Introduction This paper is a first attempt to generalize results of A Berenstein, S Fomin, and A Zelevinsky on total positivity of matrices over commutative rings to matrices over noncommutative rings The classical theory of total positivity studies matrices whose minors are all nonnegative Motivated by a surprising connection discovered by Lusztig [10, 11] between total positivity of matrices and canonical bases for quantum groups, Berenstein, Fomin, and Zelevinsky, in a series of papers [1, 2, 3, 4], systematically investigated the problem of total positivity from a representation-theoretic point of view In particular, they showed that a natural framework for the study of totally positive matrices is provided by the decomposition of a reductive group G into the disjoint union of double Bruhat cells G u,v = BuB B vb, where B and B are two opposite Borel subgroups in G, and u and v belong to the Weyl group W of G According to [1, 3, 4] there, exist families of birational parametrizations of G u,v, one for each reduced expression of the element u, v in the Coxeter group W W Every such parametrization can be thought of as a system of local coordinates in G u,v Such coordinates are called the factorization parameters associated to the reduced expression of u, v The coordinates are obtained by expressing a generic element x G as an element of the maximal torus H = B B multiplied by the product of elements of various Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 Received 19 July 2004 Revision received 10 October 2004 Communicated by Andrei Zelevinsky
2 478 A Berenstein and V Retakh one-parameter subgroups in G associated with simple roots and their negatives; the reduced expression prescribes the order of factors in this product An explicit formula for these factorization parameters as rational functions on the double Bruhat cell G u,v was given As we said, Berenstein, Fomin, and Zelevinsky came to factorization parameters first, for GL n and then for other classical groups from representation theory For the noncommutative case, our program is to go in the opposite direction; from factorization parameters for GL n to total positivity, canonical bases, and representations This paper is a beginning of the program For G = GL n F, where F is a field of characteristic zero, the explicit formulas for factorization parameters are given through the classical determinant calculus As a first step toward noncommutative representation theory and noncommutative total positivity, we generalize here the results of [3, 4] to G = GL n F where F is a noncommutative skew-field by using the quasideterminantal calculus of matrices over noncommutative rings introduced by I Gelfand and V Retakh [5, 6, 7, 8] The noncommutative point of view has many advantages Here is one Let w o W be the element of the maximal length In the commutative case the factorization parameters for x G u,v, G = GL n, u = id, v = w o are given some times as ratios ab/cd, and other times as a/b, where a, b, c, d are minors of the matrix x see [1] In the noncommutative case, for any u and v = w o, the factorization parameters can be written as f 1 g where f, g are quasiminors of x This paper contains other noncommutative formulas and constructions for GL n that are new, even for the commutative case Our results confirm the Gelfand principle: noncommutative algebra properly understood is simpler than its commutative counterpart The paper is organized as follows In Section 2, we recall some facts about quasideterminants and introduce our main tool positive quasiminors i u,v InSection 3, we study basic factorizations in GL n and its Borel subgroup Section 4 contains examples of such factorizations Section 5 is central for the paper First, we introduce noncommutative SL 2 subgroups in GL n For a generic matrix x, we define special quasiminors i u,vx, where u, v W and show that they satisfy certain Plücker relations We note that i u,vx is always positive for positive real matrices x Section 5 also contains the main result; it gives formulas for factorization coordinates for reduced double Bruhat cells For a matrix x G u,v, these coordinates are written as products of quasiminors i s,ty where the matrix y is the so-called noncommutative twist of x InSection 6, we study relations between quasiminors of x G u,wo and its corresponding twisted matrix y In this case, the quasiminors i, y in the main theorem, can be replaced by quasiminors i, x Studying twisted matrices is an important problem by itself and we present Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011
3 Noncommutative Double Bruhat Cells and Their Factorizations 479 several approaches to computations of such matrices These results are new even in the commutative case 2 Quasideterminants and quasiminors A notion of quasideterminant for matrices over a noncommutative ring was introduced in [6, 7] and developed in [8] It has been effective in many areas see, eg, the survey article [5] Here we recall a few facts about quasideterminants which will be used in this paper 21 Definition of quasideterminants Let A = a ij, i I, j J be a matrix of order n over a ring R Construct the following submatrices of A: submatrix A ij, i I, j J obtained from A by deleting its ith row and jth column; row submatrix r i obtained from ith row of A by deleting the element a ij ; column submatrix c j obtained from jth column of A by deleting the element a ij Definition 21 If n = 1, the quasideterminant A ij equals a ij Ifn>1, the quasideterminant A ij is defined if the submatrix A ij is invertible over the ring R In this case, Inductive properties of quasideterminants are given by the following proposition A ij = a ij r i A ij 1 cj 21 Proposition 22 If all quasideterminants of A ij are defined and invertible, then the quasideterminant A ij is defined and A ij = a ij a iq A ij 1 pq a pj 22 Here the sum is taken over all p I {i}, q J {j} Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 Remark 23 If each a ij is an invertible morphism V j V i in an additive category, then the quasideterminant A pq is also a morphism from the object V q to the object V p and where multiplication corresponds to the composition of morphisms This is a key point of view in determining what one can and cannot multiply together If A is an n n matrix, there exist up to n 2 quasideterminants of A By definition, an r quasiminor of a square matrix A is a quasideterminant of an r r submatrix of A
4 480 A Berenstein and V Retakh Sometimes it is convenient to adopt a more graphic notation for the quasideterminant A pq by boxing the element a pq ForA = a ij, i, j = 1,,n, we write a 11 a 1q a pn A pq = a p1 a pq a pn 23 a n1 a nq a nn Example 24 1 For a matrix A = a ij, i, j = 1, 2, there exist four quasideterminants if the corresponding entries are invertible A 11 = a 11 a 12 a 1 22 a 21, A 12 = a 12 a 11 a 1 21 a 22, A 21 = a 21 a 22 a 1 12 a 11, A 22 = a 22 a 21 a 1 11 a ForamatrixA = a ij, i, j = 1, 2, 3, there exist nine quasideterminants but we will write here only A 11 = a 11 a 12 a22 a 23 a 1 33 a 1a21 32 a 12 a32 a 33 a 1 23 a 1a31 22 a 13 a23 a 22 a 1 32 a 1a21 33 a 13 a33 a 32 a 1 22 a 1a31 23 provided all inverses are defined 25 The quasideterminant is not a generalization of the determinant over a commutative ring, but a generalization of a ratio of two determinants Proposition 25 IfA is a matrix over a commutative ring, then A pq = 1 p+q det A 26 det Apq Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 The quasideterminants of matrix A are related to the elements of the inverse matrix A 1 in the following way Proposition 26 If A is invertible and A 1 = b ij, then b 1 ij = A ji 27 if the element b ij is invertible
5 22 Elementary properties of quasideterminants Noncommutative Double Bruhat Cells and Their Factorizations 481 Here is a list of elementary properties of quasideterminants i The quasideterminant A pq does not depend on the permutation of rows and columns in the matrix A if the pth row and the qth column are not changed ii The multiplication of rows and columns Let the matrix B be constructed from the matrix A by multiplication of its ith row by a scalar λ from the left Then λ A ij if k = i, B kj = 28 A kj if k = i, λ is invertible Let the matrix C be constructed from the matrix A by multiplication of its jth column by a scalar µ from the right Then A ij µ if l = j, C il = 29 A il if l j, µ is invertible iii The addition of rows and columns Let the matrix B be constructed by adding to some row of the matrix A its kth row multiplied by a scalar λ from the left Then A ij = B ij, i = 1,,k 1, k + 1,,n, j= 1,,n 210 Let the matrix C be constructed by adding to some column of the matrix A its lth column multiplied by a scalar λ from the right Then A ij = C ij, i = 1,,n, j= 1,,l 1, l + 1,,n 211 The following homological relations play an important role in the theory Theorem 27 The following relations hold a Row homological relations: Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 A ij A il 1 sj = A il A ij 1 sl s = i 212 b Column homological relations: A kj 1 it A ij = A ij 1 kt A kj r j 213
6 482 A Berenstein and V Retakh 23 Noncommutative Sylvester formula The following noncommutative version of the famous Sylvester identity found in [6, 7] is closely related with the fundamental Heredity principle see [5, 8] Let A = a ij, i, j = 1,,nbe a matrix over a skew-field FLetk<n 1 Suppose the k k submatrix A 0 = a ij, i I 0, j J 0 is invertible For each p/ I 0, q/ J 0, construct the k + 1 k + 1 submatrix A pq = a ij, where i I 0 {p}, j J 0 {q} Set b pq = Apq pq 214 and construct the matrix B = b pq, p/ I 0, q/ J 0 We call the submatrix A 0 a pivot for matrix B Theorem 28 For s/ I 0, t/ J 0, A st = B st 215 The particular case of the theorem when I 0 = J 0 = {2,,n 1} is called the noncommutative Lewis-Carroll identity Example 29 Let n = 3, I 0 = J 0 = {2}Then A 11 equals a 11 a 12 a 21 a 22 a 12 a 13 a 21 a 22 a 22 a 23 a 22 a 23 a 31 a 32 a 32 a Quasi-Plücker coordinates and Gauss LDU-factorization Here we recall some definitions and results from [5, 8] Let A = a pq, p = 1,,k, q = 1,,n, k<n, be a matrix over a skew-field FFix 1 i, j, i 1,,i k 1 n such that i/ I = {i 1,,i k 1 }For1 s k, set Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 a 1i a 1i1 a 1ik 1 q I ija = a ki a ki1 a kik 1 1 si a 1j a 1i1 a 1ik 1 a kj a ki1 a kik 1 sj 217 Proposition 210 i q I ij A does not depend on s ii q I ij ga = qi ij A for any invertible k k matrix g over F
7 coordinate Noncommutative Double Bruhat Cells and Their Factorizations 483 We call q I ij A a left quasi-plücker coordinate of the matrix A In the commutative case, q I ij = p ji/p ii, where p α1 α k is the standard Plücker Similarly, one can introduce right quasi-plücker coordinates Consider a matrix B = b ij, i = 1,,n; j = 1,,k, k<nover a skew-field F Fix1 i, j, i 1,,i k 1 n such that j/ I = i 1,,i k 1 Alsofix1 t k and set b i1 b ik r I b i1 1 b i1 k ijb = b ik 1 1 b ik 1 k it b j1 b jk b i1 1 b i1 k b ik 1 1 b ik 1 k Proposition 211 i r I ij B does not depend on t ii r I ij Bg = ri ij B for any invertible k k matrix g over F 1 jt 218 We call r I ij B a right quasi-plücker coordinate of the matrix B To describe the Gauss decomposition we need the following notations Let A = a ij, i, j = 1,,n Set A k = a ij, i, j = k,,n, B k = a ij, i = 1,,n, j = k,,n, and C k = a ij, i = k,,n, j = 1,,n These are submatrices of sizes n k + 1 n k + 1, n n k + 1, and n k + 1 n, respectively Suppose that the quasideterminants y k = A k kk, k = 1,,n, 219 are defined and invertible Theorem y z αβ A =, 220 x βα 1 0 y n 0 1 Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 where x βα = r 1,,α 1 βα B α, 1 α<β n, z αβ = q 1,,α 1 αβ C α, 1 α<β n 221
8 484 A Berenstein and V Retakh A noncommutative analog of the Bruhat decomposition was given in [5] Example 213 For n = 2, 1 0 a a 1 11 A = a a 21 a A Positive quasiminors Most of the results in this subsection are new For a given matrix x Mat n R and I, J [1, n] = {1,2,,n}, denote by x I,J the submatrix with the I rows and J columns And, if I = J, that is, when x I,J is a square matrix, for any i I, j J, denote by x I,J i,j the quasideterminant of the submatrix x I,J with the marked position i, j We denote by i x the principal i i quasiminor of x Mat n R, that is, i x = x{1,2,,i},{1,2,,i} i,i 223 The following fact is obvious Lemma 214 For any I, J {1,2,,n} such that I = J = k and any i I, j J, there exist permutations u, v of {1,2,,n} such that I = u{1,,k}, J = v{1,,k}, i = uk, j = vk, and for any x Mat n R, k u 1 x v = xi,j i,j 224 where the permutations u and v are identified with their corresponding n n matrices Definition 215 For I, J [1, n], I = J, i I, j J, define the positive quasiminor i,j I,J as follows: i,j I,J x = 1d ii+d j J xi,j i,j, 225 Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 where d i I resp, d j J is the number of those elements of I resp, of J which are greater than i resp, than j The definition is motivated by the fact that for a commutative ring R, one has i,j I,J x = det x I,J det x I,J, 226 where I = I \{i}, J = J \{j}thatis, a positive quasiminor is a positive ratio of minors
9 Noncommutative Double Bruhat Cells and Their Factorizations 485 Let S n be the group of permutations on {1,2,,n} and k [1, n] For any permutations u, v S n, set k u,vx := i,j I,J x = 1d ii+d j J k u 1 x v, 227 where I = u{1,,k}, J = v{1,,k}, i = uk, j = vk Denote by D n = D n R the set of all diagonal n n matrices over R Clearly, positive quasiminors satisfy the relation k u,vhxh = h uk k u,vxh vk 228 for h = diagh 1,,h n,h = diagh 1,,h n D n, and k = 1,2,,n Let σ be an involutive automorphism of Mat n R defined by σx ij = x n+1 i,n+1 j 229 The following fact follows from the elementary properties of quasideterminants see, eg,[5, Proposition 124]Letw o = n, n 1,,1 be the longest permutation in S n Lemma 216 For any u, v S n and x Mat n R, i u,v σx = i wou,w ovx 230 Now we present some less obvious identities for positive quasiminors For each permutation v S n, denote by lv the number of inversions of valsofori = 1,2,,n 1, denote by s i the simple transposition i, i + 1 S n Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 Proposition 217 Let u, v S n and i [1, n 1] be such that lus i = lu + 1 and lvs i = lv + 1Then i us i,vs i = i us i,v i 1 i u,v u,vsi + i+1 i usi,v 1 i+1 u,v = i u,v u,v i+1 i+1 1 us i,v = i usi,v u,v, 1 i+1 us i,v, u,v i+1 i 1 u,vsi = i+1 u,vs i i 1, u,v i 1, u,v i+1 1 i+1 u,vs i u,v = i 1 i u,v u,vsi 231
10 486 A Berenstein and V Retakh Proof Clearly, the fourth and the fifth identities follow from the second and the third Using Lemma 214 and the Gauss factorization, it suffices to take u = v = 1, i = 1 in the first three identities, that is, work with 2 2 matrices Then the first three identities will take, respectively, the following obvious forms: x 22 = x 21 x 1 11 x x 11 x x 21 x 22, x 1 x 11 x 12 x 11 x x 21 x 22 = x 1 11 x 21 x 22, x 11 x 12 x 21 x 22 x 1 12 = x 11 x 12 x 21 x 22 x 1 11 One can prove the next proposition presenting some generalized Plücker relations Proposition 218 Let u, v S n and i [1, n 2]Iflus i s i+1 s i = lu + 3, then 232 us i+1 i+1,v = i+1 us i s i+1,v + i us i+1 s i,v i 1 i+1 usi,v u,v 233 If lvs i s i+1 s i = lv + 3, then i+1 u,vs i+1 = i+1 u,vs i s i+1 + i+1 u,v i u,vsi 1 i u,vsi+1 s i Basic factorizations in GL n F For i, j = 1,2,,n, denote by E ij the n n matrix unit in the intersection of the ith row and the jth column Then we abbreviate E i := E i,i+1 for i = 1,,n 1 The matrix units E 1,,E n 1 satisfy the relations E 2 i = 0 for i = 1,,n 1, E i E j = E j E i 31 Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 if i j 2, and E i E i±1 E i = 0 32 Let i = i 1,,i m be a sequence of indices i k {1,2,,n 1} and x = x ij, i, j = 1,,nbe an upper triangular n n matrix over a skew-field F For such an i and x,
11 we write the formal factorization Noncommutative Double Bruhat Cells and Their Factorizations 487 x = 1 + t 1 E i1 1 + t2 E i2 1 + tm E im, 33 where all t k belong to the skew-field F Let k ij be the position of the ith occurrence of the index j i in the sequence i = 1,,n 1; 1,,n 2; ; 1, 2; 1Thatis, i + 1 k ij = ni 1 + j 34 2 for 1 i<j n The following fact is obvious Lemma 31 Let i = 1,,n 1; 1,,n 2; ; 1, 2; 1 Set temporarily t ij := t kij for 1 i<j n where t k is as in factorization 33 Then the matrix entries of the product x satisfy 1 i n k n 1 x i,i+k = 1 i 1 i 2 i k n+1 i k t i1,i 1 +it i2,i 2 +i+1t i3,i 3 +i+2 t ik,i k +i+k 1 35 Remark 32 In particular, after the specialization t kij := y j for 1 i<j n in 35, for some elements y 2,,y n, we obtain x i,i+k = i<j 1 <j 2 < <j k n y j1 y j2 y jk 36 That is, each matrix entry of a so-specialized matrix x is an elementary symmetric function in y 2,,y n in the sense of [8] Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 Proposition 33 System 35 has a unique solution of the form t ij = x j i,n i+1 I,J x I1,J 1 j i+1,n i+1 37 for 1 i<j n, provided the quasiminors are defined Here I = {j i + 1,,j}, I 1 = {j i + 2,,j}, and J = {n i + 1,,n}
12 488 A Berenstein and V Retakh Proof First of all, we have the relations x i,n = t 1,i+1 t 1,i+2 t 1,n 38 for i = 1,,n 1 Therefore, t 1,i+1 = x i,n x 1 i+1,n 39 for all j = 2,,n, which verifies 37 and We define a sequence x 0,x 1,,x n 1 of matrices inductively by setting x 0 = I x m = I + t n m,n+1 m E 12 I + tn m,n+2 m E 23 I + tn m,n E m,m+1 x m 1 for m = 1,2,,n 1 Clearly, x n 1 = x Lemma 34 For all i j m + 1 n, 310 x m ij = xi, j, m ij, 311 where xi, j, m isthe n m n m submatrix of x withthe rows {i, i+1,, i+n m 1} and the columns {j; m + 2, m + 3,,n} Proof We proceed by induction on n m By definition of x m, we have a recursion for the matrix entries of x m : x m i,j for 1 i j m + 1 = t n m,i+n m x m i+1,j + xm 1 ij 312 Taking j = m + 1, we obtain Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 Therefore, t n m,i+n m = x m xm 1 i,m+1 i+1,m x m 1 ij = x m i,j x m xm 1x m i,m+1 i+1,m+1 i+1,j = x m i,j x m i+1,j x m i,m+1 x m i+1,m+1 314
13 by the above, Noncommutative Double Bruhat Cells and Their Factorizations 489 Furthermore, we use the inductive hypotheses precisely in the form 311Then, x m 1 ij = xi, j, m ij xi + 1, j, m i+1,j xi, m + 1, m i,m+1 xi + 1, m + 1, m i+1,m Using the Sylvester formula Theorem 28 with A = xi, j, m 1 and A 0 being a submatrix of A with the rows indexed by {i + 1,,i+ n m 1} and the columns indexed by {m + 2, m + 3,,n}, we obtain x m 1 ij = xi, j, m i,j xi + 1, j, m i+1,j This finishes the induction The lemma is proved xi, m + 1, m i,m+1 xi + 1, m + 1, m i+1,m+1 = x m 1 ij ij 316 Finally, using 311, 313, and the fact that x {m+2 i,,m+1},{n m,,n} = xi, m + 1, m for m = 0,1,,n 1, we obtain 37 The proposition is proved Another natural factorization of matrices is given by the following theorem Amatrixx = x ij, i, j = 1,,n, over a skew-field F is called generic if the sequence of rational functions t m,k = t m,k x, 1 m k n 1, is defined x 1,k m+1 x 1k t m,k = x m,k m+1 x mk Clearly, in terms of positive quasiminors, one has 1 x 1,k m+2 x 1,k+1, 317 x m,k m+2 x m,k+1 t m,k = m,k 1 m,k+1 {1,,m},{k m+1,,k} {1,,m},{k m+2,,k+1} 318 Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 Then define a sequence of matrices xm, k = x m,k ij, 1 m k n 1, by the inductive formula x1, n 1 = x 1 t 1,n 1 E n 1, xm, k = xm, k t m,k E k, xm + 1, n 1 = xm, m 1 t m+1,n 1 E n 1 319
14 490 A Berenstein and V Retakh In other words, xm, k 1 + ti,j E j = x, 320 i,j m,k where the order on all pairs m, k, 1 m k n 1, is defined by i, j m, k if and only if either i<mor i = m, j>k Theorem 35 a For a generic matrix x = x ij, i, j = 1,,n, over a skew-field F, x m,k ij = for all i, j such that i, j 1 m, k ie, for i<j, i<mand for i = m, j>k In particular, xn 1, n 1 is lower triangular b The entries x m,k ij are given by the following formulas For i m, 2 j k, x m,k ij = for i>m, j>k, x m,k ij = x 1,j m+1 x 1j, 322 x m 1,j m+1 x m 1,j x i,j m+1 x i,j x 1,j m x 1,j, 323 x m,j m x m,j x i,j m x i,j and x m,k ij = x ij otherwise Proof It is enough to show that matrices xm, k satisfy the following conditions: i x m,k ij = 0 for i<j, i<mand for i = m, j>k, ii x1, n 1 = x1 + E n 1 t 1,n 1, iii xm, k = xm, k E k t m,k, iv xm + 1, n 1 = xm, m1 + E n 1 t m+1,n 1 We proceed by induction over a totally ordered set of indices 1, n 1,,1, 1, 2, n 1,,2, 2,,n 1, n 1 It is easy to check that the entries of matrix x1, n 1 satisfy conditions i iv Suppose that these conditions are satisfied for matrix xm, l We consider then two cases: l>mand l = m Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011
15 Noncommutative Double Bruhat Cells and Their Factorizations 491 If l>m, then l = k + 1 for k m Define matrix xm, k by formula iii Evidently, the corresponding entries of matrices xm, k and xm, k + 1 coincide except for the entries with indices i, k for i m which are given by the formula x m,k ik = x m,k+1 ik t m,k + x m,k+1 ik For i m, the product x m,k+1 ik t m,k equals x 1,k m+1 x 1,j x 1,k m+1 x 1k x m 1,k m+1 x m 1,j x x i,k m+1 x i,j m,k m+1 x mk x 1,k m+2 x 1,k+1 x m,k m+2 x m,k According to the homological relations for quasideterminants, the last expression can be written as x 1,k m+1 x 1,j x 1,k m+1 x 1k x m 1,k m+1 x m 1,j x x i,k m+1 x i,j m,k m+1 x mk x 1,k m+2 x 1,k+1 x m,k m+2 x m,k Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 It follows that the element x m,k ik x m,k ij = = x m,k+1 ik tm, k + x m,k+1 ik+1 is zero for i = mifi>m, x 1,k m+1 x 1,k x m,k m+1 x m,k+1 x i,k m+1 x i,k+1
16 492 A Berenstein and V Retakh This follows from the Sylvester identity applied to the corresponding matrix with the pivot equal to x 1,k m+2 x 1,k 328 x m 1,k m+2 x m 1,k This shows that the entries of matrix xm, k satisfy part b of the theorem If l = m, one can check in a similar way that the entries of matrix xm + 1, n 1 satisfy part b of the theorem The theorem is proved Remark 36 It follows from the proof that matrices xm, k and elements t m,k are uniquely defined Example 37 Let n = 3Thent 1,2 = x 1 12 x 13, t 1,1 = x 1 11 x 12, x 11 x 12 t 2,2 = x 21 x 22 1 x 12 x 13 x 22 x 23, x 11 x 12 0 x 12 x 13 x 21 x 22 x1, 2 = x 22 x 23, x 12 x 13 x 31 x 32 x 32 x 33 x x 11 x 12 x 12 x 13 x 21 x1, 1 = x 21 x 22 x 22 x 23, x 11 x 12 x 12 x 13 x 31 x 31 x 32 x 32 x 33 x x 11 x 12 x 21 0 x2, 2 = x 21 x 22 x 11 x 12 x 31 x 33 x 31 x Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011
17 Noncommutative Double Bruhat Cells and Their Factorizations Examples 41 A factorization in the Borel subgroup of GL 3 F We write the formal factorization x 11 x 12 x 13 x t 12 + t 23 t 12 t 13 0 x 22 x 23 = 0 x t x x x t t 23 0 = 0 x t x assuming that all x ij,t ij are elements of a skew-field F Then we can express t ij as follows: t 13 = x 1 22 x 23, t 12 = x 1 11 x 13x 1 23 x 22, t 23 = x 1 11 x 12 x 1 11 x 13 x 1 23 x 22 = x x 12 x 13 x 22 x Remark 41 The above factorization exists and therefore is unique if and only if each of x 11, x 22, x 33, and x 23 is invertible 42 A factorization in GL 3 F We write the formal factorization over a skew-field F: x 11 x 12 x 13 x = x 21 x 22 x 23 = hx 2 t1 x 1 t2 x 2 t3 x2 t4 x1 t5 x2 t6, 43 x 31 x 32 x 33 where h t h = 0 h 2 0, x 1t = 0 1 0, x 2t = 0 1 t, 0 0 h t x 1 t = 1 t 0, x 2t = 0 t t 44 Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011
18 494 A Berenstein and V Retakh Then we can express h i and t k as follows: x 21 x x 11 x 12 x h 3 = x 31, h 2 = x 31 x 32, h 1 = x 21 x 22 x 23, x 31 x 32 x 33 t 6 = x 1 12 x 13, t 5 = x 1 11 x 12, t 4 = x 22 x 21 x 1 11 x 1 12 x23 x 22 x 1 12 x x 11 x = x 21 x 22 x 11 x 12 x 13 t 1 = x 1 x 21 x x 31 x 32, t 2 = x 1 11 x 21 x 22 x 23, x 31 x 32 x 33 1 x 11 x x 11 x 12 x t 3 = x 21 x 22 x 21 x 22 x 23 x 31 x 32 x 33 In fact, if we define a sequence of matrices 1 x 12 x 13 x 22 x 23, 45 x 5 = x x 2 t6 1, x 4 = x 5 x 1 t5 1, x 3 = x 4 x 1 t4 1, 46 then x k 1 will have exactly one more zero entry in the upper part than x k : x 11 x 12 0 x x 5 = x 21 x 22 x 23, x4 = x 21 x 22 x 23, x 31 x 32 x 33 x 31 x 32 x 33 x x 3 = x 21 x 22 0 x 31 x 32 x Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 This determines t 6, t 5, and t 4 The rest of the parameters h 1, h 2, h 3, t 1, t 2, t 3 are obtained from the equation h 1 t x 3 = hx 2 t1 x 1 t2 x 2 t3 = h 2 t 1 1 h 2 t 1 1 t 2t h 3 h 3 t1 + t 2 t 1 3 h 3 t 1 t 3
19 Noncommutative Double Bruhat Cells and Their Factorizations A factorization in the unipotent subgroup of GL 4 F We write the formal factorization 1 x 12 x 13 x 14 1 t 12 + t 23 + t 34 t 12 t 13 + t 12 t 24 + t 23 t 24 t 12 t 13 t x 23 x x 34 = 0 1 t 13 + t 24 t 13 t t t = t t t t t assuming that all x ij,t ij are elements of a skew-field F Then we can express t k as follows: t 14 = x 34, t 13 = x 24 x 1 34, t 12 = x 14 x 1 24, t 24 = x 23 x 24 x 1 34 = x 23 x 24 1 x 34, t 23 = x 13 x 14 x 1 24 x 23 x23 x 24 x x 13 x 14 x 23 x =, x 23 x 24 1 x 34 x 12 x 13 x 14 t 34 =x 12 x 13 x23 x 24 x x14 x 1 34 x23 x 24 x = 1 x 23 x x Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 Remark 42 The above factorization exists and therefore is unique if and only if x 24, x 34, and x 23 x 24 are invertible in F 1 x 34
20 496 A Berenstein and V Retakh 5 Double Bruhat cells in GL n F and their factorizations 51 Structure of GL n F Throughout this and the next section, we denote G := GL n F and will use the abbreviation for a, b Z {a, a + 1,,b} if a b, [a, b] = 51 otherwise Let U resp, U be the upper resp, lower unitriangular subgroup of GFori [1, r], we define the elementary unitriangular matrices x i t and y i t by x i t = I + te i, y i t = I + tf i 52 for i [1, n 1], where E i = E i,i+1, F i = E i+1,i are the corresponding matrix units in the notation of Section 3 Let H denote the subgroup of all diagonal matrices in G LetB resp, B be the subgroup of all upper resp, lower triangular matrices in G Clearly, B = HU, B = HU, and H = B B We denote by G 0 = B U the subset of elements x G that have a Gaussian LDUdecomposition This unique decomposition will be written as x = [x] [x] + where [x] B, but not necessarily in U For any x in the Gauss cell G 0 = B U, denote by [x] 0 the diagonal component of the Gauss LDU-factorization In particular, [x] 0 = [[x] ] 0 for any x G 0 For each i [1, n 1], let ϕ i : GL 2 F G denote the embedding corresponding to the 2 2 block at the intersection of the ith and i + 1st rows and the ith and i + 1st columns Thus, we have 1 t 1 0 x i t = ϕ i, y i t = ϕ i t 1 Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 We also set t 0 t 1 0 h i t = ϕ i H, x i t = ϕ i 0 t 1 1 t 54 for any i and any t F By definition, x i t = y i th i t 1 = h i t 1 y i t 1 55
21 Noncommutative Double Bruhat Cells and Their Factorizations 497 More generally, it is easy to see that for each i [1, n 1] and any diagonal matrix h = diagh 1,,h n H, one has Hence hx i th 1 = x i hi th 1 i+1, h 1 y i th = y i h 1 i+1 th i 56 h j sx i t = x i s ε ji ts ε ij hj s, y i th j s = h j sy i s ε ij ts ε ji 57 for any i, j [1, n 1], where ε ij = δ ij δ i,j 1 Lemma 51 i For each i [1, n 1], x i sx i t = x i s 1 ts + t 1 x i s + t ii For each i [1, n 2], x i sx i+1 t = x i+1 stx i s iii For each i [2, n 1], x i sx i 1 t = x i 1 tsx i s iv For any i, j [1, n 1] such that i j >1, x i sx j t = x j tx i s 58 Proof Part i follows from the obvious identity s t s 1 s 1 t 1 s 1 ts + t 1 s + t 1 0 = = 1 s s+ t s+ t for s, t F Part ii follows from for s, t F 59 s s s s t = 1 s st = 0 1 st 1 s Part iii follows from t 0 1 t 0 1 ts s = 0 s 1 0 = s s s s Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 for s, t F Part iv is obvious
22 498 A Berenstein and V Retakh The symmetric group S n of G is naturally embedded into G via 0 1 i, i + 1 ϕ i for i [1, n 1] We also define a representative s i of the transposition i, i + 1 by 0 1 s i = ϕ i The elements s i satisfy the braid relations in W; thus the representative w can be unambiguously defined for any w W by requiring that uv = u v whenever luv = lu + lv 52 Bruhat cells and double Bruhat cells The group G has two Bruhat decompositions, with respect to opposite Borel subgroups B and B : G = BuB = B vb 514 u S n v S n Now define the Schubert cell Uw := wu w 1 U for w S n Then the following obvious fact demonstrates that the Bruhat cells BuB and B vb behave similarly to their commutative counterparts Lemma 52 a For each u S n, BuB = UuuB = BuUu, UuU = UuuU = UuU u b For each v S n, B vb = B Uvv = B Uvv 1 1 = vu v 1 B = v 1 1 U v 1 B 516 Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 Definition 53 For any permutations u, v S n, define the double Bruhat cell G u,v by G u,v = BuB B vb In this section, we will concentrate on the following subset L u,v G u,v, which we call a reduced double Bruhat cell: L u,v = UuU B vb 517
23 Noncommutative Double Bruhat Cells and Their Factorizations 499 Remark 54 In the commutative case, the reduced double Bruhat cells are symplectic leafs of the Poisson-Lie structure on GL n C see, eg,[9] These cells also emerge in the study of total positivity see [3, Section 4] on GL n The equations defining L u,v inside G u,v are as follows Proposition 55 An element x G u,v belongs to L u,v if and only if [u 1 x] 0 = 1, or equivalently if i u,ex = 1 for each i [1, n] The maximal torus H acts freely on G u,v by left or right translations, and L u,v is a section of this action Thus, L u,v is naturally identified with G u,v /H, and all properties of G u,v can be translated in a straightforward way into the corresponding properties of L u,v A double reduced word forapairu, v S n is a reduced word for an element u, v of the group S n S n To avoid confusion, we will use the indices 1,, r for the simple reflections in the first copy of W, and 1,,rfor the second copy A double reduced word for u, v is simply a shuffle of a reduced word i for u written in the alphabet [ 1, r] we will denote such a word by i and a reduced word i for v written in the alphabet [1, r] We denote the set of double reduced words for u, v by Ru, v For any sequence i = i 1,,i m of indices from the alphabet [1, r] [ 1, r], we define the product map x i : F m G by x i t1,,t m = xi1 t1 xim tm Factorization problem for reduced double Bruhat cells In this section, we address the following factorization problem for L u,v : for any double reduced word i Ru, v, find explicit formulas for the inverse birational isomorphism x 1 i between L u,v and F m, thus expressing the factorization parameters t k in terms of the product x = x i t 1,,t m L u,v Definition 56 Let x x ι be the involutive antiautomorphism of G given by Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 x ι = J n x 1 J n 519 for any x G, where J n = diag 1, 1, 1,, 1 n We will refer to the antiautomorphism x x ι as the positive inverse in G Itis easy to see that a ι = a 1 a H, x i t ι = x i t, y i t ι = y i t 520
24 500 A Berenstein and V Retakh The following fact is a direct noncommutative analogue of [4, Theorem 16 ] Lemma 57 For any u, v S n, BuB ι = Bu 1 B, UuU ι = Uu 1 U, B vb ι = B v 1 B 521 In particular, G u,v ι = G u 1,v 1 Definition 58 For any u, v W, the twist map ψ u,v : L u,v G is defined by ψ u,v x = [ xv 1] ι x ι 1[ u 1 x ] ι The following result is a noncommutative analogue of [3, Theorem 47] Theorem 59 The twist map ψ u,v is an isomorphism between L u,v and L v,u The inverse isomorphism is ψ v,u Proof The proof essentially follows the pattern of the commutative case from [3, 4] We need the following obvious fact Lemma 510 The twist map ψ u,v satisfies ψ u,v x = [ vx ι 1] + v[ u 1 x] + ι = [ xv 1 ] ιu 1 1[ u 1 x ι 1] 523 The restriction of ψ u,v to L u,v B U isamapl u,v B U L u,v B U given by the formula ψ u,v x x + = [ x+ v 1] ι [ u 1 ] ι x 524 In particular, the twist map ψ u,e : L u,e L e,u is given by + Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 ψ u,e x = [ u 1 x ] + ι 525 And ψ e,v : L e,v L v,e is given by ψ e,v x = [ xv 1] ι 526
25 Noncommutative Double Bruhat Cells and Their Factorizations 501 The formula 523 guarantees that ψ u,v L u,v UvU B ub = L v,u, that is, ψ u,v is a well-defined map L u,v L v,u Finally, we prove that ψ v,u is the inverse of ψ u,v, that is, ψ v,u ψ u,v = id Given x L u,v, denote y = ψ u,v x By definition 522, we have y = [ xv 1] ι x ι 1[ u 1 x ] + ι 527 Or, equivalently, y ι 1 [ ] = xv 1 1x [ u 1 1, x] + x = [ xv 1] y ι 1[ u 1 x ] Since ψ v,u y = [ yu 1] ι y ι 1[ v 1 y ] ι, in order to prove that ψ v,u y = x, it suffices to show that [ ] yu 1 ι [ ] = xv 1, [ v 1 y ] ι [ + = u 1 x ], or, equivalently, [ ] yu 1 = [ xv 1] ι, [ v 1 y] + = [ u 1 x ] ι We prove the first identity 531 Denote temporarily z = [xv 1 ] ι Then523 implies that Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 yu 1 = z u 1 1[ u 1 x ι 1] u According to Lemma 57, for any x UuU, we have x ι 1 Uu 1 1 U, and furthermore, by Lemma 52a, u 1 x ι u 1 Uuu 1 1 U U U and [u 1 x ι 1 ] u 1 Uuu 1 1 Hence, u 1 1 [u 1 x ι 1 ] u 1 U Therefore, [ ] yu 1 = [ z u 1[ 1 u 1 x ι 1] u 1] = [z] = z 533
26 502 A Berenstein and V Retakh This proves the first identity in 531 Now we prove the second identity in 531 Again denote temporarily t = [u 1 x] + ι Then523 implies that v 1 y = v 1[ vx ι v t 534 1]+ According to Lemma 52b, for any x B vb, one has xv 1 B Uv and [vx ι 1 ] + Uv Hence, v 1 [vx ι 1 ] + v v 1 Uvv U Therefore, [ v 1 y] + = [ v 1[ vx ι 1] v t] = [t] = t 535 This proves the second identity in 531 Theorem 59 is proved Nowwefixapairu, v S n S n and a double reduced word i = i 1,,i m Ru, v Recall that i is a shuffle of a reduced word for u written in the alphabet [ 1, r] and a reduced word for v written in the alphabet [1, r] In particular, the length m of i is equal to lu + lv We will use the convention that s i = 1 for each i [1, n 1]Fork [1, m], denote u k = s im s im 1 s ik, u >k = s im s im 1 s ik+1, 536 v k = s i1 s i2 s ik, v <k = s i1 s i2 s ik For example, if i = 2, 1, 3, 3, 2, 1, 2, 1, 1, then, say, u 7 = s 1 s 2 and v <7 = s 1 s 3 s 2 Now we are ready to state our solution to the factorization problem Theorem 511 Let i = i 1,,i m be a double reduced word for u, v, and suppose an element x L u,v can be factored as x = x i1 t 1 x im t m, with all t k nonzero elements of F Then the factorization parameters t k are determined by the following formula: Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 i v t k = <k,u >k y 1 i v <k,u k y = v i+1 <k,u k y 1 v i+1 <k,u >k y if i k <0, i v k,u >k y 1 v i+1 <k,u >k y = i v <k,u >k y 1 v i+1 k,u >k y if i k >0, 538 where y = ψ u,v x and i = i k
27 Noncommutative Double Bruhat Cells and Their Factorizations 503 Proof First, we list some important properties of positive quasiminors Recall that in Section 25, for i [1, n], we defined the principal quasiminor i by i x = x[1,i],[1,i] i,i 539 for any x G, where x [1,i],[1,i] denotes the principal i i submatrix of x In particular, 1 x = x 11 and n x = x n,n The following fact is obvious Lemma 512 The principal quasiminors are invariant under the left multiplication by U and the right multiplication by U, that is, i x xx + = i x 540 for any x + U, x U, x G in particular, i x = i [x] 0 = [x] 0 ii Furthermore, for any u, v S n, Also i u,vx = i u 1 xv 541 i u,v x ι = n+1 i w ovw o,w ouw o x We will prove 538 by induction on lu + lv The base of the induction with u = v = e is obvious We will consider the following four cases Case 1 u = e, v = e and i is separated, that is, i 1,, i l [1, n 1] and i l+1,,i m [1, n 1] for some l, or, equivalently, u = s i1 s il and v = s il+1 s im Case 2 u e, v e and i is not separated Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 Case 3 u = e, v e Case 4 u e, v = e Consider Case 1 first Denote x := x i1 t1 xil tl, x+ := x il+1 tl+1 xim tm 543
28 504 A Berenstein and V Retakh Clearly, x L u,e, x + L e,v, and x = x x + L u,v Furthermore, the inductive hypothesis 538 for x says that t k = i e,u >k y+ 1 i v<k,u k y+ = i+1 e,u k y+ 1 i+1 e,u >k y+ 544 for k [1, l], where y + = ψ u,e x, i = i k And the inductive hypothesis 538 for x + says that t k = i 1 i+1 v k,e y v <k,e y = i 1 i+1 v<k,e y v k,e y 545 for k [l + 1, m], where y = ψ e,v x +, i = i k According to 524, 525, and 526, ψ u,v x = [ x + v 1] ι [ u 1 x ] + ι = y y Note also that j e,wy + = j e,wy y + and j w,ey = j w,ey y + for any w S n and j [1, n] Finally, taking into the account that v k = v <k = e for each k l, and u k = u >k = e for each k>l, we obtain 538 for x = x x + This finishes Case 1 Now consider Case 2 Wesaythatgiveni, apairi l,i l+1 is an inversion if i l >0 and i l+1 <0 Clearly, i has no inversions if and only if i is separated Here we will proceed by induction on the number of inversions The base of the induction is the already considered Case 1 no inversions Assume that i has an inversion i l,i l+1 = i, j, where i, j [1, n 1] Leti be obtained from i by switching i l and i l+1, that is, i has one inversion less than i According to the inductive hypothesis,538 holds for the factorization relative to i x = x i1 t1 xil 1 tl 1 x j tl xi tl+1 xil+2 tl+2 xim tm 547 Note that, according to Lemma 51, x j tl xi tl+1 = xi t l x j t l+1, 548 Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 where tl+1,t l t l,t l+1 tl t l+1,t l = tl+1 t l,t l t 1 1,tl l t l+1 tl + t l+1 + t l+1 if i j >1, if i j = 1, if i j = 1, if i = j 549
29 Noncommutative Double Bruhat Cells and Their Factorizations 505 We have to prove that each of the parameters t 1,,t l 1,t l,t l+1,t l+2,,t m in the factorization relative to i x = x i1 t1 xil 1 tl 1 xi t l x j t l+1 xil+2 tl+2 xim tm 550 is given by 538 for i Obviously, if k l, l + 1, then v i <k = vi <k, vi k = vi k, ui >k = ui >k, ui k = ui k Therefore, each t k, k = l, l + 1 in the latter decomposition is given by 538 for i It remains to prove that t l and t l+1 are both given by 538 for i Denote temporarily u = u >l, v = v <l so that taking into account that i l = i l+1 j, i l+1 = i l = i we have v i l = vi <l+1 = v, v i l+1 = v s i, u i l+1 = ui >l+1 = u, u i l = u s j Therefore,538 for i with k = l and k = l + 1 becomes with the convention y = ψ u,v x, y = v 1 yu t l = j e,ey 1 j e,s j y = j+1 e,s j y 1 j+1 e,e y, t l+1 = i s i,ey 1 i+1 e,e y = i e,ey 1 i+1 s i,ey 551 Taking into account that v i <l = v, v i l = vi <l+1 = vi l+1 = v s i, u i >l+1 = u, u i l+1 = u i >l+1 = ui l = u s j, we only have to prove that t l = i s i,s j y 1 i+1 e,s j y = i e,s j y 1 i+1 s i,s j y, 552 t l+1 = j s i,ey 1 j s i,s j y = j+1 s i,s j y 1 j+1 s i,ey 553 Consider the following four subcases 1 i j > 1 Then clearly, i s i,s j y = i s i,ey, i+1 e,s j y = i+1 e,e y, and j s i,s j y = j e,s j y, j s i,ey = j e,ey Finally, by 549, t l = t lt l+1 and t l+1 = t l All these immediately imply 552 and j = i 1 According to 549, t l = i e,s j y 1 i e,ey, t l+1 = i e,ey 1 i+1 s i,ey, t l = t lt l+1 = i e,s j y 1 i+1 s i,ey = i e,s j y 1 i+1 s i,s j y, 554 Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 which proves 552 Similarly, we obtain t l+1 = t l = j e,ey 1 j e,s j y = j s i,ey 1 j s i,s j y, 555 which proves 553
30 506 A Berenstein and V Retakh 3 j = i + 1 According to 549, t l = t l+1t l = i s i,ey 1 i+1 e,e y j e,ey 1 j e,s j y = i s i,ey 1 j e,s j y, 556 which proves 552 because i s i,ey = i s i,s i+1 y Similarly, we obtain t l+1 = t l = j+1 e,s j y 1 j+1 e,e y = j+1 s i,s j y 1 j+1 s i,ey, 557 which proves i = j According to 549, t l+1 = t l + t l+1 = i e,ey 1 i e,s i y + i s i,ey 1 e,e i+1 y = i s i,ey 1 i s i,ey i e,ey 1 i e,s i y + e,e i+1 y = i s i,ey 1 i s i,s i y by 231 This proves 553 Furthermore, according to 549, t lt l+1 = t 1 l t l+1 = i e,s i y 1 i e,ey i e,ey 1 s i+1 i,ey = i e,s i y 1 i+1 s i,ey Therefore, using the already proved formula 553, we obtain t l = i e,s i y 1 s i+1 i,ey t l 1 = i e,si y 1 s i+1 i,ey s i+1 i,ey 1 s i+1 i,s i y = i e,s i y 1 i+1 s i,s i y, which proves 552 This finishes Case Now we consider Case 3: i = i 1,,i m, where all i k >0, that is, i is a reduced word for v And let i = i m so that v = v s i and lv = lv + 1Let Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 x = x i t1 xim tm, x = x i t1 xim 1 tm 1 x i t 1 m 561 It is easy to see that xs i x i t 1 m = x 562
31 Noncommutative Double Bruhat Cells and Their Factorizations 507 Indeed, this follows from x i t 1 = x i ts i x i t 1, 563 which in turn follows from the obvious identity 1 t 0 1 t 1 t 0 1 t 1 = = t Note that x is factored along the reduced word i = i 1,,i m 1 ; i for s i,v Therefore, we can use the already proved Case 2 for the i -factorization of x Formula 538 for the factorization parameters t 1,,t m 1,t 1 m of x takes the form t k = i k v k,s i y 1 i k+1 v <k,s i y = i k v<k,s i y 1 i k+1 v k,s i y 565 for k [1, m 1], and t 1 m = i v,ey 1 i v,s i y = i+1 v,s i y 1 i+1 v,e y, 566 where y = ψ s i,v x Clearly, in order to finish Case 3, that is, to verify formula 538 for the i-factorization parameters t 1,,t m of x L e,v, it will suffice to prove that for any w S n, j [1, n], one has j w,s i y = j w,ey, 567 where y = ψ e,v xnotethat j w,s i y = j w,ey s i Thus, it will suffice to prove [ y s i ] = y 568 Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 Taking into account that xs i x i t 1 = x, 569 all we need to prove is the following fact
32 508 A Berenstein and V Retakh Lemma 513 Let v = v s i for some i such that lv = lv + 1 Then for any x L e,v and any t F, ψ e,v x x i t = [ ψ s i,v x x i t 1 s i ] 570 Proof Indeed, by Lemma 510, ψ e,v x x i t = [ x x i tv 1] ι 571 Using 563, we obtain x x i tv 1 = x x i ts i v 1 = x x i t 1 x i t 1 v 1 = x x i t 1 v 1 u for some u + U Therefore, [ x x i tv 1] = [ x x i t 1 v 1 ] u + = [ x x i t 1 v 1 ] 573 Summarizing, we obtain ψ e,v x x i t = [ x x i t 1 v 1 ] ι 574 On the other hand, by the second identity of 523 we have for any x L s i,v, [ ψ s i,v x ] ιsi 1 s [ i s i x ι 1] ] = [[ x v 1 ] + s i = [ x v 1 ] ι 575 because z = [s i x ι 1 ] + U ϕ i GL 2 and, therefore, s 1 i zs i B Thus, taking x = x x i t 1, we obtain Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 [ ψ s i,v x x i t 1 ] s i = [ x x i t 1 v ] 1 ι = ψ e,v x x i t 576 The lemma is proved This finishes Case 3 Case 4 is almost identical to Case 3 Therefore, Theorem 511 is proved
33 Remark 514 The commutative version of 538 is Noncommutative Double Bruhat Cells and Their Factorizations 509 v<k ω i,u k ω i y v<k ω i,u >k ω i y t k = v<k ω i 1,u k ω i 1 y v<k ω i+1,u k ω i+1 y v<k ω i,u k ω i y v k ω i,u >k ω i y if i k <0, if i k > Factorizations of G u,v In this section, we extend the result of Theorem 511 to factorizations in G u,v In order to do so, we first have to extend the twist ψ u,v to an isomorphism G u,v Gv,u which we will denote in the same way by ψ u,v hx = hψ u,v x 578 for any h H and any x L u,v This means that the twist ψ u,v is a left H-equivariant map G u,v G v,u Recall that for any g in the Gauss cell G 0 = B U, we denote by [g] 0 the diagonal component of the Gauss factorization Lemma 515 The general twist ψ u,v : G u,v Gv,u is given by ψ u,v g = u [ u 1 g ] [ 0 ] gv 1 ι g ι 1[ u 1 g ] ι, for any g G u,v Other formulas for ψ u,v are ψ u,v g = u [ u 1 g ] [ 0 vg ι 1] v[ u 1 g ] ι, + + ψ u,v g = u [ u 1 g ] [ ] 0 gv 1 ιu 1 1[ u 1 g ι 580 1] Also ψ u,v is symmetric: ψ u,v 1 = ψ v,u In particular, for u = v, the twist ψ v,v is an involution on G v,v Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 Proof Clearly, for any h H and x UuU, we have [ u 1 hx ] 0 = [ u 1 hu u 1 hx ] 0 = u 1 hu [u 1 hx ] 0 = u 1 hu = u 1 h 581 Therefore, taking g = hx, where h H and x L u,v, and taking into account 522 and 523, we obtain the desirable formulas
34 510 A Berenstein and V Retakh Theorem 511 admits the following obvious generalization Theorem 516 Let i = i 1,,i m be a double reduced word for u, v, and suppose an element x G u,v can be factored as x = hx i1 t 1 x im t m, with all t k nonzero elements of F, and h = diagh 1,,h n H Then the factorization parameters h 1,,h n, t 1,,t m are determined by formula 538 and the following formula: h i = u 1 i u,e x for i [1, n] 582 The following two special cases of Theorem 516 will be of particular importance: u, v = e, w 0 and u, v = w 0,e, where w 0 is the longest element in S n In these cases, Definition 58 and Theorem 59 can be simplified as follows Formula 538 now takes the following form Corollary 517 Let i = i 1,,i m be a reduced word for w S n, and let t 1,,t m be nonzero elements of F i If x = x i1 t 1 x im t m, then the factorization parameters h 1,,h n and t 1,,t m are given by h i = i e,ex = x ii 583 for i [1, n], and t k = i s i1 s ik,ey 1 i+1 s i1 s ik 1,ey = i s i1 s ik 1,ey 1 i+1 s i1 s ik,ey, where y = ψ e,w x is given by 526, and i = i k 584 ii If x = hx i1 t 1 x im t m, then the factorization parameters h 1,,h n and t 1,,t m are given by h i = w 1 i w,e x 585 Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 for i [1, n], and t k = i e,s im s ik+1 y 1 i e,s im s ik y = i+1 e,s im s ik y 1 i+1 e,s im s ik+1 y, 586 where y = ψ w,e x is given by 525
35 Noncommutative Double Bruhat Cells and Their Factorizations Other factorizations in GL n F and the maximal twist ψ wo,wo In this section, we will provide some explicit factorizations in G u,wo and G wo,v We consider a factorization of x G u,wo of the form x = x x n 1 x n 2 x 1, 61 where x G u,e and x m L e,sms m+1 s n 1 is given by x m = x m tm,m xm+1 tm,m+1 xn 1 tm,n 1 for m [1, n 1] Lemma 61 In the notation of 61, 62 t m,k = m,k [1,m],[k m+1,k] x 1 m,k+1 [1,m],[k m+2,k+1]x 63 for all 1 m k n 1 Proof The proof follows immediately from Theorem 35 Lemma 62 In the notation of 61, t ij = i,j [1,i] [n+i+1 j,n],[1,j] y 1 i,j+1 [1,i] [n+i j,n],[1,j+1]y 64 for all 1 i j<n, where y = ψ u,wo x Proof Denote by i 0 the following standard reduced word for w o : i 0 = n 1; n 2, n 1; ; 1,2,,n 1 65 Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 It is convenient to identify i 0 with the sequence of pairs n 1, n 1; n 2, n 2, n 2, n 1; ; 1, 1, 1, 2,,1, n 1 66 Let i be any reduced word for u S n Then we put i and i 0 into a separated word i = i, i 0 for the element u, w o S n S n Denote by w i,n o the longest element of the subgroup of S n generated by the simple transpositions s i,s i+1,,s n 1
36 512 A Berenstein and V Retakh pair i, j, formula: Then in the notation of 537, we have for the position k of i corresponding to the i v k = w i+1,n o s i s i+1 s j, v <k = w i+1,n o s i s i+1 s j 1, ii v k j + 1 = w i+1,n o s i s i+1 s j j + 1 = w i+1,n o i = i, iii v <k j = w i+1,n o s i s i+1 s j 1 j = w i+1,n o i = i, iv v k [1, j+1] = w i+1,n o s i s i+1 s j 1 [1, j+1] = w i+1,n o [1, j+1] = [1, i] [n+i j, n], v v <k [1, j] = w i+1,n o s i s i+1 s j 1 [1, j] = w i+1,n o [1, j] = [1, i] [n + i + 1 j, n] On the other hand, taking 538 for i = i, i 0 with i k = j yields the following t k = j v <k,ey 1 j+1 v k,ey 67 Combining this with the results of the above computations, we obtain the desired formula for t k = t ij The lemma is proved The above facts imply an immediate corollary Corollary 63 For any u S n, the twist map ψ u,wo satisfies i,j [1,i] [n+i+1 j,n],[1,j] ψ u,w o x 1 i,j+1 [1,i] [n+i j,n],[1,j+1] ψ u,w o x 68 = i,j [1,i],[j+1 i,j] x 1 i,j+1 [1,i],[j i+2,j+1] x for all 1 i j n 1 We consider a factorization of x G wo,v of the form x = h x n 1 x n 2 x 1 x +, 69 Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 where x + L e,v, h H, and x m L sm s n 1s m,e is of the form x m = x m τm,m x m 1 τm,m+1 x n 1 τm,n for m [1, n 1]
37 Noncommutative Double Bruhat Cells and Their Factorizations 513 The following result generalizes the factorization from Section 42 Proposition 64 In the notation of 69, that is, x n 1,1 h n = x n1,h n 1 = x n,1 x n,2 x n 1,2,, h 1 = 1 n 1 x 1n, 611 h m = m,n+1 m [m,n],[1,n+1 m]x 612 for m [1, n], and x m,1 x m,k+1 m τ m,k = 1 k m x k,1 x k,k+1 m for all 1 m k<n, that is, 1 h m 613 τ m,k = m,k+1 m [m,k],[1,k+1 m] x 1 m,n+1 m [m,n],[1,n+1 m]x 614 The proof is similar to the proof of Theorem 35 Example 65 Let n = 3 Then in the factorization x = h x 2 τ22 x 1 τ11 x 2 τ12 x+, 615 where h H and x + U, we have τ 11 = x ,3 123,123x, τ 12 = 1,2 12,12x 1 1,3 123,123x, τ 22 = x ,1 23,12x 616 Downloaded from imrnoxfordjournalsorg at Mass Inst of Technology on January 5, 2011 Our next result is a direct consequence of Theorem 516 Lemma 66 In the notation of 69, τ ij = j,j+1 i [1,j],[n+2 i,n] [1,j+1 i] y 1 j,n+1 i [1,j],[n+1 i,n] [1,j i]y 617 for all 1 i j<n, where y = ψ wo,e x
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