aa r X i v : . [ m a t h . G T ] S e p Bowling ball representations of braid groups
Stephen Bigelow
Abstract
In a remark in his seminal 1987 paper, Jones describes a way to definethe Burau matrix of a positive braid using a metaphor of bowling a balldown a bowling alley with braided lanes. We extend this definition toallow multiple bowling balls to be bowled simultaneously. We obtainthe Iwahori-Hecke algebra and a cabled version of the Temperley-Liebrepresentation.
The positive braid monoid B + n is the monoid generated by σ , . . . , σ n − modulothe following relations. Far commutativity: σ i σ j = σ j σ i if | i − j | > The braid relation: σ i σ j σ i = σ j σ i σ j if | i − j | = 1.Alternatively, B + n is the set of geometric braids with n strands that involve onlypositive crossings.In a remark in [3], Jones describes a definition of the (non-reduced) Buraurepresentation of the positive braid monoid using a “bowling ball” metaphor.Here is the relevant passage (except we change “ t ” to “1 − q ” and ( i, j ) to ( j, i ),to match our conventions).For positive braids there is also a mechanical interpretation of theBurau matrix: Lay the braid out flat and make it into a bowling alleywith n lanes, the lanes going over each other according to the braid.If a ball traveling along a lane has probability 1 − q of falling off thetop lane (and continuing in the lane below) at every crossing thenthe ( j, i ) entry of the (non-reduced) Burau matrix is the probabilitythat a ball bowled in the i th lane will end up in the j th.This approach was generalized to string links in [5]. Subsequent papers,for example [4], [6], and [1], have pursued the related idea of random walks onbraids and knot diagram. Our goal is to generalize the bowling ball definition toallow several balls to be bowled simultaneously. We obtain the Iwahori-Heckealgebra and a cabled version of the Temperley-Lieb representation.1hroughout the paper, we work over an arbitrary field containing an element q . The probability metaphor only makes literal sense when q is a real numberin the range [0 , q , and evenover a ring. Fix N ≥
1. Let β be a positive braid, thought of as a bowling alley with n lanes.Simultaneously bowl balls into the lanes so that each lane receives at most N balls.At each crossing, some balls may fall, according to the following rule. Sup-pose a balls arrive on the top lane of a crossing, and b arrive on a bottom. If a ≤ b then no balls will fall. If a > b then, with probability 1 − q , exactly a − b balls will fall from the top lane to join the b balls on the lane below.Use this to define a matrix ρ ( β ) whose rows and columns are indexed by n -tuples u = ( u , . . . , u n ) of integers such that 0 ≤ u i ≤ N . The ( v , u ) entryof ρ ( β ) is the probability that, if u i balls are bowled into the i th lane for all i ,then v j balls end up in the j th lane for all j . Theorem 2.1. ρ is a well-defined representation of B + n .Proof. The definition of ρ clearly respects multiplication, and the far commu-tativity relation. It remains to check the braid relation.For convenience, we can assume n = 3. Let β = σ σ σ = σ σ σ . Call the three lanes the top, middle, and bottom. The top lane crosses overboth other lanes, and the bottom lane crosses under both other lanes.We will compute the entries of the matrix ρ ( β ) in a way that does not dependon the specific word used to represent β . This will show that ρ ( β ) is well-defined. Case 0: No balls.
If no balls are bowled in to β then no balls will emerge. Case 1: One ball.
Suppose one ball is bowled into β . If it is bowled into one of the lower twolanes then the top lane plays no role, so we can simply use the probabilities fora single crossing between the lower two lanes.Now suppose the ball is bowled into the top lane. The probability that itwill end up in the top lane is q , since it must pass over two empty lanes. Theprobability that it will end up among the top two lanes is q , since it must passover the bottom lane exactly once, regardless of whether or not it falls to themiddle lane. By subtraction, the probability that it will end up in the middlelane is q − q , and the probability that it will end up in the bottom lane is 1 − q . Case 2: Two balls.
Suppose two balls are bowled into β . If they are bowled into the same lanethen they behave as a single ball, which was covered in the Case 1. If one of2hem is bowled into the bottom lane then the bottom lane plays no role, sowe can simply use the probabilities for a single crossing between the upper twolanes.Now suppose the two balls are bowled into the top and the middle lanes.The probability that the bottom lane ends up empty is q , since the two ballsmust pass over the bottom lane. The probability that one of the lower twolanes ends up empty is q , since the top ball must pass over the empty lane. Bysubtraction, the probability that the middle lane will end up empty is q − q ,and the probability that the top lane will end up empty is 1 − q . Case 3: Three balls.
Suppose three balls are bowled into β . If they are bowled into the samelane then they behave as a single ball, which was covered in Case 1. If they arebowled one into each lane then no balls will fall.Now suppose one ball is bowled into one lane and two balls are bowledinto another. We can compute the probability that any given lane ends upempty by ignoring the distinction between having one or two balls in a lane,and proceeding as one ball had been bowled into each of two lanes. Similarly,we can compute the probability that any given lane ends up with two ballsby ignoring the distinction between having one or zero balls in a lane, andproceeding as if one ball had been bowled in.Consider the six possible outcomes in the following order:(0 , , , (0 , , , (1 , , , (2 , , , (2 , , , (1 , , . Any pair of consecutive terms in this list represents either the two outcomesthat have 0 in a given position or the two outcomes that have 2 in a givenposition. This means we have computed the sum of the probabilities of any twoconsecutive outcomes in the list. It remains to compute the probability of anyone of the possible outcomes.Consider the outcome (0 , , , ,
0) then the probability of the outcome (0 , , q , since there are three crossings at which a larger number of balls must passover a smaller number without falling. If the input is any other permutation of(2 , ,
0) then the probability of the outcome (0 , ,
2) is 0, since balls cannot fallup. We can now deduce the probability of any outcome for any given input ofthree bowling balls.
Case 4: The general case.
Suppose a , b and c balls are bowled into the lanes. The only thing thatmatters about the numbers a , b and c is which equalities and inequalities holdbetween them. Thus we can reduce to one of the cases we have already covered.In every case, for any given input we can compute the probability of anygiven output. Our computation applies equally well to σ σ σ and σ σ σ , sothese have the same matrix. 3 The Iwahori-Hecke algebra
Let ρ be the representation of B + n defined in the previous section. The Iwahori-Hecke algebra H n ( q ) is the monoid algebra of formal linear combinations ofpositive braids modulo the quadratic relations( q + σ i )(1 − σ i ) = 0 . Theorem 3.1. ρ factors through H n ( q ) .Proof. Only two lanes are involved, so it suffices to treat the case n = 2. Let v be the vector corresponding to ( a, b ). We must show that v is in the kernel of( q + ρ ( σ ))(1 − ρ ( σ )) . If a = b then v is fixed by ρ ( σ ), and we are done. If a = b then the actionof ρ ( σ ) on the the vectors corresponding to ( a, b ) and ( b, a ) is the same as theBurau representation. It is well known, and easily checked, that this satisfiesthe required quadratic relation.Note that if q is invertible then the σ i are invertible in H n ( q ), with σ − i = q − ( σ i + q − . In this case, ρ is a representation of the braid group B n and not just B + n .There is another important element of the kernel in the case n ≥ N + 2. Fix k with 1 ≤ k ≤ n − N −
1. Suppose w is a permutation of { k, . . . , k + N + 1 } .Let sgn( w ) = ± β w denote the uniquepositive braid with a minimal number of crossings such that the lane at position i goes to position w ( i ) for all i = k, . . . , k + N +1. Let x k be the following elementof H n ( q ). x k = X w sgn( w ) β w . A generalization of x k appears in the definition of the Specht modules in[2]. The quotient of H n ( q ) by the elements x k is the representation of B n corresponding to the Lie group SL( N + 1), as defined in [7]. If N = 1 then it isthe Temperley-Lieb algebra.We will need the following property of x k . x k = X w ( i ) 1. Let β be a positive braid, thought of as a bowling alley with n lanes.Create a cabling of β by replacing every lane with K parallel lanes. Suppose a = ( a , . . . , a n ) is an n -tuple of integers such that 0 ≤ a i ≤ K . Bowl balls intothe lanes so that each lane gets at most one ball and, for all i , a i balls go intothe i th collection of K parallel lanes. Whenever a ball passes over an emptylane, it falls with probability 1 − q .Use this to define a matrix ρ K ( β ) whose rows and columns are indexed by n -tuples ( a , . . . , a n ) of integers such that 0 ≤ a i ≤ K . The ( b , a ) entry of ρ K ( β ) is the probability that, if a i balls are bowled into the i th set of N parallellanes for all i , then b i balls end up in the i th set of N parallel lane for all i . Theorem 4.1. ρ K is a well-defined representation of B + n .Proof. Our definition only keeps track of the number of balls in each collectionof K parallel lanes. We must check that it is not necessary to know preciselywhich lanes they are in.Consider the cabling of a single positive crossing. Suppose we bowl a ballsinto the upper K lanes of a cabled crossing, and b into the lower K lanes. Theempty lanes in the upper lanes will remain empty, and the balls in the lowerlanes will remain there. The probability that exactly c balls will fall dependsonly on the number a of occupied upper lanes and the number K − b of emptylower lanes. 5ne motivation for studying ρ K would be to use it to compute the coloredJones polynomial of a knot. Other apparently similar approaches to the coloredJones polynomial have appeared in [6] and [1]. It would be interesting to knowsomething about the limiting behavior of ρ K if we set q = e iπ/K and let K goto infinity. This may have some connection to the Kashaev conjecture.For all 0 ≤ c ≤ a ≤ N and 0 ≤ b ≤ N , let f ab ( c ) denote the probability,when a balls enter the top lane of a crossing and b balls enter the bottom, that c balls fall from the upper lane to the lower lane. We will compute a formulafor f ab ( c ). First we define some notation.If k is a non-negative integer, we define the quantum integer[ k ] = 1 − q k − q . Note that we are using the definition that involves only positive exponents of q ,not the definition that is symmetric under mapping q to q − .The q -factorial is [ k ]! = [ k ][ k − . . . [1]. If 0 ≤ r ≤ k , the Gaussian binomialis (cid:18) kr (cid:19) q = [ k ]![ r ]![ k − r ]! . These have a combinatorial interpretation as follows.An inversion of a permutation φ of { , . . . , k } is a pair such that i < j and φ ( i ) > φ ( j ). The quantum factorial [ k ]! is the sum over φ of q to the power ofthe number of inversions of φ .An inversion of a sequence ( ǫ , . . . , ǫ k ) of ones and zeros is a pair such that i < j , ǫ i = 1 and ǫ j = 0. The Gaussian binomial (cid:0) kr (cid:1) q is the sum over all suchsequences that have r ones of q to the power of the number of inversions.Finally, let (cid:18) ∞ r (cid:19) q = 1[ r ]! (1 − q ) r . If | q | < q , we can justtake it as a definition.We now give a formula for f ab ( c ). Despite appearances, it is a polynomialfunction of q . Theorem 4.2. If a, b, c are integers and ≤ c ≤ a then f ab ( c ) = (cid:0) ac (cid:1) q (cid:0) K − bc (cid:1) q (cid:0) ∞ c (cid:1) q q ( a − c )( K − b − c ) . Proof. Consider a crossing where K parallel lanes pass over K parallel lanes.Now bowl a balls into the upper collection of lanes and b into the lower. Wemust compute the probability that exactly c balls will fall.Fix a choice of c of the upper a balls, a choice of c of the lower K − b emptylanes, and a bijection φ from these balls to these empty lanes. We compute theprobability that our chosen balls fall into our chosen lanes according to φ .6ome terminology will help us to stay organized. The K upper lanes consistof our chosen c briefly-full lanes, a − c always-full lanes, and K − a irrelevant lanes. The K lower lanes consist of our chosen c briefly-empty lanes, K − b − c always-empty lanes, and b irrelevant lanes. The irrelevant lanes have no effecton the probability.Consider the crossings where an always-full lane passes over an always-emptylane. At each such crossing, a ball will pass over an empty lane, contributing afactor of q . Taken together, these crossings contribute the term q ( a − c )( K − b − c ) . Consider the crossings where an always-full lane passes over a briefly-emptylane. Such a crossing will contribute a factor of q if and only if the briefly-empty lane is still empty, having not yet met its corresponding briefly-full lane.The number of times this happens is the number of pairs of lanes consistingof a briefly-full lane to the left of an always-full lane. Taken together, thesecontribute the power of q in the definition of (cid:18) ac (cid:19) q (using the combinatorial definition, with appropriate conventions).Consider the crossings where a briefly-full lane passes over an always-emptylane. Such a crossing will contribute a factor of q if and only if the briefly-full lane is still full, having not yet met its corresponding briefly-empty lane.The number of times this happens is the number of pairs of lanes consisting ofan always-empty lane to the left of a briefly-empty lane Taken together, thesecontribute the power of q in the definition of (cid:18) K − bc (cid:19) q (using the combinatorial definition, with appropriate conventions).Consider the crossings where a ball falls from a briefly-full lane to a briefly-empty lane. Each such crossing contributes a factor of (1 − q ). Taken together,these contribute the term (1 − q ) c . Finally, consider the crossings where a briefly-full lane passes over a briefly-empty lane but no ball falls there. This will contribute a factor of q if and onlyif the briefly-full lane is still full and the briefly-empty lane is still empty. Thenumber of times this happens is the number of pairs of briefly-full lanes i and j such that i is to the left of j and φ ( i ) is to the left of φ ( j ). This contributesthe power of q in the definition of [ c ]!(using the combinatorial definition, with appropriate conventions).Now multiply the above contributions, and sum over all possible choices of c balls, c empty lanes, and φ . This gives the desired formula for f ab ( c ).7 eferences [1] Cody W. Armond. Walks along braids and the colored Jones polynomial. J. Knot Theory Ramifications , 23(2):1450007, 15, 2014.[2] Richard Dipper and Gordon James. Representations of Hecke algebras ofgeneral linear groups. Proc. London Math. Soc. (3) , 52(1):20–52, 1986.[3] V. F. R. Jones. Hecke algebra representations of braid groups and linkpolynomials. Ann. of Math. (2) , 126(2):335–388, 1987.[4] Paul Kirk, Charles Livingston, and Zhenghan Wang. The Gassner represen-tation for string links. Commun. Contemp. 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