Stephen Bigelow. The Lawrence-Krammer representation of the braid groups recently. 1. Introduction

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1 Topology Geometric Georgia International Topology Conference 2001 Studies in Advanced Mamatics AMS/IP 35, 2003 Volume Lawrence-Krammer representation braid groups recently Abstract. came to prominence when it was shown to be faithful by myself It is an action braid group on a certain homology module Krammer. ~ C) over ring Laurent polynomials in q t. In this paper we de- H2( some surfaces in ~ C representing elements homology. use se to scribe a new pro that H2( ~ C) is a free module. also show that (n 2; 2) give Temperley-Lieb algebra is image a map to relative representation at t = q 1, clarifying work Lawrence. homology Lawrence-Krammer representation is action braid group B n on certain homology module H 2 ( ~ C) over ring = Z[q 1 ; t 1 ]. It was introduced a Lawrence [Law90], except that she worked over C instead. It recently by to prominence when it was shown to be faithful for n = 4 [Kra00], n came all n [Big01], [Kra02]. Since n, a number papers have appeared which for examine certain aspects representation. See [PP01], [Son02], closely now continue in this tradition, with specic goal understing [Bud02]. connection with Temperley-Lieb algebra. paper was partly motivated by an attempt to clarify two points from This First, pairing between a \noodle" a \fork" involved an algebraic [Big01]. number between two non-compact surfaces in ~ C. Such a thing is not intersection well-dened, so I gave an indirect pro existence a certain necessarily surface corresponding to a fork. I now have an explicit description this closed which will be given in Section 3.4. In fact it is possible to dene surface, without reference to this surface by using results from [Kaw96, Appendix pairing (Thanks to Won Taek Song, whose paper [Son02] drew my attention to this.) E]. to compute matrices for representation, I tensored H 2 ( ~ C) with a Second, containing. resulting vector space contains H 2 ( ~ C), but strictly speaking, eld action B n on this vector space should not be called Lawrence-Krammer In Section 5 we give a new pro that H 2 ( ~ C) is a free -module. representation. Mamatics Subject Classication. Primary 20F36; Secondary 20C words phrases. Braid group, Temperley-Lieb algebra, linear representations. Key Lawrence-Krammer Representation Stephen Bigelow 1. Introduction This is originally due to Paoluzzi Paris [PP01], but our pro uses an explicit Research supported by Australian Research Council. c2003 American Mamatical Society International Press 51

2 surfaces representing elements a free basis H 2 ( ~ C). correct description for Lawrence-Krammer representation could be computed from this, matrices we do not do so since y are quite complicated. but addressing se two issues from [Big01] we also shed new light on work By Lawrence [Law90]. re, Lawrence-Krammer representation was used to a topological interpretation to a representation Hecke algebra. idea give to specialise t to value q 1, at which point Lawrence-Krammer representation is becomes reducible desired representation appears as a quotient. complicated methods are used in [Law90] to dene required quotient. Somewhat In Section 6 we show that it is simply image a map to a certain relative module. homology this paper n is a positive integer, D is unit disk centred at Throughout origin in complex plane, 1 < p 1 < < p n < 1 are real numbers, n = D n fp 1 ; : : :; p n g is n-times punctured disk. braid group B n is D class group D n. also assume familiarity with presentation B n mapping Artin generators i, interpretation B n as fundamental group using a conguration space in plane. research was supported by Australian Research Council. Thanks to This Matic or organisers for such an inspiring entertaining Gordana conference. briey review denition Lawrence-Krammer representation as in [Big01]. C be space unordered pairs distinct points in D n. given c 0 = fd 1 ; d 2 g be a basepoint in C, where d 1 d 2 are distinct points on disk. boundary follows. Suppose : I! C is a closed loop in C representing an element as 1 (C; c 0 ). By ignoring puncture points we can consider as a loop in space unordered pairs points in disk, hence as a braid in B 2. j be this braid in Artin generator 1. Similarly, map exponent a braid in B n+2. j 0 be exponent sum this braid in determines generators B n+2. Note that j j 0 have same parity. i = 1 2 (j0 j). Artin ~ C be connected covering space C such that 1 ( ~ C) = ker(). Fix a ~c 0 in bre over c 0. homology group H 2 ( ~ C) admits a -module choice where q t act by covering transformations. structure, f is a homeomorphism from D n to itself, representing an element Suppose see this, think 1 (C; c 0 ) as subgroup B n+2 consisting braids whose To n strs are straight. n f acts on 1 (C; c 0 ) by conjugation with a braid rst 52 STEPHEN BIGELOW 2. Lawrence-Krammer representation dene a homomorphism : 1 (C; c 0 )! hqi hti s 7! fp 1 ; : : :; p n g [ (s) dene () = q i t j : B n. f be induced map from 1 (C; c 0 ) to itself. I claim that f = :

3 rst n strs. This preserves image under, since maps to an abelian on group. follows that f has a unique lift It : ( ~ C; ~c0 )! ( ~ C; ~c0 ); ~f : H 2 ( ~ C)! H 2 ( ~ C) ~f a -module automorphism. dene Lawrence-Krammer representation to is this action B n on H 2 ( ~ C). be now introduce some relative homology modules. For > 0, let be set points fx; yg 2 C such that eir x y are within distance each or, or least one m is within distance a puncture point. ~ be preimage at this section we describe some immersed surfaces in ~ C. se will be used to In elements homology or relative homology. Each immersed surface will represent specied by giving a map from a surface to C showing that it can be lifted be ~ C. will not specify a choice lift, we will not pay much attention to to orientation. Thus resulting element homology or relative homology issues only be dened up to multiplication by a unit in. This will be sucient for will Squares triangles. describe some properly embedded surfaces 3.1. elements second homology ~ C relative to ~ C, ~ ~ C [ ~. representing will be represented by one or two embedded edges in disk. se 1 ; 2 : I! D are disjoint embeddings interior I into D n, Suppose map endpoints I to puncture points (not necessarily injectively). f map from interior I I to C given by f(x; y) = f 1 (x); 2 (y)g. A lift be f to ~ C will represent an element H2 ( ~ C; ~). Similarly, we can dene an element H 2 ( ~ C; ~ C) corresponding to a pair disjoint edges in D n with endpoints on D. we can dene an element H 2 ( ~ C; ~ C [ ~) corresponding to a pair edges Finally, have a mixture endpoints on D on puncture points. For all se which call resulting element relative homology square corresponding examples, edges 1 2. to suppose : I! D is an embedding interior I into D n, maps Now endpoints I to puncture points. Dene a map f(x; y) = ff(x); f(y)g. A lift f to ~ C will represent an element H2 ( ~ C; ~). by we can obtain an element H 2 ( ~ C; ~ C [ ~) if we allow one or both Similarly, edge to be on D. For all se examples, call resulting endpoints relative homology triangle corresponding to. element THE LAWRENCE-KRAMMER REPRESENTATION 53 that induced map in ~ C. relative homology modules H 2 ( ~ C; ~ ) are nested by inclusion. 2 ( ~ C; ~) = lim H H 2 ( ~ C; ~ );!0 2 ( ~ C; ~ C [ ~) = lim H H 2 ( ~ C; ~ C [ ~ ):!0 braid group B n acts on se, on H 2 ( ~ C; ~ C), by -module automorphisms. 3. Some surfaces our purposes. f : f(x; y) 2 I I : 0 < x < y < 1g! C

4 s s s s 7! (1 q) 2 s s s s. A genus one surface. Suppose 1 ; 2 : S 1! D n are disjoint gureeights, 3.2. each going around two puncture points, as in Figure 1. Dene a map f torus S 1 S 1 to C by f(x; y) = f 1 (x); 2 (y)g. Both meridian from torus are mapped into kernel, so f lifts to a map longitude ~f : S 1 S 1! ~ C: obtain an element [ ~ f] H 2 ( ~ C). will be useful to know image [ ~ f] in H2 ( ~ C; ~). I claim this is (1 q) 2 It square as indicated in Figure 1. To see this, homotope each gure-eight times i so as to map points 1 2 S 1 into respective -neighbourhoods two points enclosed by i. n f maps four lines f1gs 1 S 1 f1g puncture. se lines cut torus into four squares. restriction ~ f to each into se squares represents an element H 2 ( ~ C; ~). claim now follows from a comparison lifts orientations se squares, which is left to careful A genus two surface. Suppose 1 ; 2 : S 1! D n are gure-eights such passes around p i p j, 2 passes around p j p k, 1 intersects that 2 twice, as in Figure 3. B D be a disk centred at p j, containing two intersection, meeting each 1 2 in a single edge. For i = 1; 2, points let I i S 1 be interval 1 with I = [0; 1], oriented as shown in Figure 2. both T be closure that for (x; y) 2 T we have 1 (x) 6= 2 (y). can refore dene f : T! C Note f(x; y) = f 1 (x); 2 (y)g. by f 0 with an anticlockwise rotation B by an angle s about composition p j. can assume B has a rotational symmetry so that f 1 (x; y) = f 0 (y; x) centre 54 STEPHEN BIGELOW Figure 1. A torus a square. s Figure 2. disk B edges 1 (I) 2 (I). reader. (B). For convenience, assume I 1 = I 2, identify i (S 1 S 1 ) n (I 1 I 2 ): f 0 be restriction f to T = (I I). For s 2 I, let f s be

5 all (x; y) 2 (I I). Thus f 0 f 1 represent same loops, but with opposite for orientations. now build a closed genus two surface 2 by gluing toger T I two T as follows. First glue T I to T by ((x; y); 0) (x; y). n glue copies to a second copy T 0 T by ((x; y); 1) (y; x). 2 be surface so T I g : 2! C be given by gj T = gj 0 T = f gj (Tfsg) = f s. obtained. fundamental group 2 is generated by meridian longitude T 0. Each se is mapped by g into kernel, so g lifts to a map T : 2! ~ C. obtain an element [~g] H2 ( ~ C). ~g now compute image [~g] in H 2 ( ~ C; ~). For any > 0 we can assume that disk B has radius less than. n g maps T I into, so ~gj T represents element H 2 ( ~ C; ~ ). This element is (1 q) 2 times a square, by a similar an to one given for genus one surface. It remains to gure out how argument 0 is related to ~gj T. T ~gj goes from (0; 0) 2 T to (0; 0) 2 T 0. Now g is a loop in C in which pair This points switch places by an anticlockwise rotation through an angle around j Thus (g ) = qt. It follows that ~gj 0 T = qt ~gj T. p. that T 0 T inherit same orientation from 2. This is because note Also that image [~g] in H 2 ( ~ C; ~) is (1 q) 2 (1 + qt) times a square, as conclude in Figure 3. shown A genus three surface. Suppose 1 ; 2 : S 1! D n are gure-eights, 3.4. passing around p i p j, intersecting transversely at four points, as in both 4. construct a map from a genus three surface 3 into ~ C by a slight Figure procedure used above for 2. This time surface T will be modication torus with two disks removed, one for each p i p j. Two annuli are n a to glue T to anor copy T 0 T. obtain a genus three surface 3 needed map g : 3! C. a now show that g lifts to a map ~g : 3! ~ C. image longitude, boundary components T all lie in kernel. Thus gj T meridian, both ~gj T. This lift can be extended to annuli T I. Finally, dene ~gj 0 T to to lifts transformation qt applied to ~gj T. be covering compute image [~g] in H 2 ( ~ 2 C; ~). This is (1 q) (1 + qt) times a now by a similar argument to one given for genus two surface. This square square, to a parallel pair edges from p i to p j. se edges can be homotoped corresponds THE LAWRENCE-KRAMMER REPRESENTATION 55 s s s 7! (1 q) 2 (1 + qt) s s s Figure 3. A genus two surface a square. Consider path : I! 2 given by (s) = ((0; 0); s) 2 T I: ends annulus T I were attached with opposite orientations.

6 s s 7! (1 q) 2 (1 + qt)(1 t) s s. as to lie within each or. n a diagonal square in ~ C will lie in ~. so diagonal cuts square into two triangles, which have opposite orientations This dier by covering transformation t. Thus [~g] represents (1 q) 2 (1+qt)(1 t) a triangle, as shown in Figure 4. times is not immediately obvious that surfaces described in Section 3 represent It elements homology or relative homology. In Sections 5 6 we will non-trivial to prove even stronger results concerning linear independence various need such elements. Our main tool will be following intersection pairing. sets x 2 H 2 ( ~ C) y 2 H 2 ( ~ C; ~ C [ ~) let (x y) 2 Z denote stard For number. dene an intersection pairing intersection hx; yi = X i;j2z (x q i t j y)q i t j : check that se are well dened requires some elementary homology ory. To [Kaw96, Appendix E], where following properties are also proved. See is image under automorphism taking q to q 1 t to where 1. Similar identities hold for h; i 0. t that above denition h; i diers from that [Big01] in that Note entries is reversed. above is consistent with [Kaw96], with order usual denition sesquilinear. will frequently be necessary to compute intersection pairing in specic It up to multiplication by a unit in. rest this section is devoted examples, discussion how to do this. to 1 2 be edges in D with endpoints on puncture points, representing square a 2 H 2 ( ~ C; ~). 1 2 be edges in D with endpoints on D, a a b 2 H 2 ( ~ C; ~ 0 C). discuss how to compute ha; bi up to representing square a unit in. by multiplication A B be surfaces in C corresponding to a b respectively. A \ B is set pairs form fx; yg where eir intersection 56 STEPHEN BIGELOW Figure 4. A genus three surface a triangle. 4. An intersection pairing h; i: H 2 ( ~ C) H2 ( ~ C; ~ C [ ~)! by A similar denition gives a pairing h; i 0 : H 2 ( ~ C; ~) H2 ( ~ C; ~ C)! : For x 2 H 2 ( ~ C), y 2 H 2 ( ~ C; ~ C [ ~), 2 B n, 2, we have hx; yi = hx; yi; hx; yi = hx; yi = hx; yi;

7 x 2 1 \ 1 y 2 2 \ 2, or x 2 1 \ 2 y 2 2 \ 1. all edges, intersect transversely, so A intersects B transversely. ~ A Assume be lifts A B to ~ C. ~B A B do not intersect n neir will any ir lifts, so ha; bi 0 = 0. If If B intersect at one point fx; yg n choose lifts orientations so that ~ A A ~ B with positive sign at a point in bre fx; yg. n ha; bi 0 = 1. intersects suppose A B intersect in two points, fx; yg fx 0 ; y 0 g. Choose lifts Now orientations so that ~ A intersects ~ B with positive sign at a point in bre yg. i j be integers such that ~ A intersects q i t j ~ B at a point in bre fx; fx 0 ; y 0 g, let = 1 be sign that intersection. n ha; bi 0 = 1 + q i t j. remains to compute i, j. It A B be paths from fx; yg to fx 0 ; y 0 g that lie in A B respectively. ~ A ~ B be lifts that start in ~ A\ ~ B. n ~A (1) ~ B (1) are points i t j = ( 1 A B): q practice this will be easy to compute. It remains to compute. In if necessary so that x; x y; y Orient edges 1, 2, Relabel 1 2. sign intersection A with B at fx; yg is determined by sign intersection 1 with 1 [ 2 at x, sign intersection 2 with 1 [ 2 at y, wher x 2 2 or x 2 1. assumption, conventions are such that this intersection is positive. Each By following will reverse sign intersection A with B at fx 0 ; y 0 g. 1 intersects 1 [ 2 with dierent signs at x x 0, 2 intersects 1 [ 2 with dierent signs at y y 0, eir x 2 1 y 2 2, or x 2 2 y 2 1. none or two se hold n = 1. If one or three se hold n = 1. If will sometimes need to compute intersection pairings between surfaces that not squares. First consider case A is a triangle corresponding to an edge, are is a square as before, A B intersect transversely at two points. n B above discussion applies almost unchanged. In place assumption that surfaces which are neir squares nor triangles, but same ideas will between apply. suppose A B intersect at more than two points. n each point Finally, intersection contributes a monomial to pairing. can assume one se this section, we analyse Lawrence-Krammer representation using In developed in Sections 3 4. give a new pro following orem. tools THE LAWRENCE-KRAMMER REPRESENTATION 57 in bre fx 0 ; y 0 g lying in ~ A ~ B respectively. Thus q i t j ~ B (1) = ~ A (1); so x y; y 0 2 2, make assumption that x occurs before y x 0 x; before y 0 with respect to orientation. will also compute pairings occurs monomials is 1, compute ors by methods discussed above. 5. A basis orem 5.1. H 2 ( ~ C) is a free -module rank n 2.

8 is originally due to Paoluzzi Paris [PP01]. Our pro gives a more This description surfaces representing elements a free basis for H 2 ( ~ C). explicit usually given for Lawrence-Krammer representation use a basis for matrices t)h 2 ( ~ C), not for H 2 ( ~ C). This subtle distinction will be discussed at end Q(q; this section. use following basic result about H 2 ( ~ C) vector space Q(q; t) H 2 ( ~ C) has dimension n 2. natural Lemma pro uses a nite 2-complex that is homotopy equivalent to C. Various for constructing required complex can be found in [Law90], [Big01], methods [Bud02]. Since complex has dimension two, H 2 ( ~ C) is kernel [PP01], a matrix with entries in. Now Q(q; t) H 2 ( ~ C) is kernel matrix, as a matrix over Q(q; t). Thus H 2 ( ~ C) is submodule Q(q; t)h 2 ( ~ C) considered those vectors whose entries lie in. dimension Q(q; t)h 2 ( ~ C) consisting be computed from an explicit description matrix, which can be found can any papers mentioned above. in Pro orem. now prove orem 5.1. For 1 i < j n, 5.1. dene v 0 i;j 2 H 2( ~ C; ~) as follows. If j i > 2, let v 0 i;j be square corresponding we to edges [p i ; p i+1 ] [p j 1 ; p j ]. If j i = 2 i > 1, let v 0 i;j to edge [p i ; p i+1 ] an edge from p i 1 to p j whose interior lies corresponding lower half plane. v 0 1;3 be square corresponding to edges [p 1; p 2 ] in [p 2 ; p 3 ]. If j i = 1, let v 0 i;j be triangle corresponding to edge [p i; p j ]. 1 i < j n, let v i;j be an element H 2 ( ~ C) whose image in H 2 ( ~ C; ~) is For (1 q) 2 (1 + qt)(1 t)v 0 i;j (1 q) 2 (1 + qt)v 0 i;j an element exists by Section 3. will show that v i;j form a free basis Such H 2 ( ~ C). for every 1 i < j n, let x i;j 2 H 2 ( ~ C; ~ C) be square corresponding to a For vertical edges, one passing just to right p i, or passing just to pair hv 0 i;j; x i;j i 0 = 1 for 1 i < j n, hv 0 i;i+2 ; x i 1;i+1i 0 = 1 for i = 2; : : :; n 2, hv 0 i;i+2 ; x i;i+1i 0 = 1 t for i = 2; : : :; n STEPHEN BIGELOW map from H 2 ( ~ C) to Q(q; t) H 2 ( ~ C) is injective. be square if j = i + 1, if i = 1 j = 3, (1 q) 2 v 0 i;j orwise. left p j, both having endpoints on D. Lemma 5.3. following identities hold up to multiplication by a unit in. All or pairings hv 0 i 0 ;j 0; x i;ji 0 are zero. Pro. Use methods described in Section 4. For convenience, choose lifts orientations so that hv 0 i;j; x i;j i 0 = 1 for 1 i < j n.

9 q 2 t 2 qt 2 Lemma 5.3 it follows that v 0 i;j are linearly independent. Hence v i;j From linearly independent. By Lemma 5.2, y span Q(q; t) H 2 ( ~ C) as a Q(q; t)- are It remains to show that y span H 2 ( ~ C) as a -module. In or words, module. must prove following. we 1i<jn c i;j v i;j 5.5. image x n 1;n in H 2 ( ~ C; ~ C [ ~) is (1 q)(1 + qt)(1 t) Lemma a triangle. times : I! D be a straight edge from D to p n. 2 be an edge Pro. an -neighbourhood with endpoints on D, passing anticlockwise around in 1 be an edge in an -neighbourhood with endpoints on D, passing. around 2. Dene f : I I! C by f(x; y) = f 1 (x); 2 (y)g. n anticlockwise lifts to a map ~ f which represents x n 1;n. f can assume that for all s 2 I four points 1 (s), 2 (s), 1 (1 s), 1 x)g, f( 1 2 ; y)g, f(x; 1 2 )g into. se lines cut I I into eight pieces. f(x; restriction ~ f to each piece represents an element H 2 ( ~ C; ~ C [ ~ ). Each se elements is a multiple triangle corresponding to, as shown in 5. Combining se, we see that x n 1;n is (1 q)(1 + qt)(1 t) times Figure 5.6. For all 1 i < j n, image x i;j in H 2 ( ~ C; ~ C [ ~) is Lemma q) 2 times a linear combination squares. (1 Lemma 5.6, x 2;3 is a multiple (1 q) 2, hence (1 q) 2. Since h; i is By it follows that (1+qt)(1 t)c 2;3 2. Similarly, Lemma 5.5 implies that sesquilinear, q)c 2;3 2. Since is a unique factorisation domain, it follows that c 2;3 2. (1 a symmetrical argument, c 1;2 2. By THE LAWRENCE-KRAMMER REPRESENTATION 59 qt q 2 t q 1 t qt Figure 5. function representing x n 1;n breaks into triangles. Lemma 5.4. c i;j 2 Q(q; t) for 1 i < j n be such that v = X lies in H 2 ( ~ C). n ci;j 2 for all 1 i < j n. use following facts about x i;j. 2 (1 s) lie within distance each or. n f maps lines f(x; x)g, triangle corresponding to. Pro. Use a similar argument to Lemma 5.5, as suggested by Figure 6. Pro Lemma 5.4. First consider case n = 3. n hv; x 2;3 i = (1 q) 2 (1 + qt)(1 t)c 2;3 :

10 lies in H 2 ( ~ C), so we can reduce to case v = c 1;3 v 1;3. following holds still to multiplication by a unit in. up Lemma 5.5, 2 x 2;3 is a multiple (1 t)(1 q)(1 + qt). Thus (1 q)c 1;3 2. By But Lemma 5.6, (1 + qt)c 1;3 2. Thus c 1;3 2. By suppose n > 3. For i = 1; : : :; n 2 we have Now Lemma 5.6, (1 + qt)(1 t)c n 1;n 2. By Lemma 5.5, (1 q)c n 1;n 2. Thus By n 1;n 2. can now subtract terms c i;n v i;n, so reduce to case c i;n = 0 for all i = 1; : : :; n 1. n v represents an element homology c in ~ C space unordered pairs distinct points in an (n 1)-times preimage Krammer representation. re is some confusion as to exact 5.2. denition \Lawrence-Krammer representation". In an attempt to clarify situation, we now compare contrast a slightly dierent representation which will call \Krammer representation". we >< >: F j;k E E E A A A i;k + (q 2 q)f i;j + (1 q)f j;k i = j 1; qf j+1;k i = j 6= k 1; F j;i + (1 q)f j;k + (1 q)qtf i;k i = k 1 6= j; qf j;k+1 i = k; F 60 STEPHEN BIGELOW 0 B 1 C + q q q q q q + : : : q q q q q C q q q q q = (1 q)(1 q 1 ) C B B A B C E A Figure 6. x 2;4 is (1 q) 2 times a linear combination squares. Now v (c 1;2 v 1;2 + c 2;3 v 2;3 ) hv 0 1;3 ; 2x 2;3 i 0 = (1 t): Thus hv; 2 x 2;3 i = (1 t)(1 q) 2 (1 + qt)c 1;3 : hv; x 1;3 i = (1 q) 2 (1 + qt)c 1;3 : hv; x i;n i = (1 q) 2 c i;n : By Lemma 5.6, c i;n 2. Also hv; x n 1;n i = (1 q) 2 (1 + qt)(1 t)c n 1;n : punctured disk. result now follows by induction on n. This completes pro orem 5.1. representation be following action B n on a free - Krammer rank n 2 with basis ffi;j : 1 i < j ng. V module 8 i 62 fj 1; j; k 1; kg; i (F j;k ) = tq 2 F j;k i = j = k 1:

11 representation is given in [Kra00], but with dierent conventions as explained This It is also in [Big01], but with a sign error. name \Krammer representa- below. was chosen because Krammer seems to have initially found this independently tion" Lawrence without any use homology. vector spaces Q(q; t) V Q(q; t) H 2 ( ~ C) are isomorphic representations B n, by [Big01, orem 4.1]. isomorphism is given by maps F i;j to genus three surface corresponding to an edge from p i to Thus j in lower half plane. p basis elements in [Kra00] correspond to similar edges in upper half Thus Krammer's v i;j should be identied with my plane. Krammer's t is my t. Also, n 3, map : V! H 2 ( ~ C) dened as above is not an isomorphism. For see this, note that composition with natural map from H 2 ( ~ C) to To 2 ( ~ C; ~) sends V into (1 t)h2 ( ~ C; ~). However image v1;3 in H 2 ( ~ C; ~) does H lie in (1 t)h 2 ( ~ C; ~). Thus v 1;3 is not in image. not fact Paoluzzi Paris [PP01] showed that modules V H 2 ( ~ C) are In isomorphic representations B n for n 3. distinction becomes important not we specialise q t to values which are not algebraically independent, as we when in next section. will is possible to compute matrices for Lawrence-Krammer representation It respect to basis fv i;j g. will not do this since y are quite complicated. with must be conjugate to those Krammer representation when considered y matrices over Q(q; t), but not when considered as matrices over. as following conjecture would give a nice topological interpretation B n on H 2 ( ~ C; ~). respects action B n by pro [Big01, orem 4.1]. It remains This show that terms (F i;j ) form a basis for H 2 ( ~ C; ~). pro should follow to R H 2 ( ~ C)! R H2 ( ~ C; ~) : induced by natural map from homology to relative homology. Suppose 1+qt = be q has a square root, q 2 6= 1, q 3 6= 1. n image is representation 0, TL n (R) corresponding to partition (n 2; 2). THE LAWRENCE-KRAMMER REPRESENTATION 61 : F i;j 7! ( i 1 : : : 2 1 )( j 1 : : : 3 2 )v 1;2 : ( j 1 : : : 2 1 )( k 1 : : : 3 2 )F 1;2 : Krammer representation. Conjecture 5.7. Krammer representation is isomorphic to action isomorphism should be map : V! H 2 ( ~ C; ~) given by (F i;j ) = ( i 1 : : : 2 1 )( j 1 : : : 3 2 )v 0 1;2: same method as that orem 5.1. only diculty is showing that Q(q; t) H 2 ( ~ C; ~) has dimension n A representation Temperley-Lieb algebra aim this section is to prove following. orem 6.1. R be a domain containing invertible elements q t, let

12 representation in Section 6.4. desired [Law90, orem 5.1], Lawrence constructed representation Hecke In corresponding to partition (n k; k) for k = 1; : : :; bn=2c. se are algebra that factor through Temperley-Lieb algebra. This was gener- representations still furr in [Law96] to give Hecke algebra representation correspondinalised to any partition n. orem 6.1 only covers case (n 2; 2), but also a number advantages over work Lawrence. Firstly, it gives a more has description required quotient R H 2 ( ~ C). Secondly, it works elementary a fairly general ring, whereas Lawrence worked over C used matrices over gave in Section 5.2 for \Krammer representation". Finally, our pro gives we explicit isomorphism between two representations. It is to be hoped that an advantages generalise to arbitrary partitions n. se suspect that requirements on R in orem 6.1 can be weakened somewhat. I main example to keep in mind is R = Z[q 1 2 ] with t = q 1. A basis. For 1 i < j n, let v i;j, v 0 i;j x i;j be as dened in n Section if j = i + 1 or j = 3; 0 q) 2 v 0 i;j orwise: (1 5.3 still holds, so v 0 i;j are linearly independent in R H 2( ~ C; ~). Thus Lemma image is free module with basis ease notation, we will work with H instead image, since two To isomorphic. are K be eld fractions R. n K H is a vector space dimension 3)=2, contains H as an embedded submodule. It is sometimes easier to n(n Hecke algebra. now prove that H is a representation 6.2. algebra. Hecke 6.2. Hecke algebra H n (R), or simply H n, is R-algebra Definition by generators 1; 1 ; : : :; n 1 relations given i j = j i if ji jj > 1, i j i = j i j if ji jj = 1, 62 STEPHEN BIGELOW Here, TL n (R) is Temperley-Lieb algebra. will dene this, (v i;j ) = pairing h; i 0 can be extended to a map h; i 0 : (R H 2 ( ~ C; ~)) H 2 ( ~ C; ~ C)! R: f(1 q) 2 v 0 i;j : j > max(i + 1; 3)g: H be free module with basis fv 0 i;j : j > max(i + 1; 3)g: prove that a relation holds in H by proving that it holds in K H. ( i 1)( i + q) = 0. Thus H n is group algebra RB n braid group modulo relations ( i 1)( i + q) = 0: Lemma 6.3. ( i 1)( i + q) acts as zero map on H.

13 ? = x 1;n 1 q 1 x 1;n 7. Showing that 1 n 1 x i;n = x i;n 1 q 1 x i;n. Figure Since i are all conjugate, we need only check case i = n 1. Pro. do this, we nd a basis for K H consisting eigenvectors n 1, each having To 1 or q. eigenvalue elements v 0 i;j for max(i + 1; 3) < j n 2 are linearly independent eigen- n 1 with eigenvalue 1. re are (n 2)(n 5)=2 se. vectors be a circular closed curve based at p n 2 enclosing p n 1 p n. i = 1; : : :; n 4, let u i be square corresponding to edges [p i ; p i+1 ] For u n 3 be square corresponding to edge [p n 3 ; p n 2 ] a circular. curve based at p n 4 enclosing p n 3 ; : : :; p n. se are eigenvalues n 1 closed eigenvalue 1. For i; i 0 = 1; : : :; n 3, following identities hold up to with hu ; x i;n i 0 = 1 q, i i ; x 0 i ;n i 0 = 0 for i 0 6= i, hu u i are linearly independent not in span fv 0 i;j : j n 2g. Now Thus u be square corresponding to a circular closed curve based at p n 3 let hu; x n 2;n 1 i 0 = (1 q 2 )(1 + q 2 t)(1 t), hu i ; x n 2;n 1 i 0 = 0 for i = 1; : : :; n 3, hv 0 i;j; x n 2;n 1 i 0 = 0 for j n 2. can be computed using methods Section 4. rst is quite dicult, se we will discuss a less direct way to compute it later. above identities, it follows that From a linearly independent set eigenvectors n 1 with eigenvalue 1. Each v 2 B 1 is in H, as can be seen by constructing a surface in ~ C whose image in H 2 ( ~ C; ~) lies (1 q) 2 v. is i = 1; : : :; n 3, note that v 0 i;n must be an eigenvector n 1, because For function n 1 can be chosen to x setwise square in C whose lift represents 0 i;n. To nd eigenvalue, note that v 1 n 1 x i;n = x i;n 1 q 1 x i;n ; n 1 v 0 i;n; x i;n i = hv 0 i;n; 1 n 1 x i;ni = qhv 0 i;n; x i;n i: h THE LAWRENCE-KRAMMER REPRESENTATION 63 s s s s 6 multiplication by a unit in. hv 0 i;j ; x i 0 ;ni 0 = 0 for 1 i < j n 2. enclosing. following identities hold up to multiplication by a unit in. B 1 = fv i;j : max(i + 1; 3) < j n 2g [ fu 1 ; : : :; u n 3 g [ fug as shown in Figure 7, so Thus B q = fv 0 i;n : i = 1; : : :; n 3g:

14 a set eigenvectors n 1 with eigenvalue q. y are linearly independent, is since q 6= 1, y are not in span B 1. By a dimension count, B 1 [B q is basis for K H. Each element this basis is annihilated by ( n 1 1)( n 1 +q), a we are done. so to multiplication by a unit in. now describe an indirect argument to up this identity. denote (1 q 2 )(1 + q 2 t)(1 t). obtain P be an -neighbourhood horizontal edge [p n 1 ; p n ], where > 0 is Z be set points fx; yg 2 C such that eir x y are within small. each or, or at least one m lies in P. ~ Z be preimage distance Z in ~ C. x 0 2 H 2 ( ~ C; ~ C [ ~ Z) be a triangle represented by horizontal edge [p n ; 1]. claim that image x n 2;n 1 in H 2 ( ~ C; ~ C [ ~ Z) is x 0. pro is almost I to that Lemma 5.5. This time P is to be treated as one large puncture identical Since P actually contains two puncture points, it is necessary to substitute point. 2 for q in pro Lemma 5.5. q surface representing u does not meet ~ Z, so it is possible to dene an pairing hu; x 0 i. surfaces representing u x 0 intersect transversely intersection one point when projected to C. Thus hu; x 0 i is a unit in. pairing is at so hu; xi =. But is equal to, up to multiplication by a unit in, sesquilinear, we are done. so Temperley-Lieb algebra. now prove that H is a representation 6.3. Temperley-Lieb algebra TL n R be a domain containing an invertible element q with Definition 1 root q 2. Temperley-Lieb algebra TL n (R), or simply TL n, is square a given by generators 1; e 1 ; : : :; e n 1 relations R-algebra e i e j = e j e i if ji jj > 1, e e j e i = e i if ji jj = 1, i 2 1 i = ( q 2 q 1 2 )e i. e 1 map from H n to TL n given by i 7! 1 + q 2 e i. I claim this is Consider has kernel generated by well-dened It be veried by adding relations z i;j = 0 to presentation conventions. 1 can n, substituting 1 + q 2 e i for i, simplifying to obtain presentation for H z i;j are all conjugate to each or, so it suces to prove that Pro. n 2;n 1 acts as zero map on H. have z n 2;n 1 = ( n 2 n 1 n 1 + 1)( n 2 1) z ( n 1 n 2 n 2 + 1)( n 1 1): = 64 STEPHEN BIGELOW In above pro we used fact that hu; x n 2;n 1 i 0 = (1 q 2 )(1 + q 2 t)(1 t); z i;j = i j i i j j i + i + j 1 for ji jj = 1. This is [Jon87, Equation 11.6], except using slightly dierent TL n. Lemma 6.5. z i;j acts as zero map on H.

15 it suces to nd a basis for K H, each whose elements is an eigenvector Thus eir n 1 or n 2 with eigenvalue 1. B 1 be linearly independent set eigenvectors n 1 with eigenvalue as dened in pro Lemma be a circular closed curve based at p n enclosing p n 1 p n 2. i = 1; : : :; n 4, let w i be square corresponding to edges [p i ; p i+1 ] For 0. w n 3 be square corresponding to 0 a circular closed curve based p n 3 enclosing 0. w i all lie in H, are eigenvectors for n 2 with at 1. eigenvalue : S 1! D n be a gure-eight going around p n 1 p n in a regular [p n 1 ; p n ]. For i = 1; : : :; n 3, let i : I! D n be a vertical edge neighbourhood endpoints on D, passing just to right p i. See Figure 8. Dene with f i (x; y) = f i (x); (y)g. This lifts to ~ C. y i be corresponding element by 2 ( ~ C; ~ C). For i; j = 1; : : :; n 3, following holds up to multiplication by a H q 3 i = j 6= n 3 1 q 3 )(1 q 3 t 2 ) i = j = n 3 (1 fw 1 ; : : :; w n 3 g [ B 1 linearly independent. By a dimension count, it forms a basis K H. Each is in this basis is annihilated by z n 2;n 1, so we are done. vector (n 2; 2) representation. have shown that H can be considered 6.4. as a representation TL n. now dene representation S TL n to partition (n 2; 2), show that it is isomorphic to H. A corresponding introduction to representation ory Temperley-Lieb algebra is good 6.6. M be left-ideal TL n generated by e 1 e 3, let N be Definition left-ideal generated by fe 5 ; : : :; e n 1 g. representation TL n corresponding THE LAWRENCE-KRAMMER REPRESENTATION 65 s s s s Figure 8. arcs 1 used to dene y 1. f i : I S 1! C unit in. 8 < hw i ; y j i 0 = 0 i 6= j : Furr, hv; y i i 0 = 0 for all v 2 B 1. Thus [s95]. to partition (n 2; 2) is TL n -module S = M=(M \ N). This is called S(n; 2) in [s95]. It is a free R-module with basis s i;j = (e i : : : e 3 e 2 )(e j 1 : : :e 5 e 4 )(e 1 e 3 )

16 : TL n! H be unique map compatible with action such that = v 0 1;4. n (1) 0 = (q q 1 2 ) 2 j M : (N) = 0, so 0 (M \ N) = 0. refore have a map : S! H induced Thus 0. by 1 i < j n such that j > max(i + 1; 3). will show that (s i;j ) = v 0 i;j up for multiplication by a unit in. to Lemma 6.7. For j = 5; : : :; n we have ( j 1 1)v 0 1;j 1 = v0 1;j 1 : I! D be edge [p 1 ; p 2 ]. 2 : I! D be edge Pro. j 2 ; p j 1 ]. Dene f : I I! C by f(x; y) = f 1 (x); 2 (y)g. n f lifts to a [p ~ f which represents v 0 1;j 1. map can assume that j 1 2 ( 1 2 ) is within distance p j 1. n f maps line f(x; 1 2 )g into. restriction ~ f to I [0; 1 2 v 1;j 1. restriction ~ f to I [ 1 2 ; 1] represents some unit multiple v 1;j. Thus j 1 v 0 1;j 1 = v 0 1;j 1 + v 0 1;j 66 STEPHEN BIGELOW 1 i j n 1 i j n p p p p p p ppp ppp ppp, Figure 9. s i;j for j = i + 2 j > i + 2. for 1 i < j n such that j > max(i + 1; 3). re is a diagrammatic interpretation TL n, in which s i;j corresponds to diagram shown in Figure 9. (e 3 ) = q 1 2 (3 1)(v 0 1;4) = ( q 1 2 q 1 2 )v 0 1;4 : Similarly (e 1 ) = ( q 1 2 q 1 2 )v 0 1;4. Thus (e 1 e 3 ) = (q q 1 2 ) 2 v 0 1;4: 0 : M! H be given by For i = 5; : : :; n 1 we have that (e i ) = q 1 2 (i 1)(v 0 1;4) = 0: Recall that S has a basis s i;j = (e i : : : e 2 )(e j 1 : : :e 4 )(e 1 e 3 ) up to multiplication by a unit in. ] represents some unit multiple for some units ; 2.

17 remains to show that = 1. Recall that hv 0 1;j 1; x 1;j 1 i 0 = 1. Thus It j 1 v 0 1;j 1 ; x 1;j 1i 0 =. But j 1 x 1;j 1 = x 1;j 1, so h h j 1 v 0 1;j 1; x 1;j 1 i 0 = h j 1 v 0 1;j 1; j 1 x 1;j 1 i 0 Lemma 6.8. For i = 2; : : :; j 1 we have ( i 1)v 0 i 1;j = v0 i;j In case i < j 1, pro is almost identical to that Pro. lemma. Suppose i = j 1. n previous i v 0 i 1;j = v0 i 1;j + v0 i;j some ; 2. To see that = 1, use same argument as in pro for lemma. To see that is a unit in, note that it is equal to h i v 0 i 1;j ; x i;ji 0, previous surfaces representing i v 0 i 1;j x i;j intersect transversely at one point projected to C. when By se two lemmas we have (s i;j ) = v 0 i;j References Stephen J. Bigelow, Braid groups are linear, J. Amer. Math. Soc. 14 (2001), no. 2, [Big01] (electronic). MR 2002a: {486 Ryan Budney, A note on image Lawrence-Krammer representation, math.gt/ [Bud02] F. R. Jones, Hecke algebra representations braid groups link polynomials, Ann. V. Math. (2) 126 (1987), no. 2, 335{388. MR 89c:46092 Akio Kawauchi, A survey knot ory, Birkhauser Verlag, Basel, 1996, Translated [Kaw96] from 1990 Japanese original by author. MR 97k:57011 revised Daan Krammer, braid group B4 is linear, Invent. Math. 142 (2000), no. 3, 451{486. [Kra00] 2001k:20078 MR, Braid groups are linear, Ann. Math. (2), 155 (2002), no. 1, 131{156. MR 1 [Kra02] R. J. Lawrence, Homological representations Hecke algebra, Comm. Math. Phys. [Law90] (1990), no. 1, 141{191. MR 92d: Ruth J. Lawrence, Braid group representations associated with slm, J. Knot ory [Law96] 5 (1996), no. 5, 637{660. MR 98j:57011 Ramications Paoluzzi Luis Paris, A note on Lawrence-Krammer-Bigelow representation, Luisa Algebr. Geom. Topol. 2 (2002), 499{518. Taek Song, Lawrence{Krammer representation is unitary, math.gt/ , Won February B. W. stbury, representation ory Temperley-Lieb algebras, Math. Z. [s95] (1995), no. 4, 539{565. MR 96h: Mamatics Statistics, University Melbourne, Victoria Department Australia 3010, address: Department Mamatics, University California at Santa Barbara, Current California THE LAWRENCE-KRAMMER REPRESENTATION 67 hv 0 1;j 1; x 1;j 1 i 0 = 1: = Thus = 1, so ( j 1 1)v 0 1;j 1 = v0 1;j. up to multiplication by a unit in. up to multiplication by a unit in. This implies that is an isomorphism, which completes pro orem , February [Jon87] [PP01] [Son02]