ON THE ACTION OF A q-DEFORMATION OF THE TRIVIAL IDEMPOTENT ON THE GROUP ALGEBRA OF THE SYMMETRIC GROUP

Let A be the arrangement of hyperplanes consisting of the re ecting hyperplanes for the root system An 1. Let B = B(q) be the Varchenko matrix for this arrangement with all hyperplane parameters equal to q. We show that B is the matri x with rows and columns indexed by permutations with ; entry equal to qi( 1) where i( ) is the number of inversions of . Equivalently B is the matrix for left multiplication on CSn by n(q) = X 2Sn qi( ) : Clearly B commutes with the right-regular action of Sn on CSn. A general theorem of Varchenko applied in this special case shows that B is singular exactly when q is a j(j 1)st root of 1 for some j between 2 and n. In this paper we prove two results which partially solve the problem (originally posed by Varchenko) of describing the Sn-module structure of the nullspace of B in the case that B is singular. Our rst result is that ker(B) = indSn Sn 1(Lien 1)=Lien in the case that q = e2 i=n(n 1) where Lien denotes the multilinear part of the free Lie algebra with n generators. Our second result gives an elegant formula for the determinant of B restricted to the virtual Sn-module P with ch aracteristic the power sum symmetric function p (x). 2 Section 1: Introduction. Let A = fH1; : : : ; H`g be an arrangement of hyperplanes in IRn, let be the union of the hyperplanes in A and let R be the set of regions in IRnn . In [V1], Varchenko de nes a matrix B wit h rows and columns indexed by the regions in R. This matrix depends on a collection of real parameters aH , one for each hyperplane H in A. The R; S entry of B is given by BR;S = YH aH where the product on the right is over hyperplanes H which separate R and S. The matrices B de ned above are studied in papers of Varchenko [V], Schechtman/Varchenko [SV] and Brylawski/Varchenko [BV]. These matrices describe the analogue of Serre's relations for quantum Kac-Moody Lie algebras and are relevant to the study of h ypergeometric functions and the representation theory of quantum groups. In these applications, the kernel of B is of particular interest. The kernel of B depends on the values of the aH . In general, the exact relationship is di cult to describe. But the following theorem of Varchenko gives considerable information about this dependence. Theorem 1.1: (Varchenko [V]). Let notation be as above. Then det(B) = YF 1 a(F )2 p(F )q(F ) where the product is over the non-zero ats in the intersection lattice of A, (see [O]) where a(F ) denotes the product of the aH over hyperplanes H 2 A which contain F and where p(F ) and q(F ) are non-negative integers def ined in [V1]. Our work in this paper will concern a particular arrangement. Fix a positive integer n. For each i < j let Hij denote the hyperplane in IRn consisting of all (x1; : : : ; xn) such that xi = xj. Let An be the arrangement consisting of all Hij with 1 i < j n. For each H 2 An, let aH = q where q is a complex indeterminate. 3 In this case is the set of all vectors (x1; : : : ; xn) 2 IRn which have (at least) one pair of equal coordinates. The regions in R are naturally indexed by permutations in Sn according to 2 Sn indexes the region R = f(x1; : : : ; xn) : x 1 < x 2 < : : : < x ng: ( ) Via this indexing, we can think of the rows and columns of B as corresponding to permutations. We will let B ; denote the R ; R entry of B for ; 2 Sn. In this way, we will think of B as an endomorphism o f CSn. Lemma 1.2: Let n(q) 2 C[q]Sn be de ned by n(q) = X 2Sn qi( ) where i( ) is the number of inversions of . Then B, considered as an endomorphism of CSn, is the matrix for left multiplication by n(q). Proof: Consider ; entry of B for ; 2 Sn. This entry is q` where ` is the number of hyperplanes Hjk which must be crossed in going from R to R . Note that: Hjk must be crossed in going from R to R i j > k and j < k i j > k and ( 1) j < ( 1) k i ( j; k) is an inversion of 1. So the ; entry of B is equal to qi( 1) which proves the lemma. [] 4 Next we consider what Theorem 1.1 tells us in this case. The intersection lattice of A is the partition lattice n. It turns out that the exponent p(F ) is 0 unless the partition F has exactly one non-trivial block. If F has one no n-trivial block of size ` then p(F ) = (` 2)!(n `+ 1)!. So in this case, Theorem 1.1 states that det(B) = n Ỳ=2 1 q`(` 1) (ǹ)(` 2)!(n `+1)! : (1:1) It should be noted that (1.1) was rst proved by Zagier [Z]. In this paper we realize it as a special case of the Varchenko determinant factorization. There is one other feature of this arrangement which makes it particularly interesting. The symmetric group acts on the set of regions R via the right-regular representation on the index. More precisely, if 2 Sn then (R ) = R . By Lemma 1.2, this action of Sn commutes with B and so B preserves the isotypic components of this Sn-action and Sn acts on the kernel of B. Two questions motivated the work in this paper. Question 1: What is the Sn-module structure of ker(B) at values of q for which det(B) = 0? Question 2: For eachq irreducible representation S of Sn , what is the determinant of the restriction of B to the S -isotypic component of R? Question 2 was posed independently by Varchenko [V] and by Zagier [Z, page 201]. Zagier's work concerns the relevance of the operator B to a model in in nite statistics. This model, suggested by Hegstrom gives a q-deformation of the classical Bose (q = 1) and Fermi (q = 1) statistics. It was studied for q = 0 by Greenberg [Gr] and for general q by Polyakov and Biedenharn [B]. 5 In this paper we will prove a conjecture due to R. Stanley which answers a question which is equivalent to Question 2. With regard to Question 1, we will state a conjecture due to Hanlon and Varchenko which provides an elegant answer to Question 1 for a number of interesting values of q. We will prove this conjecture in one particularly interesting special case. Section 2: A factorization of n(q). In this section we state a remarkable result which was rst proved by Zagier [Z] and later re-discovered in the paper [DKKT]. To state this result we need some notation. We will use two embeddings of CSn 1 in CSn. The rst is the usual embedding, namely we think of Sn 1 as the collection of permutations xing n. More precisely, if is in Sn 1, its image under this embedding is obtained from by adding a one-cycle containing the number n. If A 2 CSn 1 we let A also denote its image in CSn under this embedding. The second embedding, which we denote , maps Sn 1 to the collection of permutations in Sn which x 1. If 2 Sn 1 is written in disjoint cycle form then ( ) is obtained from by increasing every number by 1 and then adding a one-cycle containing the number 1. For example, ((1; 3; 5)(2; 4)) = (2; 4; 6)(3; 5)(1). This de nes on Sn 1 and it extends linearly to an embedding of CSn 1 into CSn. For each n, let Tn(q) denote the sum Tn(q) = n X j=1 qj 1 n j(zj): Here zj is the j-cycle (j; j 1; : : : ; 1) in Sj so Tn(q), with permutations written out in disjoint cycle form, is Tn(q) = n X j=1 qj 1(n; n 1; : : : ; n j + 1): It is straightforward to verify that n(q) = T2(q)T3(q) : : : Tn(q): (2:1) 6 The factorization result in [Z] gives a factorization of Tn(q) which in turn gives a factorization of n(q) by (2.1). To state the result, we need to de ne elements Gj(q); Hj(q) 2 CSj which we do inductively according to the following equations: G1(q) = H1(q) = 1 Gj(q) = 1 qjzj 1 (Gj 1(q)) Hj(q) = (Hj 1(q)) 1 qj 1zj 1 : Note that Hj(q) is de ned only for those values of q with the property that 1 q` 1z` is invertible in CSj for all ` 2 f2; 3; : : : ; jg. It is straightforward to see that 1 q` 1z` is invertible in Sj if and only if 1 q`(` 1) is nonzero. Theorem 2.1: ([Z], Prop. 2) Suppose Qǹ=2 1 q`(` 1) is nonzero. Then Tn(q) = Gn(q)Hn(q), and so n(q) = G2(q) H2(q) G3(q) H3(q) : : :Gn(q) Hn(q): This result can be extended to all values of q by restating it as: Tn(q) 1 qn 1zn = (1 qnzn 1) (Tn 1(q)) : (2:2) As an example of (2.2), let n = 3. Then (2.2) is the equation (1 + q(3; 2) + q2(3; 2; 1))(1 q2(3; 2; 1) = (1 q3(2; 1))(1 + q(3; 2)): which is easily veri ed. 7 Section 3 { The Main Results. In this section we state and prove our two main results. We begin with a conjecture that is relevant to Question 1. In general, it seems that the Sn-module structure of the kernel of n(q) is quite complicated. Below we see a table which giv es the character values of Sn acting on the kernel of 4(q) for all values for which 4(q) is singular. In this table the rows are indexed by values of q and the columns by conjugacy classes in S4 (denoted by cycle type). 14 212 22 31 4 1 9 -1 1 0 -1 -1 9 1 1 0 1 e2 i=3 3 1 -1 0 1 e2 i=4 2 0 2 2 0 e2 i=6 3 -1 -1 0 1 e2 i=12 2 0 2 -1 0 The reader will note that it is di cult to pick out a pattern even in the degrees of the representations. This observation is reinforced when one examines the same data for other values of n. However, Conjecture 3.1 states that the Sn-module stru cture of ker( n(q)) has a very elegant description for certain values of q. Conjecture 3.1: Suppose that q is a root of exactly one of the factors on the right-hand side of (1.1). More precisely, suppose that q = e2 is=j(j 1) for some j 2 f2; 3; : : : ; ng and some nonnegative integer s and that qk(k 1) 6= 1 for k 6= j; 2 k n. Then (a) The right Sn-module structure of ker(Tn(q)) is ker(Tn(q)) = indSn Cj 1(qj)= indSn Cj (qj 1) where Cj 1 is the subgroup of Sn generated by the (j 1)-cycle zj 1 = (j 1; j 2; : : : ; 2; 1), and where qj denotes the linear character of Cj 1 whose value on zj 1 is qj. (b) As right Sn-module ker( n(q)) consists of n j + 1 copies of ker(Tn(q)) 8 It is interesting to note that there are examples of j(j 1)st roots of unity q (with 2 j n) for which indSn Cj (qj 1) is not even contained in indSn Cj 1(qj). If Conjecture 3.1 is correct, then this containment must hold whenever 1 qk(k 1) is nonzero for all k not equal to j. We will prove Conjecture 3.1 in what is arguably the most interesting case. We denote by Lien the representation of Sn on the multilinear part of the free Lie algebra with n generators. There is an extensive literature on this representation (se e for example [G], [R1], [R2]). We will need one fact about the Lien representation. Lemma 3.2: (see [R1], pg. 215) For each n, Lien = indSn Cn e2 i=n : Theorem 3.3: Let q = e2 i=n(n 1). Then ker n(q) = indSn Sn 1(Lien 1)=Lien: Before going ahead with the proof of Theorem 3.3 we will need a pair of technical lemmas. Lemma 3.4: Suppose j n. As an endomorphism of CSn (acting by left multiplication) Tj(q) has determinant j Ỳ=2 1 q`(` 1) n!=`(` 1) : Proof: The proof is by induction on n. The result is easy to check for n = 1 so assume the result is true for n 1. First, suppose j n 1. As an endomorphism of CSn 1; Tj(q) has determinant j Ỳ=2 1 q`(` 1) (n 1)!=`(` 1) : 9 For each i = 1; 2; : : : ; n, let Vi be the subspace of CSn spanned by all permutations which map i to n. Then Tj(q) preserves each Vi and the action of Tj(q) on Vi is clearly isomorphic to the action of Tj(q) on CSn 1. I t follows that the determinant of Tj(q) on CSn 1 is the nth power of the quantity given in (3.1) which proves Lemma 3.4 in this case. Next suppose that j = n. By (2.1) we have det (Tn(q)) = det ( n(q)) n 1 Y j=2 det Tj(q) 1 = n 1 Ỳ=2 1 q`(` 1) E` where E` = ǹ!(` 2)!(n `+ 1)! (n `) n! `(` 1)! = n! `(` 1) : Here we have combined (1.1) with the cases j < n for this lemma to get E`. This completes the induction step. [] The next lemma is a simple consequence of the Smith normal function. Lemma 3.5: Let M(z) be an N by N matrix whose entries are polynomials in z. Let z0 be a complex number such that det(M(z)) = (z z0)D f(z) where f(z0) 6= 0. Then dim(ker(M(z0))) D. The third lemma is again a standard result from representation theory. Although it may not be explicitly stated in most references, it follows easily from the de nition of induced representations. 10 Lemma 3.6: Let H be a subgroup of a nite group G and let be a linear character of H. Let be the element of CG given by: = 1 jHj X h2H (h)h: Then the right ideal of CG generated by is isomorphic, as a right G-module, to indGH( ). We are now ready to proceed with the proof of Theorem 3.3. Let q = e2 i=n(n 1). By Lemma 3.4, Tj(q) is invertible as an endomorphism of CSn for j < n. So ker( n(q)) = ker(Tn(q)) where the isomorphism is one of right Sn-modules. Hence it is su cient to show that ker (Tn(q)) = indSn Sn 1(Lien 1)=Lien: (3:2) Recall from (2.2) that Tn(q)(1 qn 1zn) = (1 qnzn 1) (Tn 1(q)): (2:2) Our rst step is to understand the kernel of the right-hand side of (2.2). Again appealing to Lemma 3.4 we have that (Tn 1(q)) is invertible as an endomorphism of CSn. So, as right Sn-modules, ker 1 qnzn 1 Tn 1(q) = ker (1 qnzn 1) : It is straightforward to check that the kernel of 1 qnzn 1 (acting as usual on CSn by left-multiplication) is the right ideal generated by = 1 (n 1) n 2 X̀=0(qnzn 1)`: So by Lemma 3.6, we have the following equalities of right Sn-modules: ker (1 qnzn 1) (Tn 1(q)) = indSn Cn 1(qn) = indSn Sn 1(Lien 1): (3:3) (The last equality from Lemma 3.2). We now consider the kernel of the left-hand side of (2.2). A similar argument involving Lemma 3.6 shows that ker 1 qn 1zn = indSn Cn(qn 1) = Lien; (3:4) 11 the last equality from Lemma 3.2. By standard linear algebra arguments, dim ker Tn(q) 1 qn 1zn dim ker Tn(q) + dim ker 1 qn 1zn (3:5) with equality if and only if ker Tn(q)(1 qn 1) = ker(Tn(q)) ker (1 qn 1) (3:6) (the last isomorphism being an isomorphism of right Sn-modules). Combining Lemmas 3.4 and 3.5 we have dim ker Tn(q) (n 1)!: This inequality together with (3.4) and (3.5) imply that dim ker Tn(q) 1 qn 1zn (n 2)! + (n 1)! = n(n 2)! (3:7) with equality if and only if (3.6) holds. But by (3.3) ker Tn(q) 1 qn 1 = indSn Sn 1(Lien 1); which has dimension n(n 2)!. This implies that there is equality in (3.7) and hence the isomorphism in (3.6) holds, i.e., indSn Sn 1(Lien 1) = ker Tn(q) Lien: [] It is interesting to note that the same representation indSn Sn 1(Lien 1)=Lien has recently appeared in another quite di erent context. Let Fn denote the space of homeomorphically irreducible trees with n labelled leaves. We think of Fn as a topological space with the trees near to each other if they di er by a small deformation (just as drawings in the plane). There is an action of Sn on Fn induced by permutations of the labels on the leaves. This action lifts to an action of Sn on H (Fn), the homology of Fn. Sarah Whitehouse [W] showed that H (Fn) = indSn Sn 1(Lien 1)=Lien: An interesting problem is to construct an Sn-equivariant isomorphism between ker(Tn(q)) and H (F). 12 We next consider Question 2. As stated, this question probably does not have a nice answer. Below we see a chart which has rows indexed by partitions of small size. The entry in row is the determinant DS (B) of the restriction of B to t he S -isotypic component of R. In this chart we let Ij denote the jth cyclotomic polynomial, i.e., the polynomial whose roots are the primitive jth roots of 1. DS (B) 1 I0 2 I2 12 I1 3 I2I3 21 I6 1I6 2 13 I1I6 4 I2 2I3I4 31 I18 1 I18 2 I9 3 22 I8 1I8 2I2 12 212 I18 1 I18 2 I9 6 14 I2 1I4I6 : 13 DS (B) 5 I2 2I3I4I5 41 I10 1 I12 2 I6 3I3 4 32 I14 1 I14 2 I5 3I10I2 12 312 I16 1 I16 2 I5 3I5 6 I20 221 I14 1 I14 2 I5I5 6 I2 12 213 I12 1 I10 2 I3 4I6 6 15 I2 1I4I6I10 6 I3 2I2 3I4I5I6 51 I15 1 I18 2 I9 3I6 4 I3 5 42 I31 1 I34 2 I14 3 I5 4I3 10I3 12I30 412 I34 1 I36 2 I16 3 I5 4I7 6I15I20 32 I18 1 I17 2 I8 3I10I3 12I15 321 I58 1 I58 2 I15 3 I3 5I15 6 I3 10I6 12I2 20 313 I36 1 I34 2 I7 3I5 4 I16 6 I20I30 23 I17 1 I18 2 I5I8 6 I3 12I3