TUTORIAL PROBLEMS 1 - SOLUTIONS RATIONAL CHEREDNIK ALGEBRAS

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1 TUTORIAL PROBLEMS 1 - SOLUTIONS RATIONAL CHEREDNIK ALGEBRAS ALGEBRAIC LIE THEORY AND REPRESENTATION THEORY, GLASGOW 014 (-1) Let A be an algebra with a filtration 0 = F 1 A F 0 A F 1 A... uch that a) F i A F j A F i+j A b) i Z F i A = A. Let u prove that the et i Z F i A/F i+1 A i an algebra. It i clearly an additive abelian group, a all the filtered degree F i A are additive abelian group. The multiplication i defined a follow: for a F i A and b F j A, their clae in [a] F i A/F i+1 A and [b] F j A/F j+1 A multiply a: [a] [b] = [ab] F i+j A/F i+j+1 A. Thi i well defined on clae becaue F i+1 1 F j A F i+j+1 A and F i 1 F j+1 A F i+j+1 A. It inherit the aociativity, ditributivity and C-bilinearity from multiplication on A. (0) Solution to Problem 0 i between Problem and 3, after we have et up the convention for it. 1. Reflection group Let W be a finite complex reflection group, h it reflection repreentation, h the dual repreentation to h, (, ) : h h C the natural pairing, S the et of all reflection in W (element W uch that rank h (1 ) = 1), α h and α h a choice of bai vector of Im(1 ), normalized o that (α, α ) =. We conitenly write x, x,... for element of h and y, y,... for element of h. (1) a) Let be a reflection on h. A rank h (1 ) = 1, (1 ) i diagonalizable, with eigenvalue 0 of multiplicity dimh 1, and one nonzero eigenvalue of multiplicity 1. Therfore, W i diagonalizable on h, with eigenvalue 1 of multiplicity dimh 1, and ome λ 1 1 of multilplicity 1. b) The action of any w W on the dual repreentation i given by w(x) = x w 1 ; in the language of the natural pairing (, ) : h h C, (w(x), y) = (x, w 1 (y)). In particular, on h i diagonalizable with eigenvalue 1 of multiplicity dimh 1, and λ 1 of multilplicity 1. c) In the natural pairing h h C, the eigenpace for λ pair with the eigenpace for λ 1, and i orthogonal to all other. To ee that, notice that if x, y are eigenvector with eigenvalue a, b, then (x, y) = (w(x), w(y)) = ab(x, y), o either ab = 1 or (x, y) = 0. The eigenpace for 1 in h i orthogonal to the eigenpace for λ 1 in h; thi tatement can be rewritten a ((x) = x) (x, α ) = 0. The formula (x) = x 1 λ (x, α )α 1

2 ALGEBRAIC LIE THEORY AND REPRESENTATION THEORY, GLASGOW 014 can now be verified by checking it on a bai of eigenvector: if (x) = x, then (α, x) = 0 o the formula become If y = α, then the LHS i and the RHS i x = x 0. (α ) = λ α α 1 λ (α, α )α = α 1 λ α = λ 1 α. The correponding formula on h i checked accordingly. d) When W i a real reflection group (finite Coxeter group), then any reflection atifie = 1, o λ = 1 and h = h. () For W = S n the ymmetric group on n letter, we can take h = C = h, with the bae y 1,... y n and x 1,... x n, and the pairing (x i, y j ) = δ ij. The action i by permuting the coordinate. The reflection are tranpoition ij which exchange the y i and y j and leave all other y k fixed. All reflection are conjugate, o c i a contant. A choice of α, α, i α ij = x i x j, α ij = y i y j, i < j. λ = 1 for all. Thi repreentation i not irreducible (y y n i a trivial ubrepreentation!), o ometime the ubrepreentation of all the vector in C n whoe coordinate um up to 0 i ued; thi i the n 1 dimenional ubpace panned by all the α ij = y i y j.. The rational Cherednik algebra and PBW Let t C, and let c C be a collection of number parametrized by reflection S, uch that c = c ww 1 for any w W. The rational Cherednik algebra H t,c (W ) aociated to thi data i the quotient of W T (h h ) (the emidirect product of the tenor algebra on h h with W ) by the relation: for all x, x h, y, y h, [x, x ] = 0, [y, y ] = 0 [y, x] = t(x, y) c (α, y)(x, α ). We begin by comparing it with the definition of SRA from the lecture. (0) In lecture we aw the definition of a ymplectic reflection algebra a the quotient of T V C[W ] by the relation [v 1, v ] = tω(v 1, v ) c ω (v 1, v ). Let u how that in a particularly nice cla of example, i coincide with the above definition of a rational Cherednik algebra given here. The etting i: let V = h h. The group W act on h, and therefore on V, and i generated by ymplectic reflection S: the rank of (1 ) on h i one if and only if the rank of (1 ) on h h i. The ymplectic form on V = h h i defined a follow: for x 1, x h, y 1, y h, In particular thi mean that: ω(x 1 + y 1, x + y ) = (x, y 1 ) (x 1, y ). ω(x 1, x ) = 0, ω(y 1, y ) = 0 ω(x, y) = ω(y, x) = (x, y). (Remember that (x, y) i the natural pairing h h C).

3 TUTORIAL PROBLEMS 1 - SOLUTIONS RATIONAL CHEREDNIK ALGEBRAS 3 Next, let u decribe ω. Let be a reflection on h (o, a ymplectic reflection on V ). The element α and α are a bai of Im(1 ) in V. The projection Proj to Im(1 ) along Ker(1 ) can be calculated from Problem 1c), a Proj h (x) = 1 1 λ (1 )(x) = 1 (x, α )α. (To ee that thi i correct: the econd and the third expreion are the ame by 1c, and they define an operator Proj on h. To ee that it i indeed a projection to Im(1 ) along Ker(1 ), firt notice that Proj(x) = 1 1 λ (1 )(x) = 0 for all x Ker(1 ). Then, notice that Proj(α ) = 1 (α, α )α = α.) Similarly, 1 Proj h (y) = 1 λ 1 (1 )(y) = 1 (α, y)α. Finally, ω (y, x) = ω(proj(y), Proj(x)) = ω( 1 (α, y)α, 1 (x, α )α ) = 1 4 (α, y)(x, α )ω(α, α ) = 1 4 (α, y)(x, α )(α, α ) = 1 (α, y)(x, α ) and ω (x 1, x ) = ω (y 1, y ) = 0. The Symplectic Reflection Algebra in thi cae become the quotient of by the following relation: T V C[W ] = T (h h) C[W ] [x, x ] = tω(x, x ) [y, y ] = tω(y, y ) [y, x] = tω(y, x) = t(x, y) c ω (x, x ) = 0 c ω (y, y ) = 0 c ω (y, x) c (α, y)(x, α ). (3) deg(x) = 1, deg(y) = 1, deg(w) = 0 give a filtration of H t,c (W ), becaue all the relation are filtered (in fact, wxw 1 = w(x) and [x, x ] = 0 are homogeneou, and the only filtered i [y, x] = t(x, y) c (α, y)(x, α ), whoe LHS i of degree and RHS i of degree 0. The eay direction of the PBW i that W S(h h ) urject onto gr(h t,c (W )), or alternatively that {x a xan n w y a yan n a 1,... b n Z 0, w W } i a panning et of H t,c (W ). Thi i eay to ee from the relation: they give u a rule for reordering x, y, w o that all the x (which commute) are written firt, then all the group element (which multiply like they did in the group), then all the y (which

4 4 ALGEBRAIC LIE THEORY AND REPRESENTATION THEORY, GLASGOW 014 commute!). Reorder here mean replace by omething in the correct order, plu term of a lower filtered degree. The hard half i to ee that uch a reordering i well defined, i.e. that the map W S(h h ) gr(h t,c (W )) i injecitve, ie that the above et i linearly independent. (4) To ee that the hard half of the PBW theorem i nontrivial, conider the following algebra given by generator and relation: L = C x, y, = 1, x = x, y = y, [y, x] = 1, (y )x = xy +. The argument a above produce a panning et x a y b, x a y b. However, the lat two relation alo give 1 = [y, x] = + x, which can be multiplied by to give x + 1 = or x = 1. The relation x = x now give x = x = 0, and the relation [y, x] = 1 become 0 = 1. So, the algebra i The Dunkl embedding We will fix t = 1 for thi part, although mot tatement work for any t. Let D(h) be the algebra of differential operator on h, generated by function x h and differential operator y, y h, with relation [x, x ] = 0, [ y, y ] = 0, [ y, x] = (x, y). Let h reg = h \ ( {α = 0}) be the open affine ubet obtained by removing the reflection hyperplane ker α = ker( 1). Function on h are C[h] = Sh, and function on h reg are obtained from function on h by inverting all α. The Dunkl embedding i the map H 1,c (W ) W D(h reg ) given by x x w w y D y = y c (α, y) (1 ). In the following problem, we will prove thi i indeed a homomorphim of algebra, and ue it to define a repreentation of H 1,c (W ) on C[h]. (5) a) There i pretty much nothing to check. b) If W = S n, and = ij, thi i the tatement that f(..., x i,..., x j,...) f(..., x j,..., x i,...) i diviible by x i x j, which i true (plugging in x i = x j give 0). In general, let u how it by induction on degree of f, auming WLOG that it i homogeneou: if f i of degree 0, thi i trivially true; if it i of degree 1, then f = x h, and the formula from 1c) give u that f (f) = x (x) i diviible by α. If it i true for polynomial f and h, then it i true for fh, a fh (fh) = (f (f))h + (f)(h (h)). Thi prove the claim by induction on degree of the polynomial. c) Every term of the Dunkl operator map C[h] back to C[h]. (6) We want to how that wd y w 1 = D w(y). Firt, w y w 1 = w(y) (check thi by pairing with x). If i a reflection, then o i ww 1 ; they are conjugate operator and have the ame eigenvalue λ = λ ww 1, and their eigenvalue are related by w: α ww 1 = w(α ). Thu, w ( ) c (α, y) (1 ) w 1 = c 1 λ (α, y) w(α ) (w(1 )w 1 ) =

5 TUTORIAL PROBLEMS 1 - SOLUTIONS RATIONAL CHEREDNIK ALGEBRAS 5 = (7) Calculate, c (α, y) (1 ww 1 ) = c ww 1 1 λ α ww 1 1 λ ww 1 = c (α, y) [D y, x] = [ y, x] (1 ). [ ] c 1 (α, y) (1 ), x c 1 λ (α, y) = (x, y) = (x, y) c (α, y) x (x) = (x, y) c (α, y)(x, α ). (α ww 1, w(y)) (1 ww 1 ) = α ww 1 ( 1 α x 1 α x x α + x α (8) It i poible but pretty painful to how [D y1, D y ] = 0. Intead, it i enough to how that [[D y1, D y ], x] = 0. To ee that, firt ue: Next, [[D y1, x], D y ] = [(x, y 1 ) The next tep i [[D y1, D y ], x] = [[D y1, x], D y ] [[D y, x], D y1 ]. c (α, y 1 )(x, α ), D y ] = ) c (α, y 1 )(x, α )[, D y ]. [, D y ] = D y D y = D (y) D y = D ((y) y ) = 1 λ 1 (α, y )D α, o [[D y1, x], D y ] = c (α, y 1 )(x, α ) 1 λ 1 (α, y )D α. Thi i ymmetric under 1, o [[D y1, D y ], x] = [[D y1, x], D y ] [[D y, x], D y1 ] = 0. (9) In the lat few problem, we have jut checked all the relation of H 1,c (W ) after the map to W D(h reg ), o we proved that the Dunkl embedding i a homomorphim of algebra H 1,c (W ) W D 1 (h reg ). It end a panning et to a linearly independent et, o it i an embedding. Together with Problem (5), we have defined a faithful repreentation of H 1,c (W ) on C[h] = Sh, where x h act by multiplication, w W diagonally, and y h by Dunkl operator. In fact, we have written down the tandard module (triv).