FROM DAHA TO EHA ANDREI NEGUȚ

Transcription

1 FROM DAHA TO EHA ANDREI NEGUȚ 1. Goals The mai purpose of this talk is two coect the two halves of our semiar. Specifically, we will follow the outlie below: Cosider the spherical double affie Hecke algebra DAHA of gl Defie the limit Idetify the geerators ad relatios of the limit with those i the elliptic Hall algebra EHA Note that the first bullet was itroduced i José s talks, although we will recall it explicitly with focus o gl. The we will use thigs from Chris talks to work out the secod bullet. Fially, the formulas we will work out i the third bullet will be compared with Mitya s talks ext week. The referece is Schiffma Vasserot [1]. 2. The spherical DAHA of gl Recall Defiitio ad Theorem of José s otes the DAHA: H = Cq, v T 1,..., T 1, X 1 ±1,..., X±1, Y 1 ±1,..., Y ±1 subject to the relatios that all X s commute, all Y s commute, ad: 1 T i vt i + v 1 = 0 T i T j = T j T i T i T i+1 T i = T i+1 T i T i+1 2 T i X j = X j T i T i Y j = Y j T i T i X i T i = X i+1 T 1 i Y i T 1 i = Y i+1 3 Y 1 X 1...X = qx 1...X Y 1 Y 1 2 X 1 Y 2 X 1 1 = T 2 1 where i, j go over all possible idices such that j / {i 1, i, i+1}. Recall the actio: 4 SL 2 Z H i which the geerators of SL 2 Z act as: T i T i : X 0 1 i X i Y i Y i X i T 1...T i 1 1 T i 1...T : 1 1 T i T i X i X i Y i T i 1...T 1 T 1...T i 1 Y i Y i 1

2 2 ANDREI NEGUȚ Fially, recall the idempotet: e = 1 []! + v σ S v lσ T σ where T σ = T i1...t ir correspods to a reduced decompositio of σ as a product of traspositios. Recall that the v factorial is defied by settig: 7 [i] ± v = v±2i 1 v ±2 1 It is easy to show that: = []! ± v = [1] ± v... [] ± v 8 e 2 = e ad et i = T i e = ve As i Defiitio of José s otes, let: SH = eh e deote the spherical subalgebra of H, which is a algebra i its ow right with uit e. As we will see i Propositio 3, the size of this subalgebra is the same as the size of Cq, v[x 1 ±1,..., X±1, Y 1 ±1,..., Y ±1 ] S. Sice the actio of SL 2 Z o H leaves e ivariat, we coclude that SL 2 Z preserves SH. 3. The geerators For ay k > 0, Schiffma Vasserot i [1] cosider the followig elemets of SH : 9 0,k = ey k Y k e They further geeralize these elemets to arbitrary a, b Z 2 \0, by lettig k = gcda, b ad defiig: 10 a = g 0,g where deote arbitrary itegers such that the matrix o the left has determiat 1. We claim that there is o ambiguity here, sice the itegers deoted are determied up to multiplyig the matrix 10 o the right with powers of the matrix 6. Sice the latter preserves both e ad the Y s, it preserves the elemets 9, ad so 10 are uiquely defied for ay a ad b. Propositio 1. For ay a, b Z, we have: 11 a,1 = [] v ey 1 X a 1 e 12 1,b = q[]+ v ex 1 Y b 1 e 13 1,b = [] v ey b 1 X 1 1 e 14 a, 1 = q[]+ v ex a 1 Y 1 1 e Proof. Sice Y i+1 = T 1 i Y i T 1 i, we ca use 8 to ifer ey i+1 e = v 2 ey i e. Iteratig this relatio gives us: 0,1 = 1 + v v 2+2 ey 1 e = [] v ey 1 e b g

3 FROM DAHA TO EHA 3 Let us ow hit this elemet with various matrices SL 2 Z to obtai 11: a [ a ] 1 1 a,1 = ,1 = [] v e Y e = [] v ey 1 X1 a e Hittig the case a = 1 with the other geerator of SL 2 Z gives us: [ 1 ] b ,b = 0 b 1 1 1,1 = [] v e Y X 1 e = [] v ey 1 X 1 Y1 b 1 e Usig formula 2.10 of [1] together with 8, we have: 15 Y 1 X 1 = qt 1...T 1 T 1...T 1 X 1 Y 1 = = ey 1 X 1 Y1 b 1 e = qv 2 2 ex 1 Y1 b e ad so the above relatio implies 12. To obtai 13 ad 14, ote that: = seds X X 1 Y 1 X1 1, Y 1 X = Therefore, formulas 11 ad 12 imply: 0 1 1,b = b,1 = [] v a, 1 = ,a = q[]+ v 1 0 thus provig 13 ad 14. Propositio 2. For ay k N, we have: 18 0,k = e Yi k e 19 k,0 = e X k i e 20 0, k = qk e 21 k,0 = qk e Proof. The followig elemet of SL 2 Z: = seds X 1 Y1 1, Y 1 Y 1 X 1 Y1 1 ey 1 X b 1e = [] v ey b 1 X 1 1 e ex 1 Y a 1 e = q[] + v ey a 1 1 X a 1 Y 1 1 e Y k i e Xi k e takes Y 1 Y 1 X1 1 X1 1 ad T i T i. The oe ca iterate 2 to show that this matrix takes ay Y i X 1 i, ad so it takes relatio 18 to 19 ad 20 to 21. However, ote that 18 is

4 4 ANDREI NEGUȚ simply the defiitio 9, so it remais to prove 20. To this ed, let us cosider the followig elemet of SL 2 Z: Γ := = takes ΓY 1 = X 1 Y 1 1 X1 1. Usig 15 ad 2, we may rewrite this as: ΓY 1 = qy 1 1 T 1...T 1 T 1...T 1 = qt T 1 1 Y 1 T 1...T 1 Because the product of T s o the left is the iverse of the product o the right, we may raise this relatio to the k th power: ΓY k 1 = q k T T 1 1 Y k T 1...T 1 Because the idempotet e satisfies et i = T i e = ve, see 8, we coclude that: 22 ΓeY1 k e = q k ey k e As [1] claims, there is a uique polyomial P with coefficiets i Cq, v such that: 23 P ey 1 e,..., ey1 k e = e Yi k e i fact, this is true if oe replaces i Y i with ay other degree k symmetric polyomial i the Y variables, ad that the same polyomial satisfies: 24 P ey 1 e,..., ey k e = e Y k i e The reaso why these two equalities hold for the same polyomial P follows from the automorphism of the sigle affie Hecke algebra that seds T i T i ad Y i Y 1 +1 i this automorphism ca be checked either by had, ote that it is closely related to Theorem of Seth s otes. Note that the degree of P i its first variable, plus twice its degree i the secod variable,... plus k times its degree i the last variable, equals k. Combiig 22, 23, 24 yields: Γ e Yi k e = q k e Y k i e which is precisely what we eeded to prove. Propositio 3. The elemets geerate SH as a algebra. Proof. Sice all the structure costats i SH are Lauret polyomials i q ad v, ad SH is free over the rig C[q ±1, v ±1 ] this was proved by Cheredik it is eough to prove the propositio i the specializatio q = v = 1. Note that: 25 SH q=v=1 = C[x ±1 1,..., x±1, y 1 ±1,..., y±1 by sedig ep X, Y e to the symmetrizatio that is, the average over all! permutatios of the variables of the Lauret polyomial P x, y. Sice the symmetrizatios of the polyomials i the right had sides of ad geerate the right had side of 25 this is a exercise, the Propositio follows. ] S

5 FROM DAHA TO EHA 5 4. The relatios I preparatio for the stable limit, let us rescale our geerators to: 26 u = vk q k v k k where k = gcda, b. For all coprime a, b, defie: θ ak,bk x k = exp v k v k u ak,bk x k k=1 Propositio 4. The elemets u SH satisfy the commutatio relatios: k=1 28 [u, u a,b ] = 0 if ab = a b, ad: 29 [u, u a,b ] = ±θ a+a,b+b ql 1v l q l v l lv v 1 if oe of the followig situatios occurs let l = gcda, b above: ab = a b ± k, gcda, b = k, gcda, b = 1, gcda + a, b + b = 1 ab = a b ± k, gcda, b = 1, gcda, b = 1, gcda + a, b + b = k Remark 1. By Pick s theorem, the lattice poits a, b, a, b that appear i 29 are those such that the triagle with vertices 0, 0, a, b, a+a, b+b has o lattice poits iside ad o the edges, with the possible exceptio of the edge 0, 0, a, b i the case of the first bullet or the edge 0, 0, a + a, b + b i the case of the secod bullet. Proof. Sice the actio of SL 2 Z o lattice poits is trasitive, it is eough to check 28 whe a = 0. I this case, relatios 18 ad 20 tell us that: 30 u 0,b = cost e i Y b i e ad u 0,b = cost e i Yi b e Because e commutes with symmetric polyomials i the Y i, 30 commute because the Y i s commute with each other. By a similar logic, we ca use the SL 2 Z actio to make a, b, a, b equal to 0, ±k, 1, 0 i the case of the first bullet, ad k, 1, 0, 1 i the case of the secod bullet. Moreover, usig oe more rotatio, we may assume k > 0. Therefore, it remais to prove: 31 [u 0,±k, u 1,0 ] = ±u 1,±k qk 1v k q k v k k 32 [u k, 1, u 0,1 ] = θ k,0 q 1vq 1 v 1 v v 1 we used the fact that θ = u v 1 v if gcda, b = 1. Let us prove the first. I the otatio of the previous Subsectio, it amouts to: [ ] 0,±k, 1,0 = ± 1,±k qk 1

6 6 ANDREI NEGUȚ Whe the sig is ± = +, relatios 12 ad 18 make this relatio is equivalet to: [ ] [ ] 33 e Yi k e, ex 1 e = e Yi k, X 1 e = q k 1eX 1 Y1 k e The first equality holds o geeral grouds because Y i k is symmetric, while the secod equality is proved i a 6 page computatio i Appedix A of [1]. Whe the sig is ± =, oe must apply the followig automorphism to 33: T i T1 1, X 1 Y 1 X 1 Y1 1, Y i Y 1 i, v v 1, q q The above is the compositio of SL Z of 17 ad the automorphism: T i T 1 1, X i Y i, Y i X i, v v 1, q q 1 that was itroduced i Theorem of José s otes. As for relatio 32, it reads: [ ] k, 1, θ k,0 q 1 0,1 = v v 1 vq 1 v where θ k,0 is defied via the geeratig series: θ k,0 1 + x k = exp v k v k v k q k v k q k e kx k k=1 k=1 Xk i e The we may use 14, 18 ad 21 to write the required relatio as: [ ] 1 qv 2 v e X1 k Y1 1, Y i e = θ k,0 q 1 which is proved i a 5 page computatio i Appedix B of [1]. Let us defie the Cq, v algebra: 5. The stable limit A = u Z /relatios 28 ad 29 SH [[x 1 ]] Mitya will discuss this algebra i more depth ext week, ad the Alexey ad Tudor s talks will idetify it with the elliptic Hall algebra. Meawhile, ote that Propositio 3 ad 4 imply that there exist surjective rig homomorphisms: 35 A φ SH, u the RHS of 26 for ay N. Our goal is to make the above ito a isomorphism by lettig. Ufortuately, this will oly be possible whe we restrict to the positive halves of the algebras i questio: A A + = Cq, v u Z 2,+ SH SH + = Cq, v Z 2,+ where Z 2 Z 2,+ = {a, b, a > 0 or a = 0, b > 0} deotes half of the lattice plae. The the goal of the remaider of this talk is to prove the followig Propositios:

7 FROM DAHA TO EHA 7 Propositio 5. There exists a morphism SH + SH + 1 give by P 1 P. Clearly, the maps φ of 35 are compatible with the morphisms i Propositio 5. Propositio 6. The iduced map: A + θ + lim SH + give by u...,,..., is a isomorphism. Proof. of Propositio 5: Recall Cheredik s basic represetatio: 36 H DiffA S where DiffA = Cq, v[x ±1 1,..., x±1, D 1 ±1,..., D± ] is the rig of q differece operators o puctured dimesioal space, whose geerators satisfy the relatios: [x i, x j ] = [D i, D j ] = 0 D i x j = q δj i xj D i Specifically, the map 36 is give by: 37 X i multiplicatio by x i 38 T i vs i + x i+1v v 1 s i 1 x i x i+1 39 Tm Ti Y i T i 1...T 1 s m 1...s 1 D 1 where s i S deotes the traspositio of i ad i+1. Note that the basic represetatio was discussed i both Seth s ad José s otes Theorem of the latter, together with the first formula after Defiitio Because of the deomiators x i x i+1, the target of the map 36 is more precisely a certai localizatio of the rig DiffA, but there s o eed to burde the otatio with detail. Whe we restrict this embeddig to the spherical Hall algebra, we obtai the compositio: 40 SH DiffA S S DiffA S which is also a embeddig. The map o the right is the projectio D σ D for all D DiffA ad σ S. Let us cosider the smaller subalgebras: SH + SH ++ = Cq, v, 0 DiffA Diff ++ A = Cq, v[x 1,..., x, D 1,..., D ] We claim that the maps 40 restrict to: 41 ψ : SH ++ Diff ++ A S Ideed, Propositios 3 ad 4 imply that the domai is geerated by 0,k ad P k,0 this kid of geeratio statemet will be discussed i more detail i Mitya s talk so it is eough to show that these elemets lad i Diff ++ A DiffA. Comparig formulas 18, 21 with 37, 39, this statemet is clear sice o egative powers of x i ad D i come up i the latter formulas.

8 8 ANDREI NEGUȚ Lemma 1. The maps ψ of 41 ca be completed to a commutig square: 42 SH ++ ψ Diff ++ A S SH ++ 1 where the dotted map o the left takes ψ 1 Diff ++ A 1 S 1 γ 1 P, ad the map γ is give by: 43 γx i = x i γx = 0 44 γd i = D i γd = 0 v for all i {1,..., 1}. Note that γ is a homomorphism. Note that the map γ would ot have bee defied if we had cosidered aythig greater tha the ++ algebras, because the with b < 0 ivolve iverse powers of Y i, ad we could ot have set D 0 i 44. Let us show how this Lemma implies Propositio 5. Cosider ay relatio: 45 0 = cost i a i,b i SH + for various choices of a i > 0 or a i 0, b i > 0. Because the sum is fiite, we may choose some k large eough so that b i + ka i 0 for all i that appear i 45. The SL 2 Z ivariace of spherical DAHAs implies that we have a relatio: 46 0 = cost i a i,b i+ka i SH ++ By Lemma 1, relatio 46 also holds with replaced by 1, ad therefore SL 2 Z ivariace implies that so does 45. Havig showed that ay relatio betwee the geerators of SH + also holds i SH + 1, this cocludes the proof of Propositio 5. Proof. of Lemma 1: Relatios 28 ad 29 imply that for ay a, b, there exists a fiite polyomial Q i the variables α 1, α 2,..., β 1, β 2,... such that: Q 0,1, 0,2,..., P 1,0, 2,0,... = for all. It s really importat that the above relatio holds i SH for all, for a fixed polyomial Q the combiatorics which establishes this fact will be discussed i more detail by Mitya i the ext talk, but it s ot hard to believe. Sice ψ, ψ 1, γ i 42 are homomorphism, to establish the commutativity of the square, it is therefore eough to prove that: γ ψ ad γ ψ ψ 1 SH ,a This is obvious for the latter, amely a,0, because its image uder ψ is q k xa i. As for the former, it is true that for ay symmetric polyomial fy 1,..., Y we have: 47 γ ψ efy 1,..., Y 1, Y e = γ ψ efy 1,..., Y 1, 0e a,0

9 FROM DAHA TO EHA 9 Oe way to see this is to chase through the defiitios ad observe that: ψ efy 1,..., Y 1, Y e = L f where the operator L f DiffA S was itroduced i Lemma of José s talk, or Defiitio 3.13 of Chris talk. The equatio 47 is merely the compatibility of L f s for ad 1 via the homomorphism γ. Alteratively, sice both sides of 47 are additive ad multiplicative i f, it is eough to check the equality whe f is the k th elemetary symmetric fuctio i Y 1,..., Y. I this case, Macdoald shows see Lemma 4.5 of [1] that: I =k i I ψ e Y i1...y ik e x i v x j v 1 = D i x i x j 1 i 1<...<i k I {1,...,} j / I ad it is clear that settig x, D 0 i the right had side produces the correspodig expressio whe is replaced by 1 up to a power of v, which is accouted for i 44. Proof. of Propositio 6: Sice the geerate SH+ by Propositio 4, ι + is surjective. To prove it is also ijective, it is eough to show that the aalogous map: A ++ ι ++ lim SH ++ is ijective we have already see the reaso for this: if there s a relatio of the form 45 i the kerel of ι +, the we could act with a elemet of SL 2 Z to tur it ito a relatio of the form 46 i the kerel of ι ++. We claim the followig: The algebra A ++ is graded by N 0 N 0, with u i degree a, b The dimesio of A ++ i I is equal to the umber of uordered collectios: 48 a 1, b 1,..., a t, b t with ai = a ad b i = b The first bullet is immediate, ad the secod bullet will be explaied by Mitya i more detail. The ituitio behid it is the followig: elemets of A ++ are liear combiatios of ordered products u a1,b 1...u at,b t with a i, b i 0, ad the secod bullet claims that we ca always use relatios 28 ad 29 to rearrgage the terms i this product such that the lattice poits a 1, b 1,..., a t, b t form a covex path. The umber of covex paths is equal to the umber of uordered collectios 48. Therefore, to prove the ijectivity of ι ++, it is eough to show that: 49 dim lim SH ++ i degree a, b = the umber i the secod bullet To prove this, we will ivoke the argumet used i the proof of Propositio 3. Sice the itegral form of SH ++ is a free module over the rig C[q ±1, v ±1 ], its rak ca be computed i the specializatio q = v = 1: dim SH ++ i degree a, b = dim C C[x 1,..., x, y 1,..., y ] S i degree a, b

10 10 ANDREI NEGUȚ where a refers to the degree i the x variables ad b refers to the degree i the y variables. The polyomial rigs i the right had side have a well-kow iverse limit, the rig of polyomials i ifiitely may variables: dim lim SH ++ i degree a, b = dim C C[x 1,..., y 1,...] sym i degree a, b All that remais is to observe that a basis of the space i the RHS is give by: Sym x a1 1 xa2 2...yb1 1 yb with a i = a, b i = b. The umber of such basis vectors is precisely the umber i the secod bullet, which appears i 49. Refereces [1] Schiffma O., Vasserot E., Cheredik algebras, W algebras ad the equivariat cohomology of the moduli space of istatos o A 2, Publ. Math. Ist. Hautes Etud. Sci., , Issue 1,