SCRIBE: JAKE LEVINSON

Transcription

1 GL n REPRESENTATION THEORY NOTES FOR SCRIBE: JAKE LEVINSON As th th last lctur, ths on s basd on John Stmbrdg s papr: A local charactrzaton of smpl-lacd crstals, Trans. Amr. Math. Soc. 355 (2003), no. 12, Rcall Last class, dfnd rgular crstals. Ths ar crstals th th proprt that If 2, thn can fll n squars : mpls If = 1, thn changs lngths of -strngs b ±1, and c rsa: and a f othr spcfc ruls (s th prous lctur). 2. Thorm for toda Toda s man rsults ar th follong to thorms: Thorm 1 (Stmbrdg). A fnt connctd rgular crstal has unqu hgh ght lmnt. Thorm 2 (Stmbrdg). If B and B ar fnt connctd rgular crstals th hgh ght lmnts u and u, and f t(u) = t(u ), thn B = B. All th hard or tas plac n th follong lmma: Lmma 1. Lt B b a rgular crstal, u a hgh ght ctor, and B th = f r f 2 f 1 u, for som, 1, 2,..., r {1,..., n 1}. Thn n fact = f r 1 f 1 u for som 1,..., r 1 {1,..., n 1}. In othr ords, t s possbl to omt th upards stp, and gt from u to ust usng donards stps f l. Proof. B nducton on r. If r = 0, u = 0 snc u s hgh ght, and 0 / B, ths s a contradcton. For r > 0, lt s sa ha... (rst of proof s n th dagram) u 1 2 r 1 r f Thr ar thr cass. Frst of all, f r =, thn =, so r don. Scond, f r 2, thn can fll n th squar to gt: r r f 1

2 2 SCRIBE: JAKE LEVINSON So, n othr ords, ha u 1 2 r 1 so b nducton, 1 2 r 2 u r. In th thrd cas, r = 1. Ths splts nto thr sub-cass: frst, r s a contradcton. Nt, to dal th r r and us th fllng n squars aom and contnu as n cas 2. Fnall, f both arros pont out of, ha to us th damond-shapd aom to pull bac to stps: u 1 r r Inductl, can pull bac thr tms n a ro: gt a ln of arros from u to 1 ; thn from u to 2 ; thn from u to. Fnall, fll n th arros Ths complts th proof. u 1 2. Corollar 1. If B s rgular, u s hgh ght, n th sam connctd componnt as u, thn = f s f 1 u. Proof. Wrt n trms of u usng both s and f s, thn lmnat s usng th lmma. Quston: Ho should thn about ths proof? Hr ar som scattrd thoughts, addd b Dad. On should compar ths to th proof that r gl n rrp V has a unqu hgh ght ctor u. Th lmma as that, for u hgh ght, th ctors f r f 2 f 1 u span V. To sho ths, t as nough to sho that f r f 2 f 1 u s n th span of ctors of th form f r 1 f 1 u. S th Nombr 21 nots for a r smlar argumnt n th L algbra sttng. A bttr a, prhaps, to undrstand th argumnt of No. 21 s through th PBW thorm. Th ( (as part of th) PBW thorm shos that U(gl n ) s spannd b monomals of th form ) ( > Ea ) ( d Eb ) < Ec. From ths, conclud that V s spannd b ctors of th ( ) ( form > Ea ) ( d Eb ) < Ec u. But such a ctor s 0 unlss all th c ar 0 (snc u ( ) s hgh ght) and n that cas s proportonal to > Ea u (snc u s a ght ctor). If had somthng l th unrsal nlopng algbra for crstals, could magn shong that ths algbra as spannd b monomals hr do all of th s frst, and thn all of th f s. Snc an annhlats u, and no that can gt to anhr n B b applng s and f s

3 GL n REPRESENTATION THEORY NOTES FOR to u, ths shos that can gt anhr from u b dong th s frst. I am not famlar th a unrsal-nlopng-algbra l obct for crstals, but n th appnd I g a proof l ths. 3. Proofs of Thorms 1 and 2 No can pro Thorm 1 abo (unqunss of hgh-ght ctors n fnt connctd rgular crstals). Proof of Thorm 1. Frst, an fnt crstal has a hgh ght lmnt (appl oprators, hch ncras th ght, untl can t anmor). Suppos u and u ar both hgh ght. B th abo corollar, = f s f 1 u. Thn s 0, a contradcton. Rmar. Ths s actl l th corrspondng proof for gl n, naml: f ou ha a hgh ght ctor, and ou no ou can rach an othr ctor b (onl) gong donards, thn thr s no room for an othr hgh ght ctor. An mportant obsraton s th follong: f B s a crstal and u s hgh ght, thn ts ght ctor s dcrasng, that s, t(u) = (λ 1,..., λ n ) th λ 1 λ n. To s ths, consdr th -strng through u: Thn s that and so f 2 u f u u t() = s t(u) = (λ 1, λ 2,..., λ +1, λ,..., λ n ), t() = t(u) + φ (0, 0,..., 1, 1,..., 0), so λ +1 = λ φ. Thus λ +1 λ. W ll us ths fact n our proof of th scond thorm. Proof of Thorm 2. Lt t(u) = t(u ) = (λ 1,..., λ n ) b th common hgh-ght ctors of th to crstals B and B. Dfn hght functons h : B Z and B Z as follos: f t(b) = (κ 1,..., κ n ), thn n h(b) = (κ λ ). In partcular, and smlarl =1 h(f b) = h(b) + 1 f f b 0, h( b) = h(b) 1 f b 0. W sho, b nducton on r, that thr s a bcton btn B r := {b B, h(b) r} and th corrspondng st B r, such that Th bcton commuts th and f as maps thn B r {0} and B r {0}, Th bcton prsrs th quantts ε (b) and φ (b) (hch masur th dstanc from b to th nds of th -strng through b. Snc our crstals ar fnt, hn r s larg nough, ths gs a bcton B B commutng th, f. (Dad sad: fntnss s not as mportant as I m mang t sound th papr ths proof coms from dosn t alas rqur t. But r assumng fntnss for smplct.) Th bas cas s r = 0. In ths cas B 0 = {u} and B 0 = {u}: th map s obousl bct, and t agrs th th s, snc u = u = 0 for an. Th maps f all map out of ths sts, so don t nd to rf anthng ls. Ths shos that th frst proprt holds. For th scond proprt, not that u s at th -nd of ts -strng, so ε (u) = 0 and φ (u) = λ λ +1. Th sam fact holds for u.

4 4 SCRIBE: JAKE LEVINSON At ths pont, th rst of th proof s ssntall an rcs: t suffcs to sho that th f arros act appropratl (snc an pont can b rachd ust usng f arros, b th Lmma); and thn t s ust a mattr of consdrng th arous as f arros can com togthr, and applng th arous aoms of rgular crstals. Proof contnud on 12/05. For th nduct cas, suppos ha α : B r B r. W sh to tnd ths bcton to r + 1. W no that r b B r+1 s f c for som and som c B r. Frst of all, obsr that f c = 0 f and onl f f α(c) = 0. To s ths, not that th frst statmnt s qualnt to φ (c) = 0, and smlarl for th scond. (Th bcton α prsrs φ b nducton). In othr ords, α nducs a bcton btn {(c, ) : c B r, {1,..., n}, f c 0} and th corrspondng st for B r. To construct α on B r, sho that f (c 1 ) = f (c 2 ) f and onl f f (α(c 1 )) = f (α(c 2 )), for all c 1, c 2 B r, 1, n 1. (W onl nd to pro th forard drcton, snc th stuaton s smmtrc btn B r and B r.) W ll also nd to chc that th n α stll prsrs ε, φ. If =, thn f c 1 = f c 2 f and onl f c 1 = c 2, so th clam follos mmdatl. Nt, f 2, thn ha th pctur d f c 1 c 2 (Rad ths l a commutat dagram assrtng an stnc statmnt: assumng th stnc of th sold lns, th rgulart aoms mpl that can fll n th dashd lns.) Appl α to th pctur: b nducton, gt b α(d) f α(c 1 ) α(c 2 ) Ths shos that f α(c 1 ) = f α(c 2 ) = b. Fnall, f = 1, thr ar four cass (basd on th orntatons of th dgs). Th most ntrstng cas s b c 1 c 2 b Hr us th last rgulart aom (gornng th damond-shapd arro stup) to pull bac to stps; thn appl α throughout. Fnall, us rgulart agan to fnd a b mappng to α(c 1 ), α(c 2 ). No dfnd α rhr; nd to sho t stll prsrs ε and φ. So lt b = f c as arlr; ant to consdr ε (b), φ (b). If =, φ (c) = φ (b) 1 and ε (c) = ε (b) + 1.

5 GL n REPRESENTATION THEORY NOTES FOR If 2, φ (c) = φ (b) and ε (c) = ε (b). As usual, th cas = 1 s th ntrstng on. For smplct, ta = + 1. Frst of all, no α commuts th, so n partcular, (b) = 0 f and onl f (α(b)) = 0. Not that n ths cas ε (b) = ε (α(b)) = 0, and can comput φ (b) from t(b) = t(c) + (0,..., 0, }{{} 1, }{{} 1, 0,..., 0) = t(α(c)) + (0,..., 0, }{{} 1, }{{} 1, 0,..., 0) So φ s prsrd. Othrs (b) 0, bra nto som mor cass. In ach cas, th rgulart aoms sho that th orntatons of th f +1 dgs b c 2 and α(b) α(c 2 ) match. Thn can comput ε (b), φ (b) from th noldg of ε (c 2 ), φ (c 2 ) and ths orntaton. Appnd: An altrnat rout to Corollar 1 (Addd b Dad) I don t no hthr th follong proof of Corollar 1 ll sm mor or lss clar, but I prsnt t as an altrnat prspct. Lt B b a fnt connctd rgular crstal, and lt u and b lmnts of B. Consdr all as to rt = g r g r 1 g 2 g 1 u for som squnc of crstal oprators g r g 1. W assgn a scor to ach lttr g. If g s an f, th scor s 0. If g s an, thn th scor s 2 h(g 1g 2 g 1 u) hr h s th hght functon (so f s ncras hght, s dcras hght and hght s alas 0). Th scor of th prsson g r g 2 g 1 u s th sum of th scors of ach lttr. Suppos that th ord g contans f at som pont. W can no ma th follong rplacmnts: If =, can dlt th par f. Ths rmos on lttr (that ) hch contrbuts postl to th scor and mantans th scor of r othr lttr; hnc t dcrass th scor. If 2, can rplac f b f. Ths changs th scor of that from 2 h to 2 h 1 (for som h) and ps all othr scors th sam. If = 1 and th dgs corrspondng to ths lttrs ar orntd as f or f, thn ma rplac f b f. As abo, ths changs th contrbuton of that from 2 h to 2 h 1. If = 1 and th corrspondng dgs ar orntd as f, thn ma rplac f b f f f. Ths rmos a lttr th scor 2 h and nsrts n lttrs th scors 2 h h h 3 = (7/8)2 h. Thus, hnr ha f, can dcras th scor. But th scor s a post ntgr, so t cannot dcras forr. So, f p mang th abo rplacmnts, ll ntuall gt to a ord hch s of th form fff f. In partcular, f u s a hght ght lmnt, ll ntuall gt to a formula = f r f 2 f 1 u, snc u s annhlatd b r.