Are Deligne-Lusztig representations Deligne-Lusztig? Except when they are complex?

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1 Are Deligne-Lusztig representtions Deligne-Lusztig? Except when they re complex?. Representtion theory Let K be (connected) compct Lie group, nd let π n irreducible representtion of K. This mens tht π is finite-dimensionl, i.e. π is homomorphism from K to GL (V ), where V is finite dimensionl (complex) vector spce. It mkes sense to tlk bout Θ π (h) = T r (π (h)), sclr-vlued function. It turns out tht this chrcterizes the representtion π up to unitry equivlence, ie. if π is nother irreducible representtion on spce V, nd it Θ π = Θ π then there is n isomorphism T : V V such tht π (g) T = T π (g) for ll g K. We cll Θ π is clled the chrcter of π. This function is clss function, ie. it is constnt on conjugcy clsses. However it need not be multiplictive, so it is not necessrily homomorphism. We hve in generl ( Θ π (k) Θ π (k ) = deg (π) Θ π kuk u ) du. So, we understnd Θ π, hence π, if we cn evlute Θ π on set of conjugcy clss representtives. If K contins mximl torus T ( subgroup isomorphic to product of S s), then every element of K lies in some conjugte of this torus. So if, we understnd which functions on T re chrcters, then we understnd the representtion theory. Exmples: K = SO 2 ( ) = { G M 2 (R) : GG T = nd det G = } {} cos θ sin θ = S = e sin θ cos θ ιθ K This hs only one-dimensionl irreducible representtions, which re z S z n for fixed ( n in Z. In terms ) of ctions, K cts cos θ sin θ on V = C by letting ct by multipliction sin θ cos θ by e inθ. Then Θ π (z) = z n in this cse. K = SU 2 (C) = {g M 2 (C) : det g = nd g g = }, which is isomorphic to SO 3 (R), which is topologiclly {( RP) 3. It is not} z 0 belin. A mximl torus looks like T = 0 z : z S =

2 2 S. The irreducible representtions of SU 2 (C) re relized in C [x, y] hom = {homogeneous polynomils in 2 vribles}. Specificlly, we ct by trnsposition: f (x, y) = f (x + by, cx + dy). c d If you fix the degree d to get V d, then this is (d + )-dimensionl irreducible representtion (bsis x d, x d y,..., y d ). The mtrix in this bsis is digonl mtrix with z i+j long the digonl. So Θ π (z) = z d + z d z d = zd+ z (d+) z z This chrcter is sum of terms of vrious weights, nd the highest weight is d. The denomintor is the difference of the eigenvlues, the Weyl denomintor. These re clss functions on SU 2 (C). Next time: Let s geometrize in the lgebro sense, working with lgebric groups over C nd use Weyl s unitry trick. Complexifying: SU 2 (C) SL 2 (), becuse su 2 (C) R C = sl 2 (C)... Some questions tht cme up. Wht s n irreducible representtion? A representtion is liner ction of group G on (complex) vector spce V. Tht is, it is homomorphism from G to GL (V ). If G is finite-dimensionl, we cn think of the codomin nd rnge s consisting of mtrices. Two bd exmples: () The rottion representtion of SO 2 (R) on R 2. (2) The sher representtion of R on R 2. x x + ty On the level of ctions, the second is t =. y y These re bd in different wys: The rottion representtion hs no simultneous eigenvectors, i.e. there no vector v such tht gv = λ (g) v for ll g G, so it s irreducible. But there re complex simultneous eigenvectors. Specificlly, cos θ sin θ = e sin θ cos θ ±i ±iθ. ±i We sy this representtion is irreducible but not bsolutely irreducible becuse we mke it reducible by chnging sclrs. This isn t problem if the field is lgebriclly closed. Also if the chrcteristic

3 is zero, it lso voids other problems. This why we work over C. On the level of mtrices, cos θ sin θ sin θ cos θ isn t digonlizble over R. But () cos θ sin θ Int i i sin θ cos θ cos θ sin θ = = i i sin θ cos θ i i e iθ 0 0 e iθ. ( On) the other hnd, the sher representtion hs n invrint vector ; but only one (up to scling). Even pssing to C doesn t give 0 more eigenvlues. Unlike the rottion representtion, where we could write C 2 = C C, i i decomposition tht is invrint under rottion (completely reducible cse), there is no wy to write in terms of the sher representtion C 2 = C something invrint, 0 so we sy tht this ction is reducible but indecomposble. On the level of mtrices, the rel number t cts by the mtrix t, 0 it s unipotent. In crude sense, reducing representtion mens conjugting it so ll the mtrices re block upper tringulr. Decomposing representtion mens conjugting it so ll the mtrices re block digonl. The representtion is irreducible, respectively indecomposble, if this cnnot be done nontrivilly. Unipotence cuses reducibility but indecomposbility. We try to void it if we wnt to reduce the study of represention theory to the study of irreducible representtions. Good news: This never hppens for complex representtions of compct groups. Any representtion of compct group is completely reducible. Some words: linerly reductive: irreducibility of representtions reductive: how mny unipotents? (i.e. few) 3

4 4 In chrcteristic zero, both notions re the sme. Eg - dditive group of rel numbers in not reductive. Lst time, we discussed exmples of representtions of compct connected Lie groups, nd I mentioned Weyl s unitrin trick: every irreducible representtion of connected compct Lie group is the restriction of unique holomorphic representtion of n ssocited complex Lie group. Good: then you cn do lgebric geometry on complex Lie groups. Exmple: K = SO 2 (R) = { g GL 2 (R) : det (g) =, g T g = } is cos θ sin θ isomorphic to S by e iθ. It s irreducible representtions one-dimensionl nd re z z n. The complex picture: sin θ cos θ G = K C = SO 2 (C). (Why is this the right nswer? Algebric groups... On the level of Lie lgebrs, so 2 (R) R C = so 2 (C), but this only identifies the complex group up to covers. so 2 (R) = { X gl 2 (R) : trx = 0, X + X T = 0 }. ( Then SO 2 (C) is belin, nd it s isomorphic to C. z (z + 2 z ) (z + 2i z ) (z + 2i z ) (z + 2 z ) ( rtionl isomorphism). We cll C n lgebric torus. Wht do the representtions of C look like? The representtion z z n is the restriction of the representtion z z n of C. These re ll the holomorphic representtions of C. but there re others (eg z z n z m ). Exmple: K = SU 2 (C) = {g GL 2 (C) : det g =, gg = }. Note tht su 2 (C) = {X gl 2 (C) : trx = 0, X + X = 0}. This is rel, not complex, vector spce. Why not just complexify? su 2 (C) R C is complex Lie lgebr of wht? This is sl 2 (C) = {X gl 2 (R) : trx = 0}. Ruth: the isomorphism from sl 2 (C) to su 2 (C) R C is X (X + ix)+ 2 (X ix) i. 2i Lst time, we tlked bout complexifiction. This is wy of pssing from (compct) connected (rel) Lie groups to complex Lie groups. As follows, given K compct connected Lie group, its (rel) Lie lgebr is k, then let g = k R C is complex Lie lgebr. For K compct, there exists connected complex Lie group G whose Lie lgebr is g. But Lie lgebrs don t see covers; so we choose G so tht π G = π K. Then G is the complexifiction of K. This obeys universl property. ) K H (complex Lie group) (smooth) (nlytic) G

5 But, for noncompct Lie groups, K G my hve kernel. For exmple, if K = SO 2 (R) = { g GL 2 (R) : gg T = }, then k = so 2 (R) = { X : X + X T = 0 } = R. Then g = k R C = so 2 (C) = { X gl 2 (C) : X + X T = 0 } = C Among ll the complex Lie groups with Lie (G) = C, we pick the one with π (G) = Z, which is G = C. More explicitly, Also, G = SO 2 (C) = { g GL 2 (C) : gg T =, det g = } K = SU 2 (C) = SU 2,C/R (R) = { g GL 2 (C) : det (g) =, gg T = } k = su 2 (C) Lst time, k R C = sl 2 (C). Among the complex connected Lie groups G with Lie lgebr sl 2 (C), we pick the one with π G = {}, nmely G = SL 2 (C) (not P GL 2 (C) ). Lst time: Weyl s unitrin trick: Every continuous irreducible representtion of K (must be compct), i.e,. continuous homomorphism from K to GL n (C), extends uniquely to n nlytic representtion of G = K C, i.e. n nlytic homomorphism from G to GL n (C). For exmple, irreducible continuous representtions of SO 2 (R) = S re of the form z z n. Irreducible nlytic representtions of SO 2 (C) = C re of the form z z n (lwys -dimensionl). Next exmple: Two specil irreducible continuous representtions of SU 2 (R) : The 2-dim representtion coming from K GL 2 (C). The 3-dimensionl djoint representtion: K cts on su 2 (C) byconjugtion, nd su 2 (C) is 3-dimensionl. The ction preserves the Killing form: X R Y tr ([X, [Y, ]] : su 2 (C) su 2 (C)) = tr (d (X) d (Y )). The Killing form of semi-simple Lie lgebr is negtive definite if nd only if the corresponding Lie group is compct. Note tht semisimple Lie groups hve finite fundmentl group. So this mens tht we hve mp to the orthogonl group. SU 2 (C) 0 (Killing form) = O (3, 0) = O (0, 3) = O 3 (R). 5

6 6 By connectedness, it mps to SO 3 (R). By Lie lgebrs, it is surjective, nd it hs kernel (the kernel of the djoint representtion is the center of the group) Z 2 (generted by ). Thus, K ± = SO 3 (R) so K = Spin 3 (R). The (d + )-dimensionl representtion is sym d (C 2 ) (there s exctly one in ech dimension). These representtions extend to SL 2 (C), s follows. The 2-d representtion extends in the obvious wy, with SL 2 (C) GL 2 (C). All others extend s symmetric powers. The three-dimensionl representtion is the djoint representtion of sl 2 (C). 2. Representtions of Complex Groups Clssicl exmples of these: SL n (simple), GL n = SL n GL (reductive). SO n, Sp 2n. We sw tht every connected, compct Lie group K hs n ssocited connected complex Lie group G = K C, nd the restriction mp from LHS=holomorphic irreducible representtions of G to RHS=continuous irreducible representtions of K is ijection. Note tht on the RHS, we hve nice structure theory: every conjugcy clss in K intersects fixed but rbitrry mximl connected belin subgroup clled torus T (becuse it is torus = (S ) r s Lie group, with r =rnk). On the LHS, this is not true. We still hve nice mximl connected belin subgroups, but they don t meet every conjugcy clss. For complex groups, mximl connected, belin, (consists of semisimple digonlizble elements) is clled n (lgebric) torus A. As {( Lie group, ) A = (C ) r for some r. For exmple, in SU 2 (C), T = z 0 0 z : z S }. In G = K C = SL 2 (C), { } { } z 0 z A = 0 z : z C but not : z C = U, which 0 is mximl, connected, belin subgroup of G but is not n lgebric torus. It turns out tht the solution to the conjugcy clss problem is to enlrge n lgebric torus to Borel subgroup. The definition {( is tht ) B = A U. } For exmple, if G = SL 2 (C), B = 0 : C, b C. Then B consists of upper tringulr mtrices. We hve B = {g G : g upper-tringulr}. The honest definition from the lgebric point of view is tht Borel subgroup B in G is mximl connected, solvble subgroup. (Solvble:

7 B [B, B] [[B, B], [B, B]]....) The geometriefinition is tht Borel subgroup B in G is miniml closed subgroup such tht G B (s spce) is projective (s n lgebric vriety). (Topologiclly: G B is compct). Borel subgroups re unique up to conjugcy. ( Going) bck to our ( exmple, ) the mp SL 2 (C) B CP defined by B C is iholomorphism. c Next exmple: SL 3 (C) c 0 e f 0 0 e Flg SL 3 = flg vriety defined by c d e f g h i B C d g, C d g C b e h Note tht Flg SL3 Gr (2, ). In generl, G B is clled the flg vriety of G. Fcts: () ll Borel subgroups re G-conjugte. (2) B is self-normlizing, mening tht the normlizer N G (B) = B. So G B {Borel subgroups} is bijective, with gb gbg. Another concrete perspective: A (mximl) flg in C n is chin of subspces 0 V V 2... C n, nd Borel in GL n (C) is the stbilizer of flg. Tht is, the stbilizer is {g GL n (C) : gv i = V i i}. The holomorphic representtions of G re relized inside the cohomology of G-equivrint line bundles over G B. (Borel-Weil-Bott). 3. Borel-Weil-Bott nd nlogues over other fields Lst time: Let K be compct, connected Lie group K, nd let G = K C (complex, connected). Inside G is B, the Borel subgroup ( G = Ad(G) B), such tht G B is projective. Obligtory: Consider P ic (G B), the group of line bundles. Actully, we just consider G-equivrint holomorphic line bundles. In prticulr, B cts on the fiber over eb = B. Tht is, we hve homomorphism from B to GL (fiber) = GL (C) (belin), nd thus, we get mp from B [B, B]. Recll tht B = A U, where A is n lgebric torus, mximl subgroup in G sisomorphic to (C ) r, nd U is the unipotent rdicl, nd it hppens tht U = [B, B]. For exmple, if K = SU 2 (C), G = SL 2 (C), nd B is the block upper tringulr, U is s on the digonl nd upper tringulr, A is block. So, given n equivrint line bundle, we obtin holomorphic homomorphism A C (liner chrcter of A). We cn differentite this homomorphism: C. Then P ic G (G B). We cn prtilly 7.

8 8 invert this mp: given n integrl λ, which loosely mens exponentible (ie exp (λ) is induced from A to C ). Think: A = C, = C, nd integrl mens it s in Z. We cn now define n ction of B on C vi B A eλ C. Explicitly, b z = e λ () z, where b = u. Then cll this C λ. Then G B C λ is line bundle over G B. Its nturl G-ction mkes it eqiuivrint. z With the erlier K = SU 2, G, B, A, if λ : z z, then G B C λ is the tutologicl bundle ( ( B z z c ) (, C c )). Theorem. (Borel-Weil) If λ is integrl nd dominnt (lies in certin cone in ), then Γ hol (G B, L λ ) is n irreducible G-module. All irreducible G-representtions rise in this wy. If λ is not dominnt, then Γ hol (G B, L λ ) = 0. Theorem 2. (Borel-Weil-Bott) If λ is integrl nd regulr (lies in certin open subset of - complexment of hyperplnes), there is n explicit integer i 0 such tht H i (G B, L λ ) = 0 unless i = i 0, nd H i 0 (G B, L λ ) is n irreducible G-module. (Dolbeult cohomology) In our exmple, = C (with λ ), integrl mens Z, dominnt mens positive, regulr mens nonzero. L λ is the dul of the tutologicl bundle, which hs two-dimensionl spce of globl holomorphic sections. We re looking for holomorphic mps G G B C λ. We ll look insted t mps f : G C such tht f (gu) = e λ () f (g). Given f, cn construct F : G G B C λ by g g B f (g). And, given F : G G B C λ, we my define f : G C by f (g) = z, where F (g) = g B z. We hve the Crtn decomposition G = HB, where B is s before nd H = SO 2 (C), by Grm-Schmidt orthogonliztion. One cn check tht H B = {±}. The mp f : G C s bove is determined by its restriction to H; nerly ny mp H C cn be extended to suitble mp G C. (Nerly: up to ± ) However, the extension is not utomticlly holomorphic. Explicit clcultion:

9 9 So ( ) = ( c c ) ( ) 2 + c 2? 0 2 +c 2 f So e λ ( 2 + c 2 0 = e λ ( 2 + c ) ) (( f c 2 +c 2 c )) = 2 + c 2. So we wnt f to pick off either the entry or the 2 entry (nd everything else is in the spn of those ( two). ) c ie f = or c, then the resulting extension is holomorphic. c Obviously, these re the only possibilities up to liner combintions. Cll these two mps X nd Y respectively. Then the G-ction on Γ (G B, L λ ) = CX CY is given by X = X + cy, nd Y = bx +dy, nd the mtrix of this ction is. So the chrcter of this representtion is Θ (g) = tr (ction of g) = + d. In prticulr, if g =, then Θ (g) = + = 2 2, 2 where the numertor is the signed sum of liner chrcters of A, the denomintor is the Weyl denomintor.