McKay Correspondence

1 McKay Correspondence To the memory of Peter Slodowy Iku Nakamura 005 November

1 Dynkin diagrams and ADE The following appear in connection with Dynkin (0) Dynkin diagram ADE (1) Finite subgroups of SL(, C), Regular polyhedra () -dim simple sing (deform-stable critical points) (3) simple Lie algebras ADE, II 1 -factors etc. (4) Partition func. of SL(, Z)-inv. conformal field thrys (5) Finite simple groups Fischer F 4, Baby monster B, Monster M, and McKay s 3rd observation on them (E 6, E 7, E 8 correspond to them)

3 A n 5 5 5... 5 5 5 D n 5 5 5... 5 5 5 E 6 5 5 5 5 5 5 E 7 5 5 5 5 5 5 5 E 8 5 5 5 5 5 5 5

4 Partition funct. of SL(, Z)-inv. conformal field theorys Dofinition: Z =Trq L 0+ L 0 where q = e π 1τ Assumptions 1. Unique vacuum( 1 state of min. energy). χ k : a A (1) 1 -character corresp. to a particle (or an operator in a physical theory) 3. Partition func. Z is a sum of χχ, (χ, χ :A (1) 1 -characters) 4. Z is SL(, Z)-inv., invariant under τ τ 1 (the parameter τ H : upper half-plane)

5 Classification of Z Cappelli, Itzykson-Zuber, A.Kato Type k + Partition function Z(q, θ, q, θ) A n n +1 n λ=1 χ λ D r 4r r 1 λ=1 χ λ 1 + χ 4r+1 λ + χ r 1, D r+1 4r r λ=1 χ λ 1 + r 1 λ=1 (χ λ χ 4r λ + χ λ χ 4r λ )+ χ r E 6 1 χ 1 + χ 7 + χ 4 + χ 8 + χ 5 + χ 11 E 7 18 χ 1 + χ 17 + χ 5 + χ 13 + χ 7 + χ 11 + χ 9 +(χ 3 + χ 15 ) χ 9 + χ 9 ( χ 3 + χ 15 ) E 8 30 χ 1 + χ 11 + χ 19 + χ 9 + χ 7 + χ 13 + χ 17 + χ 3

6 Indices of χ = Coxeter exponents of ADE Type Partition function Z and Coxeter exponents E 6 χ 1 + χ 7 + χ 4 + χ 8 + χ 5 + χ 11 1, 4, 5, 7, 8, 11 E 7 χ 1 + χ 17 + χ 5 + χ 13 + χ 7 + χ 11 + χ 9 +(χ 3 + χ 15 ) χ 9 + χ 9 ( χ 3 + χ 15 ) 1, 5, 7, 9, 11, 13, 17 E 8 χ 1 + χ 11 + χ 19 + χ 9 + χ 7 + χ 13 + χ 17 + χ 3 1, 7, 11, 13, 17, 19, 3, 9

7 (5) McKay s 3rd observation only conj. classes of involutions of Monster M. (Fischer) is one of the classes (Fischer involution). 1. {the conj. class of a b; a, b (Fischer)} = 9 classes. {order of a b; a, b (Fischer)} = {1,,, 3, 3, 4, 4, 5, 6}, the same as mult. of vertices of Ẽ8, extended E 8. 6 Ẽ 8 6 6 6 3 6 6 6 6 6 4 6 5 4 3 1

8 Today McKay correspondence (1) = (0) [Ito-N. 1999] explains McKay (1) = (0) + () by Hilbert scheme of G-orbits G-Hilb(C ) G SL() Recall: (0) Dynkin diagrams ADE (1) Finite subgroups of SL(, C) () -dim. Simple sing. and their resol. (3) Simple Lie algebras ADE

9 well known and trivial 1. From finite groups to singularities : (1) = (). From singularities to Dynkin diag. () = (0) Recall: (0) Dynkin diagrams ADE (1) Finite subgroups of SL(, C) () -dim. Simple sing. and their resol. (3) Simple Lie algebras ADE

10 well known, but nontrivial 1. From Lie algebra to singulalities (3) = () Grothendieck, Brieskorn, Slodowy. From finite groups to Dynkin (1) = (0) (McKay) 3. Explanation for McKay (1) + () = (0) + () by vector bundles and their Chern classes (Gonzalez-Sprinberg, Verdier) Today [Ito-N. 1999] explains McKay (1) = (0) + () by Hilbert scheme of G-orbits : G-Hilb(C ), G SL()

11 More recent progress Topic(4) : Kawahigashi and other (Ann. Math. ) Topic(5) : Conway, Miyamoto, Lam-Yamada-Yamauchi Hot news John McKay received 003 CRM-Feilds-PIMS prize. Reason : Discovery of McKay corresp. and McKay s observation about Monster.

1 History 1978.11 Moonshine - Character table of Monster (Fischer-Thompson) 1978.1 McKay correspondence 1979.0 Discovery of (5) - Griess constructs Monster

13 3 dim. McKay corresp. thereafter G SL(3) Consider G-Hilb(C 3 ) only. 1. Smoothness G abelian : Nakamura. Smoothness & 3dim McKay corresp. (G abelian : Ito-Nakajima, G general : Bridgeland-King-Reid) 3. Structure G abelian : Nakamura, Craw-Reid 4. Structure of M θ (variants of G-Hilb(C 3 )) G abelian : Craw-Ishii 5. Structure G non-abelian : Gomi-Nakamura-Shinoda

14 Today McKay correspondence (1) = (0) [Ito-Nakamura 1999] explain (1) = (0) + () by G-Hilb(C ) : Hilbert scheme of G-orbits G SL() Recall: (0) Dynkin diagrams ADE (1) Finite subgroups of SL(, C) () -dim. Simple sing. and their resol.

15 Review : what is McKay correpondence Exam 1 G = G(D 5 ) : a dihedral group of order 1 G : generated by σ and τ σ = ζ 6 0 0 ζ6 1, τ = 0 1 1 0

16 Repres Tr(ρ) e e σ σ τ τ 3 ρ 0 χ 0 1 1 1 1 1 1 ρ 1 χ 1 1 1 1 1 1 1 ρ χ 1 1 0 0 ρ 3 χ 3 1 1 0 0 ρ 4 χ 4 1 1 1 1 i i ρ 5 χ 5 1 1 1 1 i i

17 Let ρ nat : G SL(, C) (the natural inclusion) Then repres.ρ i and their tensor products with ρ nat ρ ρ nat = ρ 0 + ρ 1 + ρ 3, ρ 0 ρ nat = ρ 1 ρ nat = ρ, ρ 3 ρ nat = ρ + ρ 4 + ρ 5, ρ 4 ρ nat = ρ 3, ρ 5 ρ nat = ρ 3. Draw a graph as follows: D 5 ρ 0 6 6 ρ ρ 3 ρ 4 6 6 6 6 ρ 1 ρ 5 Rule : Connect ρ i and ρ j ρ i ρ nat = ρ j + Remove ρ 0 (triv. repres.).then we get a Dynkin.

18 Exam (Continued) G = G(D 5 ) : a dihedral group of order 1 G : generated by σ and τ σ = ζ 6 0 0 ζ6 1, τ = 0 1 1 0 The quotient C /G has a unique sing. Invariant polynomials of G(D 5 ) and their relation are F = x 6 + y 6,G = x y,h = xy(x 6 y 6 ) G 4 GF + H =0

19 Invariant polynomials of G(D 5 ) and their relation are G 4 GF + H =0 The equation of D 5 is usually refered to as X 4 + XY + Z =0 A unique singularity (F, G, H) =(0, 0, 0) The exceptional set for the sing.(0, 0, 0) C 1 C C 4 C 5 C 3 C 1 C C 3 C 4 6 = 6 6 6 6 C 5 D 5 C i is P 1 (a line), intersecting at most transversely

0 The exceptional set for the sing.(0, 0, 0) C 1 C C 4 C 5 C 3 C 1 C C 3 C 4 6 = 6 6 6 6 C 5 D 5 C i is P 1 (a line), intersecting at most transv. The dual grapf of it is a Dynkin diagram D 5 Rule of dual graph : C i = a vertex, C i C j = an edge Conclusion : both Dynkin diagrams are the same (McKay correspondence)

1 McKay corresp. asserts : a relationship between (1) and () (1) Resolution of sing. of C /G () Representation theory of G

Exam 3 (Continued) G = G(D 5 ) : a dihedral group of order 1 G : generated by σ and τ σ = ζ 6 0 0 ζ6 1, τ = 0 1 1 0 The quotient C /G has a unique sing. Invariant polynom. of G(D 5 ) and relation F = x 6 + y 6,G = x y,h = xy(x 6 y 6 ) G 4 GF + H =0

3 Invariant polynom. of G(D 5 ) and relation are G 4 GF + H =0 The usual eq. of D 5 : X 4 + XY + Z =0 A unique singularity (F, G, H) =(0, 0, 0) When resolve sing, take quotients H/G = (x6 y 6 ) xy etc. with Denominators/Numerators not G(D 5 )-invariant. But xy, x 6 y 6 belong to the same repres. of G(D 5 ) Therefore Resolution of sing. and Repres. theory of G relate each other.

4 3 New resolution of simple sing. We dare to regard C /G as a moduli space C /G ={a G-inv. subset consisting of 1 points} : moduli of geometric G-orbits Resolution of C /G = moduli of ring-theor. G-orbits G - Hilb(C ):={a O C - G - module of length 1} For a module M G - Hilb(C ) 0 I O C M 0 (exact) G-module generating I is almost G-irreducible Example of generators of I : F t = xy t(x 6 y 6 )

5 4 The Hilbert scheme of n points The Hilbert scheme of n points in the space X Z : n points of X n points Z is a formal sum Z = n 1 P 1 + n P + + n r P r (P i P j ) (where n = n 1 + + n r )

6 Exam 4 Assume X =C. Let P i : x = α i, I Z = the ideal of Z I Z = f Z C[x] Z is the zero locus f Z (x) Z is n points equiv. dim C C[x]/I Z = n f Z (x) =(x α 1 ) n 1 (x α ) n (x α r ) n r = x n + a 1 x n 1 + + a n 1 x + a n e.g. if z = n [0], then I Z =(x n ) Hilb n (C) = {n points of X} n 1 bij. = xn + a n j x j ; a j C j=0 = C n

7 Exam 5 Assume X =C. Then Z = n 1 P 1 + + n r P r, (formal sum), P i P j, namely, Z X }{{} n X/order forgotton= X (n) X (n) = X }{{} n X/S n X (n) is very singular, has a lot of sing. Caution X (n) is different from Hilb n (C ).

8 Assume X =C. Let X [n] =Hilb n (C ). X [n] = {an ideal I C[x, y]; dim C[x, y]/i = n} I C[x, y]; I : a vector subsp of C[x, y] = xi I,yI I, dim C[x, y]/i = n Thm 1 (Fogarty 1968) X [n] is a resolution of X (n). A natural map π : X [n] X (n) is defined, π : Z n 1 P 1 + + n r P r where Z = {P 1,,P r },n i = multiplicity of P i in Z.

9 Thm (Fogarty 1968)(revisited) The natural morphism X [N] = Hilb N (C ) π X (N) is a resol (mimimal) of sing. The map π sends G-fixed points to G-fixed points. where G SL(), N = G. Then π G-inv. :(X [N] ) G-inv. (X (N) ) G-inv. By Thm of Fogarty (X [N] ) G-inv. is nonsing. G -inv. part of Fogarty = Next theorem

30 5 The G-orbit Hilbert scheme of C Lemma 3 (X (N) ) G-inv. =C /G. Def 4 G - Hilb(C ):=(X [N] ) G-inv. G - Hilb(C ) is called the G-orbit Hilbert scheme of C. For I G - Hilb(C ), C[x, y]/i = C[G] regular repres. Thm 5 (Ito-Nakamura 1999) G - Hilb(C ) is a minimal resol. of C /G with enough information about repres. of G.

31 Thm 6 (Ito-Nakamura 1999) G - Hilb(C ) is a minimal resol. of C /G with enough information about repres. of G. This gives a new explanation for McKay corresp. Vertices of Dynkin diagram (bijective) Irred. components of except. set (bijective) (McKay corresp.) Eq. class of irred. reps( trivial) of G

3 A n... D n... D 5 E 6 E 7 E 8

33 A n... D n... D 5 E 6 E 7 E 8

34 A n... D n... D 5 E 6 E 7 E 8

35 A n... D n... D 5 E 6 E 7 E 8

36 A n... D n... D 5 E 6 E 7 E 8

37 A n... D n... D 5 E 6 E 7 E 8

38 A n... D n... D 5 E 6 E 7 E 8

39 1 1 1 1 1 A n... (n vertices) 1 1 D n... 1 1 D 5 E 6 1 E 7 3 E 8 4 1 3 1 1 4 3 1 3 6 5 4 3 (n vertices)

40 6 The exceptional set D 5 case π : G - Hilb(C ) C /G the natural,ap G - Hilb(C ) \ π 1 (0) C /G \{0} (isom.) Let E = π 1 (0) be the exceptional set. Then E = {I G - Hilb(X); I m} where m =(x, y)c[x, y] : the maximal ideal. For I E, we set V (I) =I/(mI + n) where n =(F, G, H)C[x, y]. V (I) isag-module of generators of I other than G-inv. For I G - Hilb(C ), C[x, y]/i =C[G] : Regular rep.

41 Define a subset of E by E 1 :={I E; ρ 1 V (I)} = {I G - Hilb; m I,ρ 1 V (I)} E :={I E; V (I) ρ }. We will see E 1 = {I 1 (s); s C} I 1 ( ) P 1, E = {I (s); s C } I (0) I ( ) P 1, I 1 ( ) =I (0) (intersection)

4 The exceptional set E, E 1 an irred. comp. of E V (ρ 1 )={xy},v 6 (ρ 1 )={x 6 y 6 } I 1 (s) ={xy + s(x 6 y 6 )} + n I 1 (s) is an ideal generated by xy + s(x 6 y 6 ) and n =(F, G, H). We see, dim C[x, y]/i 1 (s) =1, I 1 (s) G - Hilb(C ) ( s C)

43 As s, If E I, we have I n =(F, G, H). Hence C[x, y]/n C[x, y]/i (surjective) E Grass.(C[x, y]/n, codim 11 Grass is compact, so that the sequence I 1 (s) (s C) converges. Now we define: I 1 ( )/n = lim s I 1(s)/n To compute I 1 ( ) = McKay corresp. appears!

44 I 1 (s) ={xy + s(x 6 y 6 )} + n (s 0) = { 1 s xy +(x6 y 6 )} + n (s 0) What happens when s =? Let 1 s =0. If I 1 ( ) ={x 6 y 6 } + n, Then I 1 ( ) / X [1], Absurd. The answer : I 1 ( ) ={x 6 y 6 } + {x y, xy } + n Since I 1 (s) n, x y, xy I 1 (s) for s 0. V 3 (ρ )={x y, xy } = {x, y} {xy} = ρ nat V (ρ 1 ) This reminds us of the McKay rule : ρ = ρ nat ρ 1

45 Recall E 1 :={I E; ρ 1 V (I)} E :={I E; ρ V (I)}. We saw We will see E 1 = {I 1 (s); s C} I 1 ( ) P 1 E = {I (s); s C} I ( ) P 1

46 V 3 (ρ )={x y, xy }, V 5 (ρ 1 )={ y 5,x 5 } Similarly for t 0, we define I (t) :=(x y ty 5,xy + tx 5 )+n We see V (I (t)) ρ. Then, E = {I E; ρ V (I)} P 1 As t 0, by the same reason as I 1 ( ), I (0) (x y, xy )+n

47 I (0) = (x y, xy )+(x 6 y 6 )+n = I 1 ( ) Intersection of E 1 and E!! This comes from McKay rule because V 6 (ρ 1 ) = {x 6 y 6 } = {x, y} { y 5,x 5 } mod n = ρ nat V 5 (ρ ) mod n +(x y, xy ) This reminds us of McKay rule : ρ 1 + = ρ nat ρ D 5 ρ 0 ρ ρ 3 ρ 4 ρ 1 6 6 6 6 6 6 ρ 5

48 I 1 (s) ={(1/s)(xy) +(x 6 y 6 )+n I 1 ( ) =(x y, xy ) +(x 6 y 6 )+n V 3 (ρ )={x y, xy } = {x, y} {xy} = ρ nat V (ρ 1 ) This reminds us of McKay rule : ρ = ρ nat ρ 1 I (t) =(x y + t( y 5 ),xy + tx 5 )+n I (0) = (x y, xy )+(x 6 y 6 ) + n V 6 (ρ 1 ) {x 6 y 6 } {x, y} { y 5,x 5 } ρ nat V 5 (ρ ) mod n +(x y, xy ) This reminds us of McKay rule : ρ 1 + = ρ nat ρ Let (1/s) =0 t =0,Red part changes into Green.

49 E(ρ 1 ) E(ρ ) E(ρ 1 ) I 1 ( ) E(ρ ) I 1 (s) (s C) I (t) (t C ) I (0) V (I 1 (s)) ρ 1, V (I (t)) ρ, V (I 1 ( )) = V (I (0)) ρ 1 ρ s C, t C(s 0) where V (I) =I/(mI + n) generators of I

50 D 5 ρ 0 ρ ρ 3 ρ 4 ρ 1 6 6 6 6 6 6 ρ 5 ρ ρ nat = ρ 0 + ρ 1 + ρ 3 ρ 1 ρ nat = ρ How does the Dynkin diag. come out?

51 m V m (ρ) Eq. class 1 {x, y} ρ {xy} {x,y } ρ 1 + ρ 3 3 {x y, xy } {x 3 ± iy 3 } ρ + ρ + 4 + ρ 5 4 {y 4,x 4 } {x 3 y, xy 3 } ρ 3 5 {y 5, x 5 } {xy(x 3 ± ( iy 3 ))} ρ + ρ + 4 + ρ 5 6 {x 6 y 6 } {x 5 y, xy 5 } ρ 1 + ρ 3 7 {xy 6,x 6 y} ρ Decomposition of the coinv. alg. into repres. of G(D 5 ) Coinv. alg. : C[x, y]/n =C[x, y]/(f, G, H)

5 Degree 1 3 4 Repr. ρ ρ 1 + ρ 3 ρ + ρ 4 + ρ 5 ρ 3 Degree 7 6 5 Repr. ρ ρ 1 + ρ 3 ρ + ρ 4 + ρ 5 The quiver str. of the red part determ. the Dynkin diag. Let V 1 = {x, y} V 1 V (ρ 1 )=V 3 (ρ ), V 1 V 3 (ρ ) V 4 (ρ 3 ), V 1 V 3 (ρ 4 ) V 4 (ρ 3 ),

53 Degree 1 3 4 Repr. ρ ρ 1 + ρ 3 ρ + ρ 4 + ρ 5 ρ 3 Degree 7 6 5 Repr. ρ ρ 1 + ρ 3 ρ + ρ 4 + ρ 5 Quiver str. of Red part determ. Dynkin. Deg.3: V 1 ρ 1 = ρ, where, V 1 = {x, y} Deg.4: V 1 ρ = ρ 3, V 1 ρ 4 = ρ 3, V 1 ρ 5 = ρ 3 Deg.5: V 1 ρ 3 ρ ρ 4 ρ 5 Deg.6: V 1 ρ ρ 1, V 1 ρ 4 0, V 1 ρ 5 0 The green part is contained in I

54 Quiver str. of D 5 V (ρ 1 ) V 6 (ρ 1 ) V 3 (ρ ) V 5 (ρ ) V 4 (ρ 3 ) V 3 (ρ 4 ) V 5 (ρ 4 ) V 3 (ρ 5 ) V 5 (ρ 5 )

55 Dynkin diagram D 5 V (ρ 1 ) V 6 (ρ 1 ) V 3 (ρ ) V 5 (ρ ) V 4 (ρ 3 ) V 3 (ρ 4 ) V 5 (ρ 4 ) V 3 (ρ 5 ) V 5 (ρ 5 )

56 Compute from lower degrees V (ρ 1 ) V 6 (ρ 1 ) V 3 (ρ ) V 5 (ρ ) V 4 (ρ 3 ) V 3 (ρ 4 ) V 5 (ρ 4 ) V 3 (ρ 5 ) V 5 (ρ 5 )

57 7 Exceptional set E 6 -case G(E 6 ) SL(, C) (Binary Tetrahedral Group) σ = G(E 6 )= σ, τ, µ order 4 i, 0, τ = 0, 1, µ = 1 0, i 1, 0 where ɛ = e πi/8. ɛ7,ɛ 7, ɛ 5, ɛ G(D 4 ):= σ, τ, normal in G(E 6 ) 1 G(D 4 ) G(E 6 ) Z/3Z 1. C /G(D 4 ) 3:1 C /G(E 6 )

58 Irred. rep. of E 6 (irred. character) 1 1 τ µ µ µ 4 µ 5 ( ) 1 1 6 4 4 4 4 ρ 0 1 1 1 1 1 1 1 ρ 0 1 1 1 1 ρ 3 3 3 1 0 0 0 0 ρ 0 ω ω ω ω ρ 1 1 1 1 ω ω ω ω ρ 0 ω ω ω ω ρ 1 1 1 1 ω ω ω ω

59 Coinv. alg. of E 6 m V m 1 ρ ρ 3 3 ρ + ρ 4 ρ 1 + ρ 1 + ρ 3 5 ρ + ρ + ρ 6 ρ 3 7 ρ + ρ + ρ 8 ρ 1 + ρ 1 + ρ 3 9 ρ + ρ 10 ρ 3 11 ρ

60 V 5 (ρ ) V 7 (ρ ) V 4 (ρ 1 ) V 8 (ρ 1 ) V 5 (ρ ) V 7 (ρ ) V 6 (ρ 3 ) V 5 (ρ ) V 7 (ρ ) V 4 (ρ 1 ) V 8 (ρ 1 ) Dynkin diagram E 6

61 V 5 (ρ ) V 7 (ρ ) V 4 (ρ 1 ) V 8 (ρ 1 ) V 5 (ρ ) V 7 (ρ ) V 6 (ρ 3 ) V 5 (ρ ) V 7 (ρ ) V 4 (ρ 1 ) V 8 (ρ 1 ) Dynkin diagram E 6

6 A n... D n... D 5 E 6 E 7 E 8

63 A n... D n... D 5 E 6 E 7 E 8

64 A n... D n... D 5 E 6 E 7 E 8

65 A n... D n... D 5 E 6 E 7 E 8

66 A n... D n... D 5 E 6 E 7 E 8

67 A n... D n... D 5 E 6 E 7 E 8

68 A n... D n... D 5 E 6 E 7 E 8

69 8 Summary Need only to compute E the exceptional set. Need only to see the coinv. alg. C[x, y]/(f, G, H) G-inv. polyn. are zero in the coinv. alg. Look at coinv. alg. Forget trivial rep. in D, Ẽ. The red part of the coinv. alg. determines Dynkin. Degree 1 3 4 Repr. ρ ρ 1 + ρ 3 ρ + ρ 4 + ρ 5 ρ 3 Degree 7 6 5 Repr. ρ ρ 1 + ρ 3 ρ + ρ 4 + ρ 5

70 For I E : (except. set), we define V (I) :=I/(mI + n) Then V (I) is either irred. or ρ ρ (ρ, ρ irred.) For ρ, ρ irred rep of G, ρ ρ, define subsets of E by E(ρ) :={I E; V (I) ρ} P (ρ, ρ ):={I E; V (I) ρ ρ }

71 Thm 7 (Ito-Nakamura 1999) Let G be a finite subgroup of SL(, C). Then (1) G-Hilb(C ) is a min. resol. of C /G. () For any irred rep. ρ of G, ρ trivial, E(ρ) =P 1, The map ρ E(ρ) is McKay corresp. (3) Decomp. rule of ρ nat into irred. rep. determine the quiver str. of the coinv. algebra, from which the intersection P (ρ, ρ ) between E(ρ) come out.

7 End Thank you for your attention.