SO(n) [8] SU(2)

Transcription

1 SO(n) [8] SU(2) SO(3) SU(3) , version

2 U(n) U(n) U(n) U(n) SU(n) U(n) SU(n) Spin(2n) Sp(n) Sp(n) Sp(n) Sp(n) Sp(n) Spin(4n) Cl 2m grading U(n) Sp(n)Sp(1) E H formulation sp(n) sp(1) Sp(n)Sp(1) Sp(n)Sp(1) Sp(n)Sp(1) Kraines SU(2), SU(3) 2

3 1.1 Definition 1.1. G V C π : G GL(V ) V G G V g V π : g End(V ) Definition 1.2. (, ) V G (π(g)v, π(g)w) = (v, w) g G, v, w V (π(x)v, w) + (v, π(x)w) = 0 X g, v, w V Remark 1.1. (π(exp tx)v, π(exp tx)w) = (v, w) (π(x)v, w) + (v, π(x)w) = 0 Proposition 1.1. G Proof. V (, ) 1 (v, w) inv = (gv, gw)dg vol(g) G G Definition 1.3. V G W G GW W V, {0} Proposition 1.2. G Proof. V G W V G W W W G 3

4 Definition 1.4. G (π, V ), (π, V ) Φ : V V G Φ(π(g)v) = π (g)φ(v) g G, v V Proposition 1.3 ( ). G (π, V ), (π, V ) Φ : V V G Φ(π(g)v) = π (g)φ(v) g G, v V V V Φ = 0 V V Φ = λid λ C λ = 0 Definition 1.5. (π, V ), (π, V ) G Hom G (V, V ) Corollary 1.4. (π, V ), (π, V ) { 0 π π Hom G (V, V ) = C π π Proof of proposition. Φ : V V G ker Φ, ImageΦ G Φ 0 ker Φ = 0 ImageΦ = V Φ π π π π Φ = 0 G T : V V T 1 Φ : V V Φ Φ λ Φ λ G ker(φ λ) {0} G ker(φ λ) = V Φ = λid Example 1.1. G (π, V ) V (π(g)f)(v) = f(π(g 1 )v), f V, v V V V V V := { v v V } v + w = v + w, z v = z v v v V π(g) v = π(g)v, v V 4

5 V G V V G V Example 1.2. (π, V ), (π, V ) 1. V V (π π )(g) := π(g) π (g) G 2. V Λ k (V ) G 3. V S k (V ) G 4. Hom(V, V ) G Definition 1.6. g g C = g C C g g C Example 1.3. gl(n, R), u(n), u(p, q) p + q = n gl(n, C) Definition 1.7. G H g h G H H G Example 1.4. GL(n, R), U(n), U(p, q) GL(n, C) Proposition 1.5. Lie Lie easy Proposition 1.6 (Weyl ). G C G V C 1. g V 2. g C V 3. G V 4. G C V holomorphic 5

6 Proof. g g C π(x + 1Y ) = π(x) + 1π(Y ) g C G C G C GL(V ) π G C GL(V ) exp exp g C π End(V ) G g (π, V ) χ(g) = tr V π(g) χ(gtg 1 ) = χ(t) U(n) Lie Lie weight 1.2 S 1 = U(1) u(1) = 1R exp : 1R t e 2πit U(1) U(1) V g U(1) π(g) : V V U(1) π(g) U(1) π(g) = λ g λ g C 1 U(1) U(1) GL(1, C) π : U(1) U(1) π 1R 1R π : 1R 1R π(t) = αt α R π : U(1) U(1) π(e 2πit ) = e α2πit well-defined α Z m Z π π m : U(1) g g m U(1) m n π m π n 6

7 Proposition 1.7. U(1) 1 Z π m : U(1) g g m U(1) GL(1, C) α R 1R t αt 1R gl(1, C) α Z integral T l = S 1 S 1 Proposition 1.8. T l 1 Z l π m1,,m l (t 1,, t l ) = t m 1 1 t m l l T l V = m Z le(m 1,, m l ) E(m 1,, m l ) (t 1,, t l ) T l t m 1 1 t m l l 1.3 SU(2) π π(g)v gv SU(2) ( ) e it 0 T = { 0 t 2π} S 1 0 e it Proof. T T T g T, g / T T g T T = T 7

8 su(2) ( ) ( ) ( ) i i σ 1 =, σ 2 =, σ 3 = 0 i 1 0 i 0 t = {tσ 1 t R} su(2) exp t = T su(2) C = sl(2, C) ( ) ( ) ( ) 1 0 H =, X = (σ iσ 3 ) =, Y = ( σ iσ 3 ) = 1 0 [H, X] = 2X, [H, Y ] = 2Y, [X, Y ] = H SU(2) V = C 2 SU(2) g g SU(2) GL(C 2 ) T C 2 = E(1) E( 1) = C C E(±1) v ±1 Hv + = v +, Xv + = 0, Y v + = v, Hv = v, Xv = v +, Y v = 0 k S k (V ) k + 1 {w k 2i := v k i + v i i = 0,, k} H S k (V ) k i + ( i) = k 2i S k (V ) = i E(k 2i) = i C(w k 2i ), Hw k 2i = (k 2i)w k 2i Hw k 2i = (k 2i)w k 2i, Xw k 2i = iw k 2i+2, Y w k 2i = (k i)w k 2i 2 X E( k) X E( k + 2) X X E(k 2) X E(k) X {0} Y {0} Y E( k) Y E( k + 2) Y 8 Y E(k 2) Y E(k)

9 Lemma 1.9. S k (V ) Proof. W S k (V ) v 0 v W v = a i w k 2i v X X l v 0, X(X l v) = 0 X l v X l v = cw k Y w k 2i W = V Lemma SU(2) S k (V ) k Proof. W T W = E(i) E(i) H i T i Z k v E(k) [H, X] = 2X H(Xv) = (k +2)Xv Xv 0 Xv = 0 [H, Y ] = 2Y HY v = (k 2)Y v X(Y v) = Y Xv + Hv = kv XY 2 v = [X, Y ]Y v + Y XY v = HY v + ky v = (k 2)Y v + ky v XY l v = l(k l + 1)Y l 1 v {v, Y v, Y 2 v,, } W = {v, Y v, Y 2 v,, } Y n v = 0 Y n 1 v 0 0 = XY n v = n(k n + 1)Y n 1 v k n+1 = 0 W = {v, Y v,, Y k v} X, Y, H W = S k (V ) SU(2) SU(2) W Zv, w + v, Zw = 0 Z su(2) 1Z W t t R = 1t t R W = E(i) weight weight vector weight weight weight vector v Xv = 0 highest weight vector weight highest weight highest weight W weight 9

10 Example 1.5. S k (V ) spin-k/2 S k (V ) i E(k 2i) weight k 2i weight k highest weight Theorem SU(2) Z 0 k Z 0 k V k := S k (V ) k highest weight Example 1.6. SU(2) sl(2, C) sl(2, C) = E(2) E(0) E( 2) = C(X) C(H) C(Y ) weight weight Example 1.7. SU(2) J; C 2 (α, β) ( β, ᾱ) C 2 J 2 = 1 J(g(α, β)) = gj(α, β) SU(2) SU(2) J k : (C 2 ) k k k S k (V ) k k Ω((α, β), (α, β )) = αβ α β k k (π k, V k ) (π l, V l ) {w k, w k 2,, w k }, {v l, v l 2,, v l } H(w k v l ) = (Hw k ) v l + w k Hv l = (k + l)w k v l, X(w k v l ) = 0 w k v l highest weight vector w k v l Y V k+l V k+l V k V l X(w k v l 2 w k 2 v l ) = w k v l w k v l = 0 H(w k v l 2 w k 2 v l ) = (k + l 2)(w k v l 2 w k 2 v l ) 10

11 V k+l 2 V k V l s ( s X{ i i=0 s ( s = i i=1 s 1 ( s = j =0 j=0 H{ s i=0 ) ( 1) i w k 2i v l 2(s i) } ) ( 1) i iw k 2(i 1) v l 2(s i) + s 1 i=0 ) (s j)( 1) j+1 w k 2j v l 2(s j 1) + ( ) s (s i)( 1) i w k 2i v l 2(s i 1) i s 1 j=0 ( ) s ( 1) i w k 2i v l 2(s i) } = (k + l 2s){ i ( ) s (s i)( 1) i w k 2i v l 2(s i 1) i s i=0 ( ) s ( 1) i w k 2i v l 2(s i) } i V k+l 2s V k V l s = 0, 1,, min{k, l} dim V k+l 2s = k + l 2s + 1, dim V k V l = (k + 1)(l + 1) Theorem (π k, V k ) (π l, V l ) V k V l = V k+l V k+l 2 V k+l 4 V k l SO(3) SO(3) Spin(3) = SU(2) SO(3) SO(3) SU(2) SU(2) SO(3) SU(2) SO(3) S k (V ) k SO(3) SU(2) su(2) R 3 Ad : SU(2) SO(3) SU(2) T SO(3) cos t sin t 0 T = Ad(T ) = { sin t cos t 0 0 t < 2π} S 1 = SO(2)

12 Ad : T Ad(T ) U(1) = S t = span R {H = } SO(3) W Ad(T ) 1t weight W = i E(i), w.r.t Ad(T ) SU(2) lift T W = i E(2i) w.r.t T SO(3) SU(2) lift S 2k (V ) S 2k+1 (V ) SO(3) Proposition SO(3) Z 0 k Z 0 S 2k (V ) su(2) so(3) S 2k+1 (V ) so(3) su(2) so(3) H = 1H 2 V = C 2 H v + = 1 2 Hv + = 1 2 v +, H v = 1 2 v so(3) weight weight Proposition so(3) k/2 S k (V ) spin-k/2 SO(3) lift Spin(3) = SU(2) Spin(3) = SU(2) SO(3) S k (V ) so(3) highest weight k/2 so(n) weight Spin(n) SO(n) weight integral half-integral 12

13 1.6 SU(3) SU(3) e i2πa T = { 0 e i2πa 2 0 a 1 + a 2 + a 3 = 0} S 1 S e i2πa 3 h = a { 0 a 2 0 a 1 + a 2 + a 3 = 0} 0 0 a 3 a h R := { 0 a 2 0 a 1 + a 2 + a 3 = 0} 0 0 a 3 V h R weight weight h R V = E(α), α h R H h R v E(α) Hv = α(h)v α h R weight weight a h R = R{L 1, L 2, L 3 }/{L 1 + L 2 + L 3 = 0}, L i 0 a 2 0 = a i 0 0 a 3 weight m 1 L 1 + m 2 L 2 + m 3 L 3, m 1, m 2, m 3 Z T integral SU(3) sl(3, C) weight sl(3, C) = h C C(E ij ) C(E ji ) 1 i<j 3 1 i<j 3 13

14 E ij (i, j) 1 weight weight [E kk, E ij ] = δ ki E kj δ jk E ik = δ ki E ij δ jk E ij = (L i L j )(E kk )E ij E ij weight L i L j sl(3, C) = g 0 i<j g Li L j i<j g Lj L i L 1 L 2, L 2 L 3, L 1 L 3 L 2 L 1, L 3 L 2, L 1 L 3 sl(3, C) V weight V = E(α) v E(α) X g β HX(v) = XHv + [H, X]v = (α + β)xv Xv 0 Xv weight vector weight α + β V = E(α) weight α weight α {L i L j } i j E 12, E 23, E 13 weight V V weight v E 12 v = E 23 v = E 13 v = 0 highest weight vector weight highest weight vector highest weight vector SU(2) highest weight vector v E 21, E 31, E 3,2 V [E ij, E ji ] = E ii E jj 1 i < j 3 [E ii E jj, E ij ] = 2E ij, [E ii E jj, E ji ] = 2E ji E ij X, E ji Y, E ii E jj H E ij, E ji, E ii E jj 1 i < j 3 p sl(2, C) highest weight vector v E ij v = 0 p highest vector k (E ii E jj )v = kv highet weight (m 1 L 1 + m 2 L 2 + m 3 L 3 )(E ii E jj ) 0, 1 i < j 3 14

15 m 1 m 2 m 3 dominant weight m 1 m 2 m 3, m i Z dominant integral dominant integral (m 1, m 2, m 3 ) C(v) v E 12 v = E 23 v = E 13 v = 0 Hv = (m 1 L 1 + m 2 L 2 + m 3 L 3 )(H)v H h R Z(m 1, m 2, m 3 ) := span C {E k 21E l 32E m 31v k, l, m Z 0 } sl(3, C) Y (m 1, m 2, m 3 ) Z/Y highest weight m 1 L 1 + m 2 L 2 + m 3 L 3 Theorem SU(3) dominant integral weight (m 1, m 2, m 3 ) Z 3 mod Z(1, 1, 1), m 1 m 2 m 3 dominant integral weight highest weight Remark 1.2. mod Z(1, 1, 1) h R = R{L 1, L 2, L 3 }/{L 1 + L 2 + L 3 = 0} Example 1.8. SU(3) V = C 3 V = E(L 1 ) E(L 2 ) E(L 3 ) E(L 1 ) = C{(1, 0, 0)} highest weight vector g h h R g C weight h R Rl h R weight Example 1.9. sl(3, C) l(a 1 L 1 + a 2 L 2 + a 3 L 3 ) = aa 1 + ba 2 + ca 3, a + b + c = 0, a > b > c 15

16 l : h R R l(α) > 0 α > 0, l(β) < 0 β < 0 α > β l(α β) > 0 l(l 1 L 2 ) = a b > 0, l(l 2 L 3 ) = b c > 0, l(l 1 L 3 ) = a c > 0 L 1 L 2 > 0, L 2 L 3 > 0, L 1 L 3 > 0 L 2 L 3 L 2 L 1 L 2 L 1 L 3 L 1 L 3 L 3 L 1 L 1 L 2 L 3 L 2 sl(3, C) L 1 + L 2 + L 3 = 0 h R Rl α = (α 1,, α l ) g C g C = g 0 α g α α g α, g 0 = h C, α h R 1 α α dim R h R g α, g α, [g α, g α ] sl(2, C) W weight W = β E(β) β h R X g α α Xv = 0 weight vector highest weight vector highest weight 1 highest weight vector weight 1 SU(3) g α, g α, [g α, g α ] sl(2, C) dominant G lift integral half-integral G highest weight dominat integral dominant integral highest weight dominant integral weight 16

17 (π, V ), (π, V ) highest weight ρ, ρ highest weight vector v ρ, v ρ V V v ρ v ρ highest weight vector highest weight ρ + ρ V V highest weight ρ + ρ SU(2) Spin(n) 2 2n 2.1 Spin c (n). Definition 2.1. ±1 Spin(n), ±1 U(1), Spin c (n) := (Spin(n) U(1))/{±1}. Spin(n) Cl n U(1) Cl n = Cl n C 1 C Spin c (n) [g, z] g z Cl n C well-defined g z = 1 g = z 1 Spin(n) C = {±1} g = z 1 = ±1 (g, z) = (1, 1) or (g, z) = ( 1, 1) well-defined Spin c (n) 17

18 Spin c (n) := Spin(n) U(1) C Cl n Spin c (n) C = U(1) spin c (n) := spin(n) 1R Cl n Spin c (n) := exp spin c (n) Cl n Spin c (n).. Ad : Spin(n) g Ad(g) SO(n), Ad : Spin c (n) [g, z] Ad(g) SO(n), i : Spin(n) g [g, 1] Spin c (n), j : U(1) z [1, z] Spin c (n), l : Spin c (n) [g, z] z 2 U(1), p = Ad l : Spin c (n) [g, z] (Ad(g), z 2 ) SO(n) U(1). well-defined p, 1 Spin(n) i Spin c (n) l U(1) 1 1 U(1) j Spin c (n) Ad SO(n) 1 Spin(n) SO(n), Spin c (n) SO(n) U(1) U(1). Lemma 2.1. n 3, 1. π 1 (Spin c (n)) = Z, l : π 1 (Spin c (n)) π 1 (U(1)) = Z 2. α π 1 (Spin c (n)) = Z, β = 1 π 1 (SO(n)) = Z 2, γ π 1 (U(1)) = Z, l (α) = γ. p : π 1 (Spin c (n)) π 1 (SO(n)) π 1 (U(1)) p (α) = β + γ. Proof. π 1 (Spin(n)) π 1 (Spin c (n)) π 1 (U(1)) π 0 (Spin(n)) 18

19 α π 1 (Spin c (n)) = Z, β π 1 (SO(n)) = Z 2, γ π 1 (U(1)) = Z, l (α) = γ Ad (α) = β π 1 (SO(n)) = Z 2 Ad (α) = 0 Z 2 π 1 (U(1)) j π 1 (Spin c (n)) Ad π 1 (SO(n)) π 0 (U(1)) δ π 1 (U(1)) j (δ) = α γ = l (α) = l j (δ) = 2δ in π 1 (U(1)) γ U(m) R n e i e i R n+2 Cl n Cl n+2 Spin(n) Spin(n + 2). spin(n + 2) e n+1 e n+2 exp e n+1 e n+2 t U(1) (e n+1 e n+2 ) 2 = 1 e i e j 1 i < j n e n+1 e n+2 1. U(1) Spin(n + 2) (Spin(n) U(1))/{±1} Spin(n + 2). Lemma Spin c (n) p SO(n) U(1) = SO(n) SO(2) f Spin(n + 2) Ad i SO(n + 2) (2.1) f. Proof. 2.2 Definition 2.2. Cl n Spin c (n) n. W 2m+1 W 2m ± Spin c (n) W ([g, z], φ) ([g, z])φ = z (g)φ = zgφ W (2.2) well-defined.. 19

20 Spin(n) n : Spin(n) U(W n ) SU(W n ) n 3. Proof. det, Spin(n) U(1), Spin(n) Spin(n) R, Spin(n) R 0. det = 1. det( 2m+1 ([g, z])) = z 2m det( ± 2m([g, z])) = z 2m 1. Spin c (n) det U(1) z U(1) Spin c (n) Remark 2.1.,. U(1) z, J J([g, z] v) = [g, z] J(v) Remark 2.2. l = 2k + 1 k Z Spin c (n) W ([g, z], φ) l ([g, z])φ = z l (g)φ = z l gφ W l Spin(n) U(1) Spin c (n) = Spin(n) U(1)/±1 SO(n) 3 SO(n), U(n), SU(n), Sp(n), Sp(n)Sp(1), G 2, Spin(7) Hol 0 (M, g) Hol(M, g) G 2, Spin(7) G 2, Spin(7) 20

21 3.1 Lemma 3.1 ( ). G, H G G H π : G G f : H G F : H G f (π 1 (H, e H )) π (π 1 ( G, e G)) H F f G π G Proof. f (π 1 (H, e H )) π (π 1 ( G, e G)) h H e H ω f(ω) e G f(h) π : G G e G ω F (h) ω f(ω ) f(ω 1 ) γ f (π 1 (H, e H )) [γ] γ π [ γ] = [γ] γ γ γ ω ω F F Remark 3.1. H H H = G = Z 2 = {±1}, G = {1} f(±1) = 1 f F F (±1) = ±1 F (±1) = 1 Proposition 3.2 ([6]). (M, g) Hol(M) SO(n) F F G = Hol(M, g) F i Spin(n) Ad SO(n) 21

22 M M Proof. g ij (x) G f ij = F (g ij ) Spin(n) f ij f jk f kl = F (g ij g jk g kl ) = id f ij Spin(n) Ad g ij SO(n) g ij Spin(n) g ij g jk g ki = z ijk Z 2 [z ijk ] = w 2 (M) g ij g ij g jk g ki = id Z 2 [6] Hol(M) Hol(M) Example 3.1. SU(n) π 1 (SU(n)) = 1 SU(n) G 2 Spin(7) π 1 (G 2 ) = 1, π 1 (Spin(7)) = 1 G 2 Spin(7), Spin(7) Spin(8) G 2 Spin(7) Example 3.2. U(n) 3.2 U(n) U(n) I compatible I I 2 = 1, Iv, Iw = v, w, for v, w R 2n U(n) SO(2n) 22

23 Definition 3.1. U(n) = {A SO(2n) AI = AI} = {A GL(2n, R) Au, Av = u, v, AI = IA} (3.1) GL(n, C) = {A GL(2n, R) AI = AI} I V = (R 2n, I) V C I ± 1 (1, 0), (0, 1) V C = V 1,0 V 0,1 V 0,1 = V 1,0, V C h(u, v) := u, v V C V 1,0 h V 0,1 Proof. u V 1,0, v V 0,1 v V 1,0 h(u, v) = u, v = Iu, I v = 1 2 u, v = h(u, v) = 0 V 1,0, V 0,1 (V 1,0, h), (V 0,1, h) Proof. v V 1,0 h(v, v) = 0 v, v = 0 v + v, v + v = v, v + 2 v, v + v, v = v, v + v, v = 0 v, v = Iv, Iv = v, v = 0 v + v V, V v + v = 0 v = 0 = v (V 1,0, h) U(n) GL(n, C) Definition 3.2. U(n) := {A GL(V 1,0 ) h(au, Av) = h(u, v)} (3.2) 3.2 U(n) 3.1 U(n) 23

24 Proof. A GL(V 1,0 ) v V 1,0 v + v V A (v + v) = Av+Av A GL(2n, R) Av + Av = Av+Av V A (I(v + v)) = A ( 1v + 1v) = 1Av + A 1v = 1Av 1 Av = IA (v + v) I GL(V 1,0 ) = GL(n, C) GL(2n, R) I Av + Av, Aw + Aw = 2 Av, Aw = 2h(Av, Aw) U(n) (3.1) (3.2) Remark 3.2., ( ) A B GL(n, C) A + ib GL(2n, R) B A ( ) 0 id I = id 0 Λ 1,0 := (V 1,0 ), Λ 0,1 := (V 0,1 ) V 1,0 v h(, v) =, v Λ 0,1 V 0,1 Λ 1,0 Remark 3.3. (1, 0) (0, 1) U(n) V e 1, e 2 = Je 1,, e 2n 1, e 2n = Je 2n 1 a i := 1 2 (e 2i 1 1Je 2i 1 ), a i := 1 2 (e 2i 1 + 1Je 2i 1 ) Ja i = 1a i Ja i = 1a i V 1,0 = C{a i 1 i n}, V 0,1 = C{a i 1 i n} a i = a i, a i = a i h(a i, a j ) = δ ij/2, h(a i, a j ) = δ ij /2, h(a i, a j) = 0 24

25 Remark 3.4. a i 1 (e 2 2i 1 1Je 2i 1 ) a i a j + a j a i = δ ij V 1,0 Λ 0,1 Λ 0,1 = C{a i 1 i n}, Λ1,0 = C{a i 1 i n} Remark a i 2a i V 1,0 h(, 2a i ) = 2h(, a i ) 2a i V 0,1 = Λ 1,0 a i 1 (e 2 2i 1 1Je 2i 1 ) U(n) Definition 3.3. V 1,0 Λ 1,0 = V 1,0 V 0,1 gl(n, C) V 1,0 ɛ i ɛ i := 1 2 (e 2i 1 1Je 2i 1 ) = 1 2a i ɛ i := 1 2 (e 2i 1 + 1Je 2i 1 ) = 1 2a i V 1,0 Λ 0,1 V 0,1 ɛ i gl(n, C) ɛ i ɛ j = ɛ i (ɛ j ) = 2a i a j 1 i, j n E ij V 1,0 ɛ i ɛ j (ɛ k ) = δ kj ɛ i [ɛ i ɛ j, ɛ k ɛ l ] = ɛ j (ɛ k )ɛ i ɛ j ɛ l (ɛ i )ɛ k ɛ j = δ jk ɛ i ɛ j δ il ɛ k ɛ j ([E ij, E kl ] = δ jk E ij δ il E kj ) Remark 3.6. i j k = δ jk i i j k l = δ jk i l 25

26 u(n) gl(n, C) zij ɛ i ɛ j z ij ɛ j ɛ i u(n) 1ɛi ɛ i, 1 i n, 1(ɛi ɛ j +ɛ j ɛ i ), ɛ i ɛ j ɛ j ɛ i, 1 i < j n. u(n) R{ 1ɛ i ɛ i } i h R := R{ɛ i ɛ i 1 i n} gl(n, C) U(n) h R weight λ = (λ 1,, λ n ) weight λ i ɛ i ɛ i λ i Z weight highest weight ρ highest weight dominant integral ρ = (ρ 1,, ρ n ) Z n, ρ 1 ρ 2 ρ n highest weight U(n) GL(n, C) U(n) GL(n, C) Weyl highest weight ρ (π ρ, V ρ ) π ρ V ρ gl(n, C) =h C i<j C(ɛ i ɛ j ) i<j C(ɛ j ɛ i ) = i C(E ii ) i<j C(E ij ) i<j C(E ji ) Definition 3.4. weight λ = (λ 1,, λ n ) k j k j (0 i 1, 1, 0 n i ) = (0,, 0, 1, 0,, 0) }{{}}{{} i 1 n i 26

27 Example 3.3 ( ). V 1,0. span C {a i i = 1,, n} = span C {ɛ i i = 1,, n} a 1 ɛ 1 highest weight vector highest weight (1, 0 n 1 ) ɛ i ɛ i (ɛ 1 ) = δ i1 ɛ i ɛ i ɛ i i = 1 1 i 1 weight (1, 0,, 0) ɛ i ɛ j (ɛ 1 ) = δ 1j ɛ i = 0, 1 i < j n ɛ 1 highest weight vector a i weight (0 i 1, 1, 0 n i ). weights (1, 0,, 0) > (0, 1, 0,, 0) > > (0,, 0, 1) highest weight (1, 0 n 1 ) lowest weight (0 n 1, 1). Example 3.4 ( ).. span C {a i i = 1,, n}. a i weight (0 i 1, 1, 0 n i ). highest weight (0 n 1, 1) highest weight vector a n. Example 3.5 ((0, p)-form). Λ p V (1,0) = Λ 0,p highest weight (1 p, 0 n p ), highest weight vector a 1 a p. Example 3.6 (det ). det (det, Λ 0,n ) highest weight (1 n ). det k highest weight (k n ). Example 3.7 ((p, 0)-form). Λ p V (0,1) = Λ p,0. highest weight (0 n p, ( 1) p ). Λ p,0 Λ 0,n highest weight (0 n p, ( 1) p ) + (1 n ) = (1 n p, 0 p ) Λ p,0 Λ 0,n Λ 0,n p. Example 3.8 ( ). k.. highest weight (k, 0 n 1 ) weight vector k a 1. Example 3.9 ( ). U(n) gl(n, C) adjiont.. gl(n, C) C{I}. highest weight (1, 0 n 2, 1). 27

28 gl(n, C) V (0n ) V (1,0n 2, 1). gl(n, C) = sl(n, C) tracepart. gl(n, C) = V (1,0) Λ 1,0 = Λ 0,1 Λ 1,0 = Λ 1,1 Λ 1,1 U(n) Ω = 2 1 a i a i = 1 ɛ i ɛ i = n e 2i 1 e 2i Ω = Ω (1, 1) Λ 1,1 = Λ 1,1 0 ΩΛ 0,0 0 primitive (1, 1) highest weight(1, 0 n 2, 1) Λ p,q = Λ p,0 Λ 0,q Λ 1,1 Ω Ω Ω Λ p,q U(n) Λ p,q Remark 3.7. W highest weight vector highest weight ρ V ρ W V ρ dim V ρ = dim W V ρ = W W Remark 3.8., (π ρ, V ρ ), conjugate representation or contragradient representation weight weight ( 1). highest weight lowest weight ( 1) highest weight lowest weight weight diagram., postive root negative root, highest weight lowest weight. Proposition 3.3. highest weight ρ = (ρ 1,, ρ n ) (π ρ, V ρ ) highest weight. ( ρ n, ρ n 1,, ρ 1 ) 28 i=1

29 Remark 3.9. (π ρ, V ρ ) det highest weight ρ + k(1 n ) SU(n) (1 n ) {ρ mod Z(1 n ) ρ dominant integral } SU(n) SU(n) U(n) ˆρ i = ρ i 1 n n j=1 ˆρ 1 ˆρ 2 ˆρ n ˆρ i ˆρ j Z ρ j ˆρ 1 + ˆρ ˆρ n = 0 parametrize U(n) F U(n) F i Spin(2n) Ad SO(2n) Lemma 3.4. π 1 (SO(n)) = Z 2 n 3, π 1 (SO(2)) = Z cos t sin t 0 γ(t) = sin t cos t 0, 0 t 2π 0 0 I Proof. n = 2 n 3 1 Z 2 Spin(n) SO(n) 1 0 π 1 (Spin(n)) = 0 π 1 (SO(n)) π 0 (Z 2 ) 0 29

30 γ π 1 (SO(n)) Spin(n) Z 2 ±1 γ(t) γ(t) = cos t/2 + e 1 e 2 sin t/2 1 γ(t) π 1 (SO(n)) Z 2 G N H G = NH, N H = {e} G N H N H gng 1 N (nh)(n h ) = (nhn h 1 )(hh ) Lemma 3.5. U(n) = SU(n)U(1) U(n) U(1) SU(n) π 1 (U(n)) = Z ( ) e it 0 γ(t) =, 0 t 2π 0 I ( ) a 0 Proof. SU(n) U(n) U(1) U(n) 0 I SU(n) U(1) = {e} ( ) ( ) det A 1 0 det A 0 U(n) A A SU(n)U(1) 0 I 0 I Proposition 3.6. F U(n) F i Spin(2n) Ad SO(2n) Proof. π 1 (U(n)) U(n) SO(2n) cos t sin t 0 γ(t) = sin t cos t 0, 0 t 2π 0 0 I i (π 1 (U(n)) = Z 2 Ad (π 1 (Spin(2n))) = Corollary 3.7. U(n) Spin c (2n) 30

31 3.2.5 U(n) U(1) U(n) A i det Spin c (2n) p = Ad l (A, det A) SO(2n) U(1) Spin c (2n) SO(n) U(1) Proposition 3.8. F U(n) F f=i det Spin c (2n) p = Ad l SO(2n) U(1) F U(n) Spin c (2n) Proof. δ π 1 (U(n)) α, β, γ π 1 (Spin c (2n)), π 1 (SO(2n)), π 1 (U(1)) l (α) = γ p (α) = β+γ f (δ) = β +γ f (π 1 (U(n))) p (π 1 (Spin c (2n)) F f = p F f = i det i f F Remark f k : U(n) A (A, (det A) 2k+1 ) SO(2n) U(1), k Z f k (δ) = β+(2k+1)γ = (2k+1)β+(2k+1)γ f k (π 1 (U(n))) p (π 1 (Spin c (2n)) k Z U(n) F k f k Spin c (2n) p = Ad l SO(2n) U(1) U(n) Spin c (2n) 31

32 3.2.6 SU(n) SU(n) SU(n) Spin(2n) SU(n) or Proposition 3.9. F SU(n) F i Spin(2n) Ad SO(2n) Spin(2n) Spin c (2n) SU(n) Spin c (2n) U(n) U(n) Spin c (2n) u(n) R 1ɛi ɛ i, 1 i n, 1(ɛi ɛ j +ɛ j ɛ i ), ɛ i ɛ j ɛ j ɛ i, 1 i < j n 1ɛi ɛ i 1a i a i = 1 2 e 2i 1e 2i Cl2n 1(ɛi ɛ j + ɛ j ɛ i ) 1(a i a j + a j a i) Cl 2n ɛ i ɛ j ɛ j ɛ i a i a j a j a i Cl 2n 1a i a i = 1e 2 2i 1e 2i u(n) Remark a i a i a i a i u(n) C gl(n, C) ɛ i ɛ j a i a j. Lemma [a, b] = ab ba u(n) Cl 2n Proof. 32

33 u(n) Cl 2n G := exp u(n) Spin c (2n) Proposition G U(n) U(n) Spin c (2n) Propostion 3.8 Proof. V 1,0 Cl 2n [a i a j, a k ] = a i a ja k a k a i a j = δ jk a i Cl 2n G exp Cl 2n u(n) Ad U(n) exp ad u(n) ad Ad(exp X) = exp ad(x) ad U(n) g U(n) g = exp X X u(n) Ad Ad Ad(exp X) = id (exp X)a k (exp( X)) = a k k exp X = exp Y exp it G Spin c (2n) = Spin(2n) U(1) Ad(exp X)a k = (exp Y exp it)a k (exp it exp Y ) = (exp Y )a k (exp Y ) = a k (exp Y )a k (exp Y ) = a k (exp Y )a k(exp Y ) = a k exp Y Spin(2n) Spin(2n) exp Y = ±1 Ad(exp X) = id exp X = 1 exp it = 1 λ Spin(2n) U(1) t T = exp t G = g G Ad(g)T R{ 1a i a i} i exp T λ = exp X = gtg 1 t T λ t = λ t = λ T t = (cos θ 1 + sin θ 1 e 1 e 2 ) (cos θ n + sin θ n e 2n 1 e 2n )e 1(θ 1 + +θ n ) 33

34 θ 1 = m 1 π, θ 2 = m 2 π,..., θ n = m n π t = ( 1) m 1+ +m n e 1(m 1 + +m n)π = ( 1) 2(m 1+ +m n) = 1 λ = 1 Ad(exp X) = id exp X = id G = U(n) Spin c (2n) SO(2n) U(1) i det : U(n) SO(2n) U(1) U(n) Ad [a i a j, a k ] = a i a ja k a k a i a j = δ jk a i i : U(n) SO(2n) det det exp X = exp trx 1ɛi ɛ i u(n) 1ɛi ɛ i (a k ) = 1δ ik a i 1 u(n) 1ɛ i ɛ i 1a i a i e z z 2i 1e 2i u(1) 2 U(n) Spin c (2n) SO(2n) U(1) i det : U(n) SO(2n) U(1) Remark u(n) spin(2n) 1ɛi ɛ i 1a i a i 1/2 = 1 e 2 2i 1e 2i t = ±1 1 exp u(n) Ad : G U(n) U(n) exp ta 1a 1 = cos t/2 + e 1 e 2 sin t/2 Ad ( e it 0 0 I U(n) SO(2n) exp ta 1a 1 = (cos t/2 + e 1 e 2 sin t/2)e t 1/2 Ad ( ) e it 0 0 I Remark F k : U(n) Spin c (2n) 1ɛi ɛ i 1 e 2 2i 1e 2i + (2k + 1) 1/2 34 )

35 3.2.8 SU(n) Spin(2n) SU(n) Spin(2n) su(n) R 1(ɛi ɛ i ɛ n ɛ n ), 1 i n 1, 1(ɛi ɛ j +ɛ j ɛ i ), ɛ i ɛ j ɛ j ɛ i, 1 i < j n 1 1(ɛi ɛ i ɛ n ɛ n ) 1(a i a i a na n ) = 1 2 e 2i 1e 2i 1 2 e 2n 1e 2n 1(ɛi ɛ j + ɛ j ɛ i ) 1(a i a j + a j a i) ɛ i ɛ j ɛ j ɛ i a i a j a j a i su(n) exp su(n) Spin(2n) SU(n) = exp su(n) Spin c (2n) U(n) a k 1 a k 2 a k j vac, 1 k 1 < k 2 < < k j n, j = 0, 1,, n Spin c (2n) U(n) Λ 0,p U(n) Cl 2n N = a i a i Remark Ω = 2 1 a i a i Ω = i e 2i 1 e 2i = 2 1 i (a i a i 1/2) = 2 1(N n/2) 35

36 N u(n) [u(n), Ω] = 0 p N = p W p := C{a k 1 a k 2 a k p vac, 1 k 1 < k 2 < < k p n} u(n) wight a i a i a i 1 wight (0,..., 0, 1,..., 1, 0,..., 0, 1...) 1 p 0 n p highest weight (1 p, 0 n p ) = (1,..., 1, 0,..., 0) highest weight vector a 1a 2 a p vac Λ 0,p Proposition U(n) n p=0λ 0,p Remark N c 1 highest weight ρ = (ρ 1,, ρ n ) π ρ (c 1 ) = i ρi p i=0 1 = p W Λ 0,p. a i 1 a i l vac 2 l 2 a i 1 a i l 2 h(a i, a j ) = δ ij/2 W Λ 0,p. a i 2a i, a i 2i(a i ) Λ 0,p i(a i ) i(a i )(a i 1 a i l ) = ( 1) k 1 h(a i k, a i )a i 1 â i k a i l 1ai 1 2i(a i ). i( ), a i a i,. W Λ 0,p. 36

37 Remark e i 2 1(a i + i(a i )) Je i 2(a i i(a i )) Remark U(n) Spin(2n) u(n) spin(2n) 1ɛi ɛ i 1a i a i 1/2 u(n) W p := C{a k 1 a k 2 a k p vac, 1 k 1 < k 2 < < k p n} highest weight ((1/2) p, ( 1/2) n p ) dominant integral dominant integral Remark integral U(n) u(n) U(n) highest weight U(n) Remark highest weight integral u(n) u(n) C ɛ i ɛ i 1 2 id i = 1,, n u(n) highest weight ((1/2) n ) (det) 1/2 Λ 0,n = Λ n,0 1 W p Λ n,0 1 highest weight (1 p, 0 n p ) Λ 0,p 37

38 Proposition u(n) W p Λ 0,p Λ n,0, W n p=0(λ 0,p Λ n,0 ) W p W n p Proof. W p W p W p = Λ 0,p Λ n,0 = Λ p,0 Λ 0,n Λ n,0 = Λ 0,n p Λ n,0 = W n p Remark S S = S p = (Λ 0,p (M) K) K Λ n,0 (M) fiber S p = S n p SU(n) Λ n,0 Λ 0,n Proposition SU(n) W p Λ 0,p, Λ 0,p = Λ p,0 = Λ 0,n p Proof. U(n) SU(n) Λ 0,n p U(n) highest weight (1 n p, 0 p ) SU(n) highest weight (1 n p, 0 p ) mod Z(1 n ) Λ p,0 U(n) highest weight (0 n p, ( 1) p ) (0 n p, ( 1) p ) + (1 n ) SU(n) highest weight (1 n p, 0 p 1 ) mod Z(1 n ) 3.3 Sp(n) subsection subsection 38

39 3.3.1 Sp(n) (R 4n,,, I, J, K) 4n, compatible I, J, K I 2 = J 2 = 1, IJ = JI = K Sp(n) := {A SO(4n, R) AI = IA, AJ = JA, AK = KA} U(2n) SO(4n) Sp(n, R) Sp(n, C) Remark I 2 = J 2 = 1, IJ = JI K K = IJ (R 4n, I) J J JI = JI J 2 = 1 Sp(n) (V = R 4n,,, I, J, K) V C I, J, K I V 1,0 V 0,1 V 1,0, V 0,1 h JI = IJ J : V 1,0 V 0,1, J : V 0,1 V 1,0 V 0,1 = V 1,0 J : V 1,0 u J(ū) V 1,0 J 2 = 1, h(ju, Jv) = h(v, u) Proof. J u = v z V C J(v z) = (Jv) z J(u) = J(ū) J 2 (u) = J(J(ū)) = J 2 (u) = u J 2 = 1 h(ju, Jv) = J(ū), J( v) = J(ū), J(v) = ū, v = v, ū = h(v, u) (V 1,0, h) compatible J σ(u, v) := h(u, Jv) V 1,0 σ(u, v) = σ(ju, Jv), σ(v, Jv) > 0 v 0 (3.3) 39

40 Proof. σ(u, v) = h(u, Jv) = h(jjv, Ju) = h(v, Ju) = σ(v, u) σ(ju, Jv) = h(ju, v) = h(v, Ju) = σ(v, u) = σ(u, v), σ(v, Jv) = h(v, JJv) = h(v, v) > 0, v 0 Remark J (3.3) h(u, v) := σ(u, Jv) J compatible σ, J, h compatible V 1,0 Λ 0,1 V 1,0 Λ 1,0 V 1,0 V 0,1 Λ 1,0 Λ 0,1 Sp(n) Sp(n) : = {A GL(2n, C) JA = AJ, h(au, Av) = h(u, v)} = {A GL(2n, C) JA = AJ, σ(au, Av) = σ(u, v)} = {A GL(2n, C) h(au, Av) = h(u, v), σ(au, Av) = σ(u, v)} Sp(n, C) = {A GL(2n, C) σ(au, Av) = σ(u, v)} Sp(n, C) U(2n) = Sp(n) (V 1,0, J, h) E E ɛ α, α = n, (n 1),, 1, 1, 2, n σ(ɛ α, ɛ β ) = sign(α)δ α, β, h(ɛ α, ɛ β ) = δ α,β J(ɛ α ) = sign(α)ɛ α 40

41 3.3.2 Sp(n) E 2n compatible {ɛ α } α x αβ := ɛ α ɛ β sign(αβ)ɛ β ɛ α E E, α + β 0 (3.4) σ 2n 2 + n sp(n, C) sp(n, C) = C{x αβ α + β 0, α, β = ±1,, ±n}. gl(n, C) [x αβ, x µν ] = δ βµ x αν δ αν x µβ + sign(αβ)(δ βν x µ α δ αµ x βν ). E x αβ ɛ γ = δ βγ ɛ α sign(αβ)δ αγ ɛ β x αβ α + β 0 α, β (3.4) x αβ = sign(αβ)x β α Λ 1,0 V 1,0 ɛ α = Ω(, ɛ α ) sign(α)ɛ α x αβ = ɛ α ɛ β sign(αβ)ɛ β ɛ α = sign(β)ɛ α ɛ β sign(β)ɛ β ɛ α = sign(β)ɛ α ɛ β sp(n, C) sp(n, C) = S 2 (E). E ɛ α ɛ β (ɛ γ ) = σ(ɛ α, ɛ γ )ɛ β + σ(ɛ β, ɛ γ )ɛ α sp(n) E J S 2 (E) E E J J sp(n) sp(n, C) = S 2 (E) J J(ɛ α ɛ β ) = sign(αβ)ɛ α ɛ β { ɛα ɛ β + sign(αβ)ɛ α ɛ β 1(ɛα ɛ β sign(αβ)ɛ α ɛ β ) 41

42 sp(n) = sp(n, C) u(2n) { xαβ x βα 1(xαβ + x βα ) sp(n) = R{x αβ x βα α + β 0, α > β} R{ 1(x αβ + x βα ) α + β 0, α β} sp(n) 1hR := R{ 1x αα α 0} h R := R{x ii i = 1,, n} Sp(n) weight h R (V, π) Sp(n) h R V = V (λ) λ = (λ 1,, λ n ) weight λ i x ii weight weight ρ highest weight highest weight ρ = (ρ 1,, ρ n ) Z n, ρ 1 ρ 2 ρ 1 0 dominant integral ρ highest weight highest weight ρ V ρ π ρ [x ii, x αβ ] = (δ α,i + δ β,i δ β,i δ α,i )x αβ {x k, l 1 k l n} {x k,l 1 k < l n} x k, l (0,, 0, }{{} 1, 0,, 0, }{{} 1, 0,, 0) (1 k l n) k l x k,l (0,, 0, }{{} 1, 0,, 0, }{{} 1, 0,, 0) (1 k < l n) k l 42

43 Example E (1, 0 n 1 ) highest weight ɛ ±i weight (0 i 1, ±1, 0 n i ) ɛ 1 highest weight vector x k, l (ɛ 1 ) = δ l,1 ɛ k + δ k,1 ɛ l = 0, (1 k l n) x kl (ɛ 1 ) = δ l1 ɛ k δ k,1 ɛ l = 0 (1 k < l n) x ii (ɛ 1 ) = δ i1 ɛ i δ i,1 ɛ i = δ i1 ɛ 1 highest weight vector Example Λ 2 (E) Λ 2 0(E) C Λ 2 0(E) highest weight (1 2, 0 n 2 ) C E σ σ = 1 sign(α)ɛ α ɛ α 2 α Proof. σ(ɛ β, ɛ γ ) = 1 sign(α)(ɛ α ɛ 2 α)(ɛ β, ɛ γ ) α = 1 sign(α)(δ αβ δ γα δ γ α δ αβ ) 2 α =sign(β)δ γβ Example S 2 (E) = sp(n, C) highest weight (2, 0 n 1 ) Example Λ p (E) 0 p n σ Λ p (E) = [p/2] k=0 σk Λ p 2k 0 (E), Λ p 0(E) = ker σ Λ p 0(E) highest weight (1 p, 0 n p ) Remark U(n) Λ p,q Example Λ k 0(E) Λ l 0(E) k l highest weight highest weight Λ k,l 0 (E) highest weight (2 k, 1 l k.0 n 1 ) Λ k 0(E) Λ l 0(E) highest weight vector v k, v l v k v l weight (2 k, 1 l k.0 n 1 ) = (1 k, 0 n k ) + (1 l, 0 n l ) 43

44 3.3.3 Sp(n) Sp(n) [3] Sp(n) (π ρ, V ρ ) weight π ρ weight lowest weight ρ Sp(n) Z n 2 lowest weight ρ weight ρ highest weight (π ρ, V ρ ) highest weight ρ Sp(n) (π ρ, V ρ ) (π ρ, V ρ ) (π ρ, V ρ ) Sp(n) V ρ V ρ V ρ V ρ Example E E E E E Sp(n) V ρ Hom Sp(n) (V ρ, Vρ ) Sp(n) V ρ Vρ V ρ Vρ Hom Sp(n)(V ρ, Vρ ) = Hom Sp(n) (V ρ, V ρ ) = C Sp(n) Ω Proof. Sp(n) Ω radical ker Ω = {φ Ω(φ, ψ) = 0, ψ V ρ } Sp(n) V ρ V ρ Ω Ω ± (φ, ψ) := Ω(φ, ψ) ± Ω(ψ, φ) Sp(n) Ω + Ω Sp(n) 1 Ω = Ω ± Sp(n) Ω ± V ρ V ρ Vρ V ρ Ω + V ρ Sp(n) J Ω Sp(n) 44

45 J h(jφ, Jψ) = h(ψ, φ) ψ, φ V ρ Sp(n) E E E J even E odd E Proposition Sp(n) (π ρ, V ρ ) ρ i 0 mod 2 Sp(n) J ρ i 1 mod 2 Sp(n) J V ρ h h(jφ, Jψ) = h(ψ, φ) compatible Sp(n) Spin(4n) Sp(n) SU(2n) Proposition F Sp(n) F i Spin(4n) Ad SO(4n) Spin(4n) Spin c (4n) Sp(n) Spin c (4n) sp(n, C) Cl 4n x αβ = ɛ α ɛ β sign(αβ)ɛ β ɛ α a αa β sign(αβ)a β a α Cl 4n sp(n) W SU(2n) W = 2n p=0λ 0,p Sp(n) SU(2n) SU(2n) Sp(n) Λ 0,p = Λ p (E) σ σ = 1 2 sign(α)a α a α 45

46 σ = 1 2 sign(α)aα a α σ [σ, σ ] = N n, [N n, σ] = 2σ, [N n, σ ] = 2σ X := σ, Y := σ, H := N n = [X, Y ] sl(2, C) Proof. [σ, a β ] = 1 2 sign(α)(a α a αa β a β a αa α) = 1 2 sign(α)(a α (δ β α a β a α) (δ βα a αa β )a α) = sign(β)a β [σ, σ ] = 1 sign(β)([σ, aβ ]a β + a β [σ, a β ]) 2 = 1 sign(β)( sign(β)a β 2 a β + sign(β)a β a β ) = N n Λ(E) := p Λ p (E) sl(2, C) ker σ lowest vector σ Λ(E) Λ(E) = n l=0σ l ker σ σ ker σ highest weight vector sl(2, C) Sp(n) Λ(E) = 2n p=0 n l=0 (σ l ker σ Λ p (E)) p n Λ p 0(E) := ker σ Λ p (E) Λ p (E) Λ p (E) = [p/2] k=0 σk Λ p 2k 0 (E) σ Λ p 0(E) ( ) ( ) ( ) 2n 2n 2n + 1 2n 2p + 2 = p p 2 p 2n p

47 a 1a 2 a p vac σ Λ p 0(E) Sp(n) highest weight vector weight (1 p, 0 n p ) Λ p 0(E) V (1p,0 n p ) Λ p 0(E) Weyl dim V (1p,0n p ) V (1p,0n p ) = Λ p 0(E) Remark Weyl Weyl [8] φ Λ p 2k 0 (E) σ φ = 0 (N n)φ = (p 2k n)φ = ((n p) + 2k)φ φ sl(2, C) lowest weight vector σ weight (n p + 2k) weight sl(2, C) highest weight vector weight σ φ σ σφ σ σ 2 φ σ σ σ n p+2k φ = σ n p+k σ k φ σ 0 σ n p Λ p (E) Λ 2n p (E) Sp(n) σ k Λ p 2k 0 (E) σ n p dim Λ p (E) = dim Λ 2n p (E) σ n p Λ p (E) = Λ 2n p (E) Λ 2n p (E) = [p/2] k=0 σn p+k Λ p 2k 0 (E) Proposition Sp(n) Λ (E) Λ p (E) = [p/2] k=0 σk Λ p 2k 0 (E), Λ 2n p (E) = [p/2] k=0 σn p+k Λ p 2k 0 (E) Remark Λ p (E) Λ 2n p (E) σ n φ Λ p (E), ψ Λ 2n p (E) φ ψ = φ, ψ σ n Λ p (E) (Λ 2n p (E)) Λ 2n p (E) Λ p 0(E) p = 0,, n n p k=0 σk Λ p 0(E) sl(2, C) highest weight n p dim Λ p 0(E) n p k=0 σk Λ p 0(E) = S n p ˆ Λ p 0(E) S n p sl(2, C) highest weight n p n p + 1 Proposition W = Λ(E) sl(2, C) sp(n) W = n p=0s n p ˆ Λ p 0(E) 47

48 3.4 SO(n) Spin(n) g = spin(n) g C [e k, e l ] 1. n = 2m g C {ω k = a k a k 1/2} m k=1 {a k a l} k<l {a k a l } k<l {a k a l } k<l {a k a l } k<l {a k a l} k<l, {a k a l } k<l 2. n = 2m + 1 g C {ω k = a k a k 1/2} m k=1 {a k a l} k<l {a k a l } k<l {a k a l } k<l {a k a l } k<l {a k b} m k=1 {a k b}m k=1 {a k a l} k<l, {a k a l } k<l, {a k b} k g a k = a k, a k = a k a k a l + a k a l 1(a k a l a k a l ) g h 1 h R := 1h = span R {ω 1,, ω m } g (π, V ) h π(h) h h R, V (λ), λ (h R ) weight λ = (λ 1,, λ m ) λ i ω i weight λ λ 1 = λ 2 = = λ k = 0, λ k+1 > 0 k weight λ, λ λ > λ λ λ > 0 48

49 weight highest weight ρ ρ dominant integral ρ = (ρ 1,, ρ m ) in Z m or (Z + 1/2) m ρ 1 ρ m 1 ρ m for n = 2m, ρ 1 ρ m 1 ρ m 0 for n = 2m + 1. ρ, highest weight ρ g (π ρ, V ρ ) highest weight ρ Remark ρ (Z + 1/2) m integral integral root ρ (Z + 1/2) m g SO(n) ρ Z m integral ρ Z m ρ (Z + 1/2) m integral or half-integral ρ Z m g SO(n) Ad : Spin(n) SO(n) ρ (Z + 1/2) m Ad SO(n) Example SO(n) Spin(n) (π Ad, R n C) SO(n). π Ad ([e i, e j ])a = [[e i, e j ], a]. (3.5) C n {a k } k {a k } k for n = 2m {a k } k {a k } k {b} for n = 2m + 1. n = 2m. [ω k, a i ] = δ ki a i [ω k, a i ] = δ kia i a i a i weight vector weight (0 i 1, 1, 0 m i ) (0 i 1, 1, 0 m i ). weight 1 highest weight vector a 1 highest weight (1, 0 m 1) Example q q S q 49

50 x 2 i S q Λ p,q H q R n q spin(n) spin(n) H q ([e k, e l ], f(x)) 4x k f x l + 4x l f x k H q. (3.6) highest weight vector z q 1 = x 1 q 1x 2 highest weight (q, 0 m 1 ) dim H q = (n + q 3)! (n + 2q 2). q!(n 2)! Example n = 2m Λ p Λ p := Λ p (R 2m ) C p Λ p = Cl 2m spin(2m) ad([e k, e l ])(φ) = [[e k, e l ], φ] for φ in Λ p 0 p m 1 Λ p Λ 2m p Λ p ( ) 2m p highest weight vector a 1 a p highest weight (1 p, 0 m p ) p = m Λ m Λ m + Λm 1 ( 2 2m m ) Λ m + resp. Λm highest weight vector a 1 a m highest weight (1 m ) resp. a 1 a m 1 a m with weight (1 m 1, 1) Λ p Cl 2m ω spin(2m) ω 2 = 1 e k ω = ωe k ω : Λ p Λ 2m p ω Λ m ±1 ω(a 1 a m) = a 1 a m Λm ± ω ±1 Remark n = 4 Λ 2 = Λ 2 + Λ 2 Remark Λ p Λ 2m p ω ω ω 2 = 1 Λ p Λ 2m p V (1) V ( 1) V (1), V ( 1) Remark n = 2m = 4k + 2 Λ m = Λ m + Λ m Example n = 2m+1 Λ p Λ p (R 2m+1 ) C ( 2m+1 ) p spin(2m+1) n = 2m 0 p m Λ p Λ 2m+1 p highest weight vector a 1 a p highest weight (1 p, 0 m p ) 50

51 3.4.3 SO(n) Λ p = Λ p (R n ) C Λ p (R n ) R n (R n ) Λ p (Λ p ) Sp(n) Λ p SO(n) Λ p (R n ) SO(n) n = 2m Λ m ω Hodge : Λ p (R n ) Λ n p (R n ) 2 = ( 1) p(n p) n = 2m Λ m (R 2m ) Λ m (R 2m ) 2 = ( 1) m n = 2m = 4l 2 = 1 n = 2m = 4l = 1 n = 4l Λ 2l (R 4l ) Λ 2l ± Λ2l ± = (Λ 2l ±) n = 4l + 2 Λ 2l+1 (R 4l+2 ) Λ 2l+1 ± (Λ± 2l+1 ) Λ 2l+1 ± SO(n) Λ 2l+1 ± (Λ 2l+1 ) Λ 2l+1 Λ 2l+1 ± Spin(n) Sp(n) n = 2m + 1 2m+1 highest weight ((1/2) m ) weight ±1/2 weight highest weight ((1/2) m ) 2m+1 ( 2m+1 ) n = 2m ± highest weight ((1/2) m 1, ±1/2) weight 1/2, 1/2 weight 1. n = 4l ± 4l weight 1/2, 1/2 2l + 4l weight 1/2 1/2 weight weight 1/2 1/2 highest weight ((1/2) 2l ) + 4l ( + 4l ) 4l weight 1/2 1/2 4l ( 4l ) 51

52 2. n = 4l l+2 weight 1/2 1/2 highest weight 1/2 2l + 1 weight weight 1/2 1/2 highest weight ((1/2) 2l, 1/2) ( ± 4l+2 ) 4l+2 ± 4l+2 Remark highest weight Sp(n) SO(n) 1. n = 2m W W = m!2 m 1 2. n = 2m + 1 W W = m!2 m Proposition n = 2m + 1 2m+1 2m+1 2. n = 2m 2m 2m (a) n = 4l ± 4l ( ± 4l ). (b) n = 4l + 2 ± 4l+2 ( 4l+2 ). Cl n = Cl n,0 Cl 1 = C, Cl 2 = H, Cl 3 = H H, Cl 4 = H(2) Cl 5 = C(4) Cl 6 = R(8) Cl 7 = R(8) R(8) Cl 8 = R(16) 1. n = 8k + 2 8k+2 J : ± 8k+2 8k+2 2. n = 8k + 3 8k+3 3. n = 8k + 4 8k+4 n = 8k + 3 ± 8k+4 52

53 4. n = 8k + 6 8k+6 J : ± 8k+6 8k+6 5. n = 8k + 7 8k+7 6. n = 8k 8k+8 ± 8k+8 n = 8k + 1, 8k + 5 Cl 0,n Cl 0,1 = R R, Cl 0,2 = R(2), Cl 0,3 = C(2), Cl 0,4 = H(2) Cl 0,5 = H(2) H(2) Cl 0,6 = H(4) Cl 0,7 = C(8) Cl 0,8 = R(16) Cl 0,n Spin(n) Cl 0,n Cl n = Cl n,0 C R 0,n {e i} i Cl n = Cl n C e i 1e i Cl n,0 C = Cl n Cl n,0 v R n Cl n,0 J J(e i φ) = J( 1e i φ) = 1e ij(φ) = e i J(φ) 1. n = 8k + 1 8k+1 2. n = 8k + 2 8k+2 3. n = 8k + 4 8k+4 4. n = 8k + 5 8k+5 5. n = 8k + 6 8k+6 6. n = 8k + 8 8k+8 53 Cl 0,n

54 Remark Cl n,0 n = 8k+6, 8k+5 ± 8k+6 ± 8k+6 Cl0 n Cl n 1 e 1 e 2 e 1 e n Remark Cl r,s e 1,, e r e r+1,.e r+s e 1 e r+1 Proposition n = 8k + 1 8k+1 2. n = 8k + 2 8k+2 ± 8k+2 3. n = 8k + 3 8k+3 4. n = 8k + 4 8k+4 ± 8k+4 5. n = 8k + 5 8k+5 6. n = 8k + 6 8k+6 ± 8k+6 7. n = 8k + 7 8k+7 8. n = 8k 8k+8 ± 8k+8 54

55 Example n = 2 2 = = C C U(1) = Spin(2) e iθ (z, w) = (e iθ z, e iθ w) (z, w) ( w, z) (z, w) ( w, U(1) ± 2 V V Ω ± (e + f, e + f ) = f (e) ± f(e ) Ω T M Example u(n) spin(2n) W = W p Ω 1(2p m) Ω φ p W p ΩJφ = JΩφ = 1(2p n)jφ = 1(2(n p) p)jφ u(n) J : W p W n p u(n) spin c (n) u(n) u(n) n = 8k + 2 ± 8k+2, + 8k+2 8k+2 Λ2k 1. n = 2m Λ p = (1 p, 0 m p ) 1 p m 2, ± 2m = ((1/2) m 1, ±1/2) highest weight m 2 ρ = ( n p Λ p ) + k + + 2m + k 2m, p=1 n p, k ± Z 0 (1 m 1, 1) = 2 2m highest weight ρ m 2 ( n p Λ p ) ( k+ + 2m) ( k 2m) p=1 55

56 top highest weight { ρ n = 4l ( m 2 p=1 n pλ p ) + k + 2m + k + 2m n = 4l + 2 n = 4l n = 4l+2 k + = k 2l 1 ρ = ( n p Λ p ) + k( + 4l+2 + 4l+2 ) p=1 ρ Z 2l+1 SO(4l + 2) 2. n = 2m + 1 Λ p = (1 p, 0 m p ) 1 p m 1, 2m+1 = ((1/2) m ) highest weight ρ Proof. ρ ρ lowest weight Weyl W highest weight weight highest weight loweset weight W w 0 W ρ = w 0 (ρ) w 0 root positve root Φ + w 0 (Φ + ) = Φ + n = 2m + 1 (ρ 1,, ρ n ) (ρ 1,, ρ n ) w 0 V ρ Vρ 1. n (a) n = 4l + 2 V ρ Vρ 2l 1 ρ = ( n p Λ p ) + k( + 4l+2 + 4l+2 ) p=1 56

57 + 4l+2 + 4l+2 = Λ2l = (1 2l, 0) Λ p n p Λ p ) k(λ 2l ) 2l 1 ( p=1 top V ρ Remark top highest weight v i highest weight vector v ρ v i, v i = 0 v ρ, v ρ = 0 (b) n = 8l V ρ (c) n = 8l + 4 ρ = ( p=1 n p Λ p ) + k + + 8l+4 + k 8l+4, n p, k ± Z 0 k + +k SO(8l+4) 2. n (a) n = 2m + 1 = 8l + 1, 8l + 7 V ρ (b) n = 8k + 3, 8k + 5 ρ m 1 ρ = ( n p Λ p ) + k 2m+1 p=1 k SO(2m + 1) Remark

58 3.4.4 Cl 2m grading Λ Z 2 Cl 2m Cl 2m (φ, v) v φ t Cl 2m (e i1 e i2 e ik ) t = e ik e i2 e i1 Cl 2m = W 2m W2m = Hom(W 2m, W 2m ) W 2m Cl 2m = C(2 m ) W 2m W2m W 2m Cl 2m Cl 2m W 2m W2m (Cl 2m Cl 2m ) Cl 2m (φ 1 φ 2, v) φ 1 vφ t 2 Cl 2m. ( 2m ) ( 2m ) = ( 2m ) 2m = 2(Λ 0 + Λ 1 + Λ 2 + Λ m 1 ) + Λ m + + Λ m Remark Cl 2m Cl2m 0 Cl2m 1 Λ even Λ odd ω Λ ±1 even-odd Λ p Λ 2m p ω ω 2 = 1 Λ p Λ 2m p = V (1) V ( 1) W 2m W2m W 2m ±1 Cl 2m = ( + 2m 2m) 2m = ( + 2m 2m ) ( 2m 2m ) + 2m 2m + 2m + 2m top (1 m ) = 2 ((1/2) m ) highest weight Λ m + 2m 2m Λ m even-odd α : Cl 2n Cl 2n ±1 ω Cl 2n φ ωφω t = ωφω Cl 2n α 58

59 Proof. ω 2 = 1 n = 2m ωe i = e i ω ωe i ω = ωωe i = e i, ωe i e j ω = ωωe i e j = e i e j α even-odd W 2m W2m ω W 2m W2m W 2m W 2m Z 2 -grading m 1. n = 4l ± 4l ( ± 4l ) Cl 4l = ( + 4l 4l ) ( + 4l 4l ) = {( + 4l + 4l ) ( 4l 4l )} {( + 4l 4l ) ( 4l + 4l )} = Cl 0 4l Cl 1 4l highest weight Cl4l 0 = ( + 4l + 4l ) ( 4l 4l ) Λ 2l +, Λ 2l 2. n = 4l + 2 4l+2 ( ± 4l+2 ) Cl 4l+2 = ( + 4l+2 4l+2 ) (( + 4l+2 ) ( 4l+2 ) ) = {( + 4l+2 ( + 4l+2 ) ) ( 4l+2 ( 4l+2 ) )} {( + 4l+2 ( 4l+2 ) ) ( 4l+2 ( + 4l+2 ) )} = {( + 4l+2 4l+2 ) ( 4l+2 + 4l+2 )} {( + 4l+2 + 4l+2 ) ( 4l+2 4l+2 )} = Cl 0 4l+2 Cl 1 4l+2 Cl4l+2 1 = ( + 4l+2 + 4l+2 ) ( 4l+2 4l+2 ) Λ2l+1 +, Λ 2l U(n) Spin(n), Sp(n) U(n) SU(n) U(n) U(n) V ρ highest weight ( ρ n, ρ n 1,, ρ 1 ) 59

60 ρ n = ρ 1, ρ n 1 = ρ 2,, Proposition V ρ U(n) 1. n = 2m highest weight ρ (λ 1, λ 2,, λ m 1, λ m, λ m, λ m 1,, λ 1 ) V ρ Vρ 2. n = 2m + 1 highest weight ρ (λ 1, λ 2,, λ m 1, λ m, 0, λ m, λ m 1,, λ 1 ) V ρ Vρ Λ 0,p (1 p, 0 n p ) highest weight top Λ 0,p = (1 p, 0 n p ) n 1 ρ = (ρ 1,, ρ n ) = (ρ p ρ p+1 )Λ 0,p + ρ n Λ 0,n p=1 ρ p ρ p+1 Z 0 ρ n Z Λ 0,p Λ n,0 (Λ 0,n p ) (Λ 0,p Λ n,0 Λ 0,n p ) (Λ 0,p Λ n,0 Λ 0,n p ) (Λ 0,p Λ n,0 Λ 0,n p ) Λ 0,p + Λ n,0 + Λ 0,n p highest weight Proof. V V (V V ) (V V ) = V V V V V V V V Ω((e 1 + f 1 ) (e 2 + f 2 ), (e 3 + f 3 ) (e 4 + f 4 )) = (f 1 (e 3 ) ± f 3 (e 1 ))(f 2 (e 4 ) ± f 4 (e 4 )) V V V V Ω(f 1 e 2, e 3 f 4 ) = ±f 1 (e 3 )(f 4 (e 4 )) top 60

61 n = 2m V ρ Vρ highest weight ρ =(λ 1 λ 2 )Λ 0,1 + + (λ m 1 λ m )Λ 0,m 1 + 2λ m Λ 0,m + (λ m 1 λ m )Λ 0,m+1 + (λ m 2 λ m 1 )Λ 0,m (λ 1 λ 2 )Λ 0,2m 1 λ 1 Λ 0,2m =(λ 1 λ 2 )Λ 0,1 + + (λ m 1 λ m )Λ 0,m 1 + 2λ m Λ 0,m + (λ m 1 λ m )Λ 0,m+1 + (λ m 2 λ m 1 )Λ 0,m (λ 1 λ 2 )Λ 0,2m 1 + {(λ 1 λ 2 ) + (λ 2 λ 3 ) + + (λ m 1 λ m ) + λ m }Λ 2m,0 =(λ 1 λ 2 )(Λ 0,1 + Λ 2m,0 + Λ 0,2m 1 ) + + λ m (Λ 0,m + Λ 2m,0 + Λ 0,m ) V ρ n = 2m + 1 ρ =(λ 1 λ 2 )Λ 0,1 + + (λ m 1 λ m )Λ 0,m 1 + λ m Λ 0,m + λ m Λ 0,m+1 + (λ m 1 λ m )Λ 0,m+2 + (λ m 2 λ m 1 )Λ 0,m (λ 1 λ 2 )Λ 0,2m λ 1 Λ 0,2m+1 =(λ 1 λ 2 )Λ 0,1 + + (λ m 1 λ m )Λ 0,m 1 + λ m Λ 0,m + λ m Λ 0,m+1 + (λ m 1 λ m )Λ 0,m+2 + (λ m 2 λ m 1 )Λ 0,m (λ 1 λ 2 )Λ 0,2m {(λ 1 λ 2 ) + (λ 2 λ 3 ) + + (λ m 1 λ m ) + λ m }Λ 0,2m+1 V ρ SU(n) U(n) SU(n) Λ 0,p (Λ 0,n p ) φ Λ 0,p, ψ Λ 0,n p φ ψ = φ, ψ ɛ 1 ɛ n Λ 0,p Λ 0,n p pairing SU(n) U(n) n = 2m Λ 0,m (Λ 0,m ) n = 4m Λ 0,2m n = 4m + 2 Λ 0,2m+1 U(n) Example SU(2) Λ 0,1 = H U(2) U(2) with highest weight (k, k) SU(2) highest weight (2k) S 2k (H) 61

62 3.5 Sp(n)Sp(1) Sp(n)Sp(1) SO(4n) Sp(n)Sp(1) Sp(n)Sp(1) := (Sp(1) Sp(n))/Z 2 Sp(n)Sp(1) U(n) Sp(n) ±1 (1, 1), ( 1, 1) Z 2 semi-quternionic structure ( [4] V End(V ) R Q id Q R Q H Example R 4n = H n H H End(V ) H n v v q H n, q H (H n, Q = H) Example H n q H p o Sp(1) H n x xp 0 qp 1 0 = xp 0 qp 1 0 H n p 0 Hp 1 0 = H End(V ) p 0 idp 1 0 = id p 0 Hp 1 0 p 0Hp 1 0 H (H n, p 0 Hp 1 0 ) Q H (V, Q) f : V V f Q : Q Q f(qv) = f Q (q)f(v) Remark f Q : Q Q f id : Q Q f : V V f(qv) = qf(v) (V, Q) Q H Q = H g : H H H R R1 H g(a1) = ag(1) = a a R IH = R 3 SO(3) 62

63 Sp(1) = SU(2) SO(3) g : H H p Sp(1) g : H q g(q) = pqp 1 H (V, Q) (H n, H) f p Sp(1) f(v q) = f(v)p qp 1 f (v) := f(v)p f (v q) = f(v q)p = f(v)p qp 1 p = f(v)p q = f (v) q f A gl(n, H) f (v) = Av (H n, H) f(v) = Avp 1 GL(n, H)Sp(1) (V, Q) Q q Q qu, v = u, qv Q H Q = R IQ q Remark Iu, Iv = u, v, Ju, Jv = u, v, Ku, Kv = u, v q H qu, v = u, qv Sp(n)Sp(1) Proposition (V, Q,, ) Sp(n)Sp(1) E H formulation (V, Q,, ) Sp(n), Sp(1) (V, Q,, ) Sp(n) (E, σ E, J) E 2n σ E J compatible σ E (ɛ, ɛ ) = σ E (Jɛ, Jɛ ), σ(ɛ, Jɛ) > 0, ɛ 0 {ɛ α α = ±1,, ±n} (H, σ H, J) Sp(1) {h A A = ±} 63

64 E H 4n Sp(n)Sp(1) 1. J J (E H) R J(ɛ α h A ) = sign(αa)ɛ α h A ɛ α h A / (E H) R 2. σ E σ H (E H) R (σ E σ H )(ɛ α h A, ɛ β h B ) = sign(αa)δ α, β δ A, B 3. h E (ɛ, ɛ ) = σ E (ɛ, Jɛ ) E H h H E H h := h H h E h(ɛ α h A, ɛ β h B ) = (σ E σ H )(ɛ α h A, J(ɛ β h B )) =sign(βb)(σ E σ H )(ɛ α h A, ɛ β h B )) = δ α,β δ A,B {ɛ α h A } Sp(n)Sp(1) E H (V, Q,, ) Proposition (V, Q,, ) E H σ E σ H Remark E, H Salamon E H- formulation [4],[7] Proof. E H 2n 2 (z, w) E, (α, β) H (z, w) ( w, z), (α, β) ( β, ᾱ) ( ) zα zβ (z, w) (α, β) = wα wβ E H 2n 2 z, w n ( ) ( zα zβ w β ) wᾱ wα wβ z β zᾱ 64

65 E H ( ) z 1 z 2 w 1 w 2 ( ) ( ) z 1 z 2 w 2 w 1 w 1 w 2 z 2 z 1 ( ) z w w z z +jw H n (E H) R = H n E H = H n R C Sp(n)Sp(1) A+jB Sp(n) GL(n, H), α+jβ Sp(1) GL(n, H) (z + jw)(ᾱ βj) = (zᾱ + wβ) + j( zβ + wᾱ) ( ) ( z w A B ) ( ) ( z w ᾱ w z B Ā w z β ) β α Sp(n)Sp(1) (E H) R = H n E, H σ E ((z, w), (z, w )) = zw wz z t w w t z compatible σ E (( w, z), ( w, z )) = w z + z w = σ E ((z, w), (z, w )) σ E ((z, w), ( w, z)) = z z + w w > 0 (z, w) (0, 0) σ E σ H (σ E σ H )((z, w) (α, β), (z, w ) (α, β )) =(zw wz )(αβ βα ) =(zα)(w β ) (zβ)(w α ) (wα)(z β ) + (wβ)(z α ) 65

66 E H ( ) ( ) z 1 z 2 z 1 z 2 (σ E σ H )(, ) w 1 w 2 w 1 w 2 =z 1 w 2 + w 2 z 1 z 2 w 1 w 1 z 2 ( ) ( ) z 1 z 2 z 1 z 2 (σ E σ H )(, ) w 1 w 2 w 1 w 2 =2(z 1 w 2 z 2 w 1 ) ( ) ( ) z w z w (σ E σ H )(, ) w z w z =z z + zz + ww + w w H n sp(n) sp(1) Sp(n)Sp(1) ɛ α h A (E H) R (ɛ α h A )(ɛ β h B ) + (ɛ β h B )(ɛ α h A ) = 2sign(αA)δ α β δ A B a α = 1 2 ɛ α h +, a α = 1 2 sign(α)ɛ α h [a α, a β ] + = 0, [a α, a β ] + = 0 [a α, a β ] + = a αa β + a β a α = 1/2{(ɛ α h + )(sign(β)ɛ β h ) + (sign(β)ɛ β h )(ɛ α h + )} = δ αβ 1 J(a α) = sign(α)ɛ α h = a α, J(a α ) = sign(α)sign( α)ɛ α h + = a α 66

67 a α, a α Cl 4n spin(4n, C) sp(n, C) sp(1, C) spin(4n, C) sp(n, C) x αβ a αa β sign(αβ)a β a α = a β a α + sign(αβ)a α a β x αβ (ɛ γ ) = δ βγ ɛ α + sign(γβ)δ αγ ɛ β 1 [a αa β sign(αβ)a β a α, ɛ γ h ] 2 =[a αa β sign(αβ)a β a α, sign(γ)a γ ] =[ a β a α + sign(αβ)a α a β, sign(γ)a γ] = δ αγ sign(γ)a β sign(α)δ βγ a α 1 1 = δ αγ sign(γ) sign(β)ɛ β h + δ βγ ɛ α h = (δ β γ ɛ α + δ α γ sign( γβ)ɛ β ) h 2 x αβ spin(4n, C) Cl 4n ad E H E [a αa β sign(αβ)a β a α, a γ] sp(1, C) {y AB A + B 0, A, B = ±1} y AB (h C ) = δ BC h A + sign(bc)δ AC h B y ++ (h + ) = h +, y ++ (h ) = h y ++ = y y + (h + ) = 0, y + (h ) = 2h + y + (h + ) = 2h, y + (h ) = 0, [y +, y + ] = 4y ++, [y ++, y + ] = 2y +, [y ++, y + ] = 2y + y + sign(α)a αa α = 2σ y + sign(α)a α a α = 2σ y ++ a αa α n = N n = [σ, σ ] 67

68 [ sign(α)a αa α, 1 2 sign(β)ɛ β h ] = [ sign(α)a αa α, a β ] = sign(α)( δ αβ a α + δ αβ a α) = 2sign(β)a β 1 = sign(β)ɛ β 2h + 2 [ sign(α)a α a α, 1 = ɛ β 2h ɛ β h + ] = [ sign(α)a α a α, a β ] = 2sign(β)a β sp(1, C) spin(4n, C) ad H sp(n) sp(1) spin(4n) Proof. sp(1) 1y++, y + y +, 1(y + + y + ), 1( a α a α n), 2(σ σ), 2 1(σ + σ) spin(4n) Sp(n)Sp(1) Sp(n)Sp(1) Sp(n)Sp(1) Sp(1) Sp(n) 0 Z 2 Sp(1) Sp(n) Sp(n)Sp(1) 1 π 1 (Sp(n)Sp(1)) = Z 2 Proposition F n n Sp(n)Sp(1) F i Spin(4n) Ad SO(4n) 68

69 Proof. i (π 1 (Sp(n)Sp(1))) π 1 (Sp(n)Sp(1)) Sp(1) SU(2), Sp(n) SU(2n) ( ) ( ) e it I 0 e it 0 γ(t) = Sp(n)Sp(1), 0 t π 0 e it I 0 e it γ(t) = ( e it I 0 0 e it I ) ( e it 0 0 e it ) Sp(n) Sp(1) 0 t π γ(π) = ( 1)( 1) = 1 π 1 (Sp(n)Sp(1)) γ(π) = ( 1, 1) γ(t) γ(t) γ(t) SO(4n) ( ) ( e it I 0 z 0 e it I w ) ( ) w e it 0 z 0 e it SO(4n) I 2n ( 0 ) i (γ(t)) = cos 2t sin 2t 0 sin 2t cos 2t ( z e = 2it ) w e 2it w z [γγ ] = [γ][γ ] I 0 ( ) cos 2t sin 2t 0 sin 2t cos 2t β = 1 Z 2 = π 1 (SO(4n)) [i γ(t)] = nβ n i (π 1 (Sp(n)Sp(1))) = 0 n i (π 1 (Sp(n)Sp(1))) = Z 2 Remark n Sp(n)Sp(1) Sp(n) Sp(1) w 2 (S 2 (H) R ) = 0 see [7] S 2 (H) S 2 (H) R w 2 (M) = nw 2 (S 2 (H) R ) { 0 n = even w 2 (M) = w 2 (S 2 (H) R ) n = odd n w 2 (S 2 (H) R ) = 0 w 2 (M) = 0 n w 2 (M) = 0 w 2 (S 2 (H) R ) = 0 n n 69 I n

70 sp(n) sp(1) n Sp(n)Sp(1) Sp(n) Sp(1) Proposition Spin(4n) SO(4n) Sp(n) Sp(1) Sp(n)Sp(1) F Sp(n) Sp(1) Sp(n)Sp(1) F i Spin(4n) Ad SO(4n) F ( 1, 1) = ( 1) n Proof. Sp(n) Sp(1) Sp(n)Sp(1) i SO(4n) Lemma 3.1 F ( 1, 1) = ( 1) n n F Sp(n)Sp(1) F ( 1, 1) = 1 n F ( 1, 1) = ±1 F ( 1, 1) = 1 Sp(n)Sp(1) F ( 1, 1) = 1 n F G = exp sp(n) sp(1) Spin(4n) Spin(4n) G exp spin(4n) sp(n) sp(1) Ad Sp(n)Sp(1) SO(4n) exp ad sp(n) sp(1) so(4n) U(n) Ad G Sp(n)Sp(1) G G Sp(n)Sp(1) Sp(n) Sp(1) sp(n) sp(1) 1(a α a α a αa α ) (α = ±1,, ±n), 1( a αa α n) T 1( 2a k a k n) 1 k n α 70

71 exp(π2 1( n a k a k n/2)) = exp(π e 2k 1 e 2k ) k=1 =(cos π + e 1 e 2 sin π) (cos π + e 2n 1 e 2n sin π) = ( 1) n n 1 G G Sp(n)Sp(1) Sp(n) Sp(1) n U(n) Spin c (2n) T = exp T gtg 1 1 G 1 exp T (cos t 1 + e 1 e 2 sin t 1 )(cos t 1 e 2n+1 e 2n+2 sin t 1 ) (cos t n + e 2n 1 e 2n sin t n )(cos t n e 4n 1 e 4n sin t n ) (cos s + e 1 e 2 sin s) (cos s + e 2n 1 e 2n sin s) =(cos(t 1 + s) + e 1 e 2 sin(t 1 + s))(cos t 1 e 2n+1 e 2n+2 sin t 1 ) = 1 (cos(t n + s) + e 2n 1 e 2n sin(t n + s))(cos t n e 4n 1 e 4n sin t n ) t 1 + s = ±π, t 2 + s = ±π,, t n + s = ±π, t 1 = ±π,, t n = ±π 1 1 / G Proposition sp(n) sp(1) { Sp(n)Sp(1) Spin(4n) n = even exp sp(n) sp(1) = Sp(n) Sp(1) Spin(4n) n = odd Sp(n)Sp(1) Spin c (4n) n = even Spin(4n) Spin c (4n) n = odd n = odd Sp(n)Sp(1) F i Spin c (4n) Ad SO(4n) F F F F : π 1 (Sp(n)Sp(1)) = Z 2 π 1 (Spin c (4n)) = Z Hom Z (Z/nZ, Z) = 0 F = 0 n i = id Ad F = i = 0 71

72 Remark sp(1) 1( a α a α n), 2(σ σ), 2 1(σ + σ) U(n) shift 1 a α a α E H sp(1) U(n) Spin c (2n) Proposition F n n Sp(n)Sp(1) F i Spin c (4n) Ad SO(4n) Remark Sp(1)Sp(1) Sp(n)Sp(1) Sp(n) Sp(1) = SU(2) H Sp(1) k Z 0 k V k := S k (H) k + 1 spin-k/2 highest weight k Sp(1) H S even (H) S odd (H) Sp(n) Sp(1) Sp(n) (π ρ, V ρ ) Sp(1) (π k, V k ) V ρ V k V ρ ρ i = odd ρ i = even ρ = (ρ 1,, ρ n ) Proposition Sp(n) Sp(1) V ρ V k 1. k + ρ i = even 2. k + ρ i = odd 72

73 Sp(n)Sp(1) Sp(n) ±1 π( 1)π( 1) = id π( 1) = ±id E π( 1) = id E id id ρ i = even id ρ i = odd id Lemma Sp(n) V ρ 1. ρ i = even π ρ ( 1) = id 2. ρ i = odd π ρ ( 1) = id Proposition Sp(n) Sp(1) V ρ V k 1. k + ρ i = even Sp(n)Sp(1) 2. k + ρ i = odd Sp(n)Sp(1) Sp(n)Sp(1) Spin(4n) Sp(n)Sp(1) Sp(n) Sp(1) W Sp(n) sl(2, C) σ, σ, N n sp(1, C) 3.18 Proposition W Sp(n) Sp(1) n = odd Sp(n)Sp(1) n = even : W = n p=0v (1p,0 n p ) V n p = n p=0λ p 0(E) S n p (H) Kraines subsection Sp(n)Sp(1) Kraines 4-from 73

74 Sp(n)Sp(1) Kraines Sp(n) E, Sp(1) H E, H {ɛ α } α, {h A } A {ɛ α h A } α,a E H E H Lemma V W Λ 2 (V W ) = (Λ 2 (V ) S 2 (W )) (S 2 (V ) Λ 2 (W )) Proof. (v 1 w 1 ) (v 2 w 2 ) = 1 2 {(v 1 w 1 ) (v 2 w 2 ) (v 2 w 2 ) (v 1 w 1 )} = 1 2 (v 1 v 2 ) (w 1 w 2 ) 1 2 (v 2 v 1 ) (w 2 w 1 ) =1/4((v 1 v 2 ) + (v 2 v 1 )) ((w 1 w 2 ) (w 2 w 1 )) + 1/4((v 1 v 2 ) (v 2 v 1 )) ((w 1 w 2 ) + (w 2 w 1 )) =(v 1 v 2 ) (w 1 w 2 ) + (v 1 v 2 ) (w 1 w 2 ) [1] Λ 2 (E H) Λ 2 (E H) = Λ 2 (E) S 2 (H) S 2 (E) Λ 2 (H) =(Λ 2 0(E) S 2 (H)) (C(σ E ) S 2 (H)) S 2 (E) C(σ H ) Λ 2 (E H) S 2 (H) = sp(1, C) S 2 (H) = sp(1, C) {y ++, y +, y + } y ++ = h + h = y, y + = h + h +, y + = h h sp(1, C) Sp(1) C 2 = y AB y BA = y ++ y ++ + y + y + + y y + y + y + = 2(y ++ y ++ + y + y + ) S p (H) AB π p(y A B)π p (y BA ) 74

75 highest weight vector 2p(p + 2) Λ 2 (E H) 1 sign(α)(ɛα h + ) (ɛ α h + ) 2 = 1 2 sign(α){(ɛα ɛ α ) (h + h + ) + (ɛ α ɛ α ) (h + h + )} = 1 2 (sign(α)ɛα ɛ α ) (h + h + ) =σ E y sign(α)(ɛα h ) (ɛ α h ) = 1 2 (sign(α)ɛα ɛ α ) (h h ) =σ E y sign(α)(ɛα h + ) (ɛ α h ) = 1 2 sign(α)(ɛα ɛ α ) (h + h ) + (sign(α)ɛ α ɛ α ) (h + h ) = 1 2 sign( α)(ɛ α ɛ α ) (h + h ) + (sign(α)ɛ α ɛ α ) (h + h ) = 1 2 sign(α)(ɛα ɛ α ) (h + h ) =σ E y ++ = σ E y σ E = 1/2 sign(α)ɛ α ɛ α E Sp(n) sp(1, C) = C(σ E ) S 2 (H) Λ 2 (E H) 2-form Sp(n)Sp(1) 4-from sp(1, C) (σ E y + ) (σ E y + ) + (σ E y + ) (σ E y + ) + 2(σ E y ++ ) (σ E y ++ ) Sp(n) Sp(1) Remark form 4-form 4-from Ω := 2(σ E y + ) (σ E y + ) + 2(σ E y ++ ) (σ E y ++ ) Kraines Sp(n)Sp(1) Proof. J(ɛ α ) = sign(α)ɛ α, J(h A ) = sign(a)h A J(σ E ) = 1/2 sign(α)j(ɛ α ) J(ɛ α ) = 1/2 sign(α)ɛ α ɛ α = σ E 75

76 σ E J(y AB ) = J( sign(b)h A h B ) = sign(a)h A h B = y BA Kraines Λ p 0(E) S n p (H) Kraines v α,a = ɛ α h A v α,a v β,b + v β,b v α,a = 2g(v α,a, v β,b ) = 2sign(αA)δ α β δ A B 2-form (σ E y + ) = ( 1 sign(α)vα,+ v α,+ ) = 1 sign(α)vα,+ v α,+ = 2σ 2 2 (σ E y + ) = ( 1 sign(α)vα, v α, ) = 1 sign(α)vα, v α, = 2σ 2 2 (σ E y ++ ) =( 1 2 sign(α)vα,+ v α, ) = 1 sign(α)vα,+ v α, 1 sign(α)i(vα,+ )v α, 2 2 = 1 sign(α)vα,+ v α, 1 sign(α)sign(α) 2 2 = 1 sign(α)vα,+ v α, n = N n = [σ, σ ] 2 2-from 4-form (v β, v β, v α,+ v α,+ ) =v β, (v β, v α,+ v α,+ ) +(i(v β, )(v β, v α,+ v α,+ )) =v β, v β, (v α,+ v α,+ ) +v β, (i(v β, )(v α,+ v α,+ )) + sign(β)δ β α (v β, v α,+ ) sign(β)δ βα (v β, v α,+ ) =v β, v β, (v α,+ v α,+ ) +v β, (sign(β)δ βα v α,+ sign(β)δ β α v α,+ ) + sign(β)δ β α (v β, v α,+ ) sign(β)δ βα (v β, v α,+ ) 76

77 {(σ E y + ) (σ E y + )} = 1/4 sign(αβ)(v β, v β, v α,+ v α,+ ) = 1/4 sign(αβ){v β, v β, (v α,+ v α,+ ) +v β, (sign(β)δ βα v α,+ sign(β)δ β α v α,+ ) + sign(β)δ β α (v β, v α,+ ) sign(β)δ βα (v β, v α,+ ) } =4σ σ + 1/4 v β, ( sign(β)v β,+ sign(β)v β,+ ) + 1/4 sign(α)(v α, v α,+ ) +1/4 sign(α)(v α, v α,+ ) =4σ σ 1/2 sign(β)v β, v β,+ 1/2 sign(β)(v β, v β,+ ) =4σ σ + 1/2 sign(β)v β, v β,+ 1/2 sign(β)(v β,+ v β, ) =4σ σ 1/2 sign(β)v β,+ v β, 2n 1/2 sign(β)(v β,+ v β, ) =4σ σ + (N 2n) + (N n) = 4σ σ + 2N 3n =4σσ 4[σ, σ ] + 2N 3n = 4σσ 4N + 4n + 2N 3n = 4σσ 2N + n (v β,+ v β, v α,+ v α, ) =v β,+ (v β, v α,+ v α, ) +(i(v β,+ )(v β, v α,+ v α, )) =v β,+ v β, (v α,+ v α, ) +v β,+ (i(v β, )(v α,+ v α, )) + sign(β)(v α,+ v α, ) +sign(β)δ βα (v β, v α,+ ) =v β,+ v β, (v α,+ v α, ) +sign(β)δ βα v β,+ v α, + sign(β)(v α,+ v α, ) +sign(β)δ βα (v β, v α,+ ) {(σ E y ++ ) (σ E y ++ )} = 1 sign(αβ)(vβ,+ v β, v α,+ v α, ) 4 = 1 sign(αβ){vβ,+ v β, (v α,+ v α, ) +sign(β)δ βα v β,+ v α, 4 + sign(β)(v α,+ v α, ) +sign(β)δ βα (v β, v α,+ ) } =N(N n) + 1 sign(α)δβα v β,+ v α, + 1 sign(α)((v α,+ v α, ) +δ βα (v β, v α,+ )) 4 4 =N(N n) + 1 sign(β)vβ,+ v β, 4 + n 1 sign(α)(v α,+ v α, ) 1 sign(α)(v α,+ v α, ) 2 4 α =N(N n) 1 2 N n(n n) (N n) = (N n)2 n/2 αβ αβ 77

78 Kraines Ω = 2(N n) 2 n + 2(4σσ 2N + n) = 2(N n)(n n 2) 3n + 8σσ Λ n p 0 (E) S p (H) 0 (E) n p σ sp(1, C) lowest weight vector σ Λ n p 0 (E) S p (H) Λ n p 0 (E) σ σλ n p 0 (E) σ σ σ p Λ n p 0 (E) σ 0 Kraines φ Λ n p 0 (E) Ω φ = (2p(p + 2) 3n)φ σφ σ σφ n p + 2 Ω(σφ) = 2( p + 2)( p)σφ 3n + 8σσ σφ =2p(p 2)σφ 3n + 8σ[σ, σ]φ =2p(p 4)σφ 3n + 8σ(n N)φ =2p(p 2)σφ 3n + 8pσφ = (2p(p + 2) 3n)σφ 0 (E) S p (H) Ω 2p(p + 2) 3n 2p(p + 2) Λ n p Λ n p Remark Ω Kraines Sp(n)Sp(1) E H Λ (E H) Λ = Cl 4n = W 4n W4n ( p Λ p 0(E) S n p (H)) ( q Λ q 0(E) S n q (H)) S n p (H) S n q (H) Λ p 0(E) Λ q 0(E) Λ p 0(E) Λ q 0(E) = a,b Λ a,b 0 (E) 78