Estrutura do Grupo U(n)

Vamos considerar o caso em que G = U(1) = {a ∈ GL(n, C)|at_{a = 1}. Portanto,}

a respectiva ´algebra de Lie ´e dada por g = u(n) = {X ∈ gl(n, C)|Xt+ X = 0}. Se tomarmos a representa¸c˜ao matricial X = [Xij] ∈ g, temos que Xii ∈ iR, ou seja, Xii =

B. Informa¸c˜oes Adicionais Sobre ´Algebras de Lie 106

−(Xt)ii = −Xii, e para i 6= j temos que Xij = −(Xt)ij = −Xji. Seja a sub-´algebra

abeliana maximal t de g representada por

t =      iθ1 ° . .. ° iθn   ¯ ¯ ¯ ¯ ¯ ¯θj ∈ R   ⊂ g. (B.21)

Quando X ∈ g e se [X, H] = 0, para qualquer H ∈ t, ent˜ao devemos ter que X ∈ t. O centralizador gH de H em g est´a contido em t e, portanto, gH = t para H ∈ t

com elementos diagonais distintos. Em particular chamamos H de elemento regular se dimgH = dimt.

A forma de Killing (B.11) sobre g ´e dada por B(X, Y ) = Tr(XY ) = n X j=1 (XY )jj = n X j=1 n X k=1 XjkYkj (B.22) ⇒ B(X, Y ) = Tr(XY ) = n X j,k=1 XjkYkj = n X j,k=1 (−Xkj)(−Yjk) = Tr(XY ) (B.23) ∴ Tr(XY ) ∈ R. (B.24)

Semelhantemente, temos o seguinte resultado: B(X, X) = Tr(XX) = n X j,k=1 XjkXkj = − n X j,k=1 Xjk(Xt)kj = − n X j,k=1 |Xjk|2 (B.25) ∴ −Tr(XX) ≥ 0 e Tr(XX) = 0 ⇔ X = 0. (B.26) Logo hX, Y i ≡ −Tr(XY ), X, Y ∈ g, (B.27)

define um produto interno real e Ad(G)-invariante sobre g = u(n). Uma vez que h·, ·i ´e um produto interno U(n)-invariante (ou Ad(G)-invariante), sabemos que

h[X, Y ], Zi = −hY, [X, Z]i (B.28)

para X, Y, Z ∈ u(n). Escolhemos uma decomposi¸c˜ao g = z ⊕ g1, onde g1 = [g, g] ´e semi- simples. Tamb´em B(X, X) ≤ 0 ´e a forma de Killing sobre g, onde B(X, X) < 0 sobre g1− {0}. Para X, Y ∈ g, temos que Tr[X, Y ] = 0 nos leva a concluir que [X, Y ] ∈ su(n) e, portanto, g1 ⊂ su(n). Inversamente, dado X ∈ su(n) escrevemos X = Z + X1 ∈ z ⊕ g1. Ent˜ao Z = X − X1 ∈ su(n) tal que TrZ = 0. No entanto, Z = diag(iθ, ..., iθ), θ ∈ R, o que implica em TrZ = niθ, onde θ deve ser zero, ou seja, Z = 0. Logo X = X1 ∈ g1, ou seja, g1 = su(n).

Tamb´em podemos considerar a decomposi¸c˜ao t = z ⊕ t1, onde t1 = t ∩ g1 ´e abeliano maximal em g1. Neste caso temos

t1 =      iθ1 ° . .. ° iθn   ¯ ¯ ¯ ¯ ¯ ¯θj ∈ R, θ1+ ... + θn = 0   = {H ∈ t|TrH = 0}. (B.29) Mais explicitamente, dado H = diag(iθ1, ..., iθn) ∈ t, θ ≡

P_n

j=1θj/n e tj ≡ θj − θ, vemos

B. Informa¸c˜oes Adicionais Sobre ´Algebras de Lie 107

Agora vamos olhar as complexifica¸c˜oes:

t = z ⊕ t1 → tC= zC⊕ tC1. (B.30) Dado H = diag(H1, ..., Hn) = (X, Y ) ∈ tC = t ⊕ it, temos X = Z1 + T1 ∈ z ⊕ t1 e Y = Z2 + T2 ∈ z ⊕ t2, onde (Z1, Z2) ∈ zC e (T1, T2) ∈ tC1. Devemos considerar as representa¸c˜oes matriciais explicitas (Z1, Z2) = diag(λ, ..., λ) e (T1, T2) = diag(u1, ..., u1), onde λ, uj ∈ C e Tr(T1, T2) = 0. Portanto temos que Hj = λ + uj, tal que

n X j=1 Hj = nλ + n X j=1 uj = nλ (B.31) ⇒ λ = P_n j=1Hj n e uj = Hj + λ. (B.32)

Como j´a notado, tC

1 ⊂ {H ∈ tC|TrH = 0}. Inversamente, se H ∈ tC e TrH = 0, a decomposi¸c˜ao H = X + iY ∈ t ⊕ it, onde X = (H − Ht)/2 e Y = (−i)(H + Ht)/2, implica que TrX = TrY = 0, ou seja, X, Y ∈ t1. A ´algebra sl(n, C) realiza a complexifica¸c˜ao de su(n). Em outras palavras temos que gC

1 = su(n)C= sl(n, C) e o espa¸co das ra´ızes ´e dado por

∆(gC_{, t}C_{) = ∆(gl(n, C), t}C_{) = {α ∈ (t}C₎∗_{− {0}|(g}C₎α _{6= 0}, (B.33)}

onde

(gC)α = {X ∈ gl(n, C)|[H, X] = α(H)X, ∀H ∈ tC}. (B.34) Existe uma bije¸c˜ao α → eα de ∆(gC

1, tC1) = ∆(sl(n, C), tC1) em ∆(gC, tC) definida por e

α(Z + H1) ≡ α(H1), (B.35)

para H = Z + H1 ∈ z ⊕ t1. Portanto os espa¸cos das ra´ızes coincidem, (gC1)α ∼= (gC)eα. Ent˜ao, para α = αrs, onde r 6= s, temos que

e

αrs(H) = αrs(H1) = (Hr− λ) − (Hs− λ) = Hr− Hs. (B.36) Para a escolha ∆+ _{= {α}

rs|r < s} temos P = {eαrs|1 ≤ r < s ≤ n}. Vamos definir a

matriz Hrs = diag(0, ..., 1(r), ..., −1(s), ..., 0) ∈ it, portanto hH, Hrsi = TrHH

t

rs = Hr− Hs = eαrs(H). (B.37)

Seja a ∈ NG(t) = {a ∈ G|Ad(a)t = t}. Para 1 ≤ j ≤ n seja Hj ≡ diag(0, ..., i1(j), ..., 0) ∈ t. Logo Ad(a)Hj = aHja−1 ≡ diag(iθ1, ..., iθn) ∈ t, onde θk ∈ R. Contudo aHja−1 and

Hj tem os mesmos autovalores, o que nos leva concluir que aHja−1 ∼= Hl. Logo esta-

belecemos um isomorfismo σ : {n} → {n}, tal que aHja−1 = Hσ(j), o que implica em

σ(j) = l. Ent˜ao σ ∈ Sn, grupo sim´etrico, e depende de a, σ = σa. Portanto, existe um

mapa ² : NG(t) → Sn dado por ²(a) = σa. Se b ∈ T , ent˜ao AdH = H, para qualquer

H ∈ t. Sendo assim,

Hσab_(j) = abH_jb−1a−1 = aH_ja−1 = H_σa_(j) (B.38)

⇒ σab _{= σ}a_{, (B.39)}

ou seja, podemos definir mais um mapa e² : W ≡ NG(t)/T → Sn dado por e²(aT ) = σa. W

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