Google 翻译生成的粗略草稿，质量较低。术语依照术语表，但句子可能有误；之后会由仔细的译文取代。
§ 13。经典可拆分简单李代数
在本段中，我们明确地描述了对于经典可分割简单李代数的每个型：
这个型、它的维数和它的分裂嘉当的(I)代数
子代数；
(II) 它的同根；
(III) 博雷尔子代数及其抛物线子代数；
(IV) 其基本简单表示；
(V) 其基本简单表示的那些是
正交或辛；
(VI) 代数不变量多项式函数；
(VII)群Autg \mathfrak{g} g 0 g _0\mathfrak{g} 0 g e g _e\mathfrak{g} e g
(VIII)限制杀戮形式至嘉当子代数；
(IX) 谢瓦利阶。
1. 代数型 Al _{\boldsymbol{l}} l l ≥ \boldsymbol{l}\geq l ≥
(I) 令k k k l + 1 l+ 1 l + 1 g \mathfrak{g} g s l ( V ) \mathfrak{s}\mathfrak{l}(V) sl ( V ) ( e i ) 1 ≤ i ≤ l + 1 (e_i)_{1\leq i\leq l+1} ( e i ) 1 ≤ i ≤ l + 1 g \mathfrak{g} g g \mathfrak{g} g s l ( l + 1 , k ) \mathfrak{s}\mathfrak{l}(l+ 1, k) sl ( l + 1 , k ) g \mathfrak{g} g
回想一下（代数，II, §10, no. 3章），E i j E_{ij} E ij ( α m p ) (\alpha_{mp}) ( α m p ) α i j = 1 \alpha_{ij}= 1 α ij = 1 α m p = 0 \alpha_{mp}= 0 α m p = 0 ( m , p ) ≠ ( i , j ) (m, p)\not= (i, j) ( m , p ) = ( i , j )
E i j ( 1 ≤ i , j ≤ l + 1 , i ≠ j )
E_{ij}(1\leq i, j\leq l+ 1, i\not=j)
E ij ( 1 ≤ i , j ≤ l + 1 , i = j ) E i , i − E i + 1 , i + 1 ( 1 ≤ i ≤ l )
E_{i,i}-E_{i+1,i+1}(1\leq i\leq l)
E i , i − E i + 1 , i + 1 ( 1 ≤ i ≤ l ) 形式基共g \mathfrak{g} g
dimg = l ( l + 2 ) \mathfrak{g}=l(l+ 2) g = l ( l + 2 )
令h ^ \widehat{\mathfrak{h}} h g l ( l + 1 , k ) \mathfrak{g}\mathfrak{l}(l+ 1, k) gl ( l + 1 , k ) ( E i i ) 1 ≤ i ≤ l + 1 (E_{ii})_{1\leq i\leq l+1} ( E ii ) 1 ≤ i ≤ l + 1 h ^ \widehat{\mathfrak{h}} h ( ε ^ i ) 1 ≤ i ≤ l + 1 (\widehat{\varepsilon}_i)_{1\leq i\leq l+1} ( ε i ) 1 ≤ i ≤ l + 1 h ^ ∗ \widehat{\mathfrak{h}}^* h ∗ ( E i i ) 1 ≤ i ≤ l + 1 (E_{ii})_{1\leq i\leq l+1} ( E ii ) 1 ≤ i ≤ l + 1 h ∈ h ^ h\in \widehat{\mathfrak{h}} h ∈ h
[ h , E i j ] = ( ε ^ i ( h ) − ε ^ j ( h ) ) E i j (1)
[h, E_{ij}] = (\widehat{\varepsilon}_i(h)-\widehat{\varepsilon}_j(h))E_{ij} \tag{1}
[ h , E ij ] = ( ε i ( h ) − ε j ( h )) E ij ( 1 ) 由章。 I, §1, no. 2，公式 (5)。 令h \mathfrak{h} h h ^ \widehat{\mathfrak{h}} h ε i = ε ^ i ∣ h \varepsilon_i= \widehat{\varepsilon}_i|\mathfrak{h} ε i = ε i ∣ h h \mathfrak{h} h g \mathfrak{g} g ( g , h ) (\mathfrak{g},\mathfrak{h}) ( g , h ) ε i − ε j ( i ≠ j ) \varepsilon_i-\varepsilon_j(i\not=j) ε i − ε j ( i = j ) h ^ 0 ∗ \widehat{\mathfrak{h}}^*_0 h 0 ∗ h ^ ∗ \widehat{\mathfrak{h}}^* h ∗ ( ε ^ i ) (\widehat{\varepsilon}_i) ( ε i ) h ^ 0 ∗ \widehat{\mathfrak{h}}^*_0 h 0 ∗ h ∗ \mathfrak{h}^* h ∗ λ → λ ∣ h \lambda \rightarrow \lambda |\mathfrak{h} λ → λ ∣ h ( g , h ) (\mathfrak{g},\mathfrak{h}) ( g , h ) A l A_l A l g \mathfrak{g} g g \mathfrak{g} g A l A_l A l
在基本自同构下，g \mathfrak{g} g h ′ \mathfrak{h}' h ′ h \mathfrak{h} h e g _e\mathfrak{g} e g g \mathfrak{g} g x → s x s − 1 x \rightarrow sxs^{-1} x → s x s − 1 s ∈ S L ( V ) s\in \mathbf{S}\mathbf{L}(V) s ∈ SL ( V ) β \beta β h ′ \mathfrak{h}' h ′ g \mathfrak{g} g h β \mathfrak{h}_{\beta} h β β \beta β h β \mathfrak{h}_{\beta} h β h β \mathfrak{h}_{\beta} h β β \beta β β → h β \beta \rightarrow \mathfrak{h}_{\beta} β → h β l + 1 l+ 1 l + 1 g \mathfrak{g} g α = ε i − ε j ( i ≠ j ) \alpha =\varepsilon_i-\varepsilon_j(i\not=j) α = ε i − ε j ( i = j ) g α = k E i j \mathfrak{g}^{\alpha}=kE_{ij} g α = k E ij
[ E i j , E j i ] = E i i − E j j
[E_{ij}, E_{ji}] =E_{ii}-E_{jj}
[ E ij , E j i ] = E ii − E j j 从 α ( E i i − E j j ) = 2 \alpha (E_{ii}-E_{jj}) = 2 α ( E ii − E j j ) = 2
H α = E i i − E j j
H_{\alpha}=E_{ii}-E_{jj}
H α = E ii − E j j (III) 放入α 1 = ε 1 − ε 2 , α 2 = ε 2 − ε 3 , . . . , α l = ε l − ε l + 1 \alpha_1=\varepsilon_1-\varepsilon_2, \alpha_2=\varepsilon_2-\varepsilon_3, . . . , \alpha_l=\varepsilon_l-\varepsilon_{l+1} α 1 = ε 1 − ε 2 , α 2 = ε 2 − ε 3 , ... , α l = ε l − ε l + 1 7. I , ( α 1 , . . . , α l ) 7.I, (\alpha_1, . . . , \alpha_l) 7. I , ( α 1 , ... , α l ) i < j i < j i < j ε i − ε j \varepsilon_i-\varepsilon_j ε i − ε j b \mathfrak{b} b
V中的标志是向量子空间V的集合，与{ 0 } \{0\} { 0 } { W 1 , . . . , W l } \{W_1, . . . ,W_l\} { W 1 , ... , W l } W i W_i W i i i i
W 1 ⊂ ⋯ ⊂ W l
W_1\subset \cdots \subset W_l
W 1 ⊂ ⋯ ⊂ W l 对于例，如果V i V_i V i e 1 , . . . , e i e_1, . . . , e_i e 1 , ... , e i { V 1 , . . . , V l } \{V_1, . . . ,V_l\} { V 1 , ... , V l }
It is immediate that b \mathfrak{b} b g \mathfrak{g} g { V 1 , . . . , V l } \{V_1, . . . ,V_l\} { V 1 , ... , V l } b \mathfrak{b} b h \mathfrak{h} h E i j E_{ij} E ij i < j i < j i < j V i V_i V i b \mathfrak{b} b
现在让 δ \delta δ g \mathfrak{g} g b δ \mathfrak{b}_{\delta} b δ δ \delta δ g \mathfrak{g} g g \mathfrak{g} g b \mathfrak{b} b δ → b δ \delta \rightarrow \mathfrak{b}_{\delta} δ → b δ g \mathfrak{g} g
令 β \beta β h β \mathfrak{h}_{\beta} h β β \beta β β \beta β β \beta β ω \omega ω { W 1 , . . . , W l } \{W_1, . . . ,W_l\} { W 1 , ... , W l } W i W_i W i i i i β \beta β ω \omega ω ( l + (l+ ( l + β \beta β ( l + (l+ ( l + ( s l ( V ) , h β ) ( (\mathfrak{s}\mathfrak{l}(V),\mathfrak{h}_{\beta}) ( ( sl ( V ) , h β ) (
设γ \gamma γ γ \gamma γ g \mathfrak{g} g p γ \mathfrak{p}_{\gamma} p γ γ \gamma γ g \mathfrak{g} g p γ \mathfrak{p}_{\gamma} p γ γ \gamma γ γ = { V i 1 , . . . , V i q } \gamma =\{V_{i_1}, . . . ,V_{i_q}\} γ = { V i 1 , ... , V i q } 1 ≤ i 1 < ⋯ < i q ≤ l 1\leq i_1<\cdots < i_q\leq l 1 ≤ i 1 < ⋯ < i q ≤ l i 0 = 0 , i q + 1 = l + 1 i_0= 0, i_{q+1}=l+ 1 i 0 = 0 , i q + 1 = l + 1
i 0 + 1 , i 1 , i 1 + 1 , i 2 , . . . , i q + 1 , i q + 1
i_0+ 1, i_1, i_1+ 1, i_2, . . . , i_q+ 1, i_{q+1}
i 0 + 1 , i 1 , i 1 + 1 , i 2 , ... , i q + 1 , i q + 1 { 1 , . . . , l + 1 } \{1, . . . , l+ 1\} { 1 , ... , l + 1 } l + 1 l+ 1 l + 1 ( X a b ) 1 ≤ a , b ≤ q + 1 (X_{ab})_{1\leq a,b\leq q+1} ( X ab ) 1 ≤ a , b ≤ q + 1 p γ \mathfrak{p}_{\gamma} p γ s l ( l + 1 , k ) \mathfrak{s}\mathfrak{l}(l+ 1, k) sl ( l + 1 , k ) ( X a b ) 1 ≤ a , b ≤ q + 1 (X_{ab})_{1\leq a,b\leq q+1} ( X ab ) 1 ≤ a , b ≤ q + 1 p i 1 , . . . , i q \mathfrak{p}_{i_1,...,i_q} p i 1 , ... , i q X a b = 0 X_{ab}= 0 X ab = 0 a > b a > b a > b p i 1 , . . . , i q ⊃ b \mathfrak{p}_{i_1,...,i_q}\supset \mathfrak{b} p i 1 , ... , i q ⊃ b p i 1 , . . . , i q \mathfrak{p}_{i_1,...,i_q} p i 1 , ... , i q V i V_i V i i k < i < i k + 1 i_k< i < i_{k+1} i k < i < i k + 1 p i 1 , . . . , i q \mathfrak{p}_{i_1,...,i_q} p i 1 , ... , i q E i k + 1 , i E_{i_{k+1},i} E i k + 1 , i V i V_i V i
因而，最大标志{ V 1 , . . . , V l } \{V_1, . . . ,V_l\} { V 1 , ... , V l } 2 l 2^l 2 l 2 l 2^l 2 l b \mathfrak{b} b 2 l 2^l 2 l b ( \mathfrak{b}( b ( γ → p γ \gamma \rightarrow \mathfrak{p}_{\gamma} γ → p γ g \mathfrak{g} g p γ ⊃ p γ ′ \mathfrak{p}_{\gamma}\supset \mathfrak{p}_{\gamma'} p γ ⊃ p γ ′ γ ⊂ γ ′ \gamma \subset \gamma ' γ ⊂ γ ′
回想一下抛物线子代数 p = p i 1 , . . . , i q ( 1 ≤ i 1 < ⋯ < i q ≤ l ) \mathfrak{p}=\mathfrak{p}_{i_1,...,i_q}(1\leq i_1<\cdots < i_q\leq l) p = p i 1 , ... , i q ( 1 ≤ i 1 < ⋯ < i q ≤ l ) s \mathfrak{s} s n \mathfrak{n} n s l ( l + 1 , k ) \mathfrak{s}\mathfrak{l}(l+ 1, k) sl ( l + 1 , k ) ( X a b ) 1 ≤ a , b ≤ q + 1 (X_{ab})_{1\leq a,b\leq q+1} ( X ab ) 1 ≤ a , b ≤ q + 1 X a b = 0 X_{ab}= 0 X ab = 0 a ≠ b a\not=b a = b a ≥ b a\geq b a ≥ b p = s ⊕ n \mathfrak{p}=\mathfrak{s}\oplus \mathfrak{n} p = s ⊕ n s \mathfrak{s} s g \mathfrak{g} g n \mathfrak{n} n p \mathfrak{p} p
(IV) 对于r = 1 , 2 , . . . , l r= 1,2, . . . , l r = 1 , 2 , ... , l ϖ r = ε 1 + ⋯ + ε r \varpi_r=\varepsilon_1+\cdots +\varepsilon_r ϖ r = ε 1 + ⋯ + ε r ϖ i ( H α j ) = δ i j \varpi_i(H_{\alpha_j}) =\delta_{ij} ϖ i ( H α j ) = δ ij ϖ r \varpi_r ϖ r α r \alpha_r α r
令σ \sigma σ g \mathfrak{g} g σ \sigma σ ⋀ r σ \bigwedge^r\sigma ⋀ r σ E = ⋀ r ( V ) E =\bigwedge^r(V) E = ⋀ r ( V ) ( e 1 , . . . , e l + 1 ) (e_1, . . . , e_{l+1}) ( e 1 , ... , e l + 1 ) e i 1 ∧ ⋯ ∧ e i r e_{i_1}\wedge \cdots \wedge e_{i_r} e i 1 ∧ ⋯ ∧ e i r i 1 < ⋯ < i r i_1<\cdots < i_r i 1 < ⋯ < i r h ∈ h h\in \mathfrak{h} h ∈ h
( ⋀ r σ ) ( h ) . e i 1 ∧ ⋯ ∧ e i r = ( ε i 1 + ⋯ + ε i r ) ( h ) e i 1 ∧ ⋯ ∧ e i r
(\bigwedge^r\sigma )(h).e_{i_1}\wedge \cdots \wedge e_{i_r}= (\varepsilon_{i_1}+\cdots +\varepsilon_{i_r})(h)e_{i_1}\wedge \cdots \wedge e_{i_r}
( ⋀ r σ ) ( h ) . e i 1 ∧ ⋯ ∧ e i r = ( ε i 1 + ⋯ + ε i r ) ( h ) e i 1 ∧ ⋯ ∧ e i r 因而，每个权都是重数，1 , ϖ r 1,\varpi_r 1 , ϖ r ⋀ r σ \bigwedge^r\sigma ⋀ r σ ϖ r − μ \varpi_r-\mu ϖ r − μ μ \mu μ ϖ r \varpi_r ϖ r ⋀ r σ \bigwedge^r\sigma ⋀ r σ e 1 ∧ ⋯ ∧ e r e_1\wedge \cdots \wedge e_r e 1 ∧ ⋯ ∧ e r
{ ε 1 , . . . , ε l + 1 }
\{\varepsilon_1, . . . , \varepsilon_{l+1}\}
{ ε 1 , ... , ε l + 1 } Weyl 群下的 ϖ r \varpi_r ϖ r ε i 1 + ⋯ + ε i r \varepsilon_{i_1}+\cdots +\varepsilon_{i_r} ε i 1 + ⋯ + ε i r i 1 < ⋯ < i r i_1<\cdots < i_r i 1 < ⋯ < i r e 1 ∧ ⋯ ∧ e r e_1\wedge \cdots \wedge e_r e 1 ∧ ⋯ ∧ e r ε i 1 + ⋯ + ε i r \varepsilon_{i_1}+\cdots +\varepsilon_{i_r} ε i 1 + ⋯ + ε i r ⋀ r σ \bigwedge^r\sigma ⋀ r σ ϖ r \varpi_r ϖ r
因而，表示⋀ r σ ( 1 ≤ r ≤ l ) \bigwedge^r\sigma (1\leq r\leq l) ⋀ r σ ( 1 ≤ r ≤ l ) ⋀ r σ \bigwedge^r\sigma ⋀ r σ = ( r l + 1 ) =(^{l+1}_r) = ( r l + 1 )
(V) 我们有w 0 ( α 1 ) = − α l , w 0 ( α 2 ) = − α l − 1 , . . w_0(\alpha_1) =-\alpha_l, w_0(\alpha_2) =-\alpha_{l-1}, . . w 0 ( α 1 ) = − α l , w 0 ( α 2 ) = − α l − 1 , ..
− w 0 ( ϖ 1 ) = ϖ l , − w 0 ( ϖ 2 ) = ϖ l − 1 , . . .
-w_0(\varpi_1) =\varpi_l,-w_0(\varpi_2) =\varpi_{l-1}, . . .
− w 0 ( ϖ 1 ) = ϖ l , − w 0 ( ϖ 2 ) = ϖ l − 1 , ... 让
ω = n 1 ϖ 1 + ⋯ + n l ϖ l ( n 1 , . . . , n l ∈ N )
\omega =n_1\varpi_1+\cdots +n_l\varpi_l(n_1, . . . , n_l\in \mathbf{N})
ω = n 1 ϖ 1 + ⋯ + n l ϖ l ( n 1 , ... , n l ∈ N ) 成为主导权。那么，简单的表示与最高权 ω \omega ω
n 1 = n l , n 2 = n l − 1 , . .
n_1=n_l,n_2=n_{l-1}, . .
n 1 = n l , n 2 = n l − 1 , .. （§7, no. 5，提案 12）。特别是，如果 l l l s l ( l + 1 , k ) \mathfrak{s}\mathfrak{l}(l+ 1, k) sl ( l + 1 , k ) l l l i ≠ ( l + 1 ) / 2 i\not= (l+ 1)/2 i = ( l + 1 ) /2 ⋀ i σ \bigwedge^i\sigma ⋀ i σ ϖ ( l + 1 ) / 2 \varpi_{(l+1)/2} ϖ ( l + 1 ) /2 ( α 1 , . . . , α l ) (\alpha_1, . . . , \alpha_l) ( α 1 , ... , α l )
1 l + 1 [ l + 1 2 ( 1 + 2 + ⋯ + l − 1 2 ) + l + 1 2 ( 1 + 2 + ⋯ + l + 1 2 ) ]
\frac{1}{l + 1}[\frac{l + 1}{2}(1 + 2 +\cdots +\frac{l- 1}{2})+\frac{l + 1}{2}(1 + 2 +\cdots +\frac{l + 1}{2})]
l + 1 1 [ 2 l + 1 ( 1 + 2 + ⋯ + 2 l − 1 ) + 2 l + 1 ( 1 + 2 + ⋯ + 2 l + 1 )] = 1 + 2 + ⋯ + l − 1 2 + l + 1 4
= 1 + 2 +\cdots +\frac{l- 1}{2}+\frac{l + 1}{4}
= 1 + 2 + ⋯ + 2 l − 1 + 4 l + 1 so ⋀ ( l + 1 ) / 2 σ \bigwedge^{(l+1)/2}\sigma ⋀ ( l + 1 ) /2 σ l ≡ − 1 l\equiv -1 l ≡ − 1 l ≡ 1 l\equiv 1 l ≡ 1 ⋀ l + 1 ( V ) \bigwedge^{l+1}(V) ⋀ l + 1 ( V ) e e e
⋀ ( l + 1 ) / 2 ( V ) × ⋀ ( l + 1 ) / 2 ( V )
\bigwedge^{(l+1)/2}(V)\times \bigwedge^{(l+1)/2}(V)
⋀ ( l + 1 ) /2 ( V ) × ⋀ ( l + 1 ) /2 ( V ) 到⋀ l + 1 \bigwedge^{l+1} ⋀ l + 1 ( u , v ) → Φ ( u , v ) e (u, v) \rightarrow \Phi (u, v)e ( u , v ) → Φ ( u , v ) e Φ \Phi Φ ⋀ ( l + 1 ) / 2 ( V ) \bigwedge^{(l+1)/2}(V) ⋀ ( l + 1 ) /2 ( V ) Φ \Phi Φ g \mathfrak{g} g ( l + 1 ) / 2 (l+ 1)/2 ( l + 1 ) /2 ( l + 1 ) / 2 (l+ 1)/2 ( l + 1 ) /2
(VI) 对于所有x ∈ g x\in \mathfrak{g} x ∈ g σ ( x ) = x \sigma (x) =x σ ( x ) = x
T l + 1 + f 2 ( x ) T l − 1 + f 3 ( x ) T l − 2 + ⋯ + f l + 1 ( x )
T^{l+1}+f_2(x)T^{l-1}+f_3(x)T^{l-2}+\cdots +f_{l+1}(x)
T l + 1 + f 2 ( x ) T l − 1 + f 3 ( x ) T l − 2 + ⋯ + f l + 1 ( x ) 其中f 2 , . . . , f l + 1 f_2, . . . , f_{l+1} f 2 , ... , f l + 1 g ( \mathfrak{g}( g (
如果x = ξ 1 E 11 + ⋯ + ξ l + 1 E l + 1 l + 1 ∈ h x=\xi_1E_{11}+\cdots +\xi_{l+1}E_{l+1l+1}\in \mathfrak{h} x = ξ 1 E 11 + ⋯ + ξ l + 1 E l + 1 l + 1 ∈ h f i ( x ) f_i(x) f i ( x ) 2 , . . . , l + 1 2, . . . , l+ 1 2 , ... , l + 1 ξ 1 , . . . , ξ l + 1 \xi_1, . . . , \xi_{l+1} ξ 1 , ... , ξ l + 1 S ( h ∗ ) \mathbf{S}(\mathfrak{h}^*) S ( h ∗ ) f i ∣ h f_i|\mathfrak{h} f i ∣ h 3 3 3 f 2 , f 3 , . . . , f l + 1 f_2, f_3, . . . , f_{l+1} f 2 , f 3 , ... , f l + 1 g \mathfrak{g} g
(VII) 对于所有g ∈ G L ( l + 1 , k ) g\in \mathbf{G}\mathbf{L}(l+ 1, k) g ∈ GL ( l + 1 , k ) φ k ( g ) = φ ( g ) \varphi_k(g) =\varphi (g) φ k ( g ) = φ ( g ) x → g x g − 1 x \rightarrow gxg^{-1} x → g x g − 1 g \mathfrak{g} g G L ( l + 1 , k ) \mathbf{G}\mathbf{L}(l+ 1, k) GL ( l + 1 , k ) g \mathfrak{g} g φ \varphi φ
φ ( S L ( l + 1 , k ) ) = \varphi (\mathbf{S}\mathbf{L}(l+ 1, k)) = φ ( SL ( l + 1 , k )) = e ( g ) _e(\mathfrak{g}) e ( g )
（第VII, §3, no. 1、注2章）。设k ‾ \overline{k} k k k k
G L ( l + 1 , k ‾ ) = k ‾ ∗ . S L ( l + 1 , k ‾ )
\mathbf{G}\mathbf{L}(l+ 1,\overline{k}) = \overline{k}^*.\mathbf{S}\mathbf{L}(l+ 1,\overline{k})
GL ( l + 1 , k ) = k ∗ . SL ( l + 1 , k ) 所以φ k ‾ ( G L ( l + 1 , k ‾ ) ) = φ k ‾ ( S L ( l + 1 , k ‾ ) ) = \varphi_{\overline{k}}(\mathbf{G}\mathbf{L}(l+ 1,\overline{k})) =\varphi_{\overline{k}}(\mathbf{S}\mathbf{L}(l+ 1,\overline{k})) = φ k ( GL ( l + 1 , k )) = φ k ( SL ( l + 1 , k )) = e ( g ⊗ k k ‾ ) _e(\mathfrak{g}\otimes_k\overline{k}) e ( g ⊗ k k ) φ ( G L ( l + 1 , k ) ) ⊂ \varphi (\mathbf{G}\mathbf{L}(l+ 1, k))\subset φ ( GL ( l + 1 , k )) ⊂ 0 ( g ) _0(\mathfrak{g}) 0 ( g ) 0 ( g ) ⊂ φ ( G L ( l + 1 , k ) ) _0(\mathfrak{g})\subset \varphi (\mathbf{G}\mathbf{L}(l+ 1, k)) 0 ( g ) ⊂ φ ( GL ( l + 1 , k )) g \mathfrak{g} g
自动0 ( g ) = φ ( G L ( l + 1 , k ) ) _0(\mathfrak{g}) =\varphi (\mathbf{G}\mathbf{L}(l+ 1, k)) 0 ( g ) = φ ( GL ( l + 1 , k ))
φ \varphi φ l + 1 l+1 l + 1 G L ( l + 1 , k ) \mathbf{G}\mathbf{L}(l+ 1, k) GL ( l + 1 , k ) k ∗ k^* k ∗ 0 ( g ) _0(\mathfrak{g}) 0 ( g ) G L ( l + 1 , k ) / k ∗ = P G L ( l + 1 , k ) \mathbf{G}\mathbf{L}(l+ 1, k)/k^*=\mathbf{P}\mathbf{G}\mathbf{L}(l+ 1, k) GL ( l + 1 , k ) / k ∗ = PGL ( l + 1 , k ) φ ′ = φ ∣ S L ( l + 1 , k ) \varphi '=\varphi |\mathbf{S}\mathbf{L}(l+ 1, k) φ ′ = φ ∣ SL ( l + 1 , k ) μ l + 1 ( k ) \mu_{l+1}(k) μ l + 1 ( k ) μ l + 1 ( k ) \mu_{l+1}(k) μ l + 1 ( k ) k k k ( l + (l+ ( l + e ( g ) _e(\mathfrak{g}) e ( g ) S L ( l + 1 , k ) / μ l + 1 ( k ) = P S L ( l + 1 , k ) \mathbf{S}\mathbf{L}(l+ 1, k)/\mu_{l+1}(k) =\mathbf{P}\mathbf{S}\mathbf{L}(l+ 1, k) SL ( l + 1 , k ) / μ l + 1 ( k ) = PSL ( l + 1 , k )
1 ⟶ S L ( l + 1 , k ) ⟶ G L ( l + 1 , k ) ⟶ d e t k ∗ ⟶ \longrightarrow \mathbf{S}\mathbf{L}(l+ 1, k)\longrightarrow \mathbf{G}\mathbf{L}(l+ 1, k)\longrightarrow^{det}k^*\longrightarrow ⟶ SL ( l + 1 , k ) ⟶ GL ( l + 1 , k ) ⟶ d e t k ∗ ⟶
det下的k ∗ k^* k ∗ k ∗ l + 1 k^{*l+1} k ∗ l + 1
自动0 ( g ) / _0(\mathfrak{g})/ 0 ( g ) / e ( g ) ⟶ P G L ( l + 1 , k ) / P S L ( l + 1 , k ) _e(\mathfrak{g})\longrightarrow \mathbf{P}\mathbf{G}\mathbf{L}(l+ 1, k)/\mathbf{P}\mathbf{S}\mathbf{L}(l+ 1, k) e ( g ) ⟶ PGL ( l + 1 , k ) / PSL ( l + 1 , k )
⟶ G L ( l + 1 , k ) / k ∗ . S L ( l + 1 , k ) ⟶ k ∗ / k ∗ l + 1
\longrightarrow \mathbf{G}\mathbf{L}(l+ 1, k)/k^*.\mathbf{S}\mathbf{L}(l+ 1, k)\longrightarrow k^*/k^{*l+1}
⟶ GL ( l + 1 , k ) / k ∗ . SL ( l + 1 , k ) ⟶ k ∗ / k ∗ l + 1 如果 k k k R \mathbf{R} R l + 1 l+ 1 l + 1 0 ( g ) = _0(\mathfrak{g}) = 0 ( g ) = e ( g ) _e(\mathfrak{g}) e ( g ) l + 1 l+ 1 l + 1 0 ( g ) / _0(\mathfrak{g})/ 0 ( g ) / e ( g ) _e(\mathfrak{g}) e ( g ) Z / 2 Z \mathbf{Z}/2\mathbf{Z} Z /2 Z
与5 , f ( T Q ) 5,f(T_{\mathbf{Q}}) 5 , f ( T Q ) g \mathfrak{g} g h \mathfrak{h} h φ \varphi φ G L ( l + 1 , k ) ( \mathbf{G}\mathbf{L}(l+ 1, k) ( GL ( l + 1 , k ) ( D ′ D' D ′ S L ( l + 1 , k ) \mathbf{S}\mathbf{L}(l+ 1, k) SL ( l + 1 , k ) e ( g ) _e(\mathfrak{g}) e ( g ) f ( q ( T P ) ) ⊂ φ ( D ′ ) f(q(T_P))\subset \varphi (D') f ( q ( T P )) ⊂ φ ( D ′ ) f ( q ( T P ) ) = φ ( D ′ ) f(q(T_P)) =\varphi (D') f ( q ( T P )) = φ ( D ′ )
λ 1 \lambda_1 λ 1
d = d= d =
0 λ l + 1 \lambda_{l+1} λ l + 1
是元素 / D ′ D' D ′ i i i j j j ζ ∈ \zeta \in ζ ∈ , k ∗ , k^* , k ∗ = T Q = T_Q = T Q ζ ( ε i − ε j ) = λ i λ j − 1 \zeta (\varepsilon_i-\varepsilon_j) =\lambda_i\lambda^{-1}_j ζ ( ε i − ε j ) = λ i λ j − 1 f ( ζ ) = φ ( d ) f(\zeta ) =\varphi (d) f ( ζ ) = φ ( d ) ε = ε 1 \varepsilon =\varepsilon_1 ε = ε 1 P ( R ) / Q ( R ) P(R)/Q(R) P ( R ) / Q ( R ) l + 1 l+ 1 l + 1
ζ ( ( l + 1 ) ε ) = ζ ( ( ε 1 − ε 2 ) + ( ε 1 − ε 3 ) + ⋯ + ( ε 1 − ε l + 1 ) )
\zeta ((l+ 1)\varepsilon ) =\zeta ((\varepsilon_1-\varepsilon_2) + (\varepsilon_1-\varepsilon_3) +\cdots + (\varepsilon_1-\varepsilon_{l+1}))
ζ (( l + 1 ) ε ) = ζ (( ε 1 − ε 2 ) + ( ε 1 − ε 3 ) + ⋯ + ( ε 1 − ε l + 1 )) = λ 1 l λ 2 − 1 λ 3 − 1 . . . λ l + 1 − 1 = λ 1 l + 1
=\lambda^l_1\lambda^{-1}_2\lambda^{-1}_3. . . \lambda^{-1}_{l+1}=\lambda^{l+1}_1
= λ 1 l λ 2 − 1 λ 3 − 1 ... λ l + 1 − 1 = λ 1 l + 1 所以从 P(R) 到 k ∗ k^* k ∗ ζ \zeta ζ ζ ∈ q ( T P ) \zeta \in q(T_P) ζ ∈ q ( T P ) φ ( d ) ∈ f ( q ( T P ) ) \varphi (d)\in f(q(T_P)) φ ( d ) ∈ f ( q ( T P ))
回想一下（§5, no. 3，第 5 条命题第 2 条），对于 l = 1 l= 1 l = 1 g \mathfrak{g} g = = = 0 ( g ) _0(\mathfrak{g}) 0 ( g ) l ≥ 2 l\geq 2 l ≥ 2 g \mathfrak{g} g / / / 0 ( g ) _0(\mathfrak{g}) 0 ( g ) Z / 2 Z \mathbf{Z}/2\mathbf{Z} Z /2 Z l ≥ 2 l\geq 2 l ≥ 2 θ : x → − t x \theta :x \rightarrow -^tx θ : x → − t x s l ( l + 1 , k ) \mathfrak{s}\mathfrak{l}(l+ 1, k) sl ( l + 1 , k ) a 0 = θ ∣ h ∉ W a_0=\theta |\mathfrak{h}\notin W a 0 = θ ∣ h ∈ / W a 0 a_0 a 0 g \mathfrak{g} g / / / 0 ( g ) _0(\mathfrak{g}) 0 ( g )
杀戮形式的(VIII)限制到h \mathfrak{h} h
Φ ( ξ 1 E 11 + ⋯ + ξ l + 1 E l + 1 , l + 1 , ξ 1 ′ E 11 + ⋯ + ξ l + 1 ′ E l + 1 , l + 1 )
\Phi (\xi_1E_{11}+\cdots +\xi_{l+1}E_{l+1,l+1}, \xi '_1E_{11}+\cdots +\xi '_{l+1}E_{l+1,l+1})
Φ ( ξ 1 E 11 + ⋯ + ξ l + 1 E l + 1 , l + 1 , ξ 1 ′ E 11 + ⋯ + ξ l + 1 ′ E l + 1 , l + 1 ) = ∑ i ≠ j ( ξ i − ξ j ) ( ξ i ′ − ξ j ′ ) = ∑ i , j ( ξ i − ξ j ) ( ξ i ′ − ξ j ′ )
=\sum_{i\not=j}(\xi_i-\xi_j)(\xi '_i-\xi '_j) =\sum_{i,j}(\xi_i-\xi_j)(\xi '_i-\xi '_j)
= i = j ∑ ( ξ i − ξ j ) ( ξ i ′ − ξ j ′ ) = i , j ∑ ( ξ i − ξ j ) ( ξ i ′ − ξ j ′ ) = ( l + 1 ) ∑ i ξ i ξ i ′ + ( l + 1 ) ∑ j ξ j ξ j ′ − 2 ( ∑ i ξ i ) ( ∑ j ξ j ′
= (l+ 1)\sum_i\xi_i\xi '_i+ (l+ 1)\sum_j\xi_j\xi '_j-2(\sum_i\xi_i)(\sum_j\xi '_j
= ( l + 1 ) i ∑ ξ i ξ i ′ + ( l + 1 ) j ∑ ξ j ξ j ′ − 2 ( i ∑ ξ i ) ( j ∑ ξ j ′ = 2 ( l + 1 ) ∑ i ξ i ξ i ′
= 2(l+ 1)\sum_i\xi_i\xi '_i
= 2 ( l + 1 ) i ∑ ξ i ξ i ′ (IX) 对于 1 ≤ i < j ≤ l + 1 1\leq i < j\leq l+ 1 1 ≤ i < j ≤ l + 1
X ε i − ε j = E i j X ε j − ε i = − E j i
X_{\varepsilon_i-\varepsilon_j}=E_{ij}X_{\varepsilon_j-\varepsilon_i}=-E_{ji}
X ε i − ε j = E ij X ε j − ε i = − E j i 然后，对于所有α ∈ R \alpha \in R α ∈ R [ X α , X − α ] = − H α [X_{\alpha}, X_{-\alpha}] =-H_{\alpha} [ X α , X − α ] = − H α θ ( X α ) = X − α \theta (X_{\alpha}) =X_{-\alpha} θ ( X α ) = X − α θ \theta θ x → − t x x \rightarrow -^tx x → − t x ( X α ) α ∈ R (X_{\alpha})_{\alpha\in R} ( X α ) α ∈ R ( g , h ) (\mathfrak{g},\mathfrak{h}) ( g , h )
取k = Q k=\mathbf{Q} k = Q h ( \mathfrak{h}( h ( E i i − E i + 1 , i + 1 E_{ii}-E_{i+1,i+1} E ii − E i + 1 , i + 1 Z \mathbf{Z} Z Q ( R ∨ ) Q(R^{\vee}) Q ( R ∨ ) s l ( l + 1 , Z ) \mathfrak{s}\mathfrak{l}(l+ 1,\mathbf{Z}) sl ( l + 1 , Z ) Z \mathbf{Z} Z P ( R ∨ ) P(R^{\vee}) P ( R ∨ ) Q ( R ∨ ) Q(R^{\vee}) Q ( R ∨ ) E 11 − ( l + 1 ) − 1 ∑ E i i E_{11}-(l+ 1)^{-1}\sum E_{ii} E 11 − ( l + 1 ) − 1 ∑ E ii x + ( l + 1 ) − 1 a .1 x+ (l+ 1)^{-1}a.1 x + ( l + 1 ) − 1 a .1 x x x a ∈ Z a\in \mathbf{Z} a ∈ Z s l ( l + 1 , Z ) \mathfrak{s}\mathfrak{l}(l+1,\mathbf{Z}) sl ( l + 1 , Z ) ( g , h ) (\mathfrak{g},\mathfrak{h}) ( g , h ) Q ( R ∨ ) Q(R^{\vee}) Q ( R ∨ ) ( X α ) (X_{\alpha}) ( X α ) ⋀ r Z l + 1 \bigwedge^r\mathbf{Z}^{l+1} ⋀ r Z l + 1 ⋀ r Q l + 1 \bigwedge^r\mathbf{Q}^{l+1} ⋀ r Q l + 1 s l ( l + 1 , Z ) ( \mathfrak{s}\mathfrak{l}(l+ 1,\mathbf{Z}) ( sl ( l + 1 , Z ) (
另一方面，g l ( l + 1 , Z ) \mathfrak{g}\mathfrak{l}(l+ 1,\mathbf{Z}) gl ( l + 1 , Z ) g l ( l + 1 , Q ) \mathfrak{g}\mathfrak{l}(l+ 1,\mathbf{Q}) gl ( l + 1 , Q ) s l ( l + 1 , Q ) \mathfrak{s}\mathfrak{l}(l+ 1,\mathbf{Q}) sl ( l + 1 , Q ) g l ( l + 1 , Q ) \mathfrak{g}\mathfrak{l}(l+1,\mathbf{Q}) gl ( l + 1 , Q ) Q .1 \mathbf{Q}.1 Q .1 ( g , h ) (\mathfrak{g},\mathfrak{h}) ( g , h ) h \mathfrak{h} h P ( R ∨ ) P(R^{\vee}) P ( R ∨ ) ( X α ) (X_{\alpha}) ( X α ) g l ( l + 1 , Z ) \mathfrak{g}\mathfrak{l}(l+ 1,\mathbf{Z}) gl ( l + 1 , Z ) s l ( l + 1 , Q ) \mathfrak{s}\mathfrak{l}(l+ 1,\mathbf{Q}) sl ( l + 1 , Q ) g l ( l + 1 , Q ) \mathfrak{g}\mathfrak{l}(l+ 1,\mathbf{Q}) gl ( l + 1 , Q )
2. 代数型 Bl _{\boldsymbol{l}} l l ≥ \boldsymbol{l}\geq l ≥
(I) 令V为有限维向量空间，而Ψ \Psi Ψ x x x Ψ ( x v , v ′ ) + Ψ ( v , x v ′ ) = 0 \Psi (xv, v') +\Psi (v, xv') = 0 Ψ ( xv , v ′ ) + Ψ ( v , x v ′ ) = 0 v , v ′ ∈ V v, v'\in V v , v ′ ∈ V s l \mathfrak{s}\mathfrak{l} sl V ≠ 2 V\not= 2 V = 2 o ( Ψ ) \mathfrak{o}(\Psi ) o ( Ψ ) Ψ \Psi Ψ
假设V是奇数维数2 l + 1 ≥ 3 2l+ 1\geq 3 2 l + 1 ≥ 3 Ψ \Psi Ψ l l l Ψ \Psi Ψ x ∈ V x\in V x ∈ V Q ( x ) = 1 2 Ψ ( x , x ) Q(x) =\frac{1}{2}\Psi (x, x) Q ( x ) = 2 1 Ψ ( x , x ) F ′ F' F ′ F + F ′ F + F' F + F ′ Ψ \Psi Ψ e 0 ∈ G e_0\in G e 0 ∈ G Ψ ( e 0 , e 0 ) = − 2 \Psi (e_0, e_0) =-2 Ψ ( e 0 , e 0 ) = − 2 F ′ F' F ′ Ψ \Psi Ψ ( e i ) 1 ≤ i ≤ l (e_i)_{1\leq i\leq l} ( e i ) 1 ≤ i ≤ l ( e − i ) 1 ≤ i ≤ l (e_{-i})_{1\leq i\leq l} ( e − i ) 1 ≤ i ≤ l F ′ F' F ′
( e 1 , . . . , e l , e 0 , e − l , . . . , e − 1 )
(e_1, . . . , e_l, e_0, e_{-l}, . . . , e_{-1})
( e 1 , ... , e l , e 0 , e − l , ... , e − 1 ) 是 V 中的基；我们有
Q ( ∑ x i e i ) = − x 0 2 + ∑ i = 1 i = l x i x − i
Q(\sum x_ie_i)=-x^2_0+\sum_{i=1}^{i=l}x_ix_{-i}
Q ( ∑ x i e i ) = − x 0 2 + i = 1 ∑ i = l x i x − i 相对于该基，Ψ \Psi Ψ 2 l + 1 2l+ 1 2 l + 1
0 0 ⋯ \cdots ⋯
0 0 s s s ⋯ \cdots ⋯
S = S= S = − 2 -2 − 2 , s = ,s= , s =
s s s ⋯ \cdots ⋯
1 0 ⋯ \cdots ⋯
其中s s s l l l 6 ^6 6 g = o ( Ψ ) \mathfrak{g}=\mathfrak{o}(\Psi ) g = o ( Ψ ) a a a 2 l + 1 2l+ 1 2 l + 1 o S ( 2 l + 1 , k ) \mathfrak{o}_S(2l+ 1, k) o S ( 2 l + 1 , k ) a = − S − 1 a S ( a=-S^{-1}aS( a = − S − 1 a S ( g \mathfrak{g} g
y y y x x x
其中 x x x y y y l l l A , B , C , D A, B, C, D A , B , C , D l l l B = − s t B s , C = − s t C s B=-s^tBs, C=-s^tCs B = − s t B s , C = − s t C s D = − s t A s D=-s^tAs D = − s t A s A → s t A s A \rightarrow s^tAs A → s t A s M l ( k ) \mathbf{M}_l(k) M l ( k )
昏暗g = 2 l + l 2 + 2 l ( l − 1 ) 2 = l ( 2 l + 1 ) \mathfrak{g}= 2l+l^2+ 2\frac{l(l-1)}{2}=l(2l+ 1) g = 2 l + l 2 + 2 2 l ( l − 1 ) = l ( 2 l + 1 )
令h \mathfrak{h} h g \mathfrak{g} g g \mathfrak{g} g
6 ^6 6 ( a i j ) 1 ≤ i , j ≤ n (a_{ij})_{1\leq i,j\leq n} ( a ij ) 1 ≤ i , j ≤ n a i j a_{ij} a ij
H i = E i , i − E − i , − i ( 1 ≤ i ≤ l )
H_i=E_{i,i}-E_{-i,-i}(1\leq i\leq l)
H i = E i , i − E − i , − i ( 1 ≤ i ≤ l ) 令( ε i ) (\varepsilon_i) ( ε i ) h ∗ \mathfrak{h}^* h ∗ ( H i ) (H_i) ( H i )
X ε i = 2 E i , 0 + E 0 , − i ( 1 ≤ i ≤ l )
X_{\varepsilon_i}= 2E_{i,0}+E_{0,-i}(1\leq i\leq l)
X ε i = 2 E i , 0 + E 0 , − i ( 1 ≤ i ≤ l ) X − ε i X_{-\varepsilon_i} X − ε i − 2 E − i , 0 − E 0 , i ( 1 ≤ i ≤ l ) -2E_{-i,0}-E_{0,i}(1\leq i\leq l) − 2 E − i , 0 − E 0 , i ( 1 ≤ i ≤ l )
X ε i − ε j X_{\varepsilon_i-\varepsilon_j} X ε i − ε j E i , j − E − j , − i ( 1 ≤ i < j ≤ l ) E_{i,j}-E_{-j,-i}(1\leq i < j\leq l) E i , j − E − j , − i ( 1 ≤ i < j ≤ l )
X ε j − ε i X_{\varepsilon_j-\varepsilon_i} X ε j − ε i − E j , i + E − i , − j ( 1 ≤ i < j ≤ l ) -E_{j,i}+E_{-i,-j}(1\leq i < j\leq l) − E j , i + E − i , − j ( 1 ≤ i < j ≤ l )
X ε i + ε j X_{\varepsilon_i+\varepsilon_j} X ε i + ε j E i , − j − E j , − i ( 1 ≤ i < j ≤ l ) E_{i,-j}-E_{j,-i}(1\leq i < j\leq l) E i , − j − E j , − i ( 1 ≤ i < j ≤ l )
X − ε i − ε j X_{-\varepsilon_i-\varepsilon_j} X − ε i − ε j − E − j , i + E − i , j ( 1 ≤ i < j ≤ l ) -E_{-j,i}+E_{-i,j}(1\leq i < j\leq l) − E − j , i + E − i , j ( 1 ≤ i < j ≤ l )
很容易验证这些元素形式基是g \mathfrak{g} g h \mathfrak{h} h h ∈ h h\in \mathfrak{h} h ∈ h
[ h , X α ] = α ( h ) X α (4)
[h, X_{\alpha}] =\alpha (h)X_{\alpha} \tag{4}
[ h , X α ] = α ( h ) X α ( 4 ) 对于所有α ∈ R \alpha \in R α ∈ R ± ε i \pm \varepsilon_i ± ε i ± ε i ± ε j ( 1 ≤ i < j ≤ l ) \pm \varepsilon_i\pm \varepsilon_j(1\leq i < j\leq l) ± ε i ± ε j ( 1 ≤ i < j ≤ l ) h \mathfrak{h} h g \mathfrak{g} g g \mathfrak{g} g h \mathfrak{h} h ( g , h ) (\mathfrak{g},\mathfrak{h}) ( g , h ) ( g , h ) (\mathfrak{g},\mathfrak{h}) ( g , h ) l ≥ 2 l\geq 2 l ≥ 2 B l B_l B l l = 1 l= 1 l = 1 A 1 A_1 A 1 B 1 B_1 B 1 l = 1 l= 1 l = 1 g \mathfrak{g} g B l B_l B l
o ( Ψ ) \mathfrak{o}(\Psi ) o ( Ψ ) h \mathfrak{h} h o ( Ψ ) \mathfrak{o}(\Psi ) o ( Ψ ) O ( Ψ ) \mathbf{O}(\Psi ) O ( Ψ ) g \mathfrak{g} g h β \mathfrak{h}_{\beta} h β β \beta β h β \mathfrak{h}_{\beta} h β β \beta β
如果 l = 1 l= 1 l = 1 o ( Ψ ) \mathfrak{o}(\Psi ) o ( Ψ ) s l ( 2 , k ) \mathfrak{s}\mathfrak{l}(2, k) sl ( 2 , k ) l ≥ 2 l\geq 2 l ≥ 2
(II) 根系 R ∨ R^{\vee} R ∨
H ε i = 2 H i , H ε i − ε j = H i − H j , H ε i + ε j = H i + H j
H_{\varepsilon_i}= 2H_i, H_{\varepsilon_i-\varepsilon_j}=H_i-H_j, H_{\varepsilon_i+\varepsilon_j}=H_i+H_j
H ε i = 2 H i , H ε i − ε j = H i − H j , H ε i + ε j = H i + H j (III) 放入α 1 = ε 1 − ε 2 , . . . , α l − 1 = ε l − 1 − ε l , α l = ε l \alpha_1=\varepsilon_1-\varepsilon_2, . . . , \alpha_{l-1}=\varepsilon_{l-1}-\varepsilon_l, \alpha_l=\varepsilon_l α 1 = ε 1 − ε 2 , ... , α l − 1 = ε l − 1 − ε l , α l = ε l ( α 1 , . . . , α l ) (\alpha_1, . . . , \alpha_l) ( α 1 , ... , α l ) ε i \varepsilon_i ε i ε i ± ε j ( i < j ) \varepsilon_i\pm \varepsilon_j(i < j) ε i ± ε j ( i < j ) b \mathfrak{b} b g \mathfrak{g} g
立即验证，V中唯一不同于{ 0 } \{0\} { 0 } b \mathfrak{b} b ( e i ) (e_i) ( e i ) V 1 , . . . , V l V_1, . . . ,V_l V 1 , ... , V l V i V_i V i e 1 , . . . , e i e_1, . . . , e_i e 1 , ... , e i V − 1 , . . . , V − i : V_{-1}, . . . ,V_{-i}: V − 1 , ... , V − i : V i V_i V i V − i V_{-i} V − i e 1 , . . . , e l , e 0 , e − l , . . . , e − i − 1 e_1, . . . , e_l, e_0, e_{-l}, . . . , e_{-i-1} e 1 , ... , e l , e 0 , e − l , ... , e − i − 1 g \mathfrak{g} g b \mathfrak{b} b g \mathfrak{g} g { V 1 , . . . , V l } \{V_1, . . . ,V_l\} { V 1 , ... , V l }
如果一个标志的每个元素完全各向同性，则该标志被称为各向同性。 标志{ V 1 , . . . , V l } \{V_1, . . . ,V_l\} { V 1 , ... , V l } O ( Ψ ) \mathbf{O}(\Psi ) O ( Ψ ) g \mathfrak{g} g δ \delta δ b δ \mathfrak{b}_{\delta} b δ g \mathfrak{g} g δ \delta δ g \mathfrak{g} g δ → b δ \delta \rightarrow \mathfrak{b}_{\delta} δ → b δ g \mathfrak{g} g
令δ \delta δ p δ \mathfrak{p}_{\delta} p δ g \mathfrak{g} g δ \delta δ δ ⊂ { V 1 , . . . , V l } \delta \subset \{V_1, . . . ,V_l\} δ ⊂ { V 1 , ... , V l } p δ \mathfrak{p}_{\delta} p δ g \mathfrak{g} g b \mathfrak{b} b p δ \mathfrak{p}_{\delta} p δ ≠ { 0 } \not=\{0\} = { 0 } δ \delta δ 2 l 2^l 2 l g \mathfrak{g} g b \mathfrak{b} b g \mathfrak{g} g δ → p δ \delta \rightarrow \mathfrak{p}_{\delta} δ → p δ p δ ⊂ p δ ′ \mathfrak{p}_{\delta}\subset \mathfrak{p}_{\delta'} p δ ⊂ p δ ′ δ ⊃ δ ′ \delta \supset \delta ' δ ⊃ δ ′
（IV）基本权相应的到α 1 , . . . , α l \alpha_1, . . . , \alpha_l α 1 , ... , α l
ϖ i = ε 1 + ⋯ + ε i ( 1 ≤ i ≤ l − 1 )
\varpi_i=\varepsilon_1+\cdots +\varepsilon_i(1\leq i\leq l-1)
ϖ i = ε 1 + ⋯ + ε i ( 1 ≤ i ≤ l − 1 ) ϖ l = 1 2 ( ε 1 + ⋯ + ε l )
\varpi_l=\frac{1}{2}(\varepsilon_1+\cdots +\varepsilon_l)
ϖ l = 2 1 ( ε 1 + ⋯ + ε l ) 令σ \sigma σ g \mathfrak{g} g ⋀ r σ \bigwedge^r\sigma ⋀ r σ E = ⋀ r V E =\bigwedge^rV E = ⋀ r V h ∈ h h\in \mathfrak{h} h ∈ h
σ ( h ) . e i = ε i ( h ) e i \sigma (h).e_i=\varepsilon_i(h)e_i σ ( h ) . e i = ε i ( h ) e i 1 ≤ i ≤ l 1\leq i\leq l 1 ≤ i ≤ l σ ( h ) . e 0 = 0
\sigma (h).e_0= 0
σ ( h ) . e 0 = 0 σ ( h ) . e − i = − ε i ( h ) e − i \sigma (h).e_{-i}=-\varepsilon_i(h)e_{-i} σ ( h ) . e − i = − ε i ( h ) e − i 1 ≤ i ≤ l 1\leq i\leq l 1 ≤ i ≤ l
由此可见，对于1 ≤ r ≤ l , ε 1 + ⋯ + ε r 1\leq r\leq l,\varepsilon_1+\cdots +\varepsilon_r 1 ≤ r ≤ l , ε 1 + ⋯ + ε r ⋀ r σ \bigwedge^r\sigma ⋀ r σ ε 1 + ⋯ + ε r \varepsilon_1+\cdots +\varepsilon_r ε 1 + ⋯ + ε r e 1 ∧ ⋯ ∧ e_1\wedge \cdots \wedge e 1 ∧ ⋯ ∧ e r e_r e r 1 ≤ r ≤ l − 1 1\leq r\leq l-1 1 ≤ r ≤ l − 1 ⋀ r σ \bigwedge^r\sigma ⋀ r σ ϖ r \varpi_r ϖ r g \mathfrak{g} g ⋀ r σ \bigwedge^r\sigma ⋀ r σ 0 ≤ r ≤ 2 l + 1 0\leq r\leq 2l+ 1 0 ≤ r ≤ 2 l + 1 Φ \Phi Φ ⋀ r V × ⋀ 2 l + 1 − r V \bigwedge^rV\times \bigwedge^{2l+1-r}V ⋀ r V × ⋀ 2 l + 1 − r V
x ∧ y = Φ ( x , y ) e 1 ∧ ⋯ ∧ e l ∧ e 0 ∧ e − l ∧ ⋯ ∧ e − 1
x\wedge y=\Phi (x, y)e_1\wedge \cdots \wedge e_l\wedge e_0\wedge e_{-l}\wedge \cdots \wedge e_{-1}
x ∧ y = Φ ( x , y ) e 1 ∧ ⋯ ∧ e l ∧ e 0 ∧ e − l ∧ ⋯ ∧ e − 1 在g \mathfrak{g} g ⋀ r V \bigwedge^rV ⋀ r V ⋀ 2 l + 1 − r V \bigwedge^{2l+1-r}V ⋀ 2 l + 1 − r V ⋀ 2 l + 1 − r σ \bigwedge^{2l+1-r}\sigma ⋀ 2 l + 1 − r σ ⋀ r σ \bigwedge^r\sigma ⋀ r σ ⋀ r σ \bigwedge^r\sigma ⋀ r σ 0 ≤ r ≤ l 0\leq r\leq l 0 ≤ r ≤ l ⋀ r V \bigwedge^rV ⋀ r V e 1 ∧ ⋯ ∧ e r e_1\wedge \cdots \wedge e_r e 1 ∧ ⋯ ∧ e r T r T_r T r g \mathfrak{g} g ⋀ r V \bigwedge^rV ⋀ r V r = 0 r= 0 r = 0 r = 1 r= 1 r = 1 r = 2 r= 2 r = 2 l ≥ 2 l\geq 2 l ≥ 2 ⋀ 2 σ \bigwedge^2\sigma ⋀ 2 σ g \mathfrak{g} g l ( 2 l + 1 ) l(2l+ 1) l ( 2 l + 1 ) ε 1 + ε 2 \varepsilon_1+\varepsilon_2 ε 1 + ε 2 ⋀ 2 σ \bigwedge^2\sigma ⋀ 2 σ l = 1 l= 1 l = 1 l = 2 l= 2 l = 2
我们现在通过归纳对l l l l ≥ r ≥ 3 l\geq r\geq 3 l ≥ r ≥ 3 W ′ W' W ′ Ψ \Psi Ψ Ψ W \Psi_W Ψ W o ( Ψ W ) \mathfrak{o}(\Psi_W) o ( Ψ W ) g \mathfrak{g} g W ′ W' W ′ W < W< W < Ψ W \Psi_W Ψ W T r T_r T r w ′ ∧ w w'\wedge w w ′ ∧ w w ′ ∈ ⋀ r − k W ′ w'\in \bigwedge^{r-k}W' w ′ ∈ ⋀ r − k W ′ w ∈ ⋀ k W ( 0 ≤ k ≤ r ) w\in \bigwedge^kW (0\leq k\leq r) w ∈ ⋀ k W ( 0 ≤ k ≤ r ) T r T_r T r w ′ ∧ ⋀ k W : w'\wedge \bigwedge^kW: w ′ ∧ ⋀ k W : a ∈ o ( Ψ W ) a\in \mathfrak{o}(\Psi_W) a ∈ o ( Ψ W ) a . ( w ′ ∧ w ) = w ′ ∧ a . w a.(w'\wedge w) =w'\wedge a.w a . ( w ′ ∧ w ) = w ′ ∧ a . w p ∈ 0 , r p\in 0, r p ∈ 0 , r T r T_r T r
x = e i 1 ∧ ⋯ ∧ e i r − p ∧ e j 1 ∧ ⋯ ∧ e j p
x=e_{i_1}\wedge \cdots \wedge e_{i_{r-p}}\wedge e_{j_1}\wedge \cdots \wedge e_{j_p}
x = e i 1 ∧ ⋯ ∧ e i r − p ∧ e j 1 ∧ ⋯ ∧ e j p 对于 1 ≤ i 1 < ⋯ < i r − p ≤ l 1\leq i_1<\cdots < i_{r-p}\leq l 1 ≤ i 1 < ⋯ < i r − p ≤ l − l ≤ j 1 < ⋯ < j p ≤ 0 -l\leq j_1<\cdots < j_p\leq 0 − l ≤ j 1 < ⋯ < j p ≤ 0 p = 0 p= 0 p = 0 g l ( F ) \mathfrak{g}\mathfrak{l}(F) gl ( F ) ⋀ r F \bigwedge^rF ⋀ r F
g \mathfrak{g} g F = V l = ∑ i = 1 l k e i F = V_l=\sum_{i=1}^lke_i F = V l = ∑ i = 1 l k e i p = 1 p= 1 p = 1 q ∈ ( 1 , l ) q\in (1, l) q ∈ ( 1 , l ) q ≠ − j 1 q\not=-j_1 q = − j 1 λ ∈ 1 , r − p \lambda \in 1, r-p λ ∈ 1 , r − p q = i λ q=i_{\lambda} q = i λ p ≥ 2 p\geq 2 p ≥ 2 q ∈ 1 , l q\in 1, l q ∈ 1 , l − q ∈ { j 1 , . . . , j p } -q\in \{j_1, . . . , j_p\} − q ∈ { j 1 , ... , j p } e i e_i e i q = 1 q= 1 q = 1 W ′ = k e 1 + k e − 1 W'=ke_1+ke_{-1} W ′ = k e 1 + k e − 1 p = 1 p= 1 p = 1 x ∈ e 1 ∧ ⋀ r − 1 W x\in e_1\wedge \bigwedge^{r-1}W x ∈ e 1 ∧ ⋀ r − 1 W T r T_r T r e 1 ∧ ⋯ ∧ e r e_1\wedge \cdots \wedge e_r e 1 ∧ ⋯ ∧ e r T r T_r T r x x x p ≥ 2 p\geq 2 p ≥ 2 x ∈ e − 1 ∧ ⋀ r − 1 W x\in e_{-1}\wedge \bigwedge^{r-1}W x ∈ e − 1 ∧ ⋀ r − 1 W x ∈ e 1 ∧ e − 1 ∧ ⋀ r − 2 W x\in e_1\wedge e_{-1}\wedge \bigwedge^{r-2}W x ∈ e 1 ∧ e − 1 ∧ ⋀ r − 2 W T r T_r T r e − 1 ∧ e 2 ∧ ⋯ ∧ e r − 1 e_{-1}\wedge e_2\wedge \cdots \wedge e_{r-1} e − 1 ∧ e 2 ∧ ⋯ ∧ e r − 1 e − 1 ∧ e 1 ∧ e 2 ∧ ⋯ ∧ e r − 2 e_{-1}\wedge e_1\wedge e_2\wedge \cdots \wedge e_{r-2} e − 1 ∧ e 1 ∧ e 2 ∧ ⋯ ∧ e r − 2 T r T_r T r x x x
对于 ⋀ r σ \bigwedge^r\sigma ⋀ r σ
我们现在将确定基本表示与最高权ϖ l \varpi_l ϖ l
令V为有限维向量空间，Q为V, Ψ ,\Psi , Ψ , f 0 ,f_0 , f 0
o ( Ψ ) ⟶ g l ( V ) ⟶ V ⊗ V ∗ ⟶ V ⊗ V ⟶ C + ( Q )
\mathfrak{o}(\Psi )\longrightarrow \mathfrak{g}\mathfrak{l}(V)\longrightarrow V\otimes V^*\longrightarrow V\otimes V\longrightarrow C^+(Q)
o ( Ψ ) ⟶ gl ( V ) ⟶ V ⊗ V ∗ ⟶ V ⊗ V ⟶ C + ( Q ) 引用了 8 · 被以下引用 3 引用了 VI § 4. 。 根系的分类 , 习题 18, § 13 , 定理 1, VIII § 8 , 命题 12, VIII § 7 , 命题 2, VIII § 7 , 命题 1, VIII § 6 , 命题 3, VIII § 11 , 习题 7, § 13
被以下引用 § 13. 。经典可拆分简单李代数 , 习题 18, § 13 , 习题 19, § 13
（第一个是典范嵌入，第三个由V ∗ V^* V ∗ Ψ \Psi Ψ f = 1 2 f 0 f=\frac{1}{2}f_0 f = 2 1 f 0
(i) 如果 ( e r ) , ( e r ′ ) (e_r),(e'_r) ( e r ) , ( e r ′ ) Ψ ( e r , e s ′ ) = δ r s \Psi (e_r, e'_s) =\delta_{rs} Ψ ( e r , e s ′ ) = δ r s a ∈ o ( Ψ ) a\in \mathfrak{o}(\Psi ) a ∈ o ( Ψ ) f 0 ( a ) = f_0(a) = f 0 ( a ) = ∑ r ( a e r ) e r ′ \sum_r(ae_r)e'_r ∑ r ( a e r ) e r ′
(ii) 如果a , b ∈ o ( Ψ ) a, b\in \mathfrak{o}(\Psi ) a , b ∈ o ( Ψ ) ∑ r ( a e r ) ( b e r ′ ) = − ∑ r ( a b e r ) e r ′ \sum_r(ae_r)(be'_r) =-\sum_r(abe_r)e'_r ∑ r ( a e r ) ( b e r ′ ) = − ∑ r ( ab e r ) e r ′
(iii) 如果a ∈ o ( Ψ ) a\in \mathfrak{o}(\Psi ) a ∈ o ( Ψ ) v ∈ V v\in V v ∈ V [ f ( a ) , v ] = a v [f(a), v] =av [ f ( a ) , v ] = a v
(iv) 如果a , b ∈ o ( Ψ ) a, b\in \mathfrak{o}(\Psi ) a , b ∈ o ( Ψ ) [ f ( a ) , f ( b ) ] = f ( [ a , b ] ) [f(a), f(b)] =f([a, b]) [ f ( a ) , f ( b )] = f ([ a , b ])
(v) f ( o ( Ψ ) ) f(\mathfrak{o}(\Psi )) f ( o ( Ψ )) C + ( Q ) C^+(Q) C + ( Q )
(vi) 设 N 为左 C + ( Q ) C^+(Q) C + ( Q ) C + ( Q ) C^+(Q) C + ( Q ) k ( N ) _k(N) k ( N ) ρ \rho ρ ρ ∘ f \rho \circ f ρ ∘ f o ( Ψ ) \mathfrak{o}(\Psi ) o ( Ψ ) ρ ∘ f \rho \circ f ρ ∘ f
断言 (i) 很清楚。如果a , b ∈ o ( Ψ ) a, b\in \mathfrak{o}(\Psi ) a , b ∈ o ( Ψ ) Ψ ( x , y ) = ⟨ x , y ⟩ \Psi (x, y) =\langle x, y\rangle Ψ ( x , y ) = ⟨ x , y ⟩ : : :
∑ r ( a e r ) ( b e r ′ ) = ∑ r , s , t ⟨ a e r , e s ′ ⟩ ⟨ b e r ′ , e t ⟩ e s e t ′ = ∑ r , s , t ⟨ e r , a e s ′ ⟩ ⟨ e r ′ , b e t ⟩ e s e t ′
\sum_r(ae_r)(be'_r) =\sum_{r,s,t}\langle ae_r, e'_s\rangle \langle be'_r, e_t\rangle e_se'_t=\sum_{r,s,t}\langle e_r, ae'_s\rangle \langle e'_r, be_t\rangle e_se'_t
r ∑ ( a e r ) ( b e r ′ ) = r , s , t ∑ ⟨ a e r , e s ′ ⟩ ⟨ b e r ′ , e t ⟩ e s e t ′ = r , s , t ∑ ⟨ e r , a e s ′ ⟩ ⟨ e r ′ , b e t ⟩ e s e t ′ = ∑ s , t ⟨ a e s ′ , b e t ⟩ e s e t ′ = − ∑ s , t ⟨ e s ′ , a b e t ⟩ e s e t ′ = − ∑ t ( a b e t ) e t ′
=\sum_{s,t}\langle ae'_s, be_t\rangle e_se'_t=-\sum_{s,t}\langle e'_s, abe_t\rangle e_se'_t=-\sum_t(abe_t)e'_t
= s , t ∑ ⟨ a e s ′ , b e t ⟩ e s e t ′ = − s , t ∑ ⟨ e s ′ , ab e t ⟩ e s e t ′ = − t ∑ ( ab e t ) e t ′ 这就证明了(ii)。接下来，对于所有 v ∈ V v\in V v ∈ V
1 ′ ′ '' ′′
[ f ( a ) , v ] = ∑ ( ( a e r ) e r v − v ( a e r ) e r )
[f(a), v] =\sum((ae_r)e_rv-v(ae_r)e_r)
[ f ( a ) , v ] = ∑ (( a e r ) e r v − v ( a e r ) e r ) 2
= 1 ∑ r ( ( a e r ) e r ′ v + ( a e r ) v e r ′ − ( a e r ) v e r ′ − v ( a e r ) e r ′ ) 1\sum^r((ae_r)e'_rv+ (ae_r)ve'_r-(ae_r)ve'_r-v(ae_r)e'_r) 1 ∑ r (( a e r ) e r ′ v + ( a e r ) v e r ′ − ( a e r ) v e r ′ − v ( a e r ) e r ′ )
2
r r r 1 ′ ′ '' ′′
= ∑ ( ( a e r ) ⟨ e r , v ⟩ − ⟨ a e r , v ⟩ e r ) \sum((ae_r)\langle e_r, v\rangle - \langle ae_r, v\rangle e_r) ∑ (( a e r ) ⟨ e r , v ⟩ − ⟨ a e r , v ⟩ e r )
2
= 1 a ( ∑ r ⟨ e r ′ , v ⟩ e r ) + 1 ∑ ⟨ e r , a v ⟩ e r ′ = 1 a v + 1 a v = a v 1a(\sum^r\langle e'_r, v\rangle e_r)+1\sum\langle e_r, av\rangle e'_r=1av+1av=av 1 a ( ∑ r ⟨ e r ′ , v ⟩ e r ) + 1 ∑ ⟨ e r , a v ⟩ e r ′ = 1 a v + 1 a v = a v
2 r _r r r _r r
这就证明了(iii)。然后
[ f ( a ) , f ( b ) ] = [ f ( a ) , 1 2 ∑ ( b e r ) e r ′ ] [f(a), f(b)] =[f(a),\frac{1}{2}\sum(be_r)e'_r] [ f ( a ) , f ( b )] = [ f ( a ) , 2 1 ∑ ( b e r ) e r ′ ]
r r r = 1 2 ∑ ( [ f ( a ) , b e r ] e r ′ + ( b e r ) [ f ( a ) , e r ′ ] ) =\frac{1}{2}\sum([f(a), be_r]e'_r+ (be_r)[f(a), e'_r]) = 2 1 ∑ ([ f ( a ) , b e r ] e r ′ + ( b e r ) [ f ( a ) , e r ′ ]) = 1 2 ∑ r ( ( a b e r ) e r ′ + ( b e r ) ( a e r ′ ) ) =\frac{1}{2}\sum^r((abe_r)e'_r+ (be_r)(ae'_r)) = 2 1 ∑ r (( ab e r ) e r ′ + ( b e r ) ( a e r ′ ))
r r r = 1 2 ∑ ( ( a b e r ) e r ′ − ( b a e r ) e r ′ ) =\frac{1}{2}\sum((abe_r)e'_r-(bae_r)e'_r) = 2 1 ∑ (( ab e r ) e r ′ − ( ba e r ) e r ′ )
r r r = f ( [ a , b ] ) =f([a, b]) = f ([ a , b ])
这就证明了(iv)。为了证明(v)，我们可以通过扩展标量，假设k k k ( e r ) (e_r) ( e r ) Ψ ( e r , e s ) = δ r s \Psi (e_r, e_s) =\delta_{rs} Ψ ( e r , e s ) = δ r s e r ′ = e r e'_r=e_r e r ′ = e r i ≠ j i\not=j i = j E i j − E j i ∈ o ( Ψ ) E_{ij}-E_{ji}\in \mathfrak{o}(\Psi ) E ij − E j i ∈ o ( Ψ )
f ( E i j − E j i ) = 1 2 ( e i e j − e j e i ) = e i e j
f(E_{ij}-E_{ji}) =\frac{1}{2}(e_ie_j-e_je_i) =e_ie_j
f ( E ij − E j i ) = 2 1 ( e i e j − e j e i ) = e i e j 但是e i e j e_ie_j e i e j C + ( Q ) C^+(Q) C + ( Q )
断言 (vi) 由 (iv) 和 (v) 得出。 Q.E.D.
现在回想一下在这个数开头使用的记号。将V ~ = F + F ′ \widetilde{V} = F + F' V = F + F ′ Q ~ \widetilde{Q} Q Ψ ~ \widetilde{\Psi} Ψ Ψ \Psi Ψ V ~ \widetilde{V} V Q ~ \widetilde{Q} Q 2 l 2l 2 l V ~ \widetilde{V} V l l l C ( Q ~ ) C( \widetilde{Q}) C ( Q ) 2 2 l ( 2^{2l}( 2 2 l ( e − 1 , . . . , e − l e_{-1}, . . . , e_{-l} e − 1 , ... , e − l F ′ F' F ′ F ′ F' F ′ Ψ \Psi Ψ x ∈ F ′ x\in F' x ∈ F ′ y ∈ F y\in F y ∈ F λ ( x ) \lambda (x) λ ( x ) λ ( y ) \lambda (y) λ ( y ) x x x y y y a 1 , . . . , a k ∈ F ′ a_1, . . . , a_k\in F' a 1 , ... , a k ∈ F ′
λ ( x ) . ( a 1 ∧ ⋯ ∧ a k ) = x ∧ a 1 ∧ ⋯ ∧ a k
\lambda (x).(a_1\wedge \cdots \wedge a_k) =x\wedge a_1\wedge \cdots \wedge a_k
λ ( x ) . ( a 1 ∧ ⋯ ∧ a k ) = x ∧ a 1 ∧ ⋯ ∧ a k λ ( y ) . ( a 1 ∧ ⋯ ∧ a k ) = ∑ i = 1 k ( − 1 ) i − 1 Ψ ( a i , y ) a 1 ∧ ⋯ ∧ a i − 1 ∧ a i + 1 ∧ ⋯ ∧ a k
\lambda (y).(a_1\wedge \cdots \wedge a_k) =\sum_{i=1}^k(-1)^{i-1}\Psi (a_i, y)a_1\wedge \cdots \wedge a_{i-1}\wedge a_{i+1}\wedge \cdots \wedge a_k
λ ( y ) . ( a 1 ∧ ⋯ ∧ a k ) = i = 1 ∑ k ( − 1 ) i − 1 Ψ ( a i , y ) a 1 ∧ ⋯ ∧ a i − 1 ∧ a i + 1 ∧ ⋯ ∧ a k 很容易验证 λ ( x ) 2 = λ ( y ) 2 = 0 \lambda (x)^2=\lambda (y)^2= 0 λ ( x ) 2 = λ ( y ) 2 = 0
λ ( x ) λ ( y ) + λ ( y ) λ ( x ) = Ψ ( x , y ) .1
\lambda (x)\lambda (y) +\lambda (y)\lambda (x) =\Psi (x, y).1
λ ( x ) λ ( y ) + λ ( y ) λ ( x ) = Ψ ( x , y ) .1 由此可见（代数，IX, §9, no. 1），存在从C ( Q ~ ) C( \widetilde{Q}) C ( Q ) λ : F ∪ F ′ → \lambda : F\cup F'\rightarrow λ : F ∪ F ′ → λ \lambda λ N = 2 l N = 2^l N = 2 l C ( Q ~ ) C( \widetilde{Q}) C ( Q ) 2 l ( 2^l( 2 l ( λ \lambda λ C ( Q ~ ) C( \widetilde{Q}) C ( Q ) C ( Q ~ ) ( C( \widetilde{Q}) ( C ( Q ) (
现在考虑映射μ : v → e 0 v \mu:v \rightarrow e_0v μ : v → e 0 v V ~ \widetilde{V} V C + ( Q ) C^+(Q) C + ( Q ) v ∈ V ~ v\in \widetilde{V} v ∈ V
( e 0 v ) 2 = − e 0 2 v 2 = − Q ( e 0 ) Q ( v ) = Q ( v ) = Q ~ ( v )
(e_0v)^2=-e^2_0v^2=-Q(e_0)Q(v) = Q(v) = \widetilde{Q}(v)
( e 0 v ) 2 = − e 0 2 v 2 = − Q ( e 0 ) Q ( v ) = Q ( v ) = Q ( v ) μ \mu μ μ \mu μ C ( Q ~ ) C( \widetilde{Q}) C ( Q ) C + ( Q ) C^+(Q) C + ( Q ) C ( Q ~ ) C( \widetilde{Q}) C ( Q )
昏暗C + ( Q ) = C^+(Q) = C + ( Q ) = C ( Q ~ ) = 2 2 l C( \widetilde{Q}) = 2^{2l} C ( Q ) = 2 2 l
我们看到μ \mu μ λ ∘ μ − 1 \lambda \circ \mu^{-1} λ ∘ μ − 1 C + C^+ C + ρ = λ ∘ μ − 1 ∘ f \rho =\lambda \circ \mu^{-1}\circ f ρ = λ ∘ μ − 1 ∘ f g \mathfrak{g} g
另一方面，鉴于引理 1 (i)，我们有
f ( H i ) = 1 2 ( e i e − i − e − i e i )
f(H_i) =\frac{1}{2}(e_ie_{-i}-e_{-i}e_i)
f ( H i ) = 2 1 ( e i e − i − e − i e i ) 从e i e − i = − e 0 2 e i e − i = e 0 e i e 0 e − i e_ie_{-i}=-e^2_0e_ie_{-i}=e_0e_ie_0e_{-i} e i e − i = − e 0 2 e i e − i = e 0 e i e 0 e − i e i e − i + e − i e i = 1 e_ie_{-i}+e_{-i}e_i= 1 e i e − i + e − i e i = 1
− 1 _{-1} − 1
μ ∘ f ( H i ) = − e − i e i = − + e i e − i
\mu\circ f(H_i) =-e_{-i}e_i=-+e_ie_{-i}
μ ∘ f ( H i ) = − e − i e i = − + e i e − i 2 2 我们推断，对于 1 ≤ i 1 < ⋯ < i k ≤ l : 1\leq i_1<\cdots < i_k\leq l: 1 ≤ i 1 < ⋯ < i k ≤ l :
ρ ( H ) ( e ∧ ⋯ ∧ e ) = { − 1 2 e − i 1 ∧ ⋯ ∧ e − i k \rho (H)(e\wedge \cdots \wedge e) =\{-\frac{1}{2}e_{-i_1}\wedge \cdots \wedge e_{-i_k} ρ ( H ) ( e ∧ ⋯ ∧ e ) = { − 2 1 e − i 1 ∧ ⋯ ∧ e − i k i ∈ { i 1 , . . . , i k } i\in \{i_1, . . . , i_k\} i ∈ { i 1 , ... , i k }
i − i 1 − i k 1 2 e − i 1 ∧ ⋯ ∧ e − i k ^{i-i_1}-_{i_k}\frac{1}{2}e_{-i_1}\wedge \cdots \wedge e_{-i_k} i − i 1 − i k 2 1 e − i 1 ∧ ⋯ ∧ e − i k i ∉ { i 1 , . . . i k } i \notin \{i_1, . . . i_k\} i ∈ / { i 1 , ... i k }
h ∈ h h\in \mathfrak{h} h ∈ h ρ ( h ) ( e − i 1 ∧ ⋯ ∧ e − i k ) (5)
\rho (h)(e_{-i_1}\wedge \cdots \wedge e_{-i_k}) \tag{5}
ρ ( h ) ( e − i 1 ∧ ⋯ ∧ e − i k ) ( 5 ) = ( 1 2 ( ε 1 + ⋯ + ε l ) − ( ε i 1 + ⋯ + ε i k ) ) ( h ) ( e − i 1 ∧ ⋯ ∧ e − i k )
= (\frac{1}{2}(\varepsilon_1+\cdots +\varepsilon_l)-(\varepsilon_{i_1}+\cdots +\varepsilon_{i_k}))(h)(e_{-i_1}\wedge \cdots \wedge e_{-i_k})
= ( 2 1 ( ε 1 + ⋯ + ε l ) − ( ε i 1 + ⋯ + ε i k )) ( h ) ( e − i 1 ∧ ⋯ ∧ e − i k ) 这表明ρ \rho ρ ϖ l \varpi_l ϖ l ρ \rho ρ g \mathfrak{g} g ϖ l \varpi_l ϖ l
(V) 我们有 w 0 = − 1 w_0=-1 w 0 = − 1 g \mathfrak{g} g ϖ r \varpi_r ϖ r ( α 1 , . . . , α l ) (\alpha_1, . . . , \alpha_l) ( α 1 , ... , α l ) 1 ≤ r ≤ l − 1 : 1\leq r\leq l-1: 1 ≤ r ≤ l − 1 : ⋀ r σ \bigwedge^r\sigma ⋀ r σ Ψ \Psi Ψ ⋀ r V \bigwedge^rV ⋀ r V Ψ ( r ) \Psi_{(r)} Ψ ( r )
对于旋量表示，ϖ l \varpi_l ϖ l ( α 1 , . . . , α l ) (\alpha_1, . . . , \alpha_l) ( α 1 , ... , α l ) 1 2 ( 1 + ⋯ + l ) = l ( l + 1 ) 4 ( \frac{1}{2}(1 +\cdots +l) =\frac{l(l+1)}{4}( 2 1 ( 1 + ⋯ + l ) = 4 l ( l + 1 ) ( l ≡ 0 l\equiv 0 l ≡ 0 − 1 -1 − 1 l ≡ 1 l\equiv 1 l ≡ 1 N = ⋀ F ′ N =\bigwedge F' N = ⋀ F ′ Φ \Phi Φ x ∈ ⋀ p F ′ x\in \bigwedge^pF' x ∈ ⋀ p F ′ y ∈ ⋀ q F ′ y\in \bigwedge^qF' y ∈ ⋀ q F ′ Φ ( x , y ) = 0 \Phi (x, y) = 0 Φ ( x , y ) = 0 p + q ≠ l p+q\not=l p + q = l
x ∧ y = ( − 1 ) 2 p ( p + 1 ) Φ ( x , y ) e − 1 ∧ ⋯ ∧ e − r
x\wedge y= (-1)^{^{p(p+1)}_2}\Phi (x, y)e_{-1}\wedge \cdots \wedge e_{-r}
x ∧ y = ( − 1 ) 2 p ( p + 1 ) Φ ( x , y ) e − 1 ∧ ⋯ ∧ e − r 如果p + q = l p+q=l p + q = l Φ \Phi Φ l ≡ 0 , − 1 l\equiv 0,-1 l ≡ 0 , − 1 l ≡ 1 , 2 l\equiv 1,2 l ≡ 1 , 2
f ( X ε i ) = e 0 e i , f ( X − ε i ) = − e 0 e − i
f(X_{\varepsilon_i}) =e_0e_i,f(X_{-\varepsilon_i}) =-e_0e_{-i}
f ( X ε i ) = e 0 e i , f ( X − ε i ) = − e 0 e − i 1 ≤ i ≤ l 1\leq i\leq l 1 ≤ i ≤ l
f ( X ε i − ε j ) = 1 2 ( e i e − j − e − j e i ) = e i e − j = e 0 e i e 0 e − j
f(X_{\varepsilon_i-\varepsilon_j}) =\frac{1}{2}(e_ie_{-j}-e_{-j}e_i) =e_ie_{-j}=e_0e_ie_0e_{-j}
f ( X ε i − ε j ) = 2 1 ( e i e − j − e − j e i ) = e i e − j = e 0 e i e 0 e − j 对于1 ≤ i < j ≤ l 1\leq i < j\leq l 1 ≤ i < j ≤ l
f ( X ε j − ε i ) = − e 0 e j e 0 e − i , f ( X ε i + ε j ) = e 0 e i e 0 e j , f ( X − ε i − ε j ) = e 0 e − i e 0 e − j
f(X_{\varepsilon_j-\varepsilon_i}) =-e_0e_je_0e_{-i}, f(X_{\varepsilon_i+\varepsilon_j}) =e_0e_ie_0e_j, f(X_{-\varepsilon_i-\varepsilon_j}) =e_0e_{-i}e_0e_{-j}
f ( X ε j − ε i ) = − e 0 e j e 0 e − i , f ( X ε i + ε j ) = e 0 e i e 0 e j , f ( X − ε i − ε j ) = e 0 e − i e 0 e − j 因此
μ − 1 ∘ f ( X ε i ) = e i , μ − 1 f ( X − ε i ) = − e − i
\mu^{-1}\circ f(X_{\varepsilon_i}) =e_i,\mu^{-1}f(X_{-\varepsilon_i}) =-e_{-i}
μ − 1 ∘ f ( X ε i ) = e i , μ − 1 f ( X − ε i ) = − e − i 和
μ − 1 ∘ f ( X ± ε i ± ε j ) = c e ± i e ± j \mu^{-1}\circ f(X_{\pm\varepsilon_i\pm\varepsilon_j}) =ce_{\pm i}e_{\pm j} μ − 1 ∘ f ( X ± ε i ± ε j ) = c e ± i e ± j 1 ≤ i , j ≤ l , i ≠ j 1\leq i, j\leq l, i\not=j 1 ≤ i , j ≤ l , i = j c ∈ { 1 , − 1 } c\in \{1,-1\} c ∈ { 1 , − 1 } Φ \Phi Φ g \mathfrak{g} g
(VI) 对于x ∈ g x\in \mathfrak{g} x ∈ g σ ( x ) \sigma (x) σ ( x )
T 2 l + 1 + f 1 ( x ) T 2 l + f 2 ( x ) T 2 l − 1 + ⋯ + f 2 l + 1 ( x )
T^{2l+1}+f_1(x)T^{2l}+f_2(x)T^{2l-1}+\cdots +f_{2l+1}(x)
T 2 l + 1 + f 1 ( x ) T 2 l + f 2 ( x ) T 2 l − 1 + ⋯ + f 2 l + 1 ( x ) 其中f 1 , . . . , f 2 l + 1 f_1, . . . , f_{2l+1} f 1 , ... , f 2 l + 1 g \mathfrak{g} g
如果 x = ξ 1 H 1 + ⋯ + ξ l H l ∈ h x=\xi_1H_1+\cdots +\xi_lH_l\in \mathfrak{h} x = ξ 1 H 1 + ⋯ + ξ l H l ∈ h f i ( x ) f_i(x) f i ( x ) ξ 1 , . . . , ξ l , − ξ 1 , . . . , − ξ l \xi_1, . . . , \xi_l,-\xi_1, . . . ,-\xi_l ξ 1 , ... , ξ l , − ξ 1 , ... , − ξ l
T 2 l + 1 + f 2 ( x ) T 2 l − 1 + f 4 ( x ) T 2 l − 3 + ⋯ + f 2 l ( x ) T = T ( T 2 − ξ 1 2 ) . . . ( T 2 − ξ l 2 )
T^{2l+1}+f_2(x)T^{2l-1}+f_4(x)T^{2l-3}+\cdots +f_{2l}(x)T = T(T^2-\xi_1^2). . .(T^2-\xi^2_l)
T 2 l + 1 + f 2 ( x ) T 2 l − 1 + f 4 ( x ) T 2 l − 3 + ⋯ + f 2 l ( x ) T = T ( T 2 − ξ 1 2 ) ... ( T 2 − ξ l 2 ) 因此 f 2 , . . . , f 2 l f_2, . . . , f_{2l} f 2 , ... , f 2 l ξ 1 2 , . . . , ξ l 2 \xi_1^2, . . . , \xi_l^2 ξ 1 2 , ... , ξ l 2 S ( h ∗ ) W \mathbf{S}(\mathfrak{h}^*)^W S ( h ∗ ) W f 1 = f 3 = f 5 = ⋯ = 0 f_1=f_3=f_5=\cdots = 0 f 1 = f 3 = f 5 = ⋯ = 0 ( f 2 , f 4 , . . . , f 2 l ) (f_2, f_4, . . . , f_{2l}) ( f 2 , f 4 , ... , f 2 l ) g \mathfrak{g} g
(VII) 由于 Dynkin 图的唯一自同构是恒等，因此我们有 Aut(g \mathfrak{g} g = = = 0 ( g ) _0(\mathfrak{g}) 0 ( g )
令Σ \Sigma Σ Ψ \Psi Ψ g ∈ Σ g\in \Sigma g ∈ Σ φ ( g ) \varphi (g) φ ( g ) x → g x g − 1 x \rightarrow gxg^{-1} x → g x g − 1 g \mathfrak{g} g Σ \Sigma Σ g \mathfrak{g} g φ \varphi φ α ∈ \alpha \in α ∈ g \mathfrak{g} g = = = 0 ( g ) _0(\mathfrak{g}) 0 ( g ) s ∈ G L ( V ) s\in \mathbf{G}\mathbf{L}(V) s ∈ GL ( V ) α ( x ) = s x s − 1 \alpha (x) =sxs^{-1} α ( x ) = s x s − 1 x ∈ g x\in \mathfrak{g} x ∈ g s s s Ψ \Psi Ψ Ψ ′ \Psi ' Ψ ′ g \mathfrak{g} g Ψ ( \Psi ( Ψ ( s ∈ Σ s\in \Sigma s ∈ Σ
由于g \mathfrak{g} g φ \varphi φ k ∗ k^* k ∗ g \mathfrak{g} g = = = 0 ( g ) _0(\mathfrak{g}) 0 ( g ) Σ / k ∗ \Sigma /k^* Σ/ k ∗ Σ \Sigma Σ k ∗ k^* k ∗ S O ( Ψ ) \mathbf{S}\mathbf{O}(\Psi ) SO ( Ψ ) g \mathfrak{g} g = = = 0 ( g ) _0(\mathfrak{g}) 0 ( g ) S O ( Ψ ) \mathbf{S}\mathbf{O}(\Psi ) SO ( Ψ )
令 O 0 + ( Ψ ) \mathbf{O}^+_0(\Psi ) O 0 + ( Ψ ) Ψ ( \Psi ( Ψ ( S O ( Ψ ) / O 0 + ( Ψ ) \mathbf{S}\mathbf{O}(\Psi )/\mathbf{O}^+_0(\Psi ) SO ( Ψ ) / O 0 + ( Ψ ) e ( g ) _e(\mathfrak{g}) e ( g ) O 0 + ( Ψ ) ( \mathbf{O}^+_0(\Psi ) ( O 0 + ( Ψ ) (
(VIII) h ∗ \mathfrak{h}^* h ∗ Φ R \Phi_R Φ R
Φ R ( ξ 1 ε 1 + ⋯ + ξ l ε l , ξ ′ ε 1 + ⋯ + ξ ′ ε l ) = \Phi_R(\xi_1\varepsilon_1+\cdots +\xi_l\varepsilon_l, \xi '\varepsilon_1+\cdots +\xi '\varepsilon_l) = Φ R ( ξ 1 ε 1 + ⋯ + ξ l ε l , ξ ′ ε 1 + ⋯ + ξ ′ ε l ) = ( ξ 1 ξ ′ + ⋯ + ξ l ξ ′ ) (\xi_1\xi '+\cdots +\xi_l\xi ') ( ξ 1 ξ ′ + ⋯ + ξ l ξ ′ )
1 l 4 l − 2 ‾ 1 l
^1l\overline{4l- 2}^1l
1 l 4 l − 2 1 l （第 VI, §4, no. 5.V）。由Φ R \Phi_R Φ R h \mathfrak{h} h h ∗ \mathfrak{h}^* h ∗ H i H_i H i ( 4 l − 2 ) . ε i (4l-2).\varepsilon_i ( 4 l − 2 ) . ε i Φ R \Phi_R Φ R h \mathfrak{h} h
Φ ( ξ 1 H 1 + ⋯ + ξ l H l , ξ 1 ′ H 1 + ⋯ + ξ l ′ H l ) = ( 4 l − 2 ) ( ξ 1 ξ 1 ′ + ⋯ + ξ l ξ l ′ )
\Phi (\xi_1H_1+\cdots +\xi_lH_l, \xi '_1H_1+\cdots +\xi '_lH_l) = (4l-2)(\xi_1\xi '_1+\cdots +\xi_l\xi '_l)
Φ ( ξ 1 H 1 + ⋯ + ξ l H l , ξ 1 ′ H 1 + ⋯ + ξ l ′ H l ) = ( 4 l − 2 ) ( ξ 1 ξ 1 ′ + ⋯ + ξ l ξ l ′ ) (IX) 回想一下公式 (3) 定义的X α ( α ∈ R ) X_{\alpha}(\alpha \in R) X α ( α ∈ R ) [ X α , X − α ] = − H α [X_{\alpha}, X_{-\alpha}] =-H_{\alpha} [ X α , X − α ] = − H α α ∈ R \alpha \in R α ∈ R I + E 0 , 0 I+E_{0,0} I + E 0 , 0 M = S t M − 1 S M=S^tM^{-1}S M = S t M − 1 S
θ : g → − M − 1 t g M
\theta :g \rightarrow -M^{-1t}gM
θ : g → − M − 1 t g M 对于所有α ∈ R \alpha \in R α ∈ R g \mathfrak{g} g θ ( X α ) = X − α \theta (X_{\alpha}) =X_{-\alpha} θ ( X α ) = X − α ( X α ) (X_{\alpha}) ( X α ) ( g , h ) (\mathfrak{g},\mathfrak{h}) ( g , h )
假设k = Q k=\mathbf{Q} k = Q h \mathfrak{h} h H α H_{\alpha} H α Q ( R ∨ ) Q(R^{\vee}) Q ( R ∨ ) H i H_i H i P ( R ∨ ) P(R^{\vee}) P ( R ∨ ) h \mathfrak{h} h o S ( 2 l + 1 , Z ) \mathfrak{o}_S(2l+ 1,\mathbf{Z}) o S ( 2 l + 1 , Z ) g \mathfrak{g} g g \mathfrak{g} g P ( R ∨ ) + ∑ Z . X α P(R^{\vee}) +\sum\mathbf{Z}.X_{\alpha} P ( R ∨ ) + ∑ Z . X α ( X ± ε i ) 2 = 2 E ± i , ∓ i , ( X ± ε i ) 3 = 0 (X_{\pm\varepsilon_i})^2= 2E_{\pm i,\mp i},(X_{\pm\varepsilon_i})^3= 0 ( X ± ε i ) 2 = 2 E ± i , ∓ i , ( X ± ε i ) 3 = 0 ( X ± ε i ± ε j ) 2 = 0 (X_{\pm\varepsilon_i\pm\varepsilon_j})^2= 0 ( X ± ε i ± ε j ) 2 = 0 ( e i ) − l ≤ i ≤ l (e_i)_{-l\leq i\leq l} ( e i ) − l ≤ i ≤ l V \mathscr{V} V o S ( 2 l + 1 , Z ) \mathfrak{o}_S(2l+ 1,\mathbf{Z}) o S ( 2 l + 1 , Z ) ⋀ r V \bigwedge^rV ⋀ r V ⋀ r V \bigwedge^r\mathscr{V} ⋀ r V
现在考虑 N = ⋀ F ′ N =\bigwedge F' N = ⋀ F ′ g \mathfrak{g} g ρ \rho ρ P ( R ∨ ) P(R^{\vee}) P ( R ∨ ) Z \mathbf{Z} Z o S ( 2 l + 1 , Z ) \mathfrak{o}_S(2l+ 1,\mathbf{Z}) o S ( 2 l + 1 , Z ) ( e − i 1 ∧ ⋯ ∧ e − i k ) (e_{-i_1}\wedge \cdots \wedge e_{-i_k}) ( e − i 1 ∧ ⋯ ∧ e − i k ) 1 ≤ i 1 < ⋯ < i k ≤ l 1\leq i_1<\cdots < i_k\leq l 1 ≤ i 1 < ⋯ < i k ≤ l N \mathscr{N} N G = Q ( R ∨ ) + ∑ α ∈ R Z . X α \mathscr{G}= Q(R^{\vee}) +\sum_{\alpha\in R}\mathbf{Z}.X_{\alpha} G = Q ( R ∨ ) + ∑ α ∈ R Z . X α N \mathscr{N} N
外部乘积与e − i e_{-i} e − i e i e_i e i 1 ≤ i ≤ l 1\leq i\leq l 1 ≤ i ≤ l N \mathscr{N} N ρ ( G ) \rho (\mathscr{G}) ρ ( G ) ρ ( X α ) 2 = 0 \rho (X_{\alpha})^2= 0 ρ ( X α ) 2 = 0 α ∈ R \alpha \in R α ∈ R N \mathscr{N} N
3. 代数型 Cl _{\boldsymbol{l}} l l ≥ \boldsymbol{l}\geq l ≥
(I) 设Ψ \Psi Ψ 2 l ≥ 2 2l\geq 2 2 l ≥ 2 x x x v , v ′ ∈ V v, v'\in V v , v ′ ∈ V Ψ ( x v , v ′ ) + Ψ ( v , x v ′ ) = 0 \Psi (xv, v') +\Psi (v, xv') = 0 Ψ ( xv , v ′ ) + Ψ ( v , x v ′ ) = 0 s l ( V ) \mathfrak{s}\mathfrak{l}(V) sl ( V ) s p ( Ψ ) \mathfrak{s}\mathfrak{p}(\Psi ) sp ( Ψ ) Ψ \Psi Ψ
代数，Cap。 IX, §4, no. 2、V可以写成两个最大全各向同性子空间 F和F ′ F' F ′ Ψ \Psi Ψ ( e i ) 1 ≤ i ≤ l (e_i)_{1\leq i\leq l} ( e i ) 1 ≤ i ≤ l ( e − i ) 1 ≤ i ≤ l (e_{-i})_{1\leq i\leq l} ( e − i ) 1 ≤ i ≤ l F ′ F' F ′
( e 1 , . . . , e l , e − l , . . . , e − 1 )
(e_1, . . . , e_l, e_{-l}, . . . , e_{-1})
( e 1 , ... , e l , e − l , ... , e − 1 ) 是 V 中的基；我们说它是 V 的维特基（或辛基）。相对于此基Ψ \Psi Ψ 2 l 2l 2 l
J = − s J=-s J = − s
其中 s s s l l l
代数g = s p ( Ψ ) \mathfrak{g}=\mathfrak{s}\mathfrak{p}(\Psi ) g = sp ( Ψ ) a a a 2 l 2l 2 l s p ( 2 l , k ) \mathfrak{s}\mathfrak{p}(2l, k) sp ( 2 l , k ) a = − J − 1 t a J = J t a J ( a=-J^{-1t}aJ=J^taJ( a = − J − 1 t a J = J t a J (
a = C − s t A s
a=C-s^tAs
a = C − s t A s 其中A , B , C A, B, C A , B , C l l l B = s t B s B=s^tBs B = s t B s C = C= C = c t C s c^tCs c t C s B B B C C C
昏暗g = l 2 + 2 l ( l + 1 ) 2 = l ( 2 l + 1 ) \mathfrak{g}=l^2+ 2\frac{l(l + 1)}{2}=l(2l+ 1) g = l 2 + 2 2 l ( l + 1 ) = l ( 2 l + 1 )
令h \mathfrak{h} h g \mathfrak{g} g g \mathfrak{g} g 1 ≤ i ≤ l 1\leq i\leq l 1 ≤ i ≤ l H i = E i , i − E − i , − i H_i=E_{i,i}-E_{-i,-i} H i = E i , i − E − i , − i ( ε i ) 1 ≤ i ≤ l (\varepsilon_i)_{1\leq i\leq l} ( ε i ) 1 ≤ i ≤ l ( H i ) (H_i) ( H i ) 1 ≤ i < j ≤ l 1\leq i < j\leq l 1 ≤ i < j ≤ l
X 2 ε i X_{2\varepsilon_i} X 2 ε i E i , − i E_{i,-i} E i , − i
X − 2 ε i X_{-2\varepsilon_i} X − 2 ε i − E − i , i -E_{-i,i} − E − i , i
X ε i − ε j X_{\varepsilon_i-\varepsilon_j} X ε i − ε j E i , j − E − j , − i E_{i,j}-E_{-j,-i} E i , j − E − j , − i
X − ε i + ε j X_{-\varepsilon_i+\varepsilon_j} X − ε i + ε j − E j , i + E − i , − j -E_{j,i}+E_{-i,-j} − E j , i + E − i , − j
X ε i + ε j X_{\varepsilon_i+\varepsilon_j} X ε i + ε j E i , − j + E j , − i E_{i,-j}+E_{j,-i} E i , − j + E j , − i
X − ε i − ε j X_{-\varepsilon_i-\varepsilon_j} X − ε i − ε j − E − i , j − E − j , i -E_{-i,j}-E_{-j,i} − E − i , j − E − j , i
很容易验证，元素形式基是g \mathfrak{g} g h \mathfrak{h} h h ∈ h h\in \mathfrak{h} h ∈ h
[ h , X α ] = α ( h ) X α (7)
[h, X_{\alpha}] =\alpha (h)X_{\alpha} \tag{7}
[ h , X α ] = α ( h ) X α ( 7 ) 对于所有α ∈ R \alpha \in R α ∈ R ± 2 ε i \pm 2\varepsilon_i ± 2 ε i ± ε i ± ε j ( i < j ) \pm \varepsilon_i\pm \varepsilon_j(i < j) ± ε i ± ε j ( i < j ) h \mathfrak{h} h g \mathfrak{g} g g \mathfrak{g} g h \mathfrak{h} h ( g , h ) (\mathfrak{g},\mathfrak{h}) ( g , h ) ( g , h ) (\mathfrak{g},\mathfrak{h}) ( g , h ) l ≥ 2 l\geq 2 l ≥ 2 C l C_l C l l = 1 l= 1 l = 1 A 1 A_1 A 1 C 1 C_1 C 1 l = 1 l= 1 l = 1 g \mathfrak{g} g C l C_l C l
Every splitting Cartan subalgebra of g \mathfrak{g} g h \mathfrak{h} h S p ( Ψ ) \mathbf{S}\mathbf{p}(\Psi ) Sp ( Ψ ) h β \mathfrak{h}_{\beta} h β g \mathfrak{g} g β \beta β g \mathfrak{g} g h β \mathfrak{h}_{\beta} h β β \beta β
我们有s p ( 2 , k ) = s l ( 2 , k ) \mathfrak{s}\mathfrak{p}(2, k) =\mathfrak{s}\mathfrak{l}(2, k) sp ( 2 , k ) = sl ( 2 , k ) s p ( 4 , k ) \mathfrak{s}\mathfrak{p}(4, k) sp ( 4 , k ) o S ( 5 , k ) \mathfrak{o}_S(5, k) o S ( 5 , k ) l ≥ 2 l\geq 2 l ≥ 2
(II) 根系 R ∨ R^{\vee} R ∨
H 2 ε i = H i , H ε i − ε j = H i − H j , H ε i + ε j = H i + H j
H_{2\varepsilon_i}=H_i, H_{\varepsilon_i-\varepsilon_j}=H_i-H_j, H_{\varepsilon_i+\varepsilon_j}=H_i+H_j
H 2 ε i = H i , H ε i − ε j = H i − H j , H ε i + ε j = H i + H j (III) 放入α 1 = ε 1 − ε 2 , . . . , α l − 1 = ε l − 1 − ε l , α l = 2 ε l \alpha_1=\varepsilon_1-\varepsilon_2, . . . , \alpha_{l-1}=\varepsilon_{l-1}-\varepsilon_l, \alpha_l= 2\varepsilon_l α 1 = ε 1 − ε 2 , ... , α l − 1 = ε l − 1 − ε l , α l = 2 ε l { α 1 , . . . , α l } \{\alpha_1, . . . , \alpha_l\} { α 1 , ... , α l } 2 ε i 2\varepsilon_i 2 ε i ε i ± ε j ( i < j ) \varepsilon_i\pm \varepsilon_j(i < j) ε i ± ε j ( i < j ) b \mathfrak{b} b g \mathfrak{g} g
设δ \delta δ Ψ \Psi Ψ p δ \mathfrak{p}_{\delta} p δ g \mathfrak{g} g δ \delta δ δ → p δ \delta \rightarrow \mathfrak{p}_{\delta} δ → p δ g \mathfrak{g} g p δ ⊃ p δ ′ \mathfrak{p}_{\delta}\supset \mathfrak{p}_{\delta'} p δ ⊃ p δ ′ δ ⊂ δ ′ \delta \subset \delta ' δ ⊂ δ ′
(IV) 基本的权相应的到 α 1 , . . . , α l \alpha_1, . . . , \alpha_l α 1 , ... , α l ϖ i = ε 1 + ⋯ + ε i ( 1 ≤ i ≤ l ) \varpi_i=\varepsilon_1+\cdots +\varepsilon_i(1\leq i\leq l) ϖ i = ε 1 + ⋯ + ε i ( 1 ≤ i ≤ l )
我们将展示权ϖ r \varpi_r ϖ r σ r \sigma_r σ r ⋀ r σ \bigwedge^r\sigma ⋀ r σ σ \sigma σ g \mathfrak{g} g ⋀ σ \bigwedge\sigma ⋀ σ g \mathfrak{g} g ⋀ V \bigwedge V ⋀ V
令( e i ∗ ) (e^*_i) ( e i ∗ ) V ∗ V^* V ∗ ( e i ) (e_i) ( e i ) Ψ \Psi Ψ Γ ∗ ∈ ⋀ 2 V ∗ ( \Gamma^*\in \bigwedge^2V^*( Γ ∗ ∈ ⋀ 2 V ∗ (
Γ ∗ = − ∑ i = 1 l e i ∗ ∧ e − i ∗
\Gamma^*=-\sum_{i=1}^le^*_i\wedge e^*_{-i}
Γ ∗ = − i = 1 ∑ l e i ∗ ∧ e − i ∗ 设Ψ ∗ \Psi^* Ψ ∗ Ψ \Psi Ψ
Ψ ∗ ( e i ∗ , e j ∗ ) = 0
\Psi^*(e^*_i, e^*_j) = 0
Ψ ∗ ( e i ∗ , e j ∗ ) = 0 对于i ≠ − j i\not=-j i = − j Ψ ∗ ( e i ∗ , e − i ∗ ) = − 1 \Psi^*(e^*_i, e^*_{-i}) =-1 Ψ ∗ ( e i ∗ , e − i ∗ ) = − 1 1 ≤ i ≤ l 1\leq i\leq l 1 ≤ i ≤ l Ψ ∗ \Psi^* Ψ ∗ Γ ∈ ⋀ 2 V \Gamma \in \bigwedge^2V Γ ∈ ⋀ 2 V
Γ = ∑ i = 1 l e i ∧ e − i
\Gamma =\sum_{i=1}^le_i\wedge e_{-i}
Γ = i = 1 ∑ l e i ∧ e − i 由左外部乘积和 Γ \Gamma Γ ⋀ V \bigwedge V ⋀ V X − X_- X − − Γ ∗ : -\Gamma^*: − Γ ∗ : ⋀ V \bigwedge V ⋀ V X + X_+ X +
X − u = ( ∑ i = 1 l e i ∧ e − i ) ∧ u
X_-u=(\sum_{i=1}^le_i\wedge e_{-i})\wedge u
X − u = ( i = 1 ∑ l e i ∧ e − i ) ∧ u X + u = ( ∑ i = 1 l e i ∗ ∧ e − i ∗ ) ∧ u
X_+u=(\sum_{i=1}^le^*_i\wedge e^*_{-i})\wedge u
X + u = ( i = 1 ∑ l e i ∗ ∧ e − i ∗ ) ∧ u 对于计算X + X_+ X + X − X_- X − ⋀ V \bigwedge V ⋀ V 1 , l 1, l 1 , l ( A , B , C ) (A,B,C) ( A , B , C )
e A , B , C = e a 1 ∧ ⋯ ∧ e a m ∧ e − b 1 ∧ ⋯ ∧ e − b n ∧ e c 1 ∧ e − c 1 ∧ ⋯ ∧ e c p ∧ e − c p
e_{A,B,C}=e_{a_1}\wedge \cdots \wedge e_{a_m}\wedge e_{-b_1}\wedge \cdots \wedge e_{-b_n}\wedge e_{c_1}\wedge e_{-c_1}\wedge \cdots \wedge e_{c_p}\wedge e_{-c_p}
e A , B , C = e a 1 ∧ ⋯ ∧ e a m ∧ e − b 1 ∧ ⋯ ∧ e − b n ∧ e c 1 ∧ e − c 1 ∧ ⋯ ∧ e c p ∧ e − c p 其中( a 1 , . . . , a m ) (a_1, . . . , a_m) ( a 1 , ... , a m ) ( b 1 , . . . , b n ) , ( c 1 , . . . , c p ) (b_1, . . . , b_n),(c_1, . . . , c_p) ( b 1 , ... , b n ) , ( c 1 , ... , c p ) B , C B,C B , C ⋀ V \bigwedge V ⋀ V
X − . e A , B , C = ∑ e A , B , C ∪ { j } (8)
X_-.e_{A,B,C}=\sum e_{A,B,C\cup \{j\}} \tag{8}
X − . e A , B , C = ∑ e A , B , C ∪ { j } ( 8 ) j ∈ 1 , l , j ∉ A ∪ B ∪ C j\in 1,l ,j \notin A\cup B\cup C j ∈ 1 , l , j ∈ / A ∪ B ∪ C X + . e A , B , C = − ∑ j ∈ C e A , B , C − { j } (9)
X_+.e_{A,B,C}=-\sum_{j\in C}e_{A,B,C-\{j\}} \tag{9}
X + . e A , B , C = − j ∈ C ∑ e A , B , C − { j } ( 9 ) 设H H H ⋀ V \bigwedge V ⋀ V ( l − r ) (l-r) ( l − r ) ⋀ r V ( 0 ≤ r ≤ 2 l ) \bigwedge^rV (0\leq r\leq 2l) ⋀ r V ( 0 ≤ r ≤ 2 l )
[ X + , X − ] = − H
[X_+, X_-] =-H
[ X + , X − ] = − H [ H , X + ] = 2 X +
[H, X_+] = 2X_+
[ H , X + ] = 2 X + [ H , X − ] = − 2 X −
[H, X_-] =-2X_-
[ H , X − ] = − 2 X − In other words, the vector subspace s \mathfrak{s} s X + , X − X_+, X_- X + , X − H H H ⋀ V \bigwedge V ⋀ V s l ( 2 , k ) \mathfrak{s}\mathfrak{l}(2, k) sl ( 2 , k ) ⋀ r V \bigwedge^rV ⋀ r V l − r l-r l − r ⋀ r V \bigwedge^rV ⋀ r V E r E_r E r E r = ( ⋀ r V ) ∩ E_r= (\bigwedge^rV)\cap E r = ( ⋀ r V ) ∩ X + X_+ X + r < l r < l r < l X − X_- X − ⋀ r V \bigwedge^rV ⋀ r V r ≤ l r\leq l r ≤ l ⋀ r V \bigwedge^rV ⋀ r V
⋀ r V = E r ⊕ X − ( E r − 2 ) ⊕ X − 2 ( E r − 4 ) ⊕ ⋯
\bigwedge^rV = E_r\oplus X_-(E_{r-2})\oplus X_-^2(E_{r-4})\oplus \cdots
⋀ r V = E r ⊕ X − ( E r − 2 ) ⊕ X − 2 ( E r − 4 ) ⊕ ⋯ = E r ⊕ X − ( ⋀ r − 2 V )
= E_r\oplus X_-(\bigwedge^{r-2}V)
= E r ⊕ X − ( ⋀ r − 2 V ) 这特别表明 E r = ( r 2 l ) − ( r − 2 2 l ) E_r=(^{2l}_r)-(_{r-2}^{2l}) E r = ( r 2 l ) − ( r − 2 2 l ) 0 ≤ r ≤ l 0\leq r\leq l 0 ≤ r ≤ l
另一方面，s p ( Ψ ) \mathfrak{s}\mathfrak{p}(\Psi ) sp ( Ψ ) Γ ∗ \Gamma^* Γ ∗ σ \sigma σ Γ \Gamma Γ ⋀ 2 σ \bigwedge^2\sigma ⋀ 2 σ X + X_+ X + X − X_- X − H H H ⋀ σ ( g ) \bigwedge\sigma (g) ⋀ σ ( g ) g ∈ g g\in \mathfrak{g} g ∈ g 0 ≤ r ≤ l 0\leq r\leq l 0 ≤ r ≤ l E r E_r E r ⋀ r σ \bigwedge^r\sigma ⋀ r σ ⋀ r σ \bigwedge^r\sigma ⋀ r σ E r E_r E r ϖ r ( 1 ≤ r ≤ l ) \varpi_r(1\leq r\leq l) ϖ r ( 1 ≤ r ≤ l ) σ r \sigma_r σ r
我们注首先⋀ r σ \bigwedge^r\sigma ⋀ r σ h \mathfrak{h} h
ε i 1 + ⋯ + ε i k − ( ε j 1 + ⋯ + ε j r − k )
\varepsilon_{i_1}+\cdots +\varepsilon_{i_k}-(\varepsilon_{j_1}+\cdots +\varepsilon_{j_{r-k}})
ε i 1 + ⋯ + ε i k − ( ε j 1 + ⋯ + ε j r − k ) 其中i 1 , . . . , i k i_1, . . . , i_k i 1 , ... , i k j 1 , . . . , j r − k j_1, . . . , j_{r-k} j 1 , ... , j r − k 1 , l 1, l 1 , l ⋀ r σ \bigwedge^r\sigma ⋀ r σ
ϖ r = ε 1 + ⋯ + ε r
\varpi_r=\varepsilon_1+\cdots +\varepsilon_r
ϖ r = ε 1 + ⋯ + ε r 权 ϖ ‾ r \overline{\varpi}_r ϖ r e 1 ∧ ⋯ ∧ e r = e { 1 , … , r } , ∅ , ∅ e_1\wedge\cdots\wedge e_r=e_{\{1,\ldots,r\},\varnothing,\varnothing} e 1 ∧ ⋯ ∧ e r = e { 1 , … , r } , ∅ , ∅ e 1 ∧ ⋯ ∧ e r ∈ E r e_1\wedge\cdots\wedge e_r\in E_r e 1 ∧ ⋯ ∧ e r ∈ E r ⋀ r σ \bigwedge^r\sigma ⋀ r σ E r E_r E r
如果s ∈ S p ( Ψ ) s\in\mathbf{Sp}(\Psi) s ∈ Sp ( Ψ ) s s s ⋀ V \bigwedge V ⋀ V ⋀ V ∗ \bigwedge V^* ⋀ V ∗ Γ \Gamma Γ Γ ∗ \Gamma^* Γ ∗ X + X_+ X + X − X_- X − E r E_r E r E r E_r E r F r F_r F r e 1 ∧ ⋯ ∧ e r e_1\wedge\cdots\wedge e_r e 1 ∧ ⋯ ∧ e r S p ( Ψ ) \mathbf{Sp}(\Psi) Sp ( Ψ ) r r r V V V r r r
对于 1 ≤ r ≤ ℓ 1\leq r\leq\ell 1 ≤ r ≤ ℓ F r F_r F r r r r ⋀ r V \bigwedge^r V ⋀ r V
⋀ r V = F r + X − ( ⋀ r − 2 V ) = F r + ( ∑ i = 1 ℓ e i ∧ e − i ) ∧ ⋀ r − 2 V .
\bigwedge^r V=F_r+X_-\left(\bigwedge^{r-2}V\right)
=F_r+\left(\sum_{i=1}^{\ell}e_i\wedge e_{-i}\right)\wedge\bigwedge^{r-2}V.
⋀ r V = F r + X − ( ⋀ r − 2 V ) = F r + ( i = 1 ∑ ℓ e i ∧ e − i ) ∧ ⋀ r − 2 V . 引用了 4 引用了 命题 2, VIII § 7 , VI § 4. 。 根系的分类 , 命题 3, VIII § 11 , 习题 9, § 13
我们首先展示引理 2 如何蕴含我们的断言。由于F r ⊂ E r F_r\subset E_r F r ⊂ E r E r ∩ X − ( ⋀ r − 2 V ) = { 0 } E_r\cap X_-\left(\bigwedge^{r-2}V\right)=\{0\} E r ∩ X − ( ⋀ r − 2 V ) = { 0 } F r = E r F_r=E_r F r = E r s ∈ S p ( Ψ ) s\in\mathbf{Sp}(\Psi) s ∈ Sp ( Ψ ) End ( V ) \operatorname{End}(V) End ( V ) a ↦ s a s − 1 a\mapsto sas^{-1} a ↦ s a s − 1 g \mathfrak{g} g Aut 0 ( g ) \operatorname{Aut}_0(\mathfrak{g}) Aut 0 ( g ) g \mathfrak{g} g e 1 ∧ ⋯ ∧ e r e_1\wedge\cdots\wedge e_r e 1 ∧ ⋯ ∧ e r ⋀ r σ \bigwedge^r\sigma ⋀ r σ E r = F r E_r=F_r E r = F r e 1 ∧ ⋯ ∧ e r e_1\wedge\cdots\wedge e_r e 1 ∧ ⋯ ∧ e r S p ( Ψ ) \mathbf{Sp}(\Psi) Sp ( Ψ ) g \mathfrak{g} g E r E_r E r ϖ ‾ r \overline{\varpi}_r ϖ r 1 1 1
仍有待证明引理。 r = 1 r=1 r = 1 r ≥ 2 r\geq 2 r ≥ 2
F r − 1 ∧ V ⊂ F r + Γ ∧ ⋀ r − 2 V ,
F_{r-1}\wedge V\subset F_r+\Gamma\wedge\bigwedge^{r-2}V,
F r − 1 ∧ V ⊂ F r + Γ ∧ ⋀ r − 2 V , 或者，如果 y y y ( r − 1 ) (r-1) ( r − 1 ) x ∈ V x\in V x ∈ V
z = y ∧ x ∈ F r + Γ ∧ ⋀ r − 2 V .
z=y\wedge x\in F_r+\Gamma\wedge\bigwedge^{r-2}V.
z = y ∧ x ∈ F r + Γ ∧ ⋀ r − 2 V . 令( f i ) 1 ≤ ± i ≤ ℓ (f_i)_{1\leq\pm i\leq\ell} ( f i ) 1 ≤ ± i ≤ ℓ V V V y = f 1 ∧ ⋯ ∧ f r − 1 y=f_1\wedge\cdots\wedge f_{r-1} y = f 1 ∧ ⋯ ∧ f r − 1 x = f i x=f_i x = f i i ∉ [ 1 − r , − 1 ] i\notin[1-r,-1] i ∈ / [ 1 − r , − 1 ] r r r f 1 ∧ ⋯ ∧ f r − 1 ∧ f i f_1\wedge\cdots\wedge f_{r-1}\wedge f_i f 1 ∧ ⋯ ∧ f r − 1 ∧ f i f i f_i f i i = 1 − r i=1-r i = 1 − r Γ = ∑ j = 1 ℓ f j ∧ f − j \Gamma=\displaystyle\sum_{j=1}^{\ell}f_j\wedge f_{-j} Γ = j = 1 ∑ ℓ f j ∧ f − j
f r − 1 ∧ f 1 − r = 1 ℓ − r + 2 ( Γ − ∑ i = 1 r − 2 f i ∧ f − i + ∑ j = r ℓ ( f r − 1 ∧ f 1 − r − f j ∧ f − j ) ) ,
f_{r-1}\wedge f_{1-r}
=
\frac{1}{\ell-r+2}
\left(
\Gamma-\sum_{i=1}^{r-2}f_i\wedge f_{-i}
+\sum_{j=r}^{\ell}(f_{r-1}\wedge f_{1-r}-f_j\wedge f_{-j})
\right),
f r − 1 ∧ f 1 − r = ℓ − r + 2 1 ( Γ − i = 1 ∑ r − 2 f i ∧ f − i + j = r ∑ ℓ ( f r − 1 ∧ f 1 − r − f j ∧ f − j ) ) , z = 1 ℓ − r + 2 Γ ∧ f 1 ∧ ⋯ ∧ f r − 2 + 1 ℓ − r + 2 ∑ j = r ℓ ( f 1 ∧ ⋯ ∧ f r − 2 ) ∧ ( f r − 1 ∧ f 1 − r − f j ∧ f − j ) .
z=
\frac{1}{\ell-r+2}\Gamma\wedge f_1\wedge\cdots\wedge f_{r-2}
+
\frac{1}{\ell-r+2}\sum_{j=r}^{\ell}
(f_1\wedge\cdots\wedge f_{r-2})\wedge
(f_{r-1}\wedge f_{1-r}-f_j\wedge f_{-j}).
z = ℓ − r + 2 1 Γ ∧ f 1 ∧ ⋯ ∧ f r − 2 + ℓ − r + 2 1 j = r ∑ ℓ ( f 1 ∧ ⋯ ∧ f r − 2 ) ∧ ( f r − 1 ∧ f 1 − r − f j ∧ f − j ) . 但是
f r − 1 ∧ f 1 − r − f j ∧ f − j = ( f r − 1 + f j ) ∧ ( f 1 − r − f − j ) − f j ∧ f 1 − r + f r − 1 ∧ f − j
f_{r-1}\wedge f_{1-r}-f_j\wedge f_{-j}= (f_{r-1}+f_j)\wedge (f_{1-r}-f_{-j})-f_j\wedge f_{1-r}+f_{r-1}\wedge f_{-j}
f r − 1 ∧ f 1 − r − f j ∧ f − j = ( f r − 1 + f j ) ∧ ( f 1 − r − f − j ) − f j ∧ f 1 − r + f r − 1 ∧ f − j 我们立即验证 r r r
f 1 ∧ ⋯ ∧ f r − 2 ∧ ( f r − 1 + f j ) ∧ ( f 1 − r − f − j )
f_1\wedge \cdots \wedge f_{r-2}\wedge (f_{r-1}+f_j)\wedge (f_{1-r}-f_{-j})
f 1 ∧ ⋯ ∧ f r − 2 ∧ ( f r − 1 + f j ) ∧ ( f 1 − r − f − j ) f 1 ∧ ⋯ ∧ f r − 2 ∧ f j ∧ f 1 − r f_1\wedge \cdots \wedge f_{r-2}\wedge f_j\wedge f_{1-r} f 1 ∧ ⋯ ∧ f r − 2 ∧ f j ∧ f 1 − r f 1 ∧ ⋯ ∧ f r − 1 ∧ f − j f_1\wedge \cdots \wedge f_{r-1}\wedge f_{-j} f 1 ∧ ⋯ ∧ f r − 1 ∧ f − j
r ≤ j ≤ l r\leq j\leq l r ≤ j ≤ l z ∈ F r + Γ ∧ ⋀ r − 2 V z\in F_r+\Gamma \wedge \bigwedge^{r-2}V z ∈ F r + Γ ∧ ⋀ r − 2 V
(V) 我们有 w 0 = − 1 w_0=-1 w 0 = − 1 g \mathfrak{g} g ϖ r \varpi_r ϖ r ( α 1 , . . . , α l ) (\alpha_1, . . . , \alpha_l) ( α 1 , ... , α l )
1 + 2 + ⋯ + ( r − 1 ) + r + r + ⋯ + r +
1 + 2 +\cdots + (r-1) +r+r+\cdots +r+
1 + 2 + ⋯ + ( r − 1 ) + r + r + ⋯ + r + 2
所以σ r \sigma_r σ r r r r r r r
自 e 1 ∧ ⋯ ∧ e r e_1\wedge \cdots \wedge e_r e 1 ∧ ⋯ ∧ e r e − 1 ∧ ⋯ ∧ e − r e_{-1}\wedge \cdots \wedge e_{-r} e − 1 ∧ ⋯ ∧ e − r E r E_r E r
Ψ ( r ) ( e 1 ∧ ⋯ ∧ e r , e − 1 ∧ ⋯ ∧ e − r ) = 1
\Psi_{(r)}(e_1\wedge \cdots \wedge e_r, e_{-1}\wedge \cdots \wedge e_{-r}) = 1
Ψ ( r ) ( e 1 ∧ ⋯ ∧ e r , e − 1 ∧ ⋯ ∧ e − r ) = 1 我们看到 Ψ ( r ) \Psi_{(r)} Ψ ( r ) E r E_r E r r r r r r r σ r \sigma_r σ r
(VI) 对于所有x ∈ g x\in \mathfrak{g} x ∈ g σ ( x ) \sigma (x) σ ( x )
T 2 l + f 1 ( x ) T 2 l − 1 + ⋯ + f 2 l ( x )
T^{2l}+f_1(x)T^{2l-1}+\cdots +f_{2l}(x)
T 2 l + f 1 ( x ) T 2 l − 1 + ⋯ + f 2 l ( x ) 其中f 1 , . . . , f 2 l f_1, . . . , f_{2l} f 1 , ... , f 2 l g \mathfrak{g} g
如果 x = ξ 1 H 1 + ⋯ + ξ l H l ∈ h x=\xi_1H_1+\cdots +\xi_lH_l\in \mathfrak{h} x = ξ 1 H 1 + ⋯ + ξ l H l ∈ h f i ( x ) f_i(x) f i ( x ) ξ 1 , . . . , ξ l , − ξ 1 , . . . , − ξ l \xi_1, . . . , \xi_l,-\xi_1, . . . ,-\xi_l ξ 1 , ... , ξ l , − ξ 1 , ... , − ξ l
T 2 l + f 2 ( x ) T 2 l − 2 + ⋯ + f 2 l ( x ) = ( T 2 − ξ 1 2 ) . . . ( T 2 − ξ l 2 )
T^{2l}+f_2(x)T^{2l-2}+\cdots +f_{2l}(x) = (T^2-\xi_1^2). . .(T^2-\xi^2_l)
T 2 l + f 2 ( x ) T 2 l − 2 + ⋯ + f 2 l ( x ) = ( T 2 − ξ 1 2 ) ... ( T 2 − ξ l 2 ) 如 no. 2.VI 所示，f 1 = f 3 = f 5 = ⋯ = 0 f_1=f_3=f_5=\cdots = 0 f 1 = f 3 = f 5 = ⋯ = 0
( f 2 , f 4 , . . . , f 2 l )
(f_2, f_4, . . . , f_{2l})
( f 2 , f 4 , ... , f 2 l ) 是g \mathfrak{g} g
(VII) 由于 Dynkin 图的唯一自同构是恒等，因此我们有 Aut(g \mathfrak{g} g = = = 0 ( g ) _0(\mathfrak{g}) 0 ( g )
令Σ \Sigma Σ Ψ ( \Psi ( Ψ ( g \mathfrak{g} g x → s x s − 1 x \rightarrow sxs^{-1} x → s x s − 1 s ∈ Σ s\in \Sigma s ∈ Σ g \mathfrak{g} g = = = 0 ( g ) _0(\mathfrak{g}) 0 ( g ) Σ / k ∗ \Sigma /k^* Σ/ k ∗
对于所有s ∈ Σ s\in \Sigma s ∈ Σ μ ( s ) \mu(s) μ ( s ) s s s Σ \Sigma Σ k ∗ / k ∗ 2 k^*/k^{*2} k ∗ / k ∗ 2 s → μ ( s ) s \rightarrow \mu(s) s → μ ( s ) k ∗ 2 k^{*2} k ∗ 2 k ∗ .1 k^*.1 k ∗ .1 λ \lambda λ Σ / k ∗ \Sigma /k^* Σ/ k ∗ k ∗ / k ∗ 2 k^*/k^{*2} k ∗ / k ∗ 2 S p ( Ψ ) ∩ k ∗ = { 1 , − 1 } \mathbf{S}\mathbf{p}(\Psi )\cap k^*=\{1,-1\} Sp ( Ψ ) ∩ k ∗ = { 1 , − 1 }
1⟶ S p ( Ψ ) / { 1 , − 1 } ⟶ ι Σ / k ∗ ⟶ λ k ∗ / k ∗ 2 ⟶ 1 \longrightarrow \mathbf{S}\mathbf{p}(\Psi )/\{1,-1\}\longrightarrow^{\iota}\Sigma /k^*\longrightarrow^{\lambda}k^*/k^{*2}\longrightarrow 1 ⟶ Sp ( Ψ ) / { 1 , − 1 } ⟶ ι Σ/ k ∗ ⟶ λ k ∗ / k ∗ 2 ⟶ 1 ι \iota ι ι \iota ι ⊂ \subset ⊂ λ \lambda λ S p ( Ψ ) \mathbf{S}\mathbf{p}(\Psi ) Sp ( Ψ ) s ∈ Σ s\in \Sigma s ∈ Σ k ∗ 2 k^{*2} k ∗ 2 ν ∈ k ∗ \nu \in k^* ν ∈ k ∗ ν s ∈ S p ( Ψ ) \nu s\in \mathbf{S}\mathbf{p}(\Psi ) ν s ∈ Sp ( Ψ ) ι \iota ι = = = λ \lambda λ S p ( Ψ ) / { 1 , − 1 } \mathbf{S}\mathbf{p}(\Psi )/\{1,-1\} Sp ( Ψ ) / { 1 , − 1 } Σ / k ∗ \Sigma /k^* Σ/ k ∗ k ∗ / k ∗ 2 k^*/k^{*2} k ∗ / k ∗ 2 Σ / k ∗ \Sigma /k^* Σ/ k ∗ S p ( Ψ ) / { 1 , − 1 } \mathbf{S}\mathbf{p}(\Psi )/\{1,-1\} Sp ( Ψ ) / { 1 , − 1 } e ( g ) _e(\mathfrak{g}) e ( g ) S p ( Ψ ) / { 1 , − 1 } ( \mathbf{S}\mathbf{p}(\Psi )/\{1,-1\}( Sp ( Ψ ) / { 1 , − 1 } ( g \mathfrak{g} g / / / e ( g ) _e(\mathfrak{g}) e ( g ) k ∗ / k ∗ 2 k^*/k^{*2} k ∗ / k ∗ 2
(VIII) h ∗ \mathfrak{h}^* h ∗ Φ R \Phi_R Φ R
Φ R ( ξ 1 ε 1 + ⋯ + ξ l ε l , ξ 1 ′ ε 1 + ⋯ + ξ l ′ ε l ) = 1 4 ( l + 1 ) ( ξ 1 ξ 1 ′ + ⋯ + ξ l ξ l ′ )
\Phi_R(\xi_1\varepsilon_1+\cdots +\xi_l\varepsilon_l, \xi '_1\varepsilon_1+\cdots +\xi '_l\varepsilon_l) =\frac{1}{4(l + 1)}(\xi_1\xi '_1+\cdots +\xi_l\xi '_l)
Φ R ( ξ 1 ε 1 + ⋯ + ξ l ε l , ξ 1 ′ ε 1 + ⋯ + ξ l ′ ε l ) = 4 ( l + 1 ) 1 ( ξ 1 ξ 1 ′ + ⋯ + ξ l ξ l ′ ) （第 VI, §4, no. 6.V）。因而，Φ R \Phi_R Φ R h \mathfrak{h} h
Φ ( ξ 1 H 1 + ⋯ + ξ l H l , ξ 1 ′ H 1 + ⋯ + ξ l ′ H l ) = 4 ( l + 1 ) ( ξ 1 ξ 1 ′ + ⋯ + ξ l ξ l ′ )
\Phi (\xi_1H_1+\cdots +\xi_lH_l, \xi '_1H_1+\cdots +\xi '_lH_l) = 4(l+ 1)(\xi_1\xi '_1+\cdots +\xi_l\xi '_l)
Φ ( ξ 1 H 1 + ⋯ + ξ l H l , ξ 1 ′ H 1 + ⋯ + ξ l ′ H l ) = 4 ( l + 1 ) ( ξ 1 ξ 1 ′ + ⋯ + ξ l ξ l ′ ) (IX) 回想一下由公式( 6 ) ( α ∈ R ) (6) (\alpha \in R) ( 6 ) ( α ∈ R ) X α X_{\alpha} X α [ X α , X − α ] = − H α [X_{\alpha}, X_{-\alpha}] =-H_{\alpha} [ X α , X − α ] = − H α α ∈ R \alpha \in R α ∈ R α ∈ R \alpha \in R α ∈ R θ : a → − t a \theta :a \rightarrow -^ta θ : a → − t a g \mathfrak{g} g θ ( X α ) = X − α \theta (X_{\alpha}) =X_{-\alpha} θ ( X α ) = X − α ( X α ) α ∈ R (X_{\alpha})_{\alpha\in R} ( X α ) α ∈ R ( g , h ) (\mathfrak{g},\mathfrak{h}) ( g , h )
假设k = Q k=\mathbf{Q} k = Q h \mathfrak{h} h Q ( R ∨ ) = ∑ i = 1 l Z . H i Q(R^{\vee}) =\sum_{i=1}^l\mathbf{Z}.H_i Q ( R ∨ ) = ∑ i = 1 l Z . H i P ( R ∨ ) = Q ( R ∨ ) + 1 2 Z . ∑ i = 1 l H i P(R^{\vee}) = Q(R^{\vee}) +\frac{1}{2}\mathbf{Z}.\sum_{i=1}^lH_i P ( R ∨ ) = Q ( R ∨ ) + 2 1 Z . ∑ i = 1 l H i
no. 5.VIII 章）。我们看到Q ( R ∨ ) Q(R^{\vee}) Q ( R ∨ ) h \mathfrak{h} h Q ( R ∨ ) + ∑ α ∈ R Z . X α Q(R^{\vee}) +\sum_{\alpha\in R}\mathbf{Z}.X_{\alpha} Q ( R ∨ ) + ∑ α ∈ R Z . X α g \mathfrak{g} g s p ( 2 l , Z ) \mathfrak{s}\mathfrak{p}(2l,\mathbf{Z}) sp ( 2 l , Z )
考虑约简李代数 s p ( Ψ ) + Q .1 \mathfrak{s}\mathfrak{p}(\Psi ) +\mathbf{Q}.1 sp ( Ψ ) + Q .1 s p ( Ψ ) \mathfrak{s}\mathfrak{p}(\Psi ) sp ( Ψ ) Q .1 \mathbf{Q}.1 Q .1 P ( R ∨ ) + ∑ Z . X α P(R^{\vee}) +\sum\mathbf{Z}.X_{\alpha} P ( R ∨ ) + ∑ Z . X α
最后，为所有α ∈ R \alpha \in R α ∈ R X α 2 = 0 X_{\alpha}^2= 0 X α 2 = 0 e i e_i e i V \mathscr{V} V s p ( 2 l , Z ) \mathfrak{s}\mathfrak{p}(2l,\mathbf{Z}) sp ( 2 l , Z ) E r E_r E r E r ∩ ⋀ r V E_r\cap \bigwedge^r\mathscr{V} E r ∩ ⋀ r V
最后，E r E_r E r
P ( R ∨ ) + ∑ Z . X α
P(R^{\vee}) +\sum\mathbf{Z}.X_{\alpha}
P ( R ∨ ) + ∑ Z . X α 仅当 r r r E r ∩ ⋀ r V E_r\cap \bigwedge^r\mathscr{V} E r ∩ ⋀ r V
4. 代数型 D l \boldsymbol{D}_{\boldsymbol{l}} D l l ≥ \boldsymbol{l}\geq l ≥
(I) 令V为维数2 l ≥ 4 2l\geq 4 2 l ≥ 4 Ψ \Psi Ψ l l l F ′ F' F ′ ( e i ) 1 ≤ i ≤ l (e_i)_{1\leq i\leq l} ( e i ) 1 ≤ i ≤ l F ′ F' F ′ ( e − i ) 1 ≤ i ≤ l (e_{-i})_{1\leq i\leq l} ( e − i ) 1 ≤ i ≤ l Ψ \Psi Ψ F ′ F' F ′ e 1 , . . . , e l , e − l , . . . , e − 1 e_1, . . . , e_l, e_{-l}, . . . , e_{-1} e 1 , ... , e l , e − l , ... , e − 1 Ψ \Psi Ψ 2 l 2l 2 l g = o ( Ψ ) \mathfrak{g}=\mathfrak{o}(\Psi ) g = o ( Ψ ) 2 l 2l 2 l g g g o S ( 2 l , k ) \mathfrak{o}_S(2l, k) o S ( 2 l , k ) g = − S t g S g=-S^tgS g = − S t g S l ( 2 l − 1 ) l(2l-1) l ( 2 l − 1 ) g \mathfrak{g} g
其中A , B , C , D A, B, C, D A , B , C , D l l l B = − s t B s , C = B=-s^tBs, C= B = − s t B s , C = − s t C s -s^tCs − s t C s D = − s t A s ( s D=-s^tAs(s D = − s t A s ( s l l l
令h \mathfrak{h} h g \mathfrak{g} g g \mathfrak{g} g H i = E i , i − E − i , − i H_i=E_{i,i}-E_{-i,-i} H i = E i , i − E − i , − i 1 ≤ i ≤ l 1\leq i\leq l 1 ≤ i ≤ l ( ε i ) (\varepsilon_i) ( ε i ) h ∗ \mathfrak{h}^* h ∗ ( H i ) (H_i) ( H i ) 1 ≤ i < j ≤ l 1\leq i < j\leq l 1 ≤ i < j ≤ l
X ε i − ε j X_{\varepsilon_i-\varepsilon_j} X ε i − ε j E i , j − E − j , − i E_{i,j}-E_{-j,-i} E i , j − E − j , − i
X − ε i + ε j X_{-\varepsilon_i+\varepsilon_j} X − ε i + ε j − E j , i + E − i , − j -E_{j,i}+E_{-i,-j} − E j , i + E − i , − j
X ε i + ε j X_{\varepsilon_i+\varepsilon_j} X ε i + ε j E i , − j − E j , − i E_{i,-j}-E_{j,-i} E i , − j − E j , − i
X − ε i − ε j X_{-\varepsilon_i-\varepsilon_j} X − ε i − ε j − E − j , i + E − i , j -E_{-j,i}+E_{-i,j} − E − j , i + E − i , j
这些元素形式基是g \mathfrak{g} g h \mathfrak{h} h h ∈ h h\in \mathfrak{h} h ∈ h
[ h , X α ] = α ( h ) X α
[h, X_{\alpha}] =\alpha (h)X_{\alpha}
[ h , X α ] = α ( h ) X α 对于所有α ∈ R \alpha \in R α ∈ R ± ε i ± ε j ( i < j ) \pm \varepsilon_i\pm \varepsilon_j(i < j) ± ε i ± ε j ( i < j ) h \mathfrak{h} h g \mathfrak{g} g ( g , h ) (\mathfrak{g},\mathfrak{h}) ( g , h ) ( g , h ) (\mathfrak{g},\mathfrak{h}) ( g , h ) D l D_l D l l ≥ 3 l\geq 3 l ≥ 3 A 1 × A 1 A_1\times A_1 A 1 × A 1 D 2 D_2 D 2 l = 2 l= 2 l = 2 l = 2 l= 2 l = 2 l ≥ 3 l\geq 3 l ≥ 3 g \mathfrak{g} g D l D_l D l
g \mathfrak{g} g g \mathfrak{g} g h \mathfrak{h} h O ( Ψ ) \mathbf{O}(\Psi ) O ( Ψ ) g \mathfrak{g} g h β \mathfrak{h}_{\beta} h β β \beta β h β \mathfrak{h}_{\beta} h β β \beta β
由于代数 o S ( 4 , k ) \mathfrak{o}_S(4, k) o S ( 4 , k ) s l ( 2 , k ) × s l ( 2 , k ) \mathfrak{s}\mathfrak{l}(2, k)\times \mathfrak{s}\mathfrak{l}(2, k) sl ( 2 , k ) × sl ( 2 , k ) o S ( 6 , k ) \mathfrak{o}_S(6, k) o S ( 6 , k ) s l ( 4 , k ) \mathfrak{s}\mathfrak{l}(4, k) sl ( 4 , k ) l ≥ 3 l\geq 3 l ≥ 3 R ∨ R^{\vee} R ∨
H ε i − ε j = H i − H j , H ε i + ε j = H i + H j
H_{\varepsilon_i-\varepsilon_j}=H_i-H_j,H_{\varepsilon_i+\varepsilon_j}=H_i+H_j
H ε i − ε j = H i − H j , H ε i + ε j = H i + H j （III）放入α 1 = ε 1 − ε 2 , α 2 = ε 2 − ε 3 , . . . , α l − 1 = ε l − 1 − ε l , α l = ε l − 1 + ε l \alpha_1=\varepsilon_1-\varepsilon_2, \alpha_2=\varepsilon_2-\varepsilon_3, . . . , \alpha_{l-1}=\varepsilon_{l-1}-\varepsilon_l, \alpha_l=\varepsilon_{l-1}+\varepsilon_l α 1 = ε 1 − ε 2 , α 2 = ε 2 − ε 3 , ... , α l − 1 = ε l − 1 − ε l , α l = ε l − 1 + ε l ( α 1 , . . . , α l ) (\alpha_1, . . . , \alpha_l) ( α 1 , ... , α l ) ε i ± ε j ( i < j ) \varepsilon_i\pm \varepsilon_j(i < j) ε i ± ε j ( i < j ) b \mathfrak{b} b g \mathfrak{g} g
很容易验证，b \mathfrak{b} b V 1 , . . . , V l , V l ′ V_1, . . . ,V_l,V'_l V 1 , ... , V l , V l ′ V i V_i V i e 1 , . . . , e i e_1, . . . , e_i e 1 , ... , e i V l ′ V'_l V l ′ e 1 , . . . , e l − 1 , e − l e_1, . . . , e_{l-1}, e_{-l} e 1 , ... , e l − 1 , e − l V 1 , . . . , V l − 1 V_1, . . . ,V_{l-1} V 1 , ... , V l − 1 V − 1 , . . . , V − l + 1 V_{-1}, . . . ,V_{-l+1} V − 1 , ... , V − l + 1 V i V_i V i V − i V_{-i} V − i e 1 , . . . , e l , e − l , . . . , e − ( i + 1 ) e_1, . . . , e_l, e_{-l}, . . . , e_{-(i+1)} e 1 , ... , e l , e − l , ... , e − ( i + 1 ) a ∈ g a\in \mathfrak{g} a ∈ g V l − 1 V_{l-1} V l − 1
( λ (\lambda ( λ ) ) )
0 0 − λ y -\lambda y − λ y
0 0 D D D
其中A , B , D A, B, D A , B , D l − 1 , x l-1,x l − 1 , x y y y l − 1 l-1 l − 1 l − 1 l-1 l − 1 λ ∈ k \lambda \in k λ ∈ k a a a V l V_l V l V l ′ V'_l V l ′ b \mathfrak{b} b a ∈ g a\in \mathfrak{g} a ∈ g ( V 1 , . . . , V l − 1 ) (V_1, . . . ,V_{l-1}) ( V 1 , ... , V l − 1 ) V l V_l V l V l ′ V'_l V l ′ V l − 1 V_{l-1} V l − 1
如果一个各向同性标志是由维数 1 , . . . , l − 1 1, . . . , l-1 1 , ... , l − 1 l − 1 l-1 l − 1 δ \delta δ a ∈ g a\in \mathfrak{g} a ∈ g b δ \mathfrak{b}_{\delta} b δ δ \delta δ g \mathfrak{g} g δ → b δ \delta \rightarrow \mathfrak{b}_{\delta} δ → b δ
We say that an isotropic flag is proper if it does not contain both a subspace of dimension l l l l − 1 l-1 l − 1 δ \delta δ p δ \mathfrak{p}_{\delta} p δ a ∈ g a\in \mathfrak{g} a ∈ g δ \delta δ δ ⊂ { V 1 , . . . , V l , V l ′ } \delta \subset \{V_1, . . . ,V_l,V'_l\} δ ⊂ { V 1 , ... , V l , V l ′ } p δ \mathfrak{p}_{\delta} p δ g \mathfrak{g} g b \mathfrak{b} b p δ \mathfrak{p}_{\delta} p δ ≠ { 0 } \not=\{0\} = { 0 } δ \delta δ { V 1 , . . . , V l , V l ′ } \{V_1, . . . ,V_l,V'_l\} { V 1 , ... , V l , V l ′ } V l − 1 V_{l-1} V l − 1 2 l − 2 2^{l-2} 2 l − 2 V l V_l V l V l ′ V'_l V l ′ V l − 1 V_{l-1} V l − 1 V l V_l V l V l ′ V'_l V l ′ 2 l 2^l 2 l b \mathfrak{b} b g \mathfrak{g} g δ → p δ \delta \rightarrow \mathfrak{p}_{\delta} δ → p δ
(IV) 基本的权相应的到 α 1 , . . . , α l \alpha_1, . . . , \alpha_l α 1 , ... , α l
ϖ i = ε 1 + ε 2 + ⋯ + ε i ( 1 ≤ i ≤ l − 2 )
\varpi_i=\varepsilon_1+\varepsilon_2+\cdots +\varepsilon_i(1\leq i\leq l-2)
ϖ i = ε 1 + ε 2 + ⋯ + ε i ( 1 ≤ i ≤ l − 2 ) ϖ l − 1 = 1 2 ( ε 1 + ε 2 + . . . + ε l − 2 + ε l − 1 − ε l )
\varpi_{l-1}=\frac{1}{2}(\varepsilon_1+\varepsilon_2+. . .+\varepsilon_{l-2}+\varepsilon_{l-1}-\varepsilon_l)
ϖ l − 1 = 2 1 ( ε 1 + ε 2 + ... + ε l − 2 + ε l − 1 − ε l ) ϖ l = 1 2 ( ε 1 + ε 2 + . . . + ε l − 2 + ε l − 1 + ε l )
\varpi_l=\frac{1}{2}(\varepsilon_1+\varepsilon_2+. . .+\varepsilon_{l-2}+\varepsilon_{l-1}+\varepsilon_l)
ϖ l = 2 1 ( ε 1 + ε 2 + ... + ε l − 2 + ε l − 1 + ε l ) 令σ \sigma σ g \mathfrak{g} g ⋀ r σ \bigwedge^r\sigma ⋀ r σ E = ⋀ r ( V ) E =\bigwedge^r(V) E = ⋀ r ( V ) h ∈ h h\in \mathfrak{h} h ∈ h
σ ( h ) e i = ε i ( h ) e i , σ ( h ) e − i = − ε i ( h ) e − i
\sigma (h)e_i=\varepsilon_i(h)e_i,\sigma (h)e_{-i}=-\varepsilon_i(h)e_{-i}
σ ( h ) e i = ε i ( h ) e i , σ ( h ) e − i = − ε i ( h ) e − i 对于1 ≤ i ≤ l 1\leq i\leq l 1 ≤ i ≤ l 1 ≤ r ≤ l , ε 1 + ⋯ + ε r 1\leq r\leq l,\varepsilon_1+\cdots +\varepsilon_r 1 ≤ r ≤ l , ε 1 + ⋯ + ε r ⋀ r σ \bigwedge^r\sigma ⋀ r σ ε 1 + ⋯ + ε r \varepsilon_1+\cdots +\varepsilon_r ε 1 + ⋯ + ε r e 1 ∧ ⋯ ∧ e r e_1\wedge \cdots \wedge e_r e 1 ∧ ⋯ ∧ e r
我们将证明，对于1 ≤ r ≤ l − 2 1\leq r\leq l-2 1 ≤ r ≤ l − 2 ⋀ r σ \bigwedge^r\sigma ⋀ r σ ϖ r \varpi_r ϖ r
为此，只需证明 ⋀ r σ \bigwedge^r\sigma ⋀ r σ 1 ≤ r ≤ l − 1 1\leq r\leq l-1 1 ≤ r ≤ l − 1 ⋀ l σ \bigwedge^l\sigma ⋀ l σ T r T_r T r ⋀ r V \bigwedge^rV ⋀ r V e 1 ∧ ⋯ ∧ e r e_1\wedge \cdots \wedge e_r e 1 ∧ ⋯ ∧ e r g \mathfrak{g} g ⋀ r V \bigwedge^rV ⋀ r V r = 1 r= 1 r = 1 r = 2 r= 2 r = 2 ⋀ 2 σ \bigwedge^2\sigma ⋀ 2 σ g \mathfrak{g} g g \mathfrak{g} g l l l l − 1 ≥ r ≥ 3 l-1\geq r\geq 3 l − 1 ≥ r ≥ 3
我们现在要去确定最高权ϖ l − 1 \varpi_{l-1} ϖ l − 1 ϖ l \varpi_l ϖ l x → 1 2 Ψ ( x , x ) x \rightarrow \frac{1}{2}\Psi (x, x) x → 2 1 Ψ ( x , x ) N = ⋀ F ′ N =\bigwedge F' N = ⋀ F ′ λ \lambda λ p p p ⋀ p F ′ \bigwedge^pF' ⋀ p F ′ N + N_+ N + N − N_- N − λ \lambda λ C + ( Q ) C^+(Q) C + ( Q ) N + N_+ N + N − N_- N − C + ( Q ) C^+(Q) C + ( Q ) λ + \lambda_+ λ + λ − \lambda_- λ − C + ( Q ) ( C^+(Q) ( C + ( Q ) ( 2 l − 1 2^{l-1} 2 l − 1 ρ + = λ + ∘ f \rho_+=\lambda_+\circ f ρ + = λ + ∘ f ρ − = λ − ∘ f \rho_-=\lambda_-\circ f ρ − = λ − ∘ f g \mathfrak{g} g
f ( H i ) = 1 2 ( e i e − i − e − i e i ) = e i e − i − 1 2 = 1 2 − e − i e i
f(H_i) =\frac{1}{2}(e_ie_{-i}-e_{-i}e_i) =e_ie_{-i}-\frac{1}{2}=\frac{1}{2}-e_{-i}e_i
f ( H i ) = 2 1 ( e i e − i − e − i e i ) = e i e − i − 2 1 = 2 1 − e − i e i 我们看到，如no. 2.IV，对于h ∈ h h\in \mathfrak{h} h ∈ h 1 ≤ i 1 < ⋯ < i k ≤ l 1\leq i_1<\cdots < i_k\leq l 1 ≤ i 1 < ⋯ < i k ≤ l
λ ∘ f ( h ) ( e − i 1 ∧ ⋯ ∧ e − i k )
\lambda \circ f(h)(e_{-i_1}\wedge \cdots \wedge e_{-i_k})
λ ∘ f ( h ) ( e − i 1 ∧ ⋯ ∧ e − i k ) = ( 1 2 ( ε 1 + ⋯ + ε l ) − ( ε i 1 + ⋯ + ε i k ) ) ( h ) ( e − i 1 ∧ ⋯ ∧ e − i k )
= (\frac{1}{2}(\varepsilon_1+\cdots +\varepsilon_l)-(\varepsilon_{i_1}+\cdots +\varepsilon_{i_k}))(h)(e_{-i_1}\wedge \cdots \wedge e_{-i_k})
= ( 2 1 ( ε 1 + ⋯ + ε l ) − ( ε i 1 + ⋯ + ε i k )) ( h ) ( e − i 1 ∧ ⋯ ∧ e − i k ) 因而，ρ + \rho_+ ρ + ρ − \rho_- ρ − ϖ l \varpi_l ϖ l ϖ l − 1 \varpi_{l-1} ϖ l − 1
我们说ρ + \rho_+ ρ + ρ − \rho_- ρ − g \mathfrak{g} g ρ = λ ∘ f = ρ + ⊕ ρ − \rho =\lambda \circ f=\rho_+\oplus \rho_- ρ = λ ∘ f = ρ + ⊕ ρ − g \mathfrak{g} g
(V) 对于1 ≤ r ≤ l − 2 1\leq r\leq l-2 1 ≤ r ≤ l − 2 ⋀ r σ \bigwedge^r\sigma ⋀ r σ Ψ \Psi Ψ ⋀ r V \bigwedge^rV ⋀ r V
现在考虑 g \mathfrak{g} g ρ \rho ρ 1 ≤ i , j ≤ l , i ≠ j 1\leq i, j\leq l, i\not=j 1 ≤ i , j ≤ l , i = j
f ( X ε i − ε j ) = ± e i e − j
f(X_{\varepsilon_i-\varepsilon_j}) =\pm e_ie_{-j}
f ( X ε i − ε j ) = ± e i e − j f ( X ε i + ε j ) = ± e i e j
f(X_{\varepsilon_i+\varepsilon_j}) =\pm e_ie_j
f ( X ε i + ε j ) = ± e i e j f ( X − ε i − ε j ) = ± e − i e − j
f(X_{-\varepsilon_i-\varepsilon_j}) =\pm e_{-i}e_{-j}
f ( X − ε i − ε j ) = ± e − i e − j 由此可见，no. 2.V中引入的非简并双线性形式Φ \Phi Φ ρ ( g ) \rho (\mathfrak{g}) ρ ( g ) ρ \rho ρ l ≡ 0 , − 1 l\equiv 0,-1 l ≡ 0 , − 1 l ≡ 1 , 2 l\equiv 1,2 l ≡ 1 , 2
如果 l l l Φ \Phi Φ N + N_+ N + N − N_- N − l ≡ 0 l\equiv 0 l ≡ 0 l ≡ 2 l\equiv 2 l ≡ 2 w 0 = − 1 w_0=-1 w 0 = − 1
另一方面，如果 l l l N + N_+ N + N − N_- N − Φ \Phi Φ − w 0 ( α i ) = α i -w_0(\alpha_i) =\alpha_i − w 0 ( α i ) = α i 1 ≤ i ≤ l − 2 , − w 0 ( α l ) = α l − 1 1\leq i\leq l-2,-w_0(\alpha_l) =\alpha_{l-1} 1 ≤ i ≤ l − 2 , − w 0 ( α l ) = α l − 1 − w 0 ( α l − 1 ) = α l -w_0(\alpha_{l-1}) =\alpha_l − w 0 ( α l − 1 ) = α l − w 0 ( ϖ l ) = ϖ l − 1 -w_0(\varpi_l) =\varpi_{l-1} − w 0 ( ϖ l ) = ϖ l − 1
(VI) 对于所有x ∈ g x\in \mathfrak{g} x ∈ g σ ( x ) \sigma (x) σ ( x )
T 2 l + f 1 ( x ) T 2 l − 1 + ⋯ + f 2 l ( x )
T^{2l}+f_1(x)T^{2l-1}+\cdots +f_{2l}(x)
T 2 l + f 1 ( x ) T 2 l − 1 + ⋯ + f 2 l ( x ) 我们在no. 3中看到f 1 = f 3 = f 5 = ⋯ = 0 f_1=f_3=f_5=\cdots = 0 f 1 = f 3 = f 5 = ⋯ = 0 g \mathfrak{g} g f ~ \widetilde{f} f g \mathfrak{g} g f 2 , f 4 , . . . , f 2 l − 2 , f ~ f_2, f_4, . . . , f_{2l-2},\widetilde{f} f 2 , f 4 , ... , f 2 l − 2 , f I ( g ∗ ) I(\mathfrak{g}^*) I ( g ∗ ) f ~ 2 = ( − 1 ) l f 2 l \widetilde{f}^2= (-1)^lf_{2l} f 2 = ( − 1 ) l f 2 l
对于所有x ∈ g x\in \mathfrak{g} x ∈ g t ( S x ) = t x S = − S x ^t(Sx) =^txS=-Sx t ( S x ) = t x S = − S x S x Sx S x x x x
f 2 l ( x ) = f_{2l}(x) = f 2 l ( x ) = x x x = ( − 1 ) l = (-1)^l = ( − 1 ) l S x Sx S x = ( − 1 ) l = (-1)^l = ( − 1 ) l S x Sx S x 2 ^2 2
因此，我们可以取 f ~ ( x ) = \widetilde{f}(x) = f ( x ) = S x Sx S x
(VII) 回想一下（§5, no. 3，命题 5 的修正 1），Aut(g \mathfrak{g} g / / / 0 ( g ) _0(\mathfrak{g}) 0 ( g ) ( g , h ) (\mathfrak{g},\mathfrak{h}) ( g , h ) l ≠ 4 l\not= 4 l = 4 α 1 , . . . , α l \alpha_1, . . . , \alpha_l α 1 , ... , α l α 1 , . . . , α l − 2 \alpha_1, . . . , \alpha_{l-2} α 1 , ... , α l − 2 l = 4 l= 4 l = 4 α 1 , . . . , α 4 \alpha_1, . . . , \alpha_4 α 1 , ... , α 4 α 2 \alpha_2 α 2 S 3 \mathfrak{S}_3 S 3 α 1 \alpha_1 α 1 g \mathfrak{g} g ′ ( g ) '(\mathfrak{g}) ′ ( g ) ′ ( g ) = '(\mathfrak{g}) = ′ ( g ) = g \mathfrak{g} g l ≠ 4 l\not= 4 l = 4 g \mathfrak{g} g : : : ′ ( g ) '(\mathfrak{g}) ′ ( g ) = 3 = 3 = 3 l = 4 l= 4 l = 4
（自动′ ( g ) : '(\mathfrak{g}) : ′ ( g ) : 0 ( g ) _0(\mathfrak{g}) 0 ( g ) = 2 = 2 = 2
元素 s ∈ s\in s ∈ g \mathfrak{g} g ′ ( g ) '(\mathfrak{g}) ′ ( g ) σ ∘ s \sigma \circ s σ ∘ s σ \sigma σ ϖ 1 \varpi_1 ϖ 1 σ \sigma σ ′ ( g ) '(\mathfrak{g}) ′ ( g ) Σ / k ∗ \Sigma /k^* Σ/ k ∗ Σ \Sigma Σ Ψ \Psi Ψ
令s ∈ Σ s\in \Sigma s ∈ Σ λ ( s ) \lambda (s) λ ( s ) s s s s s s s s s = λ ( s ) l =\lambda (s)^l = λ ( s ) l s s s s s s = − λ ( s ) l =-\lambda (s)^l = − λ ( s ) l Σ \Sigma Σ Σ 0 \Sigma_0 Σ 0 Σ 0 ⊃ k ∗ \Sigma_0\supset k^* Σ 0 ⊃ k ∗ Σ 0 / k ∗ \Sigma_0/k^* Σ 0 / k ∗ ′ ( g ) = Σ / k ∗ '(\mathfrak{g}) =\Sigma /k^* ′ ( g ) = Σ/ k ∗ 0 ( g ) _0(\mathfrak{g}) 0 ( g ) k k k 0 ( g ) = _0(\mathfrak{g}) = 0 ( g ) = e ( g ) _e(\mathfrak{g}) e ( g ) Σ 0 / k ∗ \Sigma_0/k^* Σ 0 / k ∗ Σ / k ∗ \Sigma /k^* Σ/ k ∗
另一方面，如no. 3.VII，有正合序列
1 ⟶ S O ( Ψ ) / { 1 , − 1 } ⟶ Σ 0 / k ∗ ⟶ k ∗ / k ∗ 2 ⟶ 1
1\longrightarrow \mathbf{S}\mathbf{O}(\Psi )/\{1,-1\} \longrightarrow \Sigma_0/k^*\longrightarrow k^*/k^{*2}\longrightarrow 1
1 ⟶ SO ( Ψ ) / { 1 , − 1 } ⟶ Σ 0 / k ∗ ⟶ k ∗ / k ∗ 2 ⟶ 1 认同 S O ( Ψ ) / { 1 , − 1 } \mathbf{S}\mathbf{O}(\Psi )/\{1,-1\} SO ( Ψ ) / { 1 , − 1 } Σ 0 / k ∗ = \Sigma_0/k^*= Σ 0 / k ∗ = 0 ( g ) _0(\mathfrak{g}) 0 ( g ) k ∗ / k ∗ 2 k^*/k^{*2} k ∗ / k ∗ 2 e ( g ) ⊂ S O ( Ψ ) / { 1 , − 1 } _e(\mathfrak{g})\subset \mathbf{S}\mathbf{O}(\Psi )/\{1,-1\} e ( g ) ⊂ SO ( Ψ ) / { 1 , − 1 } e ( g ) _e(\mathfrak{g}) e ( g ) Ψ ( \Psi ( Ψ ( O 0 + ( Ψ ) \mathbf{O}^+_0(\Psi ) O 0 + ( Ψ ) S O ( Ψ ) / { 1 , − 1 } \mathbf{S}\mathbf{O}(\Psi )/\{1,-1\} SO ( Ψ ) / { 1 , − 1 }
(VIII) h ∗ \mathfrak{h}^* h ∗ Φ R \Phi_R Φ R
′ ′ '' ′′ ′ ′ '' ′′
Φ R ( ξ 1 ε 1 + ⋯ + ξ l ε l , ξ 1 ε 1 + ⋯ + ξ l ε l ) = ( ξ 1 ξ + ⋯ + ξ l ξ )
\Phi_R(\xi_1\varepsilon_1+\cdots +\xi_l\varepsilon_l, \xi_1\varepsilon_1+\cdots +\xi_l\varepsilon_l) =(\xi_1\xi +\cdots +\xi_l\xi )
Φ R ( ξ 1 ε 1 + ⋯ + ξ l ε l , ξ 1 ε 1 + ⋯ + ξ l ε l ) = ( ξ 1 ξ + ⋯ + ξ l ξ ) 4 ( l − 1 ) ‾ 1 l
\overline{4(l- 1)}^1l
4 ( l − 1 ) 1 l （第 VI, §4, no. 8.V）。因而，杀戮形式到h \mathfrak{h} h
Φ ( ξ 1 H 1 + ⋯ + ξ l H l , ξ 1 ′ H 1 + ⋯ + ξ l ′ H l ) = 4 ( l − 1 ) ( ξ 1 ξ 1 ′ + ⋯ + ξ l ξ l ′ )
\Phi (\xi_1H_1+\cdots +\xi_lH_l, \xi '_1H_1+\cdots +\xi '_lH_l) = 4(l-1)(\xi_1\xi '_1+\cdots +\xi_l\xi '_l)
Φ ( ξ 1 H 1 + ⋯ + ξ l H l , ξ 1 ′ H 1 + ⋯ + ξ l ′ H l ) = 4 ( l − 1 ) ( ξ 1 ξ 1 ′ + ⋯ + ξ l ξ l ′ ) (IX) 回想一下公式 (11) 定义的X α ( α ∈ R ) X_{\alpha}(\alpha \in R) X α ( α ∈ R ) [ X α , X − α ] = − H α [X_{\alpha}, X_{-\alpha}] =-H_{\alpha} [ X α , X − α ] = − H α α ∈ R \alpha \in R α ∈ R α ∈ R \alpha \in R α ∈ R θ : a → − t a \theta :a \rightarrow -^ta θ : a → − t a g \mathfrak{g} g θ ( X α ) = X − α \theta (X_{\alpha}) =X_{-\alpha} θ ( X α ) = X − α ( X α ) α ∈ R (X_{\alpha})_{\alpha\in R} ( X α ) α ∈ R ( g , h ) (\mathfrak{g},\mathfrak{h}) ( g , h )
假设k = Q k=\mathbf{Q} k = Q l l l h \mathfrak{h} h l l l H i H_i H i H \mathscr{H} H g \mathfrak{g} g o S ( 2 l , Z ) \mathfrak{o}_S(2l,\mathbf{Z}) o S ( 2 l , Z ) g \mathfrak{g} g H + ∑ Z . X α \mathscr{H}+\sum\mathbf{Z}.X_{\alpha} H + ∑ Z . X α X α 2 = 0 X_{\alpha}^2= 0 X α 2 = 0 α ∈ R \alpha \in R α ∈ R ( e i ) (e_i) ( e i ) V \mathscr{V} V o S ( 2 l , Z ) \mathfrak{o}_S(2l,\mathbf{Z}) o S ( 2 l , Z ) ⋀ r V \bigwedge^rV ⋀ r V ⋀ r V \bigwedge^r\mathscr{V} ⋀ r V
另一方面，如果我们将 P ( R ∨ ) = Z . 1 2 ∑ i = 1 l H i + H P(R^{\vee}) =\mathbf{Z}.\frac{1}{2}\sum_{i=1}^lH_i+\mathscr{H} P ( R ∨ ) = Z . 2 1 ∑ i = 1 l H i + H G = P ( R ∨ ) + ∑ Z . X α \mathscr{G}= P(R^{\vee}) +\sum\mathbf{Z}.X_{\alpha} G = P ( R ∨ ) + ∑ Z . X α r r r ⋀ r V \bigwedge^rV ⋀ r V ⋀ r V \bigwedge^r\mathscr{V} ⋀ r V
考虑约简李代数 o ( Ψ ) + Q .1 \mathfrak{o}(\Psi ) +\mathbf{Q}.1 o ( Ψ ) + Q .1 G ~ = ( o ( Ψ ) + Q .1 ) ∩ g l ( 2 l , Z ) \widetilde{\mathscr{G}}= (\mathfrak{o}(\Psi ) +\mathbf{Q}.1)\cap \mathfrak{g}\mathfrak{l}(2l,\mathbf{Z}) G = ( o ( Ψ ) + Q .1 ) ∩ gl ( 2 l , Z ) G \mathscr{G} G G ~ \widetilde{\mathscr{G}} G o ( Ψ ) \mathfrak{o}(\Psi ) o ( Ψ ) Q .1 \mathbf{Q}.1 Q .1
最后，我们在no. 2中看到，由e − i 1 ∧ ⋯ ∧ e − i 2 k e_{-i_1}\wedge \cdots \wedge e_{-i_{2k}} e − i 1 ∧ ⋯ ∧ e − i 2 k e − i 1 ∧ ⋯ ∧ e − i 2 k + 1 e_{-i_1}\wedge \cdots \wedge e_{-i_{2k+1}} e − i 1 ∧ ⋯ ∧ e − i 2 k + 1 N + \mathscr{N}_+ N + N − \mathscr{N}_- N − Q ( R ∨ ) + ∑ α ∈ R Z . X α Q(R^{\vee}) +\sum_{\alpha\in R}\mathbf{Z}.X_{\alpha} Q ( R ∨ ) + ∑ α ∈ R Z . X α N + \mathscr{N}_+ N + N − \mathscr{N}_- N − o S ( 2 l , Z ) \mathfrak{o}_S(2l,\mathbf{Z}) o S ( 2 l , Z )
表1
如果相应的简单表示是正交的（分别为辛，分别为既不正交也不辛），我们就将每个基数权数 1（分别为 − 1 , 0 -1,0 − 1 , 0 A l , B l , C l , D l A_l,B_l,C_l,D_l A l , B l , C l , D l E 6 , E 7 , E 8 , F 4 , G 2 E_6,E_7,E_8,F_4,G_2 E 6 , E 7 , E 8 , F 4 , G 2
A l ( l ≥ 1 ) B l ( l ≥ 2 )
A_l(l\geq 1)B_l(l\geq 2)
A l ( l ≥ 1 ) B l ( l ≥ 2 ) { 0 r ≠ l + 1 2 ϖ r \{0r\not=\frac{l+1}{2}\varpi_r { 0 r = 2 l + 1 ϖ r r ≠ l r\not=l r = l
( − 1 ) r r = l + 1 2 ϖ l ( − 1 ) l ( l + 1 ) / 2
(-1)^rr=\frac{l+1}{2}\varpi_l(-1)^{l(l+1)/2}
( − 1 ) r r = 2 l + 1 ϖ l ( − 1 ) l ( l + 1 ) /2 C l ( l ≥ 2 ) D l ( l ≥ 2 )
C_l(l\geq 2)D_l(l\geq 2)
C l ( l ≥ 2 ) D l ( l ≥ 2 ) ϖ r ( − 1 ) r ϖ r \varpi_r(-1)^r\varpi_r ϖ r ( − 1 ) r ϖ r r ≠ l − 1 , l r\not=l-1, l r = l − 1 , l
{ 0 \{0 { 0 l l l
如果 l l l ϖ l \varpi_l ϖ l ϖ l − 1 ( − 1 ) l / 2 \varpi_{l-1}(-1)^{l/2} ϖ l − 1 ( − 1 ) l /2
E 6 E 7 E 8 F 4 G 2
E_6E_7E_8F_4G_2
E 6 E 7 E 8 F 4 G 2 ϖ 1 \varpi_1 ϖ 1 ϖ 1 \varpi_1 ϖ 1 ϖ 1 \varpi_1 ϖ 1 ϖ 1 \varpi_1 ϖ 1 ϖ 1 \varpi_1 ϖ 1
ϖ 2 \varpi_2 ϖ 2 ϖ 2 − 1 ϖ 2 \varpi_2-1\varpi_2 ϖ 2 − 1 ϖ 2 ϖ 2 \varpi_2 ϖ 2 ϖ 2 \varpi_2 ϖ 2
ϖ 3 \varpi_3 ϖ 3 ϖ 3 \varpi_3 ϖ 3 ϖ 3 \varpi_3 ϖ 3 ϖ 3 \varpi_3 ϖ 3
ϖ 4 \varpi_4 ϖ 4 ϖ 4 \varpi_4 ϖ 4 ϖ 4 \varpi_4 ϖ 4 ϖ 4 \varpi_4 ϖ 4
ϖ 5 \varpi_5 ϖ 5 ϖ 5 − 1 ϖ 5 \varpi_5-1\varpi_5 ϖ 5 − 1 ϖ 5
ϖ 6 \varpi_6 ϖ 6 ϖ 6 \varpi_6 ϖ 6 ϖ 6 \varpi_6 ϖ 6
ϖ 7 − 1 ϖ 7 \varpi_7-1\varpi_7 ϖ 7 − 1 ϖ 7
ϖ 8 \varpi_8 ϖ 8
表2
我们将相应的简单表示的每个基本权维数联系起来，通过Th计算。 2 个§9。
A l ( l ≥ 1 ) B l ( l ≥ 2 )
A_l(l\geq 1)B_l(l\geq 2)
A l ( l ≥ 1 ) B l ( l ≥ 2 ) ( l + 1 ) ( 2 l + 1 )
(l+ 1)(2l+ 1)
( l + 1 ) ( 2 l + 1 ) ϖ r ( 1 ≤ r ≤ l ) ϖ r ( 1 ≤ r ≤ l − 1 )
\varpi_r(1\leq r\leq l)\varpi_r(1\leq r\leq l-1)
ϖ r ( 1 ≤ r ≤ l ) ϖ r ( 1 ≤ r ≤ l − 1 ) C l ( l ≥ 2 ) D l ( l ≥ 2 )
C_l(l\geq 2)D_l(l\geq 2)
C l ( l ≥ 2 ) D l ( l ≥ 2 ) ( 2 l ) ( 2 l ) ( 2 l )
(2l)(2l)(2l)
( 2 l ) ( 2 l ) ( 2 l ) ϖ r ( 1 ≤ r ≤ l ) − ϖ r ( 1 ≤ r ≤ l − 2 )
\varpi_r(1\leq r\leq l)-\varpi_r(1\leq r\leq l-2)
ϖ r ( 1 ≤ r ≤ l ) − ϖ r ( 1 ≤ r ≤ l − 2 ) ϖ l − 1 2 l − 1
\varpi_{l-1}2^{l-1}
ϖ l − 1 2 l − 1 ϖ l 2 l − 1
\varpi_l2^{l-1}
ϖ l 2 l − 1 ϖ 1 27 = 3 3 ϖ 1 133 = 7.19
\varpi_127 = 3^3\varpi_1133 = 7.19
ϖ 1 27 = 3 3 ϖ 1 133 = 7.19 ϖ 2 78 = 2.3.13 ϖ 2 912 = 2 4 .3.19
\varpi_278 = 2.3.13\varpi_2912 = 2^4.3.19
ϖ 2 78 = 2.3.13 ϖ 2 912 = 2 4 .3.19 ϖ 3 351 = 3 3 .13 ϖ 3 8645 = 5.7.13.19
\varpi_3351 = 3^3.13\varpi_38645 = 5.7.13.19
ϖ 3 351 = 3 3 .13 ϖ 3 8645 = 5.7.13.19 ϖ 4 2925 = 3 2 .5 2 .13 ϖ 4 365750 = 2.5 3 .7.11.19
\varpi_42925 = 3^2.5^2.13\varpi_4365750 = 2.5^3.7.11.19
ϖ 4 2925 = 3 2 . 5 2 .13 ϖ 4 365750 = 2. 5 3 .7.11.19 ϖ 5 351 = 3 3 .13 ϖ 5 27664 = 2 4 .7.13.19
\varpi_5351 = 3^3.13\varpi_527664 = 2^4.7.13.19
ϖ 5 351 = 3 3 .13 ϖ 5 27664 = 2 4 .7.13.19 ϖ 6 27 = 3 3 ϖ 6 1539 = 3 4 .19
\varpi_627 = 3^3\varpi_61539 = 3^4.19
ϖ 6 27 = 3 3 ϖ 6 1539 = 3 4 .19 ϖ 7 56 = 2 3 .7
\varpi_756 = 2^3.7
ϖ 7 56 = 2 3 .7 ϖ 1 3875 = 5 3 .31 ϖ 1 52 = 2 2 .13
\varpi_13875 = 5^3.31\varpi_152 = 2^2.13
ϖ 1 3875 = 5 3 .31 ϖ 1 52 = 2 2 .13 ϖ 2 147250 = 2.5 3 .19.31 ϖ 2 1274 = 2.7 2 .13
\varpi_2147250 = 2.5^3.19.31\varpi_21274 = 2.7^2.13
ϖ 2 147250 = 2. 5 3 .19.31 ϖ 2 1274 = 2. 7 2 .13 ϖ 3 6696000 = 2 6 .3 3 .5 3 .31 ϖ 3 273 = 3.7.13
\varpi_36696000 = 2^6.3^3.5^3.31\varpi_3273 = 3.7.13
ϖ 3 6696000 = 2 6 . 3 3 . 5 3 .31 ϖ 3 273 = 3.7.13 ϖ 4 6899079264 = 2 5 .3.7 2 .11 2 .17.23.31 ϖ 4 26 = 2.13
\varpi_46899079264 = 2^5.3.7^2.11^2.17.23.31\varpi_426 = 2.13
ϖ 4 6899079264 = 2 5 .3. 7 2 .1 1 2 .17.23.31 ϖ 4 26 = 2.13 ϖ 5 146325270 = 2.3.5.7 2 .13 2 .19.31
\varpi_5146325270 = 2.3.5.7^2.13^2.19.31
ϖ 5 146325270 = 2.3.5. 7 2 .1 3 2 .19.31 ϖ 6 2450240 = 2 6 .5.13.19.31
\varpi_62450240 = 2^6.5.13.19.31
ϖ 6 2450240 = 2 6 .5.13.19.31 ϖ 7 30380 = 2 2 .5.7 2 .31 G 2
\varpi_730380 = 2^2.5.7^2.31G_2
ϖ 7 30380 = 2 2 .5. 7 2 .31 G 2 ϖ 8 248 = 2 3 .31 ϖ 1 \varpi_8248 = 2^3.31\varpi_1 ϖ 8 248 = 2 3 .31 ϖ 1
ϖ 2 14 = 2.7
\varpi_214 = 2.7
ϖ 2 14 = 2.7 习题
参见习题为 § 13 。