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。
§ 1。 根系
在本段中，k k k k = R k = \mathbf{R} k = R
1. 定义 / 根系
令 V V V k k k R R R V V V V V V α ∈ R \alpha \in R α ∈ R α ≠ 0 \alpha \neq 0 α = 0 V V V s s s s ( α ) = − α s(\alpha) = -\alpha s ( α ) = − α s ( R ) = R s(R) = R s ( R ) = R
被以下引用 1 被以下引用 定义 1
令G G G V V V R R R R R R V V V G G G R R R s , s ′ s, s' s , s ′ V V V s ( α ) = s ′ ( α ) = − α s(\alpha) = s'(\alpha) = -\alpha s ( α ) = s ′ ( α ) = − α s ( R ) = R s(R) = R s ( R ) = R s ′ ( R ) = R s'(R) = R s ′ ( R ) = R t = s s ′ t = ss' t = s s ′ G G G m m m
t ( α ) = α and t ( x ) ≡ x m o d . k α for all x ∈ V .
t(\alpha) = \alpha \quad \text{and} \quad t(x) \equiv x \mod. k\alpha \quad \text{for all } x \in V.
t ( α ) = α and t ( x ) ≡ x mod . k α for all x ∈ V . 因此，在V V V f f f
t ( x ) = x + f ( x ) α for all x ∈ V
t(x) = x + f(x)\alpha \quad \text{for all } x \in V
t ( x ) = x + f ( x ) α for all x ∈ V 和f ( α ) = 0 f(\alpha) = 0 f ( α ) = 0 n n n
t n ( x ) = x + n f ( x ) α for all x ∈ V .
t^n(x) = x + nf(x)\alpha \text{ for all } x \in V.
t n ( x ) = x + n f ( x ) α for all x ∈ V . 令n n n m m m m f ( x ) = 0 mf(x) = 0 m f ( x ) = 0 x ∈ V x \in V x ∈ V f = 0 f = 0 f = 0 t = 1 t = 1 t = 1 s = s ′ s = s' s = s ′
令 V V V k k k R R R V V V R R R V V V
引用了 2 引用了 V § 2. 。反思 , 引理 1
(RS_I) R R R V V V α ∈ R \alpha \in R α ∈ R V V V V ∗ V^* V ∗ α ∨ \alpha^\vee α ∨ ⟨ α , α ∨ ⟩ = 2 \langle \alpha, \alpha^\vee \rangle = 2 ⟨ α , α ∨ ⟩ = 2 s α , α ∨ s_{\alpha, \alpha^\vee} s α , α ∨ R R R
(RS_{\text{III}}) 对于所有α ∈ R \alpha \in \mathbf{R} α ∈ R α ∼ ( R ) ⊆ Z \alpha^{\sim}(R) \subseteq \mathbf{Z} α ∼ ( R ) ⊆ Z
通过引理 1，反射s α , α ∼ s_{\alpha,\alpha^{\sim}} s α , α ∼ α ∼ \alpha^{\sim} α ∼ α \alpha α s α , α ∼ = s α s_{\alpha,\alpha^{\sim}} = s_{\alpha} s α , α ∼ = s α x ∈ V x \in V x ∈ V s α ( x ) = x − ⟨ α ∼ , x ⟩ α s_{\alpha}(x) = x - \langle \alpha^{\sim}, x \rangle \alpha s α ( x ) = x − ⟨ α ∼ , x ⟩ α
R 的元素称为根（考虑到系统）。 V的维数称为系统的秩。
使 R 保持稳定的 V 的自同构称为 R 的自同构。它们形式有限群用 A(R) 表示。由s α s_{\alpha} s α
令 R 为 V 中的根系。令( x ∣ y ) (x|y) ( x ∣ y ) V ∗ V^* V ∗ α ∈ R \alpha \in R α ∈ R α \alpha α
α ∼ = 2 α ( α ∣ α ) .
\alpha^{\sim} = \frac{2\alpha}{(\alpha|\alpha)}.
α ∼ = ( α ∣ α ) 2 α . 引用了 1 · 被以下引用 5 引用了 V § 2. 。反思
被以下引用 § 1. 。 根系 , 推论 , 命题 1 , 命题 4 , 引理 4, VIII § 2
这是从第 7 章公式 (4) 得出的。 V, § 2, no. 3。
令V Q V_Q V Q V Q ∗ V_Q^* V Q ∗ α \alpha α α ∼ \alpha^{\sim} α ∼ V ∗ V^* V ∗ V Q V_Q V Q V Q ∗ V_Q^* V Q ∗ V ∗ V^* V ∗ V × V ∗ V \times V^* V × V ∗ V Q × V Q ∗ V_Q \times V_Q^* V Q × V Q ∗ V Q , V Q ∗ V_Q, V_Q^* V Q , V Q ∗ V Q V_Q V Q
引用了 3 · 被以下引用 5 引用了 A II § 8. 。 向量空间中标量的域的限制 , 引理 2 , 注 1
被以下引用 注 2 , 命题 2 , 命题 3 , 命题 9, VI § 2 , 注 1, VIII § 2
如果 k = R k = \mathbf{R} k = R α ∼ \alpha^{\sim} α ∼ V ∗ V^* V ∗ α ∼ \alpha^{\sim} α ∼ V ∗ V^* V ∗ k = Q k = \mathbf{Q} k = Q V Q V_Q V Q α ∼ \alpha^{\sim} α ∼ Q \mathbf{Q} Q α ~ \tilde{\alpha} α ~ E ∗ E^* E ∗ E ∗ E^* E ∗ α \alpha α α ~ \tilde{\alpha} α ~ α ~ \tilde{\alpha} α ~ E ∗ E^* E ∗ i : E ⊗ Q k → V i : E \otimes_Q k \to V i : E ⊗ Q k → V t i : V ∗ → E ∗ ⊗ Q k ^t i : V^* \to E^* \otimes_Q k t i : V ∗ → E ∗ ⊗ Q k t i ^t i t i t i ^t i t i α ~ \tilde{\alpha} α ~ t i ^t i t i t i ^t i t i E ⊗ k E \otimes k E ⊗ k V ∗ V^* V ∗ E ∗ ⊗ k E^* \otimes k E ∗ ⊗ k α ∼ \alpha^{\sim} α ∼ α ~ \tilde{\alpha} α ~ V Q ∗ V_Q^* V Q ∗ E ∗ E^* E ∗ V Q V_Q V Q V Q ∗ V_Q^* V Q ∗ V ∗ V^* V ∗ V × V ∗ V \times V^* V × V ∗ V Q × V Q ∗ V_Q \times V_Q^* V Q × V Q ∗ E × E ∗ E \times E^* E × E ∗
V ∗ V^* V ∗ α ~ \alpha \tilde{} α ~ α ∈ R \alpha \in R α ∈ R α ~ = α \alpha \tilde{} = \alpha α ~ = α
引用了 1 · 被以下引用 1 引用了 命题 1
被以下引用 推论
提案 1 的 α ~ \alpha \tilde{} α ~ s α , α ~ s_{\alpha, \alpha \tilde{}} s α , α ~ V V V R R R t ( s α , α ~ ) − 1 {}^t(s_{\alpha, \alpha \tilde{}})^{-1} t ( s α , α ~ ) − 1 R ~ R \tilde{} R ~ α ~ \alpha \tilde{} α ~ t ( s α , α ~ ) − 1 = s α ~ , α {}^t(s_{\alpha, \alpha \tilde{}})^{-1} = s_{\alpha \tilde{}, \alpha} t ( s α , α ~ ) − 1 = s α ~ , α R ~ R \tilde{} R ~ α ~ = α \alpha \tilde{} = \alpha α ~ = α ⟨ α ~ , β ⟩ ∈ Z \langle \alpha \tilde{}, \beta \rangle \in \mathbf{Z} ⟨ α ~ , β ⟩ ∈ Z α ~ ∈ R ~ \alpha \tilde{} \in R \tilde{} α ~ ∈ R ~ β ∈ R \beta \in R β ∈ R R ~ R \tilde{} R ~
集合 R ~ R \tilde{} R ~ R R R α ↦ α ~ \alpha \mapsto \alpha \tilde{} α ↦ α ~ R R R R ~ R \tilde{} R ~ R R R 到 R ~ R \tilde{} R ~ α , β \alpha, \beta α , β R R R α + β ∈ R \alpha + \beta \in R α + β ∈ R ( α + β ) ~ ≠ α ~ + β ~ (\alpha + \beta) \tilde{} \neq \alpha \tilde{} + \beta \tilde{} ( α + β ) ~ = α ~ + β ~
由于s α ( α ) = − α s_{\alpha}(\alpha) = -\alpha s α ( α ) = − α − R = R -R = R − R = R ( − α ) ~ = − α ~ (-\alpha) \tilde{} = -\alpha \tilde{} ( − α ) ~ = − α ~ − 1 ∈ A ( R ) -1 \in A(R) − 1 ∈ A ( R ) − 1 ∈ W ( R ) -1 \in W(R) − 1 ∈ W ( R )
等式 t ( s α , α ~ ) − 1 = s α ~ , α {}^t(s_{\alpha, \alpha \tilde{}})^{-1} = s_{\alpha \tilde{}, \alpha} t ( s α , α ~ ) − 1 = s α ~ , α W ( R ) W(R) W ( R ) W ( R ~ ) W(R \tilde{}) W ( R ~ ) u ↦ t u − 1 u \mapsto {}^t u^{-1} u ↦ t u − 1 W ( R ) W(R) W ( R ) V V V V ∗ V^* V ∗ A ( R ) A(R) A ( R )
对于x , y ∈ V x, y \in V x , y ∈ V
( x ∣ y ) = ∑ α ∈ R ⟨ α ~ , x ⟩ ⟨ α ~ , y ⟩ .
(x|y) = \sum_{\alpha \in R} \langle \alpha \tilde{}, x \rangle \langle \alpha \tilde{}, y \rangle.
( x ∣ y ) = α ∈ R ∑ ⟨ α ~ , x ⟩ ⟨ α ~ , y ⟩ . 那么 ( x ∣ y ) (x|y) ( x ∣ y ) *是*
V V V 上的非简并对称双线性形式，*在*
A ( R ) A(R) A ( R ) 下不变。
对于 x , y ∈ V Q x, y \in V_{\mathbf{Q}} x , y ∈ V Q 我们有 ( x ∣ y ) ∈ Q (x|y) \in \mathbf{Q} ( x ∣ y ) ∈ Q 。
规范扩张 ( x ∣ y ) (x|y) ( x ∣ y ) 至 V R = V Q ⊗ Q R
V_{\mathbf{R}} = V_{\mathbf{Q}} \otimes_{\mathbf{Q}} \mathbf{R}
V R = V Q ⊗ Q R 是非简并且积极的 。
引用了 1 · 被以下引用 5 引用了 命题 1
被以下引用 § 1. 。 根系 , 推论 , 命题 4 , 命题 7 , 注 2, VIII § 2
显然( x ∣ y ) (x|y) ( x ∣ y ) V V V g ∈ A ( R ) g \in A(R) g ∈ A ( R )
( g ( x ) ∣ g ( y ) ) = ∑ α ∈ R ⟨ t g ( α ~ ) , x ⟩ ⟨ t g ( α ~ ) , y ⟩ = ( x ∣ y )
(g(x)|g(y)) = \sum_{\alpha \in R} \langle {}^t g(\alpha \tilde{}), x \rangle \langle {}^t g(\alpha \tilde{}), y \rangle = (x|y)
( g ( x ) ∣ g ( y )) = α ∈ R ∑ ⟨ t g ( α ~ ) , x ⟩ ⟨ t g ( α ~ ) , y ⟩ = ( x ∣ y ) 从
t g ) ( R ~ ) = R ~ {}^t g)(R \tilde{}) = R \tilde{} t g ) ( R ~ ) = R ~ 开始。如果
x , y ∈ V Q x, y \in V_{\mathbf{Q}} x , y ∈ V Q ，则
( x ∣ y ) ∈ Q (x|y) \in \mathbf{Q} ( x ∣ y ) ∈ Q 乘以(RS_{III})。根据命题 1，如果
z ∈ V R z \in V_{\mathbf{R}} z ∈ V R ，则
( z ∣ z ) = ∑ α ∈ R ⟨ α ~ , z ⟩ 2 ≥ 0 (z|z) = \sum_{\alpha \in R} \langle \alpha \tilde{}, z \rangle^2 \geq 0 ( z ∣ z ) = ∑ α ∈ R ⟨ α ~ , z ⟩ 2 ≥ 0 ，如果
z ≠ 0 z \neq 0 z = 0 则
( z ∣ z ) > 0 (z|z) > 0 ( z ∣ z ) > 0 ，因此
( x ∣ y ) (x|y) ( x ∣ y ) 到
V R V_{\mathbf{R}} V R 的规范扩张是正的且非简并的。因此，
( x ∣ y ) (x|y) ( x ∣ y ) 到
V Q V_{\mathbf{Q}} V Q 的限制是非简并的，因此
V V V 上的形式
( x ∣ y ) (x|y) ( x ∣ y ) 是非简并的。
(i) *令X X X R \mathbf{R} R V X V_X V X X X X V V V V X ′ V'_X V X ′ V ∗ V^* V ∗ α − \alpha^- α − α ∈ X \alpha \in X α ∈ X V V V V X V_X V X V X ′ V'_X V X ′ V ∗ V^* V ∗ V X ′ V'_X V X ′ V X V_X V X V X ′ V'_X V X ′ V X V_X V X
引用了 3 · 被以下引用 1 引用了 引理 2 , 注 2 , 命题 3
被以下引用 命题 5
(ii) R ∩ V X R \cap V_X R ∩ V X V X V_X V X R ∩ V X R \cap V_X R ∩ V X α ↦ α − \alpha \mapsto \alpha^- α ↦ α − R ∩ V X R \cap V_X R ∩ V X
通过*注2*，我们可以假设k = R k = \mathbf{R} k = R V V V V ∗ V^* V ∗ α ∈ R \alpha \in R α ∈ R α − = 2 α ( α ∣ α ) \alpha^- = \frac{2\alpha}{(\alpha|\alpha)} α − = ( α ∣ α ) 2 α V V V
设V 1 V_1 V 1 V V V V 2 V_2 V 2 R ∩ V 1 R \cap V_1 R ∩ V 1 R ∩ V 1 R \cap V_1 R ∩ V 1 V 2 V_2 V 2
引用了 3 · 被以下引用 2 引用了 命题 2 , 引理 2 , 命题 3
被以下引用 推论 4 , 命题 24
这是从适用于 X = R ∩ V 1 X = R \cap V_1 X = R ∩ V 1
对于α , β ∈ R \alpha, \beta \in R α , β ∈ R
⟨ α , β − ⟩ = n ( α , β ) .
\langle \alpha, \beta^- \rangle = n(\alpha, \beta).
⟨ α , β − ⟩ = n ( α , β ) . (1)
然后
n ( α , α ) = 2
n(\alpha, \alpha) = 2
n ( α , α ) = 2 (2)
n ( − α , β ) = n ( α , − β ) = − n ( α , β ) .
n(-\alpha, \beta) = n(\alpha, -\beta) = -n(\alpha, \beta).
n ( − α , β ) = n ( α , − β ) = − n ( α , β ) . (3)
通过（RSIII），
n ( α , β ) ∈ Z .
n(\alpha, \beta) \in \mathbf{Z}.
n ( α , β ) ∈ Z . (4)
截至 n ( α , β ) n(\alpha, \beta) n ( α , β )
s β ( α ) = α − n ( α , β ) β .
s_\beta(\alpha) = \alpha - n(\alpha, \beta)\beta.
s β ( α ) = α − n ( α , β ) β . (5)
公式 (1) 和命题 2 意味着
n ( α , β ) = n ( β − , α − ) .
n(\alpha, \beta) = n(\beta^-, \alpha^-).
n ( α , β ) = n ( β − , α − ) . (6)
令( x ∣ y ) (x|y) ( x ∣ y ) V V V W ( R ) W(R) W ( R )
n ( α , β ) = 2 ( α ∣ β ) ( β ∣ β ) .
n(\alpha, \beta) = \frac{2(\alpha|\beta)}{(\beta|\beta)}.
n ( α , β ) = ( β ∣ β ) 2 ( α ∣ β ) . (7)
由此可见
n ( α , β ) = 0 ⟺ n ( β , α ) = 0 ⟺ ( α ∣ β ) = 0 ⟺ s α and s β commute.
n(\alpha, \beta) = 0 \iff n(\beta, \alpha) = 0 \iff (\alpha|\beta) = 0 \iff s_\alpha \text{ and } s_\beta \text{ commute.}
n ( α , β ) = 0 ⟺ n ( β , α ) = 0 ⟺ ( α ∣ β ) = 0 ⟺ s α and s β commute. (8)
如果( α ∣ β ) ≠ 0 (\alpha|\beta) \neq 0 ( α ∣ β ) = 0
n ( β , α ) n ( α , β ) = ( β ∣ β ) ( α ∣ α ) .
\frac{n(\beta, \alpha)}{n(\alpha, \beta)} = \frac{(\beta|\beta)}{(\alpha|\alpha)}.
n ( α , β ) n ( β , α ) = ( α ∣ α ) ( β ∣ β ) . (9)
2. 直和根系
令 V 为向量空间超过 k k k ( V i ) 1 ≤ i ≤ r (V_i)_{1 \leq i \leq r} ( V i ) 1 ≤ i ≤ r V ∗ V^* V ∗ V i ∗ V_i^* V i ∗ i i i R i R_i R i V i V_i V i R = ⋃ i R i R = \bigcup_i R_i R = ⋃ i R i R − = ⋃ i R i − R^- = \bigcup_i R_i^- R − = ⋃ i R i − R − R^- R − i i i R i R_i R i R i − R_i^- R i − R i R_i R i 直和。让α ∈ R i \alpha \in R_i α ∈ R i j ≠ i j \neq i j = i α − \alpha^- α − V j V_j V j s α s_\alpha s α V j V_j V j k α ⊆ V i k\alpha \subseteq V_i k α ⊆ V i s α s_\alpha s α V i V_i V i W ( R ) W(R) W ( R ) 可以与*∏ i = 1 r W ( R i ) \prod_{i=1}^r W(R_i) ∏ i = 1 r W ( R i )
如果 R ≠ ∅ R \neq \varnothing R = ∅
令 V 为向量空间超过 k，即向量空间* V 1 , … , V r V_1, \ldots, V_r V 1 , … , V r 令 R 为 V 中的根系。把 R i = R ∩ V i R_i = R \cap V_i R i = R ∩ V i 以下三个条件是等效的 ：
引用了 2 · 被以下引用 3 引用了 命题 3, V § 2 , 命题 4
被以下引用 习题 12, § 4 , 推论 , 推论 5
(i) **V i V_i V i 在 W ( R ) W(R) W ( R ) R ⊆ V 1 ∪ V 2 ∪ ⋯ ∪ V r R \subseteq V_1 \cup V_2 \cup \cdots \cup V_r R ⊆ V 1 ∪ V 2 ∪ ⋯ ∪ V r 对于所有 i i i R i R_i R i 是 V i V_i V i R i R_i R i ⟹ \Longrightarrow ⟹
(i) ⟹ \Longrightarrow ⟹ V i V_i V i W ( R ) W(R) W ( R ) α ∈ R \alpha \in R α ∈ R α − \alpha^- α − V i V_i V i k α k\alpha k α V i V_i V i k α k\alpha k α α ∈ V 1 ∪ V 2 ∪ ⋯ ∪ V r \alpha \in V_1 \cup V_2 \cup \cdots \cup V_r α ∈ V 1 ∪ V 2 ∪ ⋯ ∪ V r
(ii) ⟹ \Longrightarrow ⟹ i i i R i R_i R i V i V_i V i R i R_i R i V i V_i V i R i R_i R i
令 R 为 V 中的根系。以下条件是等效的*： (i) R 是不可约的 ； (ii) ** W ( R ) W(R) W ( R ) 很简单 ； (iii) **W ( R ) W(R) W ( R ) ⟺ \Longleftrightarrow ⟺ ⟺ \Longleftrightarrow ⟺
引用了 3 · 被以下引用 4 引用了 命题 2（马施克）, V Appendix 0 , 命题 5 , 命题 1, V § 2
被以下引用 命题 11 , 命题 12 , 命题 13 , 命题 2, VI § 2
每个根系* R in V 是一个家族的直和 ( R i ) i ∈ I (R_i)_{i \in I} ( R i ) i ∈ I 不可约的根系，在指标集合 的双射范围内是唯一的。
R i R_i R i R ′ , R ′ ′ R', R'' R ′ , R ′′ R ′ < R' < R ′ < R R R R ′ ′ < R'' < R ′′ < R R R R ′ R' R ′ R ′ ′ R'' R ′′ R R R R ′ R' R ′ R ′ ′ R'' R ′′ R i R_i R i R ′ R' R ′ R ′ ′ R'' R ′′ V ′ , V ′ ′ , V i ′ , V i ′ ′ V', V'', V'_i, V''_i V ′ , V ′′ , V i ′ , V i ′′ R ′ , R ′ ′ , R ′ ∩ R i , R ′ ′ ∩ R i R', R'', R' \cap R_i, R'' \cap R_i R ′ , R ′′ , R ′ ∩ R i , R ′′ ∩ R i V V V V ′ + V ′ ′ V' + V'' V ′ + V ′′ V i ′ + V i ′ ′ V'_i + V''_i V i ′ + V i ′′ R i ⊆ R ′ ∪ R ′ ′ R_i \subseteq R' \cup R'' R i ⊆ R ′ ∪ R ′′ R i R_i R i R ′ ∩ R i R' \cap R_i R ′ ∩ R i R ′ ′ ∩ R i R'' \cap R_i R ′′ ∩ R i R ′ ∩ R i = ∅ R' \cap R_i = \varnothing R ′ ∩ R i = ∅ R ′ ′ ∩ R i = ∅ R'' \cap R_i = \varnothing R ′′ ∩ R i = ∅
R i R_i R i R R R λ i \lambda_i λ i λ i R i \lambda_i R_i λ i R i V V V λ i − 1 R i \lambda_i^{-1} R_i λ i − 1 R i W ( R ) W(R) W ( R )
令R R R V V V ( R i ) (R_i) ( R i ) R i R_i R i B B B V V V V i V_i V i V V V B ′ B' B ′ V V V W ( R ) W(R) W ( R ) V i V_i V i B ′ B' B ′ i i i B B B B ′ B' B ′ V i V_i V i
引用了 2 · 被以下引用 2 引用了 命题 3 , 命题 1, V § 2
被以下引用 § 1. 。 根系 , 注
如果v i ∈ V i , v j ∈ V j , i ≠ j v_i \in V_i, v_j \in V_j, i \neq j v i ∈ V i , v j ∈ V j , i = j w ∈ W ( R j ) w \in W(R_j) w ∈ W ( R j )
B ′ ( v i , w ( v j ) ) = B ′ ( v i , v j ) ,
B'(v_i, w(v_j)) = B'(v_i, v_j),
B ′ ( v i , w ( v j )) = B ′ ( v i , v j ) , 这表明
w ( v j ) − v j w(v_j) - v_j w ( v j ) − v j 关于
B ′ B' B ′ 与
v i v_i v i 正交。由于
V j V_j V j 对于
W ( R j ) W(R_j) W ( R j ) 是不可约的，它是由
w ( v j ) − v j w(v_j) - v_j w ( v j ) − v j 生成的，因此它与
V i V_i V i 正交。
The fact that the restrictions of B B B B ′ B' B ′ V i V_i V i
3. 关系两个之间根
回想一下，*我们从现在开始假设*k = R k = \mathbf{R} k = R 注 2 中指示的方法，将定义和结果扩展到一般情况的任务留给读者。）
以下，R R R V V V V V V W ( R ) W(R) W ( R ) ( x , y ) ↦ ( x ∣ y ) (x, y) \mapsto (x|y) ( x , y ) ↦ ( x ∣ y )
让α , β ∈ R \alpha, \beta \in R α , β ∈ R
n ( α , β ) n ( β , α ) = 4 cos 2 ( α , β ^ ) ≤ 4.
n(\alpha, \beta)n(\beta, \alpha) = 4 \cos^2 (\widehat{\alpha, \beta}) \leq 4.
n ( α , β ) n ( β , α ) = 4 cos 2 ( α , β ) ≤ 4. 因此，整数n ( α , β ) n ( β , α ) n(\alpha, \beta)n(\beta, \alpha) n ( α , β ) n ( β , α ) α \alpha α β \beta β
1) n ( α , β ) = n ( β , α ) = 0 ; n(\alpha, \beta) = n(\beta, \alpha) = 0; n ( α , β ) = n ( β , α ) = 0 ;
( α , β ‾ ) = π 2 ; s α s β of order 2 ;
(\overline{\alpha, \beta}) = \frac{\pi}{2}; \quad s_{\alpha}s_{\beta} \text{ of order } 2;
( α , β ) = 2 π ; s α s β of order 2 ; n ( α , β ) = n ( β , α ) = 1 ; n(\alpha, \beta) = n(\beta, \alpha) = 1; n ( α , β ) = n ( β , α ) = 1 ; ( α , β ‾ ) = π 3 ; s α s β of order 3 ;
(\overline{\alpha, \beta}) = \frac{\pi}{3}; \quad s_{\alpha}s_{\beta} \text{ of order } 3;
( α , β ) = 3 π ; s α s β of order 3 ; ∥ α ∥ = ∥ β ∥ ;
\| \alpha \| = \| \beta \|;
∥ α ∥ = ∥ β ∥ ; n ( α , β ) = n ( β , α ) = − 1 ; n(\alpha, \beta) = n(\beta, \alpha) = -1; n ( α , β ) = n ( β , α ) = − 1 ; ( α , β ‾ ) = 2 π 3 ; s α s β of order 3 ;
(\overline{\alpha, \beta}) = \frac{2\pi}{3}; \quad s_{\alpha}s_{\beta} \text{ of order } 3;
( α , β ) = 3 2 π ; s α s β of order 3 ; ∥ α ∥ = ∥ β ∥ ;
\| \alpha \| = \| \beta \|;
∥ α ∥ = ∥ β ∥ ; n ( α , β ) = 1 , n ( β , α ) = 2 ; n(\alpha, \beta) = 1,\ n(\beta, \alpha) = 2; n ( α , β ) = 1 , n ( β , α ) = 2 ; ( α , β ‾ ) = π 4 ; s α s β of order 4 ;
(\overline{\alpha, \beta}) = \frac{\pi}{4}; \quad s_{\alpha}s_{\beta} \text{ of order } 4;
( α , β ) = 4 π ; s α s β of order 4 ; ∥ β ∥ = 2 ∥ α ∥ ;
\| \beta \| = \sqrt{2}\ \| \alpha \|;
∥ β ∥ = 2 ∥ α ∥ ; n ( α , β ) = − 1 , n ( β , α ) = − 2 ; n(\alpha, \beta) = -1,\ n(\beta, \alpha) = -2; n ( α , β ) = − 1 , n ( β , α ) = − 2 ; ( α , β ‾ ) = 3 π 4 ; s α s β of order 4 ;
(\overline{\alpha, \beta}) = \frac{3\pi}{4}; \quad s_{\alpha}s_{\beta} \text{ of order } 4;
( α , β ) = 4 3 π ; s α s β of order 4 ; ∥ β ∥ = 2 ∥ α ∥ ;
\| \beta \| = \sqrt{2}\ \| \alpha \|;
∥ β ∥ = 2 ∥ α ∥ ; n ( α , β ) = 1 , n ( β , α ) = 3 ; n(\alpha, \beta) = 1,\ n(\beta, \alpha) = 3; n ( α , β ) = 1 , n ( β , α ) = 3 ; ( α , β ‾ ) = π 6 ; s α s β of order 6 ;
(\overline{\alpha, \beta}) = \frac{\pi}{6}; \quad s_{\alpha}s_{\beta} \text{ of order } 6;
( α , β ) = 6 π ; s α s β of order 6 ; ∥ β ∥ = 3 ∥ α ∥ ;
\| \beta \| = \sqrt{3}\ \| \alpha \|;
∥ β ∥ = 3 ∥ α ∥ ; n ( α , β ) = − 1 , n ( β , α ) = − 3 ; n(\alpha, \beta) = -1,\ n(\beta, \alpha) = -3; n ( α , β ) = − 1 , n ( β , α ) = − 3 ; ( α , β ‾ ) = 5 π 6 ; s α s β of order 6 ;
(\overline{\alpha, \beta}) = \frac{5\pi}{6}; \quad s_{\alpha}s_{\beta} \text{ of order } 6;
( α , β ) = 6 5 π ; s α s β of order 6 ; ∥ β ∥ = 3 ∥ α ∥ ;
\| \beta \| = \sqrt{3}\ \| \alpha \|;
∥ β ∥ = 3 ∥ α ∥ ; n ( α , β ) = n ( β , α ) = 2 ; n(\alpha, \beta) = n(\beta, \alpha) = 2; n ( α , β ) = n ( β , α ) = 2 ; α = β ;
\alpha = \beta;
α = β ; n ( α , β ) = n ( β , α ) = − 2 ; n(\alpha, \beta) = n(\beta, \alpha) = -2; n ( α , β ) = n ( β , α ) = − 2 ; α = − β ;
\alpha = -\beta;
α = − β ; n ( α , β ) = 1 , n ( β , α ) = 4 ; n(\alpha, \beta) = 1,\ n(\beta, \alpha) = 4; n ( α , β ) = 1 , n ( β , α ) = 4 ; β = 2 α ;
\beta = 2\alpha;
β = 2 α ; n ( α , β ) = − 1 , n ( β , α ) = − 4 ; n(\alpha, \beta) = -1,\ n(\beta, \alpha) = -4; n ( α , β ) = − 1 , n ( β , α ) = − 4 ; β = − 2 α .
\beta = -2\alpha.
β = − 2 α . 尤其：
(i) *如果两个根成比例，则比例的因子只能是± 1 , ± 1 2 , ± 2 \pm 1, \pm \frac{1}{2}, \pm 2 ± 1 , ± 2 1 , ± 2
被以下引用 1 被以下引用 命题 13
(ii) 如果α \alpha α β \beta β ∥ α ∥ ≤ ∥ β ∥ \| \alpha \| \leq \| \beta \| ∥ α ∥ ≤ ∥ β ∥ n ( α , β ) n(\alpha, \beta) n ( α , β )
如果根 α ∈ R \alpha \in \mathbf{R} α ∈ R 1 2 α ∉ R \frac{1}{2}\alpha \notin \mathbf{R} 2 1 α ∈ / R α \alpha α
令α , β \alpha, \beta α , β
被以下引用 2 被以下引用 推论 , 推论 3
(i) 如果 n ( α , β ) > 0 n(\alpha, \beta) > 0 n ( α , β ) > 0 α − β \alpha - \beta α − β α = β \alpha = \beta α = β (ii) 如果 n ( α , β ) < 0 n(\alpha, \beta) < 0 n ( α , β ) < 0 α + β \alpha + \beta α + β α = − β \alpha = -\beta α = − β
如果n ( α , β ) > 0 n(\alpha, \beta) > 0 n ( α , β ) > 0
1）n ( α , β ) = 1 n(\alpha, \beta) = 1 n ( α , β ) = 1 α − β − s β ( α ) ∈ R \alpha - \beta - s_{\beta}(\alpha) \in \mathbf{R} α − β − s β ( α ) ∈ R n ( β , α ) = 1 n(\beta, \alpha) = 1 n ( β , α ) = 1 β − α = s α ( β ) ∈ R \beta - \alpha = s_{\alpha}(\beta) \in \mathbf{R} β − α = s α ( β ) ∈ R α − β ∈ R \alpha - \beta \in \mathbf{R} α − β ∈ R β = α \beta = \alpha β = α
通过将 β \beta β − β -\beta − β
(i) 如果 ( α ∣ β ) > 0 (\alpha|\beta) > 0 ( α ∣ β ) > 0 α − β \alpha - \beta α − β α = β \alpha = \beta α = β 如果 ( α ∣ β ) < 0 (\alpha|\beta) < 0 ( α ∣ β ) < 0 α + β \alpha + \beta α + β α = − β \alpha = -\beta α = − β 如果 α − β ∉ R ∪ { 0 } \alpha - \beta \notin \mathbf{R} \cup \{0\} α − β ∈ / R ∪ { 0 } α + β ∉ R ∪ { 0 } \alpha + \beta \notin \mathbf{R} \cup \{0\} α + β ∈ / R ∪ { 0 } ( α ∣ β ) = 0 (\alpha|\beta) = 0 ( α ∣ β ) = 0
断言 (i) 和 (ii) 从 Th 开始。 1 和 no. 1 的公式 (7)。 断言 (iii) 由 (i) 和 (ii) 得出。
有可能是α + β ∈ R \alpha + \beta \in \mathbf{R} α + β ∈ R ( α ∣ β ) = 0 (\alpha|\beta) = 0 ( α ∣ β ) = 0 α − β ∉ R ∪ { 0 } \alpha - \beta \notin \mathbf{R} \cup \{0\} α − β ∈ / R ∪ { 0 } α + β ∉ R ∪ { 0 } \alpha + \beta \notin \mathbf{R} \cup \{0\} α + β ∈ / R ∪ { 0 } α \alpha α β \beta β
令α \alpha α β \beta β
引用了 1 · 被以下引用 8 引用了 推论
被以下引用 注 3, IX § 4 , 习题 2, § 1 , 推论 , 命题 10 , 推论 3 , 命题 23 , 命题 3, VIII § 2 , 命题 1, VIII § 4
(i) 集合 I of 整数 j 使得 β + j α \beta + j\alpha β + j α 是根是一个区间 [ − q , p ] [ -q, p ] [ − q , p ] in Z \mathbf{Z} Z 包含 0。 (ii) 令 S 为 β + j α \beta + j\alpha β + j α 对于 j ∈ I j \in \mathrm{I} j ∈ I 然后 ,
s α ( S ) = S and s α ( β + p α ) = β − q α .
s_{\alpha}(S) = S \quad \text{and} \quad s_{\alpha}(\beta + p\alpha) = \beta - q\alpha.
s α ( S ) = S and s α ( β + p α ) = β − q α . （三）
p − q = − n ( β , α ) p - q = -n(\beta, \alpha) p − q = − n ( β , α ) 。
显然，0 ∈ I 0 \in \mathrm{I} 0 ∈ I p p p − q -q − q [ − q , p ] [ -q, p ] [ − q , p ] [ − q , p ] [ -q, p ] [ − q , p ] r , s r, s r , s s > r + 1 , s ∈ I , r ∈ I , r + k ∉ I s > r + 1, s \in \mathrm{I}, r \in \mathrm{I}, r + k \notin \mathrm{I} s > r + 1 , s ∈ I , r ∈ I , r + k ∈ / I 1 ≤ k ≤ s − r − 1 1 \leq k \leq s - r - 1 1 ≤ k ≤ s − r − 1 ( α ∣ β + s α ) ≤ 0 , ( α ∣ β + r α ) ≥ 0 (\alpha|\beta + s\alpha) \leq 0, (\alpha|\beta + r\alpha) \geq 0 ( α ∣ β + s α ) ≤ 0 , ( α ∣ β + r α ) ≥ 0
( α ∣ β + s α ) ≤ 0 > ( α ∣ β + r α ) .
(\alpha|\beta + s\alpha) \leq 0 > (\alpha|\beta + r\alpha).
( α ∣ β + s α ) ≤ 0 > ( α ∣ β + r α ) . 这就证明了(i)。
我们有 s α ( β + j α ) = β − n ( β , α ) α − j α = β + j ′ α s_{\alpha}(\beta + j\alpha) = \beta - n(\beta, \alpha)\alpha - j\alpha = \beta + j'\alpha s α ( β + j α ) = β − n ( β , α ) α − j α = β + j ′ α j ′ = − j − n ( β , α ) j' = -j - n(\beta, \alpha) j ′ = − j − n ( β , α ) s α ( S ) ⊆ S s_{\alpha}(S) \subseteq S s α ( S ) ⊆ S s α ( S ) = S s_{\alpha}(S) = S s α ( S ) = S j ↦ − j − n ( β , α ) j \mapsto -j - n(\beta, \alpha) j ↦ − j − n ( β , α ) j = p j = p j = p j ′ = − q j' = -q j ′ = − q − q = − p − n ( β , α ) -q = -p - n(\beta, \alpha) − q = − p − n ( β , α )
集合 S 被称为β \beta β α \alpha α β − q α \beta - q\alpha β − q α β + p α \beta + p\alpha β + p α p + q p + q p + q
设S是根的α \alpha α γ \gamma γ − n ( γ , α ) -n(\gamma, \alpha) − n ( γ , α )
引用了 1 · 被以下引用 1 引用了 命题 9
被以下引用 注 3, VIII § 2
第一个断言源自第9条第(iii)条，应用于β = γ \beta = \gamma β = γ q = 0 q = 0 q = 0
另一方面，由于γ \gamma γ α \alpha α ∣ n ( γ , α ) ∣ ≤ 3 |n(\gamma, \alpha)| \leq 3 ∣ n ( γ , α ) ∣ ≤ 3
令α , β \alpha, \beta α , β β + α \beta + \alpha β + α p , q p, q p , q
引用了 1 · 被以下引用 1 引用了 命题 9
被以下引用 引理 4, VIII § 2
( β + α ∣ β + α ) ( β ∣ β ) = q + 1 p .
\frac{(\beta + \alpha|\beta + \alpha)}{(\beta|\beta)} = \frac{q + 1}{p}.
( β ∣ β ) ( β + α ∣ β + α ) = p q + 1 . 令 S 为由 β , γ \beta, \gamma β , γ α \alpha α l l l ≥ 1 \geq 1 ≥ 1 β + α \beta + \alpha β + α
1）l = 1 l = 1 l = 1 β = γ , q = 0 , p = 1 , ( β + α ∣ β + α ) = ( β ∣ β ) \beta = \gamma, q = 0, p = 1, (\beta + \alpha|\beta + \alpha) = (\beta|\beta) β = γ , q = 0 , p = 1 , ( β + α ∣ β + α ) = ( β ∣ β ) l = 2 , β = γ l = 2, \beta = \gamma l = 2 , β = γ q = 0 , p = 2 , ( β + α ∣ β + α ) = 1 2 ( β ∣ β ) q = 0, p = 2, (\beta + \alpha|\beta + \alpha) = \frac{1}{2}(\beta|\beta) q = 0 , p = 2 , ( β + α ∣ β + α ) = 2 1 ( β ∣ β ) l = 2 , β = γ + α l = 2, \beta = \gamma + \alpha l = 2 , β = γ + α q = 1 , p = 1 , ( β + α ∣ β + α ) = 2 ( β ∣ β ) q = 1, p = 1, (\beta + \alpha|\beta + \alpha) = 2(\beta|\beta) q = 1 , p = 1 , ( β + α ∣ β + α ) = 2 ( β ∣ β )
4）l = 3 , β = γ l = 3, \beta = \gamma l = 3 , β = γ q = 0 , p = 3 , ( β + α ∣ β + α ) = 1 3 ( β ∣ β ) q = 0, p = 3, (\beta + \alpha|\beta + \alpha) = \frac{1}{3}(\beta|\beta) q = 0 , p = 3 , ( β + α ∣ β + α ) = 3 1 ( β ∣ β ) l = 3 , β = γ + α l = 3, \beta = \gamma + \alpha l = 3 , β = γ + α q = 1 , p = 2 , ( β + α ∣ β + α ) = ( β ∣ β ) q = 1, p = 2, (\beta + \alpha|\beta + \alpha) = (\beta|\beta) q = 1 , p = 2 , ( β + α ∣ β + α ) = ( β ∣ β ) l = 3 , β = γ + 2 α l = 3, \beta = \gamma + 2\alpha l = 3 , β = γ + 2 α q = 2 , p = 1 , ( β + α ∣ β + α ) = 3 ( β ∣ β ) q = 2, p = 1, (\beta + \alpha|\beta + \alpha) = 3(\beta|\beta) q = 2 , p = 1 , ( β + α ∣ β + α ) = 3 ( β ∣ β )
在每种情况下，都满足要证明的公式。
假设 R 是不可约的。令α \alpha α β \beta β ∥ α ∥ = ∥ β ∥ \| \alpha \| = \| \beta \| ∥ α ∥ = ∥ β ∥ g ∈ W ( R ) g \in W(R) g ∈ W ( R ) g ( α ) = β g(\alpha) = \beta g ( α ) = β
引用了 1 · 被以下引用 2 引用了 推论
被以下引用 习题 14, § 1 , 命题 13
α \alpha α W ( R ) W(R) W ( R ) g ∈ W ( R ) g \in W(R) g ∈ W ( R ) ( g ( α ) ∣ β ) ≠ 0 (g(\alpha)|\beta) \neq 0 ( g ( α ) ∣ β ) = 0 ( α ∣ β ) ≠ 0 (\alpha|\beta) \neq 0 ( α ∣ β ) = 0 n ( α , β ) = n ( β , α ) n(\alpha, \beta) = n(\beta, \alpha) n ( α , β ) = n ( β , α ) β \beta β s β ( β ) = − β s_\beta(\beta) = -\beta s β ( β ) = − β n ( α , β ) > 0 n(\alpha, \beta) > 0 n ( α , β ) > 0 α = β \alpha = \beta α = β n ( α , β ) = n ( β , α ) = 1 n(\alpha, \beta) = n(\beta, \alpha) = 1 n ( α , β ) = n ( β , α ) = 1
s α s β s α ( β ) = s α s β ( β − α ) = s α ( − β − α + β ) = α .
s_\alpha s_\beta s_\alpha(\beta) = s_\alpha s_\beta(\beta - \alpha) = s_\alpha(-\beta - \alpha + \beta) = \alpha.
s α s β s α ( β ) = s α s β ( β − α ) = s α ( − β − α + β ) = α .
4. 约化根系
如果系统中的每一个根都不可整除（no. 3），则称根系是“约简”的。
假设 R 是不可约且可约化的。
引用了 1 · 被以下引用 1 引用了 推论
被以下引用 命题 13
(i) α ∈ R , β ∈ R \alpha \in R, \beta \in R α ∈ R , β ∈ R ( β ∣ β ) ( α ∣ α ) \frac{(\beta|\beta)}{(\alpha|\alpha)} ( α ∣ α ) ( β ∣ β ) 1 2 \frac{1}{2} 2 1 1 3 \frac{1}{3} 3 1
(ii) α ∈ R \alpha \in R α ∈ R ( α ∣ α ) (\alpha|\alpha) ( α ∣ α )
由于 R 是不可约的，根的变换乘以 W ( R ) W(R) W ( R ) α , β \alpha, \beta α , β β ′ \beta' β ′ ( α ∣ β ′ ) ≠ 0 (\alpha|\beta') \neq 0 ( α ∣ β ′ ) = 0 ( β ′ ∣ β ′ ) = ( β ∣ β ) (\beta'|\beta') = (\beta|\beta) ( β ′ ∣ β ′ ) = ( β ∣ β ) ( β ′ ∣ β ′ ) ( α ∣ α ) \frac{(\beta'|\beta')}{(\alpha|\alpha)} ( α ∣ α ) ( β ′ ∣ β ′ ) 1 2 \frac{1}{2} 2 1 1 3 \frac{1}{3} 3 1 ( x ∣ y ) (x|y) ( x ∣ y ) ( α ∣ α ) = 1 (\alpha|\alpha) = 1 ( α ∣ α ) = 1 ( β ∣ β ) (\beta|\beta) ( β ∣ β ) β ∈ R \beta \in R β ∈ R β ∈ R , γ ∈ R \beta \in R, \gamma \in R β ∈ R , γ ∈ R ( γ ∣ γ ) ( β ∣ β ) = 3 2 \frac{(\gamma|\gamma)}{(\beta|\beta)} = \frac{3}{2} ( β ∣ β ) ( γ ∣ γ ) = 2 3
(i) 不可分割的根的集合R 0 R_0 R 0 W ( R 0 ) = W ( R ) W(R_0) = W(R) W ( R 0 ) = W ( R )
(ii) 令 A 为根 α \alpha α ( α ∣ α ) (\alpha|\alpha) ( α ∣ α ) λ \lambda λ
(iii) 令 B 为 β ∈ R \beta \in R β ∈ R ( β ∣ β ) = 2 λ (\beta|\beta) = 2\lambda ( β ∣ β ) = 2 λ B ≠ ∅ B \neq \varnothing B = ∅ R 0 = A ∪ B R_0 = A \cup B R 0 = A ∪ B R = A ∪ B ∪ 2 A R = A \cup B \cup 2A R = A ∪ B ∪ 2 A
如果α ∈ R − R 0 \alpha \in R - R_0 α ∈ R − R 0 1 2 α ∈ R \frac{1}{2} \alpha \in R 2 1 α ∈ R 1 2 ( 1 2 α ) ∉ R \frac{1}{2} (\frac{1}{2} \alpha) \notin R 2 1 ( 2 1 α ) ∈ / R 1 2 α ∈ R 0 \frac{1}{2} \alpha \in R_0 2 1 α ∈ R 0 R 0 R_0 R 0 1 _1 1 α ∈ R \alpha \in R α ∈ R s α , α − ( R 0 ) = R 0 s_{\alpha, \alpha^-}(R_0) = R_0 s α , α − ( R 0 ) = R 0 R 0 R_0 R 0 I I _{II} I I I I I _{III} I I I α ∈ R − R 0 \alpha \in R - R_0 α ∈ R − R 0 1 2 α ∈ R 0 \frac{1}{2} \alpha \in R_0 2 1 α ∈ R 0 s α = s α / 2 s_\alpha = s_{\alpha/2} s α = s α /2 W ( R ) = W ( R 0 ) W(R) = W(R_0) W ( R ) = W ( R 0 ) R 0 R_0 R 0
由于R R R α ∈ R 0 \alpha \in R_0 α ∈ R 0 2 α ∈ R 2\alpha \in R 2 α ∈ R R 0 R_0 R 0 dim V ≥ 2 \dim V \geq 2 dim V ≥ 2 α \alpha α β ∈ R 0 \beta \in R_0 β ∈ R 0 n ( β , α ) ≠ 0 n(\beta, \alpha) \neq 0 n ( β , α ) = 0 β \beta β α \alpha α β \beta β − β -\beta − β n ( β , α ) > 0 n(\beta, \alpha) > 0 n ( β , α ) > 0 1 2 n ( β , α ) = n ( β , 2 α ) ∈ Z \frac{1}{2} n(\beta, \alpha) = n(\beta, 2\alpha) \in \mathbf{Z} 2 1 n ( β , α ) = n ( β , 2 α ) ∈ Z n ( β , α ) ∈ 2 Z n(\beta, \alpha) \in 2\mathbf{Z} n ( β , α ) ∈ 2 Z n ( β , α ) = 2 n(\beta, \alpha) = 2 n ( β , α ) = 2 ( β ∣ β ) = 2 ( α ∣ α ) (\beta|\beta) = 2(\alpha|\alpha) ( β ∣ β ) = 2 ( α ∣ α ) R 0 R_0 R 0 γ ∈ R 0 \gamma \in R_0 γ ∈ R 0 ( γ ∣ γ ) = ( α ∣ α ) (\gamma|\gamma) = (\alpha|\alpha) ( γ ∣ γ ) = ( α ∣ α ) ( γ ∣ γ ) = 2 ( α ∣ α ) (\gamma|\gamma) = 2(\alpha|\alpha) ( γ ∣ γ ) = 2 ( α ∣ α ) γ ∈ R − R 0 \gamma \in R - R_0 γ ∈ R − R 0 1 2 γ \frac{1}{2} \gamma 2 1 γ R 0 R_0 R 0 ( 1 2 γ ∣ 1 2 γ ) = ( α ∣ α ) (\frac{1}{2} \gamma|\frac{1}{2} \gamma) = (\alpha|\alpha) ( 2 1 γ ∣ 2 1 γ ) = ( α ∣ α ) λ = ( α ∣ α ) \lambda = (\alpha|\alpha) λ = ( α ∣ α ) B ≠ ∅ B \neq \varnothing B = ∅ R 0 = A ∪ B R_0 = A \cup B R 0 = A ∪ B R ⊆ A ∪ B ∪ 2 A R \subseteq A \cup B \cup 2A R ⊆ A ∪ B ∪ 2 A γ ∈ A \gamma \in A γ ∈ A g ∈ W ( R ) g \in W(R) g ∈ W ( R ) γ = g ( α ) \gamma = g(\alpha) γ = g ( α ) 2 γ = g ( 2 α ) ∈ R 2\gamma = g(2\alpha) \in R 2 γ = g ( 2 α ) ∈ R 2 A ⊆ R 2A \subseteq R 2 A ⊆ R R = A ∪ B ∪ 2 A R = A \cup B \cup 2A R = A ∪ B ∪ 2 A γ , γ ′ \gamma, \gamma' γ , γ ′ A A A
n ( 2 γ , γ ′ ) = 2 n ( γ , γ ′ ) = 4 n ( γ , 2 γ ′ ) ∈ 4 Z , and ∣ n ( γ , γ ′ ) ∣ ≤ 1
n(2\gamma, \gamma') = 2n(\gamma, \gamma') = 4n(\gamma, 2\gamma') \in 4\mathbf{Z}, \quad \text{and} \quad |n(\gamma, \gamma')| \leq 1
n ( 2 γ , γ ′ ) = 2 n ( γ , γ ′ ) = 4 n ( γ , 2 γ ′ ) ∈ 4 Z , and ∣ n ( γ , γ ′ ) ∣ ≤ 1 由于γ \gamma γ γ ′ \gamma' γ ′ n ( γ , γ ′ ) = 0 n(\gamma, \gamma') = 0 n ( γ , γ ′ ) = 0 ( γ ∣ γ ′ ) = 0 (\gamma|\gamma') = 0 ( γ ∣ γ ′ ) = 0
假设R R R ( α ∣ α ) (\alpha|\alpha) ( α ∣ α ) λ \lambda λ 2 λ 2\lambda 2 λ α ∈ R \alpha \in R α ∈ R A A A α \alpha α ( α ∣ α ) = λ (\alpha|\alpha) = \lambda ( α ∣ α ) = λ A A A R 1 = R ∪ 2 A R_1 = R \cup 2A R 1 = R ∪ 2 A R R R R 1 R_1 R 1
显然R 1 R_1 R 1 1 _1 1 I I I _{III} I I I α , β ∈ R 1 \alpha, \beta \in R_1 α , β ∈ R 1 ⟨ α − , β ⟩ ∈ Z \langle \alpha^-, \beta \rangle \in \mathbf{Z} ⟨ α − , β ⟩ ∈ Z α ∈ R \alpha \in R α ∈ R α ∈ A \alpha \in A α ∈ A ( 2 α ) − = 1 2 α − (2\alpha)^- = \frac{1}{2} \alpha^- ( 2 α ) − = 2 1 α − α , β ∈ 2 A \alpha, \beta \in 2A α , β ∈ 2 A β ∈ R \beta \in R β ∈ R α = 2 γ \alpha = 2\gamma α = 2 γ γ ∈ A \gamma \in A γ ∈ A
1) 如果γ = ± β \gamma = \pm \beta γ = ± β ⟨ α − , β ⟩ = ± 1 2 ⟨ γ − , γ ⟩ = ± 1 \langle \alpha^-, \beta \rangle = \pm \frac{1}{2} \langle \gamma^-, \gamma \rangle = \pm 1 ⟨ α − , β ⟩ = ± 2 1 ⟨ γ − , γ ⟩ = ± 1 γ \gamma γ β \beta β β ∈ A \beta \in A β ∈ A A A A ⟨ γ − , β ⟩ = 0 \langle \gamma^-, \beta \rangle = 0 ⟨ γ − , β ⟩ = 0 ⟨ α − , β ⟩ = 0 \langle \alpha^-, \beta \rangle = 0 ⟨ α − , β ⟩ = 0 β ∈ R − A \beta \in R - A β ∈ R − A ( β ∣ β ) = 2 λ = 2 ( γ ∣ γ ) (\beta|\beta) = 2\lambda = 2(\gamma|\gamma) ( β ∣ β ) = 2 λ = 2 ( γ ∣ γ ) ⟨ β , γ − ⟩ \langle \beta, \gamma^- \rangle ⟨ β , γ − ⟩ ⟨ β , α − ⟩ = 1 2 ⟨ β , γ − ⟩ ∈ Z \langle \beta, \alpha^- \rangle = \frac{1}{2} \langle \beta, \gamma^- \rangle \in \mathbf{Z} ⟨ β , α − ⟩ = 2 1 ⟨ β , γ − ⟩ ∈ Z
这样R 1 R_1 R 1 V V V
5. 房间和基根系
对于所有α ∈ R \alpha \in R α ∈ R L α L_\alpha L α V V V s α s_\alpha s α V V V L α L_\alpha L α R R R ( x ∣ y ) (x|y) ( x ∣ y ) V → V ∗ V \to V^* V → V ∗ α ∈ R \alpha \in R α ∈ R α \alpha α 2 α − ( α − ∣ α − ) \frac{2\alpha^-}{(\alpha^-|\alpha^-)} ( α − ∣ α − ) 2 α − L α L_\alpha L α L α − L_{\alpha^-} L α − R R R R − R^- R − C C C R R R R − R^- R − C − C^- C − C − C^- C − C C C ( x ∣ y ) (x|y) ( x ∣ y )
(i) 群 W ( R ) W(R) W ( R ) 简单传递作用于室的集合 。
引用了 6 · 被以下引用 14 引用了 V § 1. 。 超平面，腔室和刻面 , IV § 1. 。考克塞特群 , 命题 7, V § 3 , 定理 1, V § 3 , 定理 2, V § 3 , 注 3
被以下引用 IX § 7. 。 不可约表示已连接紧李群 , § 1. 。 根系 , 推论 , 注 4 , 注 8 , 推论 , 推论 , 推论 3 , 推论 5 , 定理 3 , VI § 3. 。指数不变量 , 命题 5, VIII § 11 , 注, VIII § 3 , 引理 10, VIII § 4
(ii) 设 C C C 为一个房间。那么 C ‾ \overline{C} C W ( R ) W(R) W ( R )
(iii) C C C 是一个开单纯锥体 （第 V, § 1, no. 6 章）。
(iv) 设 L 1 , L 2 , … , L l L_1, L_2, \ldots, L_l L 1 , L 2 , … , L l 为 C C C 对于所有 i i i α i \alpha_i α i 使得 L i = L α i L_i = L_{\alpha_i} L i = L α i 并且使得 α i \alpha_i α i 与 L i L_i L i 与 C C C
(v) 集合 B ( C ) = { α 1 , … , α l } B(C) = \{ \alpha_1, \ldots, \alpha_l \} B ( C ) = { α 1 , … , α l } 是 V V V
(vi) C C C 是 x ∈ V x \in V x ∈ V 的集合，使得 ⟨ α i , x ⟩ > 0 \langle \alpha_i, x \rangle > 0 ⟨ α i , x ⟩ > 0 对于所有 i i i （或者，等效地，集合属于 x ∈ V x \in V x ∈ V 使得 ( x ∣ α i ) > 0 (x|\alpha_i) > 0 ( x ∣ α i ) > 0 对于所有 i i i
(vii) 设 S S S s α i s_{\alpha_i} s α i 对 ( W ( R ) , S ) (W(R), S) ( W ( R ) , S ) 是 Coxeter 系统 （第 IV, § 1, no. 3 章）。
断言 (i) 和 (vii) 源自第 1 章。 V, § 3, no. 2，周四。 1. 断言 (ii) 源自第 1 章。 V, § 3, no. 3，周四。 2. 断言 (iv) 清楚。 根α i \alpha_i α i L i L_i L i α i − \alpha_i^- α i − 2 α i / ( α i ∣ α i ) 2\alpha_i / (\alpha_i|\alpha_i) 2 α i / ( α i ∣ α i ) W ( R ) W(R) W ( R )
设B为R的基。R（相对于B）的嘉当矩阵为矩阵( n ( α , β ) ) α , β ∈ B (n(\alpha, \beta))_{\alpha, \beta \in B} ( n ( α , β ) ) α , β ∈ B
对于所有 α ∈ B \alpha \in B α ∈ B n ( α , α ) = 2 n(\alpha, \alpha) = 2 n ( α , α ) = 2 α , β ∈ B \alpha, \beta \in B α , β ∈ B
n ( α , β ) = 2 ( α ∣ β ) ( β ∣ β ) = − 2 ∥ α ∥ ∥ β ∥ cos π m ( α , β ) ,
n(\alpha, \beta) = 2 \frac{(\alpha|\beta)}{(\beta|\beta)} = -2 \frac{\|\alpha\|}{\|\beta\|} \cos \frac{\pi}{m(\alpha, \beta)},
n ( α , β ) = 2 ( β ∣ β ) ( α ∣ β ) = − 2 ∥ β ∥ ∥ α ∥ cos m ( α , β ) π , 其中m ( α , β ) m(\alpha, \beta) m ( α , β ) s α s β s_\alpha s_\beta s α s β α ≠ β \alpha \neq \beta α = β n ( α , β ) = 0 , − 1 , − 2 n(\alpha, \beta) = 0, -1, -2 n ( α , β ) = 0 , − 1 , − 2 − 3 -3 − 3
让 C 成为B = B ( C ) B = B(C) B = B ( C ) L α L_\alpha L α C ′ C' C ′ C ′ C' C ′ W ( R ) W(R) W ( R ) L α L_\alpha L α α \alpha α β \beta β α \alpha α β \beta β α = ± β \alpha = \pm \beta α = ± β α = − β \alpha = -\beta α = − β α = s β ( β ) \alpha = s_\beta(\beta) α = s β ( β )
令R 1 R_1 R 1 R 2 R_2 R 2 V 1 V_1 V 1 V 2 V_2 V 2 B 1 B_1 B 1 B 2 B_2 B 2 R 1 R_1 R 1 R 2 R_2 R 2 f : B 1 → B 2 f : B_1 \to B_2 f : B 1 → B 2 R 1 R_1 R 1 R 2 R_2 R 2 α ∈ B 1 \alpha \in B_1 α ∈ B 1 F : V 1 → V 2 F : V_1 \to V_2 F : V 1 → V 2 R 1 R_1 R 1 R 2 R_2 R 2 α \alpha α f ( α ) f(\alpha) f ( α )
引用了 2 · 被以下引用 1 引用了 命题 15 , 定理 2
被以下引用 VI § 4. 。 根系的分类
令F F F V 1 V_1 V 1 V 2 V_2 V 2 α ∈ B 1 \alpha \in B_1 α ∈ B 1 α \alpha α f ( α ) f(\alpha) f ( α ) F F F s α s_\alpha s α s f ( α ) s_{f(\alpha)} s f ( α ) W ( R 1 ) W(R_1) W ( R 1 ) W ( R 2 ) W(R_2) W ( R 2 ) R 1 R_1 R 1 R 2 R_2 R 2
若α ∈ R \alpha \in R α ∈ R t ∈ A ( R ) t \in A(R) t ∈ A ( R ) t s α t − 1 = s t ( α ) ts_\alpha t^{-1} = s_{t(\alpha)} t s α t − 1 = s t ( α ) W ( R ) W(R) W ( R ) s α s_\alpha s α W ( R ) W(R) W ( R ) A ( R ) A(R) A ( R ) R R R A ( R ) A(R) A ( R ) R R R W ( R ) W(R) W ( R ) A ( R ) A(R) A ( R ) g 1 g 2 g_1 g_2 g 1 g 2 g 1 ∈ W ( R ) g_1 \in W(R) g 1 ∈ W ( R ) g 2 ∈ G g_2 \in G g 2 ∈ G
6. 积极根
令C C C R R R B ( C ) = { α 1 , … , α l } B(C) = \{ \alpha_1, \ldots, \alpha_l \} B ( C ) = { α 1 , … , α l } R R R V V V V ∗ V^* V ∗ C C C V V V V ∗ V^* V ∗ ≥ 0 \geq 0 ≥ 0 α i \alpha_i α i α i ∗ \alpha_i^* α i ∗ ≥ 0 \geq 0 ≥ 0 C C C B ( C ) B(C) B ( C ) C ∗ C^* C ∗ W ( R ) W(R) W ( R ) V V V V ∗ V^* V ∗ V ∗ V^* V ∗ ≥ 0 \geq 0 ≥ 0 C C C ≥ 0 \geq 0 ≥ 0 C ∗ C^* C ∗ ≥ 0 \geq 0 ≥ 0 V V V x x x ≥ 0 \geq 0 ≥ 0 y ∈ C y \in C y ∈ C ( x ∣ y ) ≥ 0 (x|y) \geq 0 ( x ∣ y ) ≥ 0
C ‾ \overline{C} C C C C ≥ 0 \geq 0 ≥ 0 C C C ≥ 0 \geq 0 ≥ 0 C ‾ \overline{C} C A 2 , B 2 , G 2 A_2, B_2, G_2 A 2 , B 2 , G 2
如果α \alpha α α ∈ R \alpha \in R α ∈ R L α L_\alpha L α C ∗ C^* C ∗ α \alpha α C ∗ C^* C ∗ > 0 > 0 > 0 C ∗ C^* C ∗ < 0 < 0 < 0 α \alpha α B ( C ) B(C) B ( C ) V V V P P P α \alpha α P P P s γ s_{\gamma} s γ γ ∈ B ( C ) \gamma \in B(C) γ ∈ B ( C ) W ( R ) W(R) W ( R ) α \alpha α w ( β ) w(\beta) w ( β ) w ∈ W ( R ) w \in W(R) w ∈ W ( R ) β ∈ B ( C ) \beta \in B(C) β ∈ B ( C ) α ∈ P \alpha \in P α ∈ P
用根的 R + ( C ) R_+(C) R + ( C ) C C C
R = R + ( C ) ∪ ( − R + ( C ) )
R = R_+(C) \cup (-R_+(C))
R = R + ( C ) ∪ ( − R + ( C )) 是划分 /
R R R 。
根据no. 5的第15条命题，可以选择C C C α ∈ B ( C ) \alpha \in B(C) α ∈ B ( C )
γ = ∑ β ∈ B ( C ) n β β with n β ∈ Z .
\gamma = \sum_{\beta \in B(C)} n_{\beta} \beta \quad \text{with} \quad n_{\beta} \in \mathbf{Z}.
γ = β ∈ B ( C ) ∑ n β β with n β ∈ Z . 因此，如果
γ \gamma γ 与
α \alpha α 成正比，则
γ = n α α \gamma = n_{\alpha} \alpha γ = n α α ，这就证明了推论。
现在令S S S α ∈ B ( C ) \alpha \in B(C) α ∈ B ( C ) s α s_{\alpha} s α T T T W W W S S S α ∈ B ( C ) \alpha \in B(C) α ∈ B ( C ) w ∈ W w \in W w ∈ W T T T t = w s α w − 1 t = ws_{\alpha}w^{-1} t = w s α w − 1 β = w ( α ) \beta = w(\alpha) β = w ( α ) s β s_{\beta} s β β \beta β w ∈ W w \in W w ∈ W α = w − 1 ( β ) ∈ B ( C ) \alpha = w^{-1}(\beta) \in B(C) α = w − 1 ( β ) ∈ B ( C ) s β = w s α w − 1 ∈ T s_{\beta} = ws_{\alpha}w^{-1} \in T s β = w s α w − 1 ∈ T { ± 1 } × T \{\pm 1\} \times T { ± 1 } × T 双射 ψ \psi ψ β \beta β ( ε , s β ) (\varepsilon, s_{\beta}) ( ε , s β ) ε = + 1 \varepsilon = +1 ε = + 1 β \beta β ε = − 1 \varepsilon = -1 ε = − 1 β \beta β
另一方面，( W , S ) (W, S) ( W , S ) w w w S S S W W W q q q T T T T w T_w T w q q q w = s 1 … s q w = s_1 \ldots s_q w = s 1 … s q s i ∈ S s_i \in S s i ∈ S
t i = s 1 … s i − 1 s i s i − 1 … s 1
t_i = s_1 \ldots s_{i-1} s_i s_{i-1} \ldots s_1
t i = s 1 … s i − 1 s i s i − 1 … s 1 （
1 ≤ i ≤ q 1 \leq i \leq q 1 ≤ i ≤ q ），然后
T w = { t 1 , … , t q } T_w = \{t_1, \ldots, t_q\} T w = { t 1 , … , t q } 。回想一下，我们还在 § 1 数
η ( w , t ) \eta(w, t) η ( w , t ) （对于
w ∈ W w \in W w ∈ W 和
t ∈ T t \in T t ∈ T ）中定义了no. 4，如果
t ∉ T w t \notin T_w t ∈ / T w 则等于
+ 1 +1 + 1 ，如果
t ∈ T w t \in T_w t ∈ T w 则等于
− 1 -1 − 1 。最后，回想一下，如果我们通过公式从集合
{ ± 1 } × T \{\pm 1\} \times T { ± 1 } × T 定义映射
U w U_w U w 到自身
U w ( ε , t ) = ( ε η ( w − 1 , t ) , w t w − 1 ) ,
U_w(\varepsilon, t) = (\varepsilon \eta(w^{-1}, t), wt w^{-1}),
U w ( ε , t ) = ( ε η ( w − 1 , t ) , w t w − 1 ) , 映射
w ↦ U w w \mapsto U_w w ↦ U w 是集合
{ ± 1 } × T \{\pm 1\} \times T { ± 1 } × T 的置换从
W W W 到群的*同态*（第IV, § 1, no. 4、引理1）。
(i) 我们有ψ ( w ( α ) ) = U w ( ψ ( α ) ) \psi(w(\alpha)) = U_w(\psi(\alpha)) ψ ( w ( α )) = U w ( ψ ( α )) (ii) 假设 α \alpha α w ( α ) w(\alpha) w ( α )
η ( w − 1 , s α ) = − 1 ,
\eta(w^{-1}, s_{\alpha}) = -1,
η ( w − 1 , s α ) = − 1 , 在其他字如果 s α ∈ T w − 1 s_{\alpha} \in T_{w^{-1}} s α ∈ T w − 1 。
(iii) 当且仅当室 C 和 w ( C ) w(C) w ( C ) L α L_\alpha L α η ( w , s α ) = − 1 \eta(w, s_\alpha) = -1 η ( w , s α ) = − 1 T w T_w T w w ( C ) w(C) w ( C )
设β ∈ B ( C ) \beta \in B(C) β ∈ B ( C ) s = s β s = s_\beta s = s β T s = { s } T_s = \{ s \} T s = { s }
U s ( ε , t ) = { ( ε , s t s − 1 ) if t ≠ s ( − ε , s ) if t = s .
U_s(\varepsilon, t) = \begin{cases}
(\varepsilon, st s^{-1}) & \text{if } t \neq s \\
(-\varepsilon, s) & \text{if } t = s.
\end{cases}
U s ( ε , t ) = { ( ε , s t s − 1 ) ( − ε , s ) if t = s if t = s . 另一方面，令 ρ = ∑ γ ∈ B ( C ) n γ ( ρ ) γ \rho = \sum_{\gamma \in B(C)} n_\gamma(\rho) \gamma ρ = ∑ γ ∈ B ( C ) n γ ( ρ ) γ
s ( ρ ) = ∑ γ ∈ B ( C ) n γ ( s ( ρ ) ) γ .
s(\rho) = \sum_{\gamma \in B(C)} n_\gamma(s(\rho)) \gamma.
s ( ρ ) = γ ∈ B ( C ) ∑ n γ ( s ( ρ )) γ . 如果ρ ≠ β \rho \neq \beta ρ = β γ ∈ B ( C ) \gamma \in B(C) γ ∈ B ( C ) γ ≠ β \gamma \neq \beta γ = β n γ ( ρ ) > 0 n_\gamma(\rho) > 0 n γ ( ρ ) > 0 n γ ( s ( ρ ) ) = n γ ( ρ ) > 0 n_\gamma(s(\rho)) = n_\gamma(\rho) > 0 n γ ( s ( ρ )) = n γ ( ρ ) > 0 s ( ρ ) s(\rho) s ( ρ )
ψ ( s ( ε , ρ ) ) = { ( ε , s s ρ s − 1 ) if ρ ≠ β ( − ε , s ) if ρ = β .
\psi(s(\varepsilon, \rho)) = \begin{cases}
(\varepsilon, ss_\rho s^{-1}) & \text{if } \rho \neq \beta \\
(-\varepsilon, s) & \text{if } \rho = \beta.
\end{cases}
ψ ( s ( ε , ρ )) = { ( ε , s s ρ s − 1 ) ( − ε , s ) if ρ = β if ρ = β . (12) 和 (13) 的比较现在表明，对于所有根 γ \gamma γ s ∈ S s \in S s ∈ S U s ( ψ ( γ ) ) = ψ ( s ( γ ) ) U_s(\psi(\gamma)) = \psi(s(\gamma)) U s ( ψ ( γ )) = ψ ( s ( γ ))
另一方面，w ( α ) w(\alpha) w ( α )
ψ ( w ( α ) ) = ( − 1 , w s α w − 1 ) ,
\psi(w(\alpha)) = (-1, ws_\alpha w^{-1}),
ψ ( w ( α )) = ( − 1 , w s α w − 1 ) , 或者，通过 (i)，U w ( ψ ( α ) ) = ( − 1 , w s α w − 1 ) U_w(\psi(\alpha)) = (-1, ws_\alpha w^{-1}) U w ( ψ ( α )) = ( − 1 , w s α w − 1 ) α \alpha α ψ ( α ) = ( + 1 , s α ) \psi(\alpha) = (+1, s_\alpha) ψ ( α ) = ( + 1 , s α ) U w ( ψ ( α ) ) = ( η ( w − 1 , s α ) , w s α w − 1 ) U_w(\psi(\alpha)) = (\eta(w^{-1}, s_\alpha), ws_\alpha w^{-1}) U w ( ψ ( α )) = ( η ( w − 1 , s α ) , w s α w − 1 )
最后，根据(ii)，η ( w , s α ) = − 1 \eta(w, s_\alpha) = -1 η ( w , s α ) = − 1 α \alpha α w − 1 ( α ) w^{-1}(\alpha) w − 1 ( α ) x ∈ C x \in C x ∈ C ( α ∣ x ) . ( w − 1 ( α ) ∣ x ) = ( α ∣ x ) . ( w ∣ x ) < 0 (\alpha|x).(w^{-1}(\alpha)|x) = (\alpha|x).(w|x) < 0 ( α ∣ x ) . ( w − 1 ( α ) ∣ x ) = ( α ∣ x ) . ( w ∣ x ) < 0
让β ∈ B ( C ) \beta \in B(C) β ∈ B ( C ) s β s_\beta s β β \beta β
被以下引用 2 被以下引用 推出 2 , 命题 2, VI § 3
我们约化立即转向R减少的情况。在这种情况下，我们的断言由 (ii) 和 T s β = s β T_{s_\beta} = s_\beta T s β = s β
假设 R 减小。令w ∈ W w \in W w ∈ W w w w w = s 1 … s q w = s_1 \ldots s_q w = s 1 … s q w w w α 1 , … , α q \alpha_1, \ldots, \alpha_q α 1 , … , α q B ( C ) B(C) B ( C ) s 1 , … , s q s_1, \ldots, s_q s 1 , … , s q
θ i = s q s q − 1 … s i + 1 ( α i ) , i = 1 , … , q .
\theta_i = s_q s_{q-1} \ldots s_{i+1}(\alpha_i), \quad i = 1, \ldots, q.
θ i = s q s q − 1 … s i + 1 ( α i ) , i = 1 , … , q . 引用了 2 · 被以下引用 4 引用了 IV § 1. 。考克塞特群 , 推出 1
被以下引用 习题 12, § 4 , 推论 3 , 命题 33 , 推论 2, VIII § 6
根θ i \theta_i θ i > 0 > 0 > 0 w ( θ i ) < 0 w(\theta_i) < 0 w ( θ i ) < 0 α > 0 \alpha > 0 α > 0 w ( α ) < 0 w(\alpha) < 0 w ( α ) < 0 θ i \theta_i θ i
令X为α > 0 \alpha > 0 α > 0 w ( α ) < 0 w(\alpha) < 0 w ( α ) < 0
Card ( X ) = Card ( T w − 1 ) = l ( w − 1 ) = l ( w ) = q .
\operatorname{Card}(X) = \operatorname{Card}(T_{w^{-1}}) = l(w^{-1}) = l(w) = q.
Card ( X ) = Card ( T w − 1 ) = l ( w − 1 ) = l ( w ) = q . 另一方面，如果 α ∈ X \alpha \in X α ∈ X i ∈ { 1 , q } i \in \{1, q\} i ∈ { 1 , q }
s i + 1 … s q ( α ) > 0 and s i s i + 1 … s q ( α ) < 0.
s_{i+1} \ldots s_q(\alpha) > 0 \quad \text{and} \quad s_i s_{i+1} \ldots s_q(\alpha) < 0.
s i + 1 … s q ( α ) > 0 and s i s i + 1 … s q ( α ) < 0. 由科. 1，这意味着s i + 1 … s q ( α ) = α i s_{i+1} \ldots s_q(\alpha) = \alpha_i s i + 1 … s q ( α ) = α i α = θ i \alpha = \theta_i α = θ i θ i \theta_i θ i Card ( X ) = q \operatorname{Card}(X) = q Card ( X ) = q θ i \theta_i θ i
假设R减小。 W中存在唯一的最长元素w 0 w_0 w 0 w 0 w_0 w 0 − C -C − C w ∈ W w \in W w ∈ W w 0 2 = 1 w_0^2 = 1 w 0 2 = 1 l ( w w 0 ) = l ( w 0 ) − l ( w ) l(ww_0) = l(w_0) - l(w) l ( w w 0 ) = l ( w 0 ) − l ( w )
引用了 2 · 被以下引用 2 引用了 命题 17 , 推出 2
被以下引用 VI § 2. 。艾芬·韦尔群 , 注 2, VIII § 7
显然− C -C − C w 0 w_0 w 0 − C -C − C w 0 ( α ) < 0 w_0(\alpha) < 0 w 0 ( α ) < 0 α \alpha α w 0 2 ( C ) = C w_0^2(C) = C w 0 2 ( C ) = C w 0 2 = 1 w_0^2 = 1 w 0 2 = 1 w ∈ W w \in W w ∈ W l ( w ) l(w) l ( w ) l ( w w 0 ) l(ww_0) l ( w w 0 ) w ( C ) w(C) w ( C ) w w 0 ( C ) = − w ( C ) ww_0(C) = -w(C) w w 0 ( C ) = − w ( C ) w ( C ) w(C) w ( C ) − w ( C ) -w(C) − w ( C ) l ( w ) + l ( w 0 ) l(w) + l(w_0) l ( w ) + l ( w 0 ) l ( w 0 ) l(w_0) l ( w 0 )
(i) x ∈ C ‾ x \in \overline{C} x ∈ C x ≥ s α ( x ) x \geq s_\alpha(x) x ≥ s α ( x ) α ∈ B ( C ) \alpha \in B(C) α ∈ B ( C ) x ≥ w ( x ) x \geq w(x) x ≥ w ( x ) w ∈ W w \in W w ∈ W
由于s α ( x ) = x − ⟨ x , α ∨ ⟩ α s_\alpha(x) = x - \langle x, \alpha^\vee \rangle \alpha s α ( x ) = x − ⟨ x , α ∨ ⟩ α C ‾ \overline{C} C x ∈ V x \in V x ∈ V ⟨ x , α ∨ ⟩ ≥ 0 \langle x, \alpha^\vee \rangle \geq 0 ⟨ x , α ∨ ⟩ ≥ 0 α ∈ B ( C ) \alpha \in B(C) α ∈ B ( C ) ⟹ \Longrightarrow ⟹ ⟹ \Longrightarrow ⟹ x ∈ C ‾ x \in \overline{C} x ∈ C w ∈ W w \in W w ∈ W w w w l ( w ) l(w) l ( w ) l ( w ) = 0 l(w) = 0 l ( w ) = 0 l ( w ) ≥ 1 l(w) \geq 1 l ( w ) ≥ 1 w w w w = w ′ s α w = w' s_\alpha w = w ′ s α α ∈ B ( C ) \alpha \in B(C) α ∈ B ( C ) l ( w ′ ) = l ( w ) − 1 l(w') = l(w) - 1 l ( w ′ ) = l ( w ) − 1
x − w ( x ) = x − w ′ ( x ) + w ′ ( x − s α ( x ) ) .
x - w(x) = x - w'(x) + w'(x - s_\alpha(x)).
x − w ( x ) = x − w ′ ( x ) + w ′ ( x − s α ( x )) . 归纳假设表明x − w ′ ( x ) x - w'(x) x − w ′ ( x )
w ′ ( x − s α ( x ) ) = w ( s α ( x ) − x ) = − ⟨ x , α ∨ ⟩ w ( α ) .
w'(x - s_\alpha(x)) = w(s_\alpha(x) - x) = -\langle x, \alpha^\vee \rangle w(\alpha).
w ′ ( x − s α ( x )) = w ( s α ( x ) − x ) = − ⟨ x , α ∨ ⟩ w ( α ) . 现在s α ∈ T w − 1 s_\alpha \in T_{w^{-1}} s α ∈ T w − 1 w ( α ) < 0 w(\alpha) < 0 w ( α ) < 0
元素x ∈ C x \in C x ∈ C x > w ( x ) x > w(x) x > w ( x ) w ∈ W w \in W w ∈ W w ≠ 1 w \neq 1 w = 1
被以下引用 1 被以下引用 定义 2, VI § 3
对于室 C C C ( β i ) 1 ≤ i ≤ n (\beta_i)_{1 \leq i \leq n} ( β i ) 1 ≤ i ≤ n β 1 + β 2 + ⋯ + β n \beta_1 + \beta_2 + \cdots + \beta_n β 1 + β 2 + ⋯ + β n π ∈ S n \pi \in S_n π ∈ S n i ∈ { 1 , 2 , … , n } i \in \{1, 2, \ldots, n\} i ∈ { 1 , 2 , … , n } β π ( 1 ) + β π ( 2 ) + ⋯ + β π ( i ) \beta_{\pi(1)} + \beta_{\pi(2)} + \cdots + \beta_{\pi(i)} β π ( 1 ) + β π ( 2 ) + ⋯ + β π ( i )
引用了 1 · 被以下引用 8 引用了 推论
被以下引用 引理 3 , 推论 1 , 推论 2 , 推论 3 , 命题 20 , 推论 4, VIII § 12 , 命题 9, VIII § 3 , 推论, VIII § 4
我们通过归纳对n n n n ≤ 2 n \leq 2 n ≤ 2 β = β 1 + ⋯ + β n \beta = \beta_1 + \cdots + \beta_n β = β 1 + ⋯ + β n ∑ i = 1 n ( β ∣ β i ) = ( β ∣ β ) > 0 \sum_{i=1}^n (\beta|\beta_i) = (\beta|\beta) > 0 ∑ i = 1 n ( β ∣ β i ) = ( β ∣ β ) > 0 k k k ( β ∣ β k ) > 0 (\beta|\beta_k) > 0 ( β ∣ β k ) > 0 β = β k \beta = \beta_k β = β k β i > 0 \beta_i > 0 β i > 0 n = 1 n = 1 n = 1 i i i β − β k \beta - \beta_k β − β k β − β k = ∑ i ≠ k β i \beta - \beta_k = \sum_{i \neq k} \beta_i β − β k = ∑ i = k β i
让α ∈ R + ( C ) \alpha \in R_+(C) α ∈ R + ( C ) α ∈ B ( C ) \alpha \in B(C) α ∈ B ( C ) α \alpha α
引用了 2 · 被以下引用 1 引用了 定理 3 , 命题 19
被以下引用 VI § 4. 。 根系的分类
如果α \alpha α α ∈ B ( C ) \alpha \in B(C) α ∈ B ( C ) α ∉ B ( C ) \alpha \notin B(C) α ∈ / B ( C ) k k k n ≥ 2 n \geq 2 n ≥ 2 α = ∑ k = 1 n β k \alpha = \sum_{k=1}^n \beta_k α = ∑ k = 1 n β k β k ∈ B ( C ) \beta_k \in B(C) β k ∈ B ( C )
如有必要，对β k \beta_k β k ∑ k = 1 n − 1 \sum_{k=1}^{n-1} ∑ k = 1 n − 1 α \alpha α ∑ k = 1 n − 1 β k \sum_{k=1}^{n-1} \beta_k ∑ k = 1 n − 1 β k β n \beta_n β n
1) φ ( − α ) = − φ ( α ) \varphi(-\alpha) = -\varphi(\alpha) φ ( − α ) = − φ ( α ) α ∈ R \alpha \in R α ∈ R α ∈ R , β ∈ R \alpha \in R, \beta \in R α ∈ R , β ∈ R α + β ∈ R \alpha + \beta \in R α + β ∈ R φ ( α + β ) = φ ( α ) + φ ( β ) \varphi(\alpha + \beta) = \varphi(\alpha) + \varphi(\beta) φ ( α + β ) = φ ( α ) + φ ( β )
令Q Q Q R R R V V V Q Q Q Γ \Gamma Γ φ \varphi φ
设B B B R R R ψ \psi ψ Q Q Q Γ \Gamma Γ φ \varphi φ B B B α \alpha α B B B ψ ( α ) = φ ( α ) \psi(\alpha) = \varphi(\alpha) ψ ( α ) = φ ( α ) α = β 1 + ⋯ + β m \alpha = \beta_1 + \cdots + \beta_m α = β 1 + ⋯ + β m β i ∈ B \beta_i \in B β i ∈ B i i i β 1 + ⋯ + β h ∈ R \beta_1 + \cdots + \beta_h \in R β 1 + ⋯ + β h ∈ R h h h m m m ψ ( α ) = φ ( α ) \psi(\alpha) = \varphi(\alpha) ψ ( α ) = φ ( α ) m = 1 m = 1 m = 1
ψ ( β 1 + ⋯ + β m − 1 ) = φ ( β 1 + ⋯ + β m − 1 ) ,
\psi(\beta_1 + \cdots + \beta_{m-1}) = \varphi(\beta_1 + \cdots + \beta_{m-1}),
ψ ( β 1 + ⋯ + β m − 1 ) = φ ( β 1 + ⋯ + β m − 1 ) , 我们有ψ ( β m ) = φ ( β m ) \psi(\beta_m) = \varphi(\beta_m) ψ ( β m ) = φ ( β m ) ψ ( α ) = φ ( α ) \psi(\alpha) = \varphi(\alpha) ψ ( α ) = φ ( α )
对于R R R α = ∑ β ∈ B ( C ) n β β \alpha = \sum_{\beta \in B(C)} n_\beta \beta α = ∑ β ∈ B ( C ) n β β β ∈ B ( C ) \beta \in B(C) β ∈ B ( C ) Y ( α ) Y(\alpha) Y ( α ) n β ≠ 0 n_\beta \neq 0 n β = 0 B ( C ) B(C) B ( C ) W ( R ) W(R) W ( R ) s α i s_{\alpha_i} s α i
b) 令Y Y Y B ( C ) B(C) B ( C ) ∑ β ∈ Y β \sum_{\beta \in Y} \beta ∑ β ∈ Y β R \mathbf{R} R
为了证明a)，我们可以假设α \alpha α C a r d ( Y ( α ) ) \mathrm{Card}(Y(\alpha)) Card ( Y ( α )) C a r d ( Y ( α ) ) = 1 \mathrm{Card}(Y(\alpha)) = 1 Card ( Y ( α )) = 1 β ∈ B ( C ) \beta \in B(C) β ∈ B ( C ) α − β ∈ R \alpha - \beta \in \mathbf{R} α − β ∈ R p p p ⩾ 0 \geqslant 0 ⩾ 0 γ = α − p β ∈ R \gamma = \alpha - p\beta \in \mathbf{R} γ = α − pβ ∈ R γ − β ∉ R \gamma - \beta \notin \mathbf{R} γ − β ∈ / R γ + p β ∈ R \gamma + p\beta \in \mathbf{R} γ + pβ ∈ R ( γ ∣ β ) ≠ 0 (\gamma|\beta) \neq 0 ( γ ∣ β ) = 0 β \beta β Y ( γ ) Y(\gamma) Y ( γ ) Y ( α ) = Y ( γ ) ∪ { β } Y(\alpha) = Y(\gamma) \cup \{\beta\} Y ( α ) = Y ( γ ) ∪ { β } Y ( γ ) Y(\gamma) Y ( γ ) Y ( α ) Y(\alpha) Y ( α )
现在令Y Y Y B ( C ) B(C) B ( C ) C a r d ( Y ) \mathrm{Card}(Y) Card ( Y ) ∑ β ∈ Y β \sum_{\beta \in Y} \beta ∑ β ∈ Y β C a r d ( Y ) ⩽ 1 \mathrm{Card}(Y) \leqslant 1 Card ( Y ) ⩽ 1 C a r d ( Y ) ⩾ 2 \mathrm{Card}(Y) \geqslant 2 Card ( Y ) ⩾ 2 X X X Y Y Y β \beta β Y − { β } Y - \{\beta\} Y − { β } β \beta β α = ∑ γ ∈ Y − { β } γ ∈ R \alpha = \sum_{\gamma \in Y - \{\beta\}} \gamma \in \mathbf{R} α = ∑ γ ∈ Y − { β } γ ∈ R ( α ∣ β ) < 0 (\alpha|\beta) < 0 ( α ∣ β ) < 0 α + β ∈ R \alpha + \beta \in \mathbf{R} α + β ∈ R
7. 集合根关闭
(i) 如果条件 α ∈ P , β ∈ P , α + β ∈ R \alpha \in P, \beta \in P, \alpha + \beta \in \mathbf{R} α ∈ P , β ∈ P , α + β ∈ R α + β ∈ P \alpha + \beta \in P α + β ∈ P P P P
(ii) 如果 P P P P ∪ ( − P ) = R P \cup (-P) = \mathbf{R} P ∪ ( − P ) = R P P P
(iii) 如果 P = − P P = -P P = − P P P P
令C C C R \mathbf{R} R P P P R \mathbf{R} R R + ( C ) R_+(C) R + ( C ) Σ = B ( C ) ∩ ( − P ) \Sigma = B(C) \cap (-P) Σ = B ( C ) ∩ ( − P ) Q Q Q Σ \Sigma Σ P = R + ( C ) ∪ Q P = R_+(C) \cup Q P = R + ( C ) ∪ Q
引用了 1 · 被以下引用 3 引用了 命题 19
被以下引用 推论 , 命题 20 , 注, VIII § 3
只要证明P ∩ ( − R + ( C ) ) = Q P \cap (-R_+(C)) = Q P ∩ ( − R + ( C )) = Q − α ∈ Q -\alpha \in Q − α ∈ Q α \alpha α n n n Σ \Sigma Σ n n n − α ∈ P -\alpha \in P − α ∈ P n = 1 n = 1 n = 1 n > 1 n > 1 n > 1 α = β + γ \alpha = \beta + \gamma α = β + γ γ ∈ Σ \gamma \in \Sigma γ ∈ Σ β \beta β n − 1 n - 1 n − 1 Σ \Sigma Σ − β ∈ P -\beta \in P − β ∈ P − γ ∈ P -\gamma \in P − γ ∈ P P P P − α ∈ P -\alpha \in P − α ∈ P Q ⊆ P ∩ ( − R + ( C ) ) Q \subseteq P \cap (-R_+(C)) Q ⊆ P ∩ ( − R + ( C )) − α ∈ P ∩ ( − R + ( C ) ) -\alpha \in P \cap (-R_+(C)) − α ∈ P ∩ ( − R + ( C )) α \alpha α p p p B ( C ) B(C) B ( C ) p p p − α ∈ Q -\alpha \in Q − α ∈ Q p = 1 p = 1 p = 1 p > 1 p > 1 p > 1 γ ∈ B ( C ) \gamma \in B(C) γ ∈ B ( C ) β \beta β α = β + γ \alpha = \beta + \gamma α = β + γ p − 1 p - 1 p − 1 B ( C ) B(C) B ( C ) − γ = β + ( − α ) -\gamma = \beta + (-\alpha) − γ = β + ( − α ) P P P − γ ∈ P -\gamma \in P − γ ∈ P γ ∈ Σ \gamma \in \Sigma γ ∈ Σ P P P − β = γ + ( − α ) -\beta = \gamma + (-\alpha) − β = γ + ( − α ) − β ∈ P -\beta \in P − β ∈ P − β ∈ Q -\beta \in Q − β ∈ Q − α = − β − γ ∈ Q -\alpha = -\beta - \gamma \in Q − α = − β − γ ∈ Q P ∩ ( − R + ( C ) ) ⊆ Q P \cap (-R_+(C)) \subseteq Q P ∩ ( − R + ( C )) ⊆ Q
(i) P P P 是抛物线 ； (ii) P P P R R R C C C P ⊇ R + ( C ) P \supseteq R_+(C) P ⊇ R + ( C ) 存在R R R C C C B ( C ) B(C) B ( C ) Σ \Sigma Σ P P P R + ( C ) R_+(C) R + ( C ) Q Q Q Σ \Sigma Σ 。
(ii) ⟹ \Longrightarrow ⟹
(iii)⟹ \Longrightarrow ⟹ P ∪ ( − P ) = R P \cup (-P) = R P ∪ ( − P ) = R α , β ∈ P \alpha, \beta \in P α , β ∈ P α + β ∈ R \alpha + \beta \in R α + β ∈ R α + β ∈ P \alpha + \beta \in P α + β ∈ P α + β \alpha + \beta α + β α + β \alpha + \beta α + β α + β = ∑ γ ∈ B ( C ) n γ γ \alpha + \beta = \sum_{\gamma \in B(C)} n_\gamma \gamma α + β = ∑ γ ∈ B ( C ) n γ γ n γ ≤ 0 n_\gamma \leq 0 n γ ≤ 0 α \alpha α β \beta β B ( C ) − Σ B(C) - \Sigma B ( C ) − Σ γ \gamma γ ≥ 0 \geq 0 ≥ 0 n γ = 0 n_\gamma = 0 n γ = 0 γ ∈ B ( C ) − Σ \gamma \in B(C) - \Sigma γ ∈ B ( C ) − Σ α + β ∈ Q ⊆ P \alpha + \beta \in Q \subseteq P α + β ∈ Q ⊆ P
(i) ⟹ \Longrightarrow ⟹ P P P C C C C a r d ( P ∩ R + ( C ) ) \mathrm{Card}(P \cap R_+(C)) Card ( P ∩ R + ( C )) α ∈ B ( C ) \alpha \in B(C) α ∈ B ( C ) α ∈ P \alpha \in P α ∈ P − α ∈ P -\alpha \in P − α ∈ P β ∈ P ∩ R + ( C ) \beta \in P \cap R_+(C) β ∈ P ∩ R + ( C ) β \beta β α \alpha α β = 2 α \beta = 2\alpha β = 2 α α = 2 α + ( − α ) ∈ P \alpha = 2\alpha + (-\alpha) \in P α = 2 α + ( − α ) ∈ P P P P s α ( β ) ∈ R + ( C ) s_\alpha(\beta) \in R_+(C) s α ( β ) ∈ R + ( C ) C ′ = s α ( C ) C' = s_\alpha(C) C ′ = s α ( C ) β = s α ( s α ( β ) ) ∈ s α ( R + ( C ) ) = R + ( C ) \beta = s_\alpha(s_\alpha(\beta)) \in s_\alpha(R_+(C)) = R_+(C) β = s α ( s α ( β )) ∈ s α ( R + ( C )) = R + ( C ) − α ∈ P ∩ R + ( C ′ ) -\alpha \in P \cap R_+(C') − α ∈ P ∩ R + ( C ′ ) C a r d ( P ∩ R + ( C ′ ) ) > C a r d ( P ∩ R + ( C ) ) \mathrm{Card}(P \cap R_+(C')) > \mathrm{Card}(P \cap R_+(C)) Card ( P ∩ R + ( C ′ )) > Card ( P ∩ R + ( C )) α ∈ P \alpha \in P α ∈ P B ( C ) ⊆ P B(C) \subseteq P B ( C ) ⊆ P R + ( C ) ⊆ P R_+(C) \subseteq P R + ( C ) ⊆ P P P P
(i) 存在一个室C C C P = R + ( C ) P = R_+(C) P = R + ( C ) ； (ii) P P P { P , − P } \{P, -P\} { P , − P } R R R
腔室C C C P = R + ( C ) P = R_+(C) P = R + ( C ) 。
如果P = R + ( C ) P = R_+(C) P = R + ( C ) C C C x ∗ ∈ V ∗ x^* \in V^* x ∗ ∈ V ∗ ⟨ x ∗ , x ⟩ > 0 \langle x^*, x \rangle > 0 ⟨ x ∗ , x ⟩ > 0 x ∈ P x \in P x ∈ P C C C
假设V V V R R R P P P C C C R R R P = R + ( C ) P = R_+(C) P = R + ( C )
引用了 1 · 被以下引用 2 引用了 推论 1
被以下引用 命题 22 , VI § 4. 。 根系的分类
事实上，P P P
这个推论特别适用于当考虑的阶是总计时，条件在R R R ( e i ) 1 ≤ i ≤ n (e_i)_{1 \leq i \leq n} ( e i ) 1 ≤ i ≤ n V V V V V V x = ∑ i ξ i e i x = \sum_i \xi_i e_i x = ∑ i ξ i e i ≥ 0 \geq 0 ≥ 0 ξ i \xi_i ξ i ξ i > 0 \xi_i > 0 ξ i > 0 i i i ξ i ≠ 0 \xi_i \neq 0 ξ i = 0
(i) B 的元素线性无关； (ii) R 的每个根都是 B 的元素的线性组合，其中系数要么全正，要么全负； (iii) B 的每个根都是不可分的。
我们已经知道条件是必要的（no. 5，Th.2，和no. 6，Th.3）。假设满足条件 (i), (ii), (iii)。令 P 为根的集合，即 B 的元素的线性组合，系数为 ≥ 0 \geq 0 ≥ 0 P = R + ( C ) P = R_+(C) P = R + ( C ) B ′ = B ( C ) B' = B(C) B ′ = B ( C ) X ′ X' X ′ B ′ B' B ′
B ⊆ P ⊆ X and B ′ ⊆ P ⊆ X ′ ,
B \subseteq P \subseteq X \quad \text{and} \quad B' \subseteq P \subseteq X',
B ⊆ P ⊆ X and B ′ ⊆ P ⊆ X ′ , 这表明X和X ′ X' X ′ B ′ B' B ′ X ′ X' X ′ B = B ′ B = B' B = B ′
这立即来自Cor。 3 和科。提案 4。
我们称R ′ R' R ′ B ′ B' B ′
设 B 为 R 的基，B 的A 1 , A 2 , … , A r A_1, A_2, \ldots, A_r A 1 , A 2 , … , A r A = A 1 ∪ A 2 ∪ ⋯ ∪ A r A = A_1 \cup A_2 \cup \cdots \cup A_r A = A 1 ∪ A 2 ∪ ⋯ ∪ A r α \alpha α A i A_i A i
引用了 3 · 被以下引用 1 引用了 推论 4 , 命题 5 , 定理 2
被以下引用 命题 25
令E 1 , … , E r , E E_1, \ldots, E_r, E E 1 , … , E r , E A 1 , … , A r , A A_1, \ldots, A_r, A A 1 , … , A r , A E = V E = V E = V E i E_i E i W ( R ) W(R) W ( R ) R ∩ E i R \cap E_i R ∩ E i
我们采用命题20的假设和记号。令V 1 V_1 V 1 Σ \Sigma Σ P ∩ ( − P ) = Q ∪ ( − Q ) = V 1 ∩ R P \cap (-P) = Q \cup (-Q) = V_1 \cap R P ∩ ( − P ) = Q ∪ ( − Q ) = V 1 ∩ R V 1 V_1 V 1 Σ \Sigma Σ
引用了 3 · 被以下引用 1 引用了 命题 20 , 推论 4 , 定理 3
被以下引用 命题 21
我们有P ∩ ( − P ) = ( R + ( C ) ∪ Q ) ∩ ( ( − R + ( C ) ) ∪ ( − Q ) ) = Q ∪ ( − Q ) P \cap (-P) = (R_+(C) \cup Q) \cap ((-R_+(C)) \cup (-Q)) = Q \cup (-Q) P ∩ ( − P ) = ( R + ( C ) ∪ Q ) ∩ (( − R + ( C )) ∪ ( − Q )) = Q ∪ ( − Q ) Q ∪ ( − Q ) = V 1 ∩ R Q \cup (-Q) = V_1 \cap R Q ∪ ( − Q ) = V 1 ∩ R Σ \Sigma Σ V 1 ∩ R V_1 \cap R V 1 ∩ R
令C（分别C ′ C' C ′ Σ \Sigma Σ Σ ′ \Sigma' Σ ′ B ( C ) B(C) B ( C ) B ( C ′ ) B(C') B ( C ′ ) Q ′ Q' Q ′ Σ \Sigma Σ Σ ′ \Sigma' Σ ′ P = Q ∪ R + ( C ) P = Q \cup R_+(C) P = Q ∪ R + ( C ) P ′ = Q ′ ∪ R + ( C ′ ) P' = Q' \cup R_+(C') P ′ = Q ′ ∪ R + ( C ′ ) P ′ P' P ′ C ′ C' C ′ Σ \Sigma Σ Σ ′ \Sigma' Σ ′
引用了 2 · 被以下引用 1 引用了 推论 1 , 推论 6
被以下引用 推论
我们约化立即处理案件P = P ′ P = P' P = P ′ V 1 V_1 V 1 P ∩ ( − P ) P \cap (-P) P ∩ ( − P ) Σ \Sigma Σ Σ ′ \Sigma' Σ ′ V 1 V_1 V 1 R 1 = P ∩ ( − P ) R_1 = P \cap (-P) R 1 = P ∩ ( − P ) g 1 ∈ W ( R 1 ) g_1 \in W(R_1) g 1 ∈ W ( R 1 ) g 1 ( Σ ) = Σ ′ g_1(\Sigma) = \Sigma' g 1 ( Σ ) = Σ ′ g 1 g_1 g 1 W ( R ) W(R) W ( R ) g g g s σ s_\sigma s σ σ ∈ Σ \sigma \in \Sigma σ ∈ Σ γ = ∑ β ∈ B ( C ) c β β \gamma = \sum_{\beta \in B(C)} c_\beta \beta γ = ∑ β ∈ B ( C ) c β β P − R 1 P - R_1 P − R 1 c β > 0 c_\beta > 0 c β > 0 β ∈ B ( C ) − Σ \beta \in B(C) - \Sigma β ∈ B ( C ) − Σ σ ∈ Σ \sigma \in \Sigma σ ∈ Σ s σ ( γ ) − γ ∈ V 1 s_\sigma(\gamma) - \gamma \in V_1 s σ ( γ ) − γ ∈ V 1 s σ ( γ ) s_\sigma(\gamma) s σ ( γ ) B ( C ) B(C) B ( C ) > 0 > 0 > 0 s σ ( γ ) ∈ R + ( C ) s_\sigma(\gamma) \in R_+(C) s σ ( γ ) ∈ R + ( C ) s σ ( γ ) ∈ P − R 1 s_\sigma(\gamma) \in P - R_1 s σ ( γ ) ∈ P − R 1 P − R 1 P - R_1 P − R 1 s σ , σ ∈ Σ s_\sigma, \sigma \in \Sigma s σ , σ ∈ Σ g g g g ( P ) = P g(P) = P g ( P ) = P P = P ′ P = P' P = P ′ Σ = Σ ′ \Sigma = \Sigma' Σ = Σ ′ Q = Q ′ Q = Q' Q = Q ′ R + ( C ) = P − Q = P − Q ′ = R + ( C ′ ) R_+(C) = P - Q = P - Q' = R_+(C') R + ( C ) = P − Q = P − Q ′ = R + ( C ′ ) C = C ′ C = C' C = C ′
令P、P ′ P' P ′ R + ( C ) ⊆ P R_+(C) \subseteq P R + ( C ) ⊆ P R + ( C ) ⊆ P ′ R_+(C) \subseteq P' R + ( C ) ⊆ P ′ P = P ′ P = P' P = P ′
引用了 3 引用了 引理 3 , 命题 21 , 定理 2
这是从引理 3和第21条命题得出的，因为W ( R ) W(R) W ( R )
1）鉴于Cor。的 Th。 1、no. 3、假设α ∈ P , β ∈ P , ( α ∣ β ) < 0 \alpha \in P, \beta \in P, (\alpha|\beta) < 0 α ∈ P , β ∈ P , ( α ∣ β ) < 0 α + β ∈ P \alpha + \beta \in P α + β ∈ P
2）我们证明P的元素中没有和α 1 + ⋯ + α q \alpha_1 + \cdots + \alpha_q α 1 + ⋯ + α q q ⩾ 1 q \geqslant 1 q ⩾ 1 q q q q = 1 q = 1 q = 1 q ⩾ 2 q \geqslant 2 q ⩾ 2 α 1 + ⋯ + α q = 0 \alpha_1 + \cdots + \alpha_q = 0 α 1 + ⋯ + α q = 0
− α 1 = α 2 + ⋯ + α q ,
-\alpha_1 = \alpha_2 + \cdots + \alpha_q,
− α 1 = α 2 + ⋯ + α q , 所以( − α 1 ∣ α 2 + ⋯ + α q ) > 0 (-\alpha_1|\alpha_2 + \cdots + \alpha_q) > 0 ( − α 1 ∣ α 2 + ⋯ + α q ) > 0 j ∈ { 2 , q } j \in \{2, q\} j ∈ { 2 , q } ( α 1 ∣ α j ) < 0 (\alpha_1|\alpha_j) < 0 ( α 1 ∣ α j ) < 0 α 1 + α j ∈ P \alpha_1 + \alpha_j \in P α 1 + α j ∈ P ( α 1 + α j ) + ∑ i ≠ 1 , j α i = 0 (\alpha_1 + \alpha_j) + \sum_{i \neq 1, j} \alpha_i = 0 ( α 1 + α j ) + ∑ i = 1 , j α i = 0
3) 我们证明V V V γ \gamma γ ( γ ∣ α ) ⩾ 0 (\gamma|\alpha) \geqslant 0 ( γ ∣ α ) ⩾ 0 α ∈ P \alpha \in P α ∈ P P P P α 1 , α 2 , … \alpha_1, \alpha_2, \ldots α 1 , α 2 , …
β i = α 1 + ⋯ + α i ∈ P
\beta_i = \alpha_1 + \cdots + \alpha_i \in P
β i = α 1 + ⋯ + α i ∈ P 对于所有
i i i ；将存在两个不同整数
i , j i, j i , j 使得
β i = β j \beta_i = \beta_j β i = β j ，这与2)的结果相矛盾。
4）为了证明命题，只要证明V V V ( α k ) 1 ≤ k ≤ l (\alpha_k)_{1 \leq k \leq l} ( α k ) 1 ≤ k ≤ l P P P > 0 > 0 > 0 l = dim V l = \dim V l = dim V < l < l < l α ∈ P \alpha \in P α ∈ P γ ∈ V \gamma \in V γ ∈ V γ ≠ 0 \gamma \neq 0 γ = 0 ( γ ∣ α ) ⩾ 0 (\gamma|\alpha) \geqslant 0 ( γ ∣ α ) ⩾ 0 L L L γ \gamma γ R ∩ L R \cap L R ∩ L L L L V ′ V' V ′ R ∩ L R \cap L R ∩ L V ′ V' V ′ P ∩ L P \cap L P ∩ L R ∩ L R \cap L R ∩ L V ′ V' V ′ ( β 1 , … , β l ′ ) (\beta_1, \ldots, \beta_{l'}) ( β 1 , … , β l ′ ) P ∩ L P \cap L P ∩ L > 0 > 0 > 0 V V V l ′ + 1 l' + 1 l ′ + 1 γ , β 1 , … , β l ′ \gamma, \beta_1, \ldots, \beta_{l'} γ , β 1 , … , β l ′ L L L
(i) P 是封闭且对称的； (ii) P 是闭集，且 P 在 V 1 V_1 V 1 （三）Γ ∩ R = P \Gamma \cap R = P Γ ∩ R = P
假设满足这些条件。对于任何α ∈ P \alpha \in P α ∈ P α 1 ~ \alpha_1 \tilde{} α 1 ~ α ~ \alpha \tilde{} α ~ V 1 V_1 V 1 α ↦ α 1 ~ \alpha \mapsto \alpha_1 \tilde{} α ↦ α 1 ~ P P P P ~ P \tilde{} P ~
(iii) ⟹ \Longrightarrow ⟹
(i) ⟹ \Longrightarrow ⟹ P P P P P P V 1 V_1 V 1 α , β ∈ P \alpha, \beta \in P α , β ∈ P s α ( β ) ∈ P s_\alpha(\beta) \in P s α ( β ) ∈ P α \alpha α β \beta β s α ( β ) = β − n ( β , α ) α s_\alpha(\beta) = \beta - n(\beta, \alpha)\alpha s α ( β ) = β − n ( β , α ) α β − p α ∈ R \beta - p\alpha \in R β − p α ∈ R n ( β , α ) n(\beta, \alpha) n ( β , α ) p p p
β − n ( β , α ) α ∈ P
\beta - n(\beta, \alpha)\alpha \in P
β − n ( β , α ) α ∈ P 因为
P P P 是封闭且对称的。因此，
s α , α 1 ( P ) = P s_{\alpha, \alpha_1}(P) = P s α , α 1 ( P ) = P 和
P P P 满足(RS_{II})。显然
P P P 满足(RS_{III})。这样，
P P P 满足(ii)，同时我们也证明了命题的最后一个断言。
(ii) ⟹ \Longrightarrow ⟹ Γ ∩ R = P \Gamma \cap R = P Γ ∩ R = P P ⊆ Γ ∩ R P \subseteq \Gamma \cap R P ⊆ Γ ∩ R β ∈ Γ ∩ R \beta \in \Gamma \cap R β ∈ Γ ∩ R β ∈ Γ \beta \in \Gamma β ∈ Γ P = − P P = -P P = − P β = α 1 + α 2 + ⋯ + α k \beta = \alpha_1 + \alpha_2 + \cdots + \alpha_k β = α 1 + α 2 + ⋯ + α k α 1 , … , α k ∈ P \alpha_1, \ldots, \alpha_k \in P α 1 , … , α k ∈ P β ∈ P \beta \in P β ∈ P k = 1 k = 1 k = 1 k k k
0 < ( β ∣ β ) = ∑ i = 1 k ( β ∣ α i ) ,
0 < (\beta|\beta) = \sum_{i=1}^{k} (\beta|\alpha_i),
0 < ( β ∣ β ) = i = 1 ∑ k ( β ∣ α i ) , 所以
( β ∣ α i ) > 0 (\beta|\alpha_i) > 0 ( β ∣ α i ) > 0 对于一些指标
i i i 。如果
β = α i \beta = \alpha_i β = α i ，则
β ∈ P \beta \in P β ∈ P 。否则，
β − α i ∈ R \beta - \alpha_i \in R β − α i ∈ R （Th.1的Cor.，no. 3），因此
β − α i ∈ P \beta - \alpha_i \in P β − α i ∈ P 乘以归纳假设，因此
β ∈ P \beta \in P β ∈ P ，因为
P P P 是封闭的。
23号提案的条件可以用V 1 = V V_1 = V V 1 = V P ≠ R P \neq R P = R R R R G 2 G_2 G 2 A 2 A_2 A 2 P P P
设R ′ R' R ′ R R R V V V R ′ R' R ′ V ′ V' V ′ B ′ B' B ′ R ′ R' R ′
引用了 1 引用了 推论
(i) 存在R R R B ′ B' B ′ (ii) R ′ R' R ′ R R R B ′ B' B ′
断言 (ii) 很清楚。我们证明（i）。令 ( ε 1 , ε 2 , … , ε l ) (\varepsilon_1, \varepsilon_2, \ldots, \varepsilon_l) ( ε 1 , ε 2 , … , ε l ) V V V B ′ = ( ε p + 1 , ε p + 2 , … , ε l ) B' = (\varepsilon_{p+1}, \varepsilon_{p+2}, \ldots, \varepsilon_l) B ′ = ( ε p + 1 , ε p + 2 , … , ε l ) V V V C C C R R R B ′ B' B ′ R + ( C ) R_+(C) R + ( C ) B ′ ⊆ B ( C ) B' \subseteq B(C) B ′ ⊆ B ( C )
8. 最高根
假设R R R C C C R R R B ( C ) = { α 1 , … , α l } B(C) = \{\alpha_1, \ldots, \alpha_l\} B ( C ) = { α 1 , … , α l }
引用了 3 · 被以下引用 3 引用了 推论 , 推论 5 , 定理 3
被以下引用 命题 31 , 命题 5, VI § 2 , VI § 4. 。 根系的分类
(i) 存在根 α ~ = ∑ i = 1 l n i α i \tilde{\alpha} = \sum_{i=1}^{l} n_i \alpha_i α ~ = ∑ i = 1 l n i α i ∑ i = 1 l p i α i \sum_{i=1}^{l} p_i \alpha_i ∑ i = 1 l p i α i n 1 ≥ p 1 , n 2 ≥ p 2 , … , n l ≥ p l n_1 \geq p_1, n_2 \geq p_2, \ldots, n_l \geq p_l n 1 ≥ p 1 , n 2 ≥ p 2 , … , n l ≥ p l R R R C C C (ii) 我们有α ~ ∈ C ‾ \tilde{\alpha} \in \overline{C} α ~ ∈ C (iii) 每根 α \alpha α ( α ~ ∣ α ~ ) ≥ ( α ∣ α ) (\tilde{\alpha}|\tilde{\alpha}) \geq (\alpha|\alpha) ( α ~ ∣ α ~ ) ≥ ( α ∣ α ) (iv) 对于每个与 α ~ \tilde{\alpha} α ~ α ′ \alpha' α ′ n ( α ′ , α ~ ) = 0 n(\alpha', \tilde{\alpha}) = 0 n ( α ′ , α ~ ) = 0 1 1 1
1) 令 α = ∑ i = 1 l n i α i \alpha = \sum_{i=1}^{l} n_i \alpha_i α = ∑ i = 1 l n i α i β = ∑ i = 1 l p i α i \beta = \sum_{i=1}^{l} p_i \alpha_i β = ∑ i = 1 l p i α i C C C α = β \alpha = \beta α = β
2) 如果( α ∣ α i ) < 0 (\alpha|\alpha_i) < 0 ( α ∣ α i ) < 0 i i i α + α i ∈ R \alpha + \alpha_i \in R α + α i ∈ R α = − α i \alpha = -\alpha_i α = − α i α \alpha α ( α ∣ α i ) ≥ 0 (\alpha|\alpha_i) \geq 0 ( α ∣ α i ) ≥ 0 i i i
3）如果α < 0 \alpha < 0 α < 0 α < − α \alpha < -\alpha α < − α n i ≥ 0 n_i \geq 0 n i ≥ 0 i i i J J J i i i n i > 0 n_i > 0 n i > 0 J ′ J' J ′ { 1 , 2 , … , l } \{1, 2, \ldots, l\} { 1 , 2 , … , l } J J J J ≠ ∅ J \neq \varnothing J = ∅ J ′ J' J ′ i ∈ J i \in J i ∈ J i ′ ∈ J ′ i' \in J' i ′ ∈ J ′ ( α i ∣ α i ′ ) < 0 (\alpha_i|\alpha_{i'}) < 0 ( α i ∣ α i ′ ) < 0
( α ∣ α i ′ ) = ∑ i ∈ I n i ( α i ∣ α i ′ ) < 0 (\alpha | \alpha_{i'}) = \sum_{i \in I} n_i (\alpha_i | \alpha_{i'}) < 0 ( α ∣ α i ′ ) = i ∈ I ∑ n i ( α i ∣ α i ′ ) < 0 因为只要 j j j k k k ( α j ∣ α k ) ≤ 0 (\alpha_j | \alpha_k) \leq 0 ( α j ∣ α k ) ≤ 0 i i i J ′ = ∅ J' = \varnothing J ′ = ∅ n i > 0 n_i > 0 n i > 0
4) 对于所有i i i ( β ∣ α i ) ≥ 0 (\beta | \alpha_i) \geq 0 ( β ∣ α i ) ≥ 0 β ≠ 0 \beta \neq 0 β = 0 i i i ( β ∣ α i ) = 0 (\beta | \alpha_i) = 0 ( β ∣ α i ) = 0
( β ∣ α ) = ∑ i n i ( β ∣ α i ) > 0. (\beta | \alpha) = \sum_i n_i (\beta | \alpha_i) > 0. ( β ∣ α ) = i ∑ n i ( β ∣ α i ) > 0. 如果γ = α − β ∈ R \gamma = \alpha - \beta \in \mathbf{R} γ = α − β ∈ R α > β \alpha > \beta α > β β > α \beta > \alpha β > α α \alpha α β \beta β α = β \alpha = \beta α = β
5) 到 2)，α ~ ∈ C ‾ \tilde{\alpha} \in \overline{C} α ~ ∈ C α ′ ∈ R \alpha' \in \mathbf{R} α ′ ∈ R ( α ′ ∣ α ′ ) ≤ ( α ~ ∣ α ~ ) (\alpha' | \alpha') \leq (\tilde{\alpha} | \tilde{\alpha}) ( α ′ ∣ α ′ ) ≤ ( α ~ ∣ α ~ ) C ‾ \overline{C} C W ( R ) W(\mathbf{R}) W ( R ) α ′ ∈ C ‾ \alpha' \in \overline{C} α ′ ∈ C α ~ − α ′ ≥ 0 \tilde{\alpha} - \alpha' \geq 0 α ~ − α ′ ≥ 0 x ∈ C ‾ x \in \overline{C} x ∈ C ( α ~ − α ′ ∣ x ) ≥ 0 (\tilde{\alpha} - \alpha' | x) \geq 0 ( α ~ − α ′ ∣ x ) ≥ 0 ( α ~ − α ′ ∣ α ~ ) ≥ 0 (\tilde{\alpha} - \alpha' | \tilde{\alpha}) \geq 0 ( α ~ − α ′ ∣ α ~ ) ≥ 0 ( α ~ − α ′ ∣ α ′ ) ≥ 0 (\tilde{\alpha} - \alpha' | \alpha') \geq 0 ( α ~ − α ′ ∣ α ′ ) ≥ 0 ( α ~ ∣ α ~ ) ≥ ( α ′ ∣ α ~ ) ≥ ( α ′ ∣ α ′ ) (\tilde{\alpha} | \tilde{\alpha}) \geq (\alpha' | \tilde{\alpha}) \geq (\alpha' | \alpha') ( α ~ ∣ α ~ ) ≥ ( α ′ ∣ α ~ ) ≥ ( α ′ ∣ α ′ ) α ′ \alpha' α ′ α ~ \tilde{\alpha} α ~ n ( α ′ , α ~ ) n(\alpha', \tilde{\alpha}) n ( α ′ , α ~ ) α ′ ≥ 0 \alpha' \geq 0 α ′ ≥ 0 ( α ~ ∣ α ′ ) ≥ 0 (\tilde{\alpha} | \alpha') \geq 0 ( α ~ ∣ α ′ ) ≥ 0 n ( α ′ , α ~ ) ≥ 0 n(\alpha', \tilde{\alpha}) \geq 0 n ( α ′ , α ~ ) ≥ 0 n ( α ′ , α ~ ) n(\alpha', \tilde{\alpha}) n ( α ′ , α ~ )
9. 权，激进权
让l = dim V l = \dim V l = dim V R \mathbf{R} R V V V Q ( R ) Q(\mathbf{R}) Q ( R ) Q ( R ) Q(\mathbf{R}) Q ( R ) R \mathbf{R} R Q ( R ) Q(\mathbf{R}) Q ( R ) l l l V V V R \mathbf{R} R Q ( R ) Q(\mathbf{R}) Q ( R )
类似地，群Q ( R ∗ ) Q(\mathbf{R}^*) Q ( R ∗ ) l l l V ∗ V^* V ∗
x ∈ V x \in V x ∈ V y ∗ ∈ Q ( R ∗ ) y^* \in Q(\mathbf{R}^*) y ∗ ∈ Q ( R ∗ ) y ∗ ∈ R ∗ y^* \in \mathbf{R}^* y ∗ ∈ R ∗ ⟨ x , y ∗ ⟩ ∈ Z \langle x, y^* \rangle \in \mathbf{Z} ⟨ x , y ∗ ⟩ ∈ Z V V V Q ( R ) Q(\mathbf{R}) Q ( R ) G G G B ′ B' B ′ R ∗ \mathbf{R}^* R ∗ V V V B ′ B' B ′ G G G
让x ∈ V x \in V x ∈ V
(i) ⟨ x , y ∗ ⟩ ∈ Z \langle x, y^* \rangle \in \mathbf{Z} ⟨ x , y ∗ ⟩ ∈ Z y ∗ ∈ Q ( R ∗ ) y^* \in Q(\mathbf{R}^*) y ∗ ∈ Q ( R ∗ ) ⟨ x , y ∗ ⟩ ∈ Z \langle x, y^* \rangle \in \mathbf{Z} ⟨ x , y ∗ ⟩ ∈ Z y ∗ ∈ B ′ y^* \in B' y ∗ ∈ B ′ x x x B ′ B' B ′ Z \mathbf{Z} Z
由此我们推断，基对B ′ B' B ′ G G G R ⊆ G \mathbf{R} \subseteq G R ⊆ G Q ( R ) ⊆ G Q(\mathbf{R}) \subseteq G Q ( R ) ⊆ G
26号提案的群G G G P ( R ) P(\mathbf{R}) P ( R ) R \mathbf{R} R R ∗ \mathbf{R}^* R ∗ P ( R ∗ ) P(\mathbf{R}^*) P ( R ∗ )
P ( R ) / Q ( R ) , P ( R ~ ) / Q ( R ~ )
P(R)/Q(R), \quad P(\tilde{R})/Q(\tilde{R})
P ( R ) / Q ( R ) , P ( R ~ ) / Q ( R ~ ) 有限群在 Q / Z \mathbf{Q}/\mathbf{Z} Q / Z R ~ \tilde{R} R ~
如果 R 是根系 R i R_i R i R i R_i R i R i R_i R i
令R 1 R_1 R 1 R 1 R_1 R 1 Q 1 Q_1 Q 1 s α s_\alpha s α α ∈ R 1 \alpha \in R_1 α ∈ R 1 W 1 W_1 W 1 p ∈ P ( R ) p \in P(R) p ∈ P ( R ) w ∈ W 1 w \in W_1 w ∈ W 1 p − w ( p ) ∈ Q 1 p - w(p) \in Q_1 p − w ( p ) ∈ Q 1
引用了 1 · 被以下引用 1 引用了 命题 16
被以下引用 注 3, IX § 5
如果 w = s α w = s_\alpha w = s α α ∈ R 1 \alpha \in R_1 α ∈ R 1
p − w ( p ) = ⟨ p , α ∨ ⟩ α ∈ Z α ⊆ Q 1 .
p - w(p) = \langle p, \alpha^\vee \rangle \alpha \in \mathbf{Z}\alpha \subseteq Q_1.
p − w ( p ) = ⟨ p , α ∨ ⟩ α ∈ Z α ⊆ Q 1 . 如果w = s α 1 s α 2 … s α r w = s_{\alpha_1} s_{\alpha_2} \ldots s_{\alpha_r} w = s α 1 s α 2 … s α r α 1 , … , α r ∈ R 1 \alpha_1, \ldots, \alpha_r \in R_1 α 1 , … , α r ∈ R 1 p − w ( p ) ∈ Q 1 p - w(p) \in Q_1 p − w ( p ) ∈ Q 1
群 A(R) 使 P(R) 和 Q(R) 保持不变，因此作用于商 P(R)/Q(R)。根据第 27 条命题，群 W(R) 对 P(R)/Q(R) 的作用微不足道。 经过到商，我们看到 商群 A(R)/W(R)（cf. 命题 16，no. 5）*规范地作用于* P(R)/Q(R)。
10. 基本权，主导权
假设 R 被*减少*。令C为R的一个室，并令B为R的相应的基。由于R被减少，B ∨ = { α ∨ } α ∈ B B^\vee = \{ \alpha^\vee \}_{\alpha \in B} B ∨ = { α ∨ } α ∈ B R ∨ R^\vee R ∨ B ∨ B^\vee B ∨ ( ω ‾ α ) α ∈ B (\overline{\omega}_\alpha)_{\alpha \in B} ( ω α ) α ∈ B ( α 1 , … , α l ) (\alpha_1, \ldots, \alpha_l) ( α 1 , … , α l ) ( ω ‾ 1 , … , ω ‾ l ) (\overline{\omega}_1, \ldots, \overline{\omega}_l) ( ω 1 , … , ω l )
让x ∈ V x \in V x ∈ V x ∈ C x \in C x ∈ C ⟨ x , α ∨ ⟩ > 0 \langle x, \alpha^\vee \rangle > 0 ⟨ x , α ∨ ⟩ > 0 α ∈ B \alpha \in B α ∈ B ≥ 0 \geq 0 ≥ 0 ω ‾ α \overline{\omega}_\alpha ω α
对于固定的α \alpha α n ( α , β ) = ⟨ α , β ∨ ⟩ n(\alpha, \beta) = \langle \alpha, \beta^\vee \rangle n ( α , β ) = ⟨ α , β ∨ ⟩ α \alpha α ( ω ‾ β ) β ∈ B (\overline{\omega}_\beta)_{\beta \in B} ( ω β ) β ∈ B
α = ∑ β ∈ B n ( α , β ) ω ‾ β .
\alpha = \sum_{\beta \in B} n(\alpha, \beta) \overline{\omega}_\beta.
α = β ∈ B ∑ n ( α , β ) ω β . 因此嘉当矩阵是转置矩阵典范嵌入
Q ( R ) → P ( R )
Q(R) \to P(R)
Q ( R ) → P ( R ) 关于Z \mathbf{Z} Z ( ω ‾ α ) α ∈ B (\overline{\omega}_\alpha)_{\alpha \in B} ( ω α ) α ∈ B
如果权 ω ‾ \overline{\omega} ω C ‾ \overline{C} C ( ω ‾ α ) α ∈ B (\overline{\omega}_\alpha)_{\alpha \in B} ( ω α ) α ∈ B ≥ 0 \geq 0 ≥ 0 g ( ω ‾ ) ≤ ω ‾ g(\overline{\omega}) \leq \overline{\omega} g ( ω ) ≤ ω g ∈ W ( R ) g \in W(R) g ∈ W ( R ) C ‾ \overline{C} C W ( R ) W(R) W ( R ) ω ‾ \overline{\omega} ω ω ‾ ′ \overline{\omega}' ω ′ ω ‾ ′ \overline{\omega}' ω ′ ω ‾ \overline{\omega} ω W ( R ) W(R) W ( R )
我们有
⟨ ω ‾ α , β ∨ ⟩ = ( ω ‾ α ∣ 2 β ( β ∣ β ) ) = δ α β
\langle \overline{\omega}_\alpha, \beta^\vee \rangle = (\overline{\omega}_\alpha | \frac{2\beta}{(\beta|\beta)}) = \delta_{\alpha\beta}
⟨ ω α , β ∨ ⟩ = ( ω α ∣ ( β ∣ β ) 2 β ) = δ α β α , β ∈ B \alpha, \beta \in B α , β ∈ B δ α β \delta_{\alpha\beta} δ α β s β ( ω ‾ α ) = ω ‾ α − δ α β β and ( ω ‾ α ∣ β ) = 1 2 ( β ∣ β ) δ α β .
s_\beta(\overline{\omega}_\alpha) = \overline{\omega}_\alpha - \delta_{\alpha\beta}\beta \quad \text{and} \quad (\overline{\omega}_\alpha |\beta) = \frac{1}{2}(\beta|\beta)\delta_{\alpha\beta}.
s β ( ω α ) = ω α − δ α β β and ( ω α ∣ β ) = 2 1 ( β ∣ β ) δ α β . ω ‾ α \overline{\omega}_\alpha ω α 与 β \beta β 正交，对于 β ≠ α \beta \neq \alpha β = α R α R\alpha R α 是 1 2 α \frac{1}{2}\alpha 2 1 α ω ‾ α ∈ C ‾ \overline{\omega}_\alpha \in \overline{C} ω α ∈ C ( ω ‾ α ∣ ω ‾ β ) ⩾ 0 (\overline{\omega}_\alpha |\overline{\omega}_\beta) \geqslant 0 ( ω α ∣ ω β ) ⩾ 0 α , β ∈ B \alpha, \beta \in B α , β ∈ B ( ω ‾ α , ω ‾ β ) (\overline{\omega}_\alpha, \overline{\omega}_\beta) ( ω α , ω β ) ω ‾ ∈ V \overline{\omega} \in V ω ∈ V α ∈ B \alpha \in B α ∈ B 2 ( ω ‾ ∣ α ) / ( ∣ α ∣ α ) 2(\overline{\omega}|\alpha)/(|\alpha|\alpha) 2 ( ω ∣ α ) / ( ∣ α ∣ α ) ⩾ 0 \geqslant 0 ⩾ 0
Let* B B B be a basis of R R R B ′ B' B ′ a subset of B B B V ′ V' V ′ the vector subspace of V V V generated by B ′ B' B ′ R ′ = R ∩ V ′ R' = R \cap V' R ′ = R ∩ V ′ (which is a root system in V ′ V' V ′ R ′ ~ R'\tilde{} R ′ ~ the inverse root system (which is identified with the canonical image of R ′ R' R ′ in R ′ ~ R'\tilde{} R ′ ~ V 1 V_1 V 1 the orthogonal complement of R ′ ~ R'\tilde{} R ′ ~ in V V V and p p p the projection of V V V onto V ′ V' V ′ parallel to V 1 V_1 V 1 然后， Q ( R ′ ) = Q ( R ) ∩ V ′ Q(R') = Q(R) \cap V' Q ( R ′ ) = Q ( R ) ∩ V ′ P ( R ′ ) = p ( P ( R ) ) P(R') = p(P(R)) P ( R ′ ) = p ( P ( R )) The set of dominant weights of R ′ R' R ′ is the image under p p p of the set of dominant weights of R R R
引用了 2 · 被以下引用 1 引用了 命题 20 , 推论 4
被以下引用 VI § 4. 。 根系的分类
事实上，Q ( R ) Q(R) Q ( R ) V V V B B B Q ( R ′ ) Q(R') Q ( R ′ ) V ′ V' V ′ B ′ B' B ′ Q ( R ′ ) = Q ( R ) ∩ V ′ Q(R') = Q(R) \cap V' Q ( R ′ ) = Q ( R ) ∩ V ′ ω ‾ ∈ P ( R ) \overline{\omega} \in P(R) ω ∈ P ( R ) α ∈ R ′ \alpha \in R' α ∈ R ′ ⟨ p ( ω ‾ ) , α ∨ ⟩ = ⟨ ω ‾ , α ∨ ⟩ ∈ Z \langle p(\overline{\omega}), \alpha^\vee \rangle = \langle \overline{\omega}, \alpha^\vee \rangle \in \mathbf{Z} ⟨ p ( ω ) , α ∨ ⟩ = ⟨ ω , α ∨ ⟩ ∈ Z p ( ω ‾ ) ∈ P ( R ′ ) p(\overline{\omega}) \in P(R') p ( ω ) ∈ P ( R ′ ) p ( P ( R ) ) ⊆ P ( R ′ ) p(P(R)) \subseteq P(R') p ( P ( R )) ⊆ P ( R ′ ) ω ‾ ′ ∈ P ( R ′ ) \overline{\omega}' \in P(R') ω ′ ∈ P ( R ′ ) ω ‾ ′ \overline{\omega}' ω ′ ω ‾ \overline{\omega} ω V ∗ V^* V ∗ ( B − B ′ ) ~ (B - B')\tilde{} ( B − B ′ ) ~ α ∈ B \alpha \in B α ∈ B ⟨ ω ‾ , α ∨ ⟩ ∈ Z \langle \overline{\omega}, \alpha^\vee \rangle \in \mathbf{Z} ⟨ ω , α ∨ ⟩ ∈ Z ω ‾ ∈ P ( R ) \overline{\omega} \in P(R) ω ∈ P ( R ) ω ‾ ′ = p ( ω ‾ ) \overline{\omega}' = p(\overline{\omega}) ω ′ = p ( ω ) P ( R ′ ) ⊆ p ( P ( R ) ) P(R') \subseteq p(P(R)) P ( R ′ ) ⊆ p ( P ( R )) P ( R ′ ) = p ( P ( R ) ) P(R') = p(P(R)) P ( R ′ ) = p ( P ( R ))
(i) ρ = ∑ α ∈ B ω ‾ α \rho = \sum_{\alpha \in B} \overline{\omega}_\alpha ρ = ∑ α ∈ B ω α 这是 C C C s α ( ρ ) = ρ − α s_\alpha(\rho) = \rho - \alpha s α ( ρ ) = ρ − α 对于所有人 α ∈ B \alpha \in B α ∈ B ( 2 ρ ∣ α ) = ( α ∣ α ) (2\rho|\alpha) = (\alpha|\alpha) ( 2 ρ ∣ α ) = ( α ∣ α ) 对于所有人 α ∈ B \alpha \in B α ∈ B
由于R R R s α ( R + ( C ) − { α } ) = R + ( C ) − { α } s_\alpha(R_+(C) - \{\alpha\}) = R_+(C) - \{\alpha\} s α ( R + ( C ) − { α }) = R + ( C ) − { α } s α ( α ) = − α s_\alpha(\alpha) = -\alpha s α ( α ) = − α α ∈ B \alpha \in B α ∈ B s α ( 2 ρ ) = 2 ρ − 2 α s_\alpha(2\rho) = 2\rho - 2\alpha s α ( 2 ρ ) = 2 ρ − 2 α s α ( ρ ) = ρ − ⟨ ρ , α ∨ ⟩ s_\alpha(\rho) = \rho - \langle \rho, \alpha^\vee \rangle s α ( ρ ) = ρ − ⟨ ρ , α ∨ ⟩
⟨ ρ , α ∨ ⟩ = 1 = ⟨ ∑ β ∈ B ω ‾ β , α ∨ ⟩ .
\langle \rho, \alpha^\vee \rangle = 1 = \langle \sum_{\beta \in B} \overline{\omega}_\beta, \alpha^\vee \rangle.
⟨ ρ , α ∨ ⟩ = 1 = ⟨ β ∈ B ∑ ω β , α ∨ ⟩ . 因而，
ρ = ∑ β ω ‾ β \rho = \sum_{\beta} \overline{\omega}_\beta ρ = ∑ β ω β ，进而
ρ ∈ C \rho \in C ρ ∈ C 。最后，(iii) 相当于
⟨ ρ , α ∨ ⟩ = 1 \langle \rho, \alpha^\vee \rangle = 1 ⟨ ρ , α ∨ ⟩ = 1 。
令 σ \sigma σ > 0 > 0 > 0 R − R^- R − B − B^- B − α ∈ V \alpha \in V α ∈ V α \alpha α B B B ⟨ α , σ ⟩ \langle \alpha, \sigma \rangle ⟨ α , σ ⟩ α ∈ R \alpha \in R α ∈ R 1 2 ∑ β ∈ R + ( C ) n ( α , β ) \frac{1}{2} \sum_{\beta \in R_+ (\mathbf{C})} n(\alpha, \beta) 2 1 ∑ β ∈ R + ( C ) n ( α , β )
被以下引用 3 被以下引用 命题 31 , 引理 2, VIII § 7 , 命题 12, VIII § 7
交换上面R R R R − R^- R − α ∈ B \alpha \in B α ∈ B ⟨ α , σ ⟩ = 1 \langle \alpha, \sigma \rangle = 1 ⟨ α , σ ⟩ = 1
11. 考克塞特转型
令C C C R R R { α 1 , … , α l } \{ \alpha_1, \ldots, \alpha_l \} { α 1 , … , α l } R R R c = s α 1 ⋯ s α l c = s_{\alpha_1} \cdots s_{\alpha_l} c = s α 1 ⋯ s α l W W W c c c W W W C C C i ↦ α i i \mapsto \alpha_i i ↦ α i h h h W W W R R R
假设R R R m m m h − 1 h - 1 h − 1 h h h exp ( 2 i π m h ) \exp(\frac{2i\pi m}{h}) exp ( h 2 iπ m ) c c c
引用了 1 · 被以下引用 1 引用了 V § 6. 。考克塞特转型
被以下引用 VI § 4. 。 根系的分类
特别是，m m m W W W
我们首先证明引理：
我们知道（no. 6，Th.3）{ α 1 , … , α l } \{ \alpha_1, \ldots, \alpha_l \} { α 1 , … , α l } R R R V V V Q ( R ) Q(R) Q ( R ) w w w Q ( R ) Q(R) Q ( R ) { α 1 , … , α l } \{ \alpha_1, \ldots, \alpha_l \} { α 1 , … , α l }
令P P P c c c P P P z = exp ( 2 i π h ) z = \exp(\frac{2i\pi}{h}) z = exp ( h 2 iπ ) h h h P P P z z z Q \mathbf{Q} Q P P P h h h Q \mathbf{Q} Q P P P
令 R + R_+ R + C C C
n 1 + ⋯ + n l = 1 2 ∑ α ∈ R + n ( β , α )
n_1 + \cdots + n_l = \frac{1}{2} \sum_{\alpha \in R_+} n(\beta, \alpha)
n 1 + ⋯ + n l = 2 1 α ∈ R + ∑ n ( β , α ) = 1 + 1 2 ∑ α ∈ R + , α ≠ β n ( β , α ) = 1 + ∑ α ∈ R + , α ≠ β ( α ∣ β ) ( α ∣ α )
= 1 + \frac{1}{2} \sum_{\alpha \in R_+, \alpha \neq \beta} n(\beta, \alpha) = 1 + \sum_{\alpha \in R_+, \alpha \neq \beta} \frac{(\alpha | \beta)}{(\alpha | \alpha)}
= 1 + 2 1 α ∈ R + , α = β ∑ n ( β , α ) = 1 + α ∈ R + , α = β ∑ ( α ∣ α ) ( α ∣ β ) 根据no. 8，第 25 号提案 (iv)，对于所有 α ∈ R + \alpha \in R_+ α ∈ R + α ≠ β \alpha \neq \beta α = β n ( α , β ) = 0 n(\alpha, \beta) = 0 n ( α , β ) = 0 n ( α , β ) 2 = n ( α , β ) n(\alpha, \beta)^2 = n(\alpha, \beta) n ( α , β ) 2 = n ( α , β ) 4 ( α ∣ β ) 2 ( β ∣ β ) 2 = 2 ( α ∣ β ) ( β ∣ β ) \frac{4(\alpha | \beta)^2}{(\beta | \beta)^2} = \frac{2(\alpha | \beta)}{(\beta | \beta)} ( β ∣ β ) 2 4 ( α ∣ β ) 2 = ( β ∣ β ) 2 ( α ∣ β )
n 1 + ⋯ + n l + 1 = 2 + 2 ∑ α ∈ R + , α ≠ β ( α ∣ β ) 2 ( α ∣ α ) ( β ∣ β ) = 2 ∑ α ∈ R + ( α ∣ β ) 2 ( α ∣ α ) ( β ∣ β ) = ( β ∣ β ) − 1 ∑ α ∈ R ( α ∥ α ∥ ∣ β ) 2 .
n_1 + \cdots + n_l + 1 = 2 + 2 \sum_{\alpha \in R_+, \alpha \neq \beta} \frac{(\alpha | \beta)^2}{(\alpha | \alpha)(\beta | \beta)}
= 2 \sum_{\alpha \in R_+} \frac{(\alpha | \beta)^2}{(\alpha | \alpha)(\beta | \beta)} = (\beta | \beta)^{-1} \sum_{\alpha \in R} \left( \frac{\alpha}{\| \alpha \|} | \beta \right)^2.
n 1 + ⋯ + n l + 1 = 2 + 2 α ∈ R + , α = β ∑ ( α ∣ α ) ( β ∣ β ) ( α ∣ β ) 2 = 2 α ∈ R + ∑ ( α ∣ α ) ( β ∣ β ) ( α ∣ β ) 2 = ( β ∣ β ) − 1 α ∈ R ∑ ( ∥ α ∥ α ∣ β ) 2 . 由章。 V, § 6, no. 2，科。的 Th。 1、
∑ α ∈ R ( α ∥ α ∥ ∣ β ) 2 = h ( β ∣ β )
\sum_{\alpha \in R} \left( \frac{\alpha}{\| \alpha \|} | \beta \right)^2 = h(\beta | \beta)
α ∈ R ∑ ( ∥ α ∥ α ∣ β ) 2 = h ( β ∣ β ) 所以n 1 + ⋯ + n l + 1 = h n_1 + \cdots + n_l + 1 = h n 1 + ⋯ + n l + 1 = h
假设 R 是不可约的，并且所有根具有相同的长度。让α ∈ R \alpha \in R α ∈ R α \alpha α 4 h − 6 4h - 6 4 h − 6
引用了 1 引用了 推论, V § 6
令R ′ R' R ′ α \alpha α
( α ∣ α ) 2 + ( α ∣ − α ) 2 + ∑ β ∈ R ′ ( α ∣ β ) 2 = h ( α ∣ α ) 2 ,
(\alpha | \alpha)^2 + (\alpha | -\alpha)^2 + \sum_{\beta \in R'} (\alpha | \beta)^2 = h(\alpha | \alpha)^2,
( α ∣ α ) 2 + ( α ∣ − α ) 2 + β ∈ R ′ ∑ ( α ∣ β ) 2 = h ( α ∣ α ) 2 , 那就是
∑ β ∈ R ′ ( α ∣ β ) 2 = ( h − 2 ) ( α ∣ α ) 2 .
\sum_{\beta \in R'} (\alpha | \beta)^2 = (h - 2)(\alpha | \alpha)^2.
β ∈ R ′ ∑ ( α ∣ β ) 2 = ( h − 2 ) ( α ∣ α ) 2 . 如果β ∈ R ′ \beta \in R' β ∈ R ′ ( α ∣ β ) = ± 1 2 ( α ∣ α ) (\alpha | \beta) = \pm \frac{1}{2} (\alpha | \alpha) ( α ∣ β ) = ± 2 1 ( α ∣ α )
1 4 Card R ′ = h − 2 , Card R ′ = 4 h − 8 ,
\frac{1}{4} \operatorname{Card} R' = h - 2, \quad \operatorname{Card} R' = 4h - 8,
4 1 Card R ′ = h − 2 , Card R ′ = 4 h − 8 , 根与α \alpha α Card R ′ + 2 = 4 h − 6 \operatorname{Card} R' + 2 = 4h - 6 Card R ′ + 2 = 4 h − 6
假设 R 是不可约且可约化的。放入s α i = s i s_{\alpha_i} = s_i s α i = s i Γ \Gamma Γ c = s 1 … s l c = s_1 \ldots s_l c = s 1 … s l
引用了 4 · 被以下引用 4 引用了 IV § 1. 。考克塞特群 , 推论 2, IV § 1 , 推出 2 , V § 6. 。考克塞特转型
被以下引用 习题 1, § 6 , 习题 11, § 1 , 习题 22, § 1 , 习题 5, § 5
(i) 令 θ i = s l s l − 1 … s i + 1 ( α i ) \theta_i = s_l s_{l-1} \ldots s_{i+1} (\alpha_i) θ i = s l s l − 1 … s i + 1 ( α i ) i = 1 , … , l i = 1, \ldots, l i = 1 , … , l θ i > 0 , c ( θ i ) < 0 \theta_i > 0, c(\theta_i) < 0 θ i > 0 , c ( θ i ) < 0 (ii) 如果 α \alpha α > 0 > 0 > 0 c ( α ) < 0 c(\alpha) < 0 c ( α ) < 0 α \alpha α θ i \theta_i θ i (iii) 家族( θ i ) 1 ≤ i ≤ l (\theta_i)_{1 \leq i \leq l} ( θ i ) 1 ≤ i ≤ l (iv) 令Ω i \Omega_i Ω i Γ \Gamma Γ θ i \theta_i θ i Ω i \Omega_i Ω i Γ \Gamma Γ
首先观察( s 1 , … , s l ) (s_1, \ldots, s_l) ( s 1 , … , s l ) c c c s i s_i s i l − 1 l - 1 l − 1 X = S − { j } X = S - \{ j \} X = S − { j } c ∈ W X c \in W_X c ∈ W X
应用科。 no. 6到c c c
令Q i Q_i Q i α j , j > i \alpha_j, j > i α j , j > i Q i Q_i Q i s j , j > i s_j, j > i s j , j > i s j ( α i ) = α i s_j (\alpha_i) = \alpha_i s j ( α i ) = α i Q i Q_i Q i j > i j > i j > i
θ i = s l … s i + 1 ( α i ) = α i mod. Q i .
\theta_i = s_l \ldots s_{i+1} (\alpha_i) = \alpha_i \text{ mod. } Q_i.
θ i = s l … s i + 1 ( α i ) = α i mod. Q i . 在其他字中，存在整数c i j c_{ij} c ij
θ i = α i + ∑ j > i c i j α j .
\theta_i = \alpha_i + \sum_{j > i} c_{ij} \alpha_j.
θ i = α i + j > i ∑ c ij α j . 部分 (iii) 紧随其后。
最后，令α \alpha α ∑ k = 0 h − 1 c k ( α ) \sum_{k=0}^{h-1} c^k(\alpha) ∑ k = 0 h − 1 c k ( α ) c c c c k ( α ) c^k(\alpha) c k ( α ) k k k c k ( α ) > 0 c^k(\alpha) > 0 c k ( α ) > 0 c k + 1 ( α ) < 0 c^{k+1}(\alpha) < 0 c k + 1 ( α ) < 0 c k ( α ) c^k(\alpha) c k ( α ) θ i \theta_i θ i R R R Γ \Gamma Γ Ω i \Omega_i Ω i V ⊗ C V \otimes \mathbf{C} V ⊗ C ( x ∣ y ) (x|y) ( x ∣ y ) R R R Γ \Gamma Γ h h h l l l R R R h l hl h l
12. 规范双线性形式
我们已经看到（no. 1，命题 3）对称的双线性形式
( x , y ) ↦ B R ( x , y ) = ∑ α ∈ R ⟨ α ∗ , x ⟩ ⟨ α ∗ , y ⟩
(x, y) \mapsto B_R(x, y) = \sum_{\alpha \in R} \langle \alpha^*, x \rangle \langle \alpha^*, y \rangle
( x , y ) ↦ B R ( x , y ) = α ∈ R ∑ ⟨ α ∗ , x ⟩ ⟨ α ∗ , y ⟩ V V V A ( R ) A(R) A ( R ) R R R R ∗ R^* R ∗ V ∗ V^* V ∗ ( x ∗ , y ∗ ) ↦ B R ∗ ( x ∗ , y ∗ ) = ∑ α ∈ R ⟨ α , x ∗ ⟩ ⟨ α , y ∗ ⟩ (x^*, y^*) \mapsto B_{R^*}(x^*, y^*) = \sum_{\alpha \in R} \langle \alpha, x^* \rangle \langle \alpha, y^* \rangle ( x ∗ , y ∗ ) ↦ B R ∗ ( x ∗ , y ∗ ) = ∑ α ∈ R ⟨ α , x ∗ ⟩ ⟨ α , y ∗ ⟩ A ( R ) A(R) A ( R ) The inverse form of B R ∗ B_{R^*} B R ∗ B R B_R B R V V V V ∗ V^* V ∗ canonical bilinear form on V V V V ∗ V^* V ∗ Φ R \Phi_R Φ R Φ R ∗ \Phi_{R^*} Φ R ∗ A ( R ) A(R) A ( R ) σ \sigma σ B R ∗ B_{R^*} B R ∗ V V V V ∗ V^* V ∗ x ∈ V x \in V x ∈ V y ∈ V y \in V y ∈ V
Φ R ( x , y ) = B R ∗ ( σ ( x ) , σ ( y ) ) = ∑ α ∈ R ⟨ α , σ ( x ) ⟩ ⟨ α , σ ( y ) ⟩ .
\Phi_R(x, y) = B_{R^*}(\sigma(x), \sigma(y)) = \sum_{\alpha \in R} \langle \alpha, \sigma(x) \rangle \langle \alpha, \sigma(y) \rangle.
Φ R ( x , y ) = B R ∗ ( σ ( x ) , σ ( y )) = α ∈ R ∑ ⟨ α , σ ( x )⟩ ⟨ α , σ ( y )⟩ . ⟨ α , σ ( x ) ⟩ = B R ∗ ( σ ( α ) , σ ( x ) ) = Φ R ( α , x ) \langle \alpha, \sigma(x) \rangle = B_{R^*}(\sigma(\alpha), \sigma(x)) = \Phi_R(\alpha, x) ⟨ α , σ ( x )⟩ = B R ∗ ( σ ( α ) , σ ( x )) = Φ R ( α , x ) Φ R ( x , y ) = ∑ α ∈ R Φ R ( α , x ) Φ ( α , y ) .
\Phi_R(x, y) = \sum_{\alpha \in R} \Phi_R(\alpha, x) \Phi(\alpha, y).
Φ R ( x , y ) = α ∈ R ∑ Φ R ( α , x ) Φ ( α , y ) . Φ R \Phi_R Φ R W ( R ) W(R) W ( R ) 对于 β ∈ R \beta \in R β ∈ R
Φ R ( β , β ) = ∑ α ∈ R Φ R ( α , β ) 2 = 1 4 Φ R ( β , β ) 2 ∑ α ∈ R n ( α , β ) 2 ,
\Phi_R(\beta, \beta) = \sum_{\alpha \in R} \Phi_R(\alpha, \beta)^2 = \frac{1}{4} \Phi_R(\beta, \beta)^2 \sum_{\alpha \in R} n(\alpha, \beta)^2,
Φ R ( β , β ) = α ∈ R ∑ Φ R ( α , β ) 2 = 4 1 Φ R ( β , β ) 2 α ∈ R ∑ n ( α , β ) 2 , 4 Φ R ( β , β ) − 1 = ∑ α ∈ R n ( α , β ) 2 .
4 \Phi_R(\beta, \beta)^{-1} = \sum_{\alpha \in R} n(\alpha, \beta)^2.
4 Φ R ( β , β ) − 1 = α ∈ R ∑ n ( α , β ) 2 . x , y ∈ V x, y \in V x , y ∈ V B R ( x , y ) = ∑ α ∈ R Φ R ( 2 α Φ R ( α , α ) , x ) ( 2 α Φ R ( α , α ) , y ) = 4 ∑ α ∈ R Φ R ( α , x ) Φ R ( α , y ) Φ R ( α , α ) − 2 .
B_R(x, y) = \sum_{\alpha \in R} \Phi_R \left( \frac{2\alpha}{\Phi_R(\alpha, \alpha)}, x \right) \left( \frac{2\alpha}{\Phi_R(\alpha, \alpha)}, y \right)
= 4 \sum_{\alpha \in R} \Phi_R(\alpha, x) \Phi_R(\alpha, y) \Phi_R(\alpha, \alpha)^{-2}.
B R ( x , y ) = α ∈ R ∑ Φ R ( Φ R ( α , α ) 2 α , x ) ( Φ R ( α , α ) 2 α , y ) = 4 α ∈ R ∑ Φ R ( α , x ) Φ R ( α , y ) Φ R ( α , α ) − 2 . 由此可见，如果 R R R γ ( R ) > 0 \gamma(R) > 0 γ ( R ) > 0
∑ α ∈ R Φ R ( α , x ) Φ R ( α , y ) Φ R ( α , α ) − 2 = γ ( R ) Φ R ( x , y ) .
\sum_{\alpha \in R} \Phi_R(\alpha, x) \Phi_R(\alpha, y) \Phi_R(\alpha, \alpha)^{-2} = \gamma(R) \Phi_R(x, y).
α ∈ R ∑ Φ R ( α , x ) Φ R ( α , y ) Φ R ( α , α ) − 2 = γ ( R ) Φ R ( x , y ) . 到γ ( R ) \gamma(R) γ ( R ) B R ( x , y ) = 4 γ ( R ) Φ R ( x , y ) B_R(x, y) = 4\gamma(R) \Phi_R(x, y) B R ( x , y ) = 4 γ ( R ) Φ R ( x , y )
Φ R ∼ ( x ∗ , y ∗ ) = ( 4 γ ( R ) ) − 1 B R ∼ ( x ∗ , y ∗ )
\Phi_{R^{\sim}}(x^*, y^*) = (4\gamma(R))^{-1} B_{R^{\sim}}(x^*, y^*)
Φ R ∼ ( x ∗ , y ∗ ) = ( 4 γ ( R ) ) − 1 B R ∼ ( x ∗ , y ∗ ) 对于x ∗ , y ∗ ∈ V ∗ x^*, y^* \in V^* x ∗ , y ∗ ∈ V ∗ γ ( R ) = γ ( R ∼ ) \gamma(R) = \gamma(R^{\sim}) γ ( R ) = γ ( R ∼ ) β ∈ R \beta \in R β ∈ R
Φ R ∼ ( β ∼ , β ∼ ) = ( 4 γ ( R ) ) − 1 ∑ α ∈ R ⟨ β ∼ , α ⟩ 2 = γ ( R ) − 1 ∑ α ∈ R Φ R ( α , β ) 2 Φ R ( β , β ) 2
\Phi_{R^{\sim}}(\beta^{\sim}, \beta^{\sim}) = (4\gamma(R))^{-1} \sum_{\alpha \in R} \langle \beta^{\sim}, \alpha \rangle^2
= \gamma(R)^{-1} \sum_{\alpha \in R} \frac{\Phi_R(\alpha, \beta)^2}{\Phi_R(\beta, \beta)^2}
Φ R ∼ ( β ∼ , β ∼ ) = ( 4 γ ( R ) ) − 1 α ∈ R ∑ ⟨ β ∼ , α ⟩ 2 = γ ( R ) − 1 α ∈ R ∑ Φ R ( β , β ) 2 Φ R ( α , β ) 2 因此，根据 (16)，
Φ R ∼ ( β ∼ , β ∼ ) = γ ( R ) − 1 Φ R ( β , β ) − 2 Φ R ( β , β )
\Phi_{R^{\sim}}(\beta^{\sim}, \beta^{\sim}) = \gamma(R)^{-1} \Phi_R(\beta, \beta)^{-2} \Phi_R(\beta, \beta)
Φ R ∼ ( β ∼ , β ∼ ) = γ ( R ) − 1 Φ R ( β , β ) − 2 Φ R ( β , β ) 或者最后
Φ R ( β , β ) Φ R ∼ ( β ∼ , β ∼ ) = γ ( R ) − 1 .
\Phi_R(\beta, \beta) \Phi_{R^{\sim}}(\beta^{\sim}, \beta^{\sim}) = \gamma(R)^{-1}.
Φ R ( β , β ) Φ R ∼ ( β ∼ , β ∼ ) = γ ( R ) − 1 . 此外，如果R R R Φ R \Phi_R Φ R λ \lambda λ
γ ( R ) = λ − 4 .
\gamma(R) = \lambda^{-4}.
γ ( R ) = λ − 4 . 此外，如果 h h h W W W
h Φ R ( x , x ) = ∑ α ∈ R ( x , α λ ) 2 for all x ∈ V .
h \Phi_R(x, x) = \sum_{\alpha \in R} \left( x, \frac{\alpha}{\lambda} \right)^2 \text{ for all } x \in V.
h Φ R ( x , x ) = α ∈ R ∑ ( x , λ α ) 2 for all x ∈ V . 与(16)比较，我们得出
λ = h − 1 / 2 or γ ( R ) = h 2 .
\lambda = h^{-1/2} \quad \text{or} \quad \gamma(R) = h^2.
λ = h − 1/2 or γ ( R ) = h 2 . 最后，公式(19)表明R ∼ R^{\sim} R ∼ Φ R ∼ \Phi_{R^{\sim}} Φ R ∼ λ \lambda λ
习题
下面考虑的所有根系都是相对于真实的向量空间。我们用 ( x ∣ y ) (x|y) ( x ∣ y )
参见习题为 § 1 。
