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。考克塞特群
在这个截面中，W表示群乘法写成，单位元素1，S表示W的生成元的集合，使得S = S^{-1}和1\notin S。W的每个元素都是乘积 S 的元素的有限序列。从 no. 3 开始，我们假设 S 的每个元素都有阶 2。
1. 长度并减少分解
设 w \in W。w 的长度（相对于 S），用 l_S(w) 或简称为 l(w) 表示，是最小的整数 q \geq 0，使得 w 是 S 的 q 元素的序列的乘积。 w 的约化分解 （相对于 S）是 S 的元素中的任意序列 s = (s_1, \ldots, s_q)，使得 w = s_1 \cdots s_q 且 q = l(w)。
因此 1 是长度 0 的唯一元素，S 由长度 1 的元素组成。
令 w 和 w' 在 W 中。我们有公式：
l ( w w ′ ) ≤ l ( w ) + l ( w ′ ) , (1)
l(ww') \leq l(w) + l(w'), \tag{1}
l ( w w ′ ) ≤ l ( w ) + l ( w ′ ) , ( 1 ) l ( w − 1 ) = l ( w ) , (2)
l(w^{-1}) = l(w), \tag{2}
l ( w − 1 ) = l ( w ) , ( 2 ) ∣ l ( w ) − l ( w ′ ) ∣ ≤ l ( w w ′ − 1 ) . (3)
|l(w) - l(w')| \leq l(w{w'}^{-1}). \tag{3}
∣ l ( w ) − l ( w ′ ) ∣ ≤ l ( w w ′ − 1 ) . ( 3 )
令 (s_1, \ldots, s_p) 和 (s'_1, \ldots, s'_q) 分别减少 w 和 w' 的分解。因此 l(w) = p 且 l(w') = q。由于 ww' = s_1 \ldots s_p s'_1 \ldots s'_q，我们有 l(ww') \leq p + q, 证明 (1)。由于 S = S^{-1} 且 w^{-1} = s_p^{-1} \ldots s_1^{-1}，我们有 l(w^{-1}) \leq p = l(w)。 替换 w 除以 w^{-1} 给出相反的不等式 l(w) \leq l(w^{-1}), 证明 (2)。 (1) 和 (2) 中的替换 w 除以 ww'^{-1} 得到关系
l ( w ) − l ( w ′ ) ≤ l ( w w ′ − 1 ) , (4)
l(w) - l(w') \leq l(w{w'}^{-1}), \tag{4}
l ( w ) − l ( w ′ ) ≤ l ( w w ′ − 1 ) , ( 4 ) l ( w w ′ − 1 ) = l ( w ′ w − 1 ) . (5)
l(w{w'}^{-1}) = l(w'w^{-1}). \tag{5}
l ( w w ′ − 1 ) = l ( w ′ w − 1 ) . ( 5 ) 交换(4)中的w w w w ′ w' w ′ l ( w ′ ) − l ( w ) ≤ l ( w w ′ − 1 ) l(w') - l(w) \leq l(w{w'}^{-1}) l ( w ′ ) − l ( w ) ≤ l ( w w ′ − 1 )
令s = ( s 1 , … , s p ) s = (s_1, \ldots, s_p) s = ( s 1 , … , s p ) s ′ = ( s 1 ′ , … , s q ′ ) s' = (s'_1, \ldots, s'_q) s ′ = ( s 1 ′ , … , s q ′ ) S S S w = s 1 … s p w = s_1 \ldots s_p w = s 1 … s p w ′ = s 1 ′ … s q ′ w' = s'_1 \ldots s'_q w ′ = s 1 ′ … s q ′ ( s 1 , … , s p , s 1 ′ , … , s q ′ ) (s_1, \ldots, s_p, s'_1, \ldots, s'_q) ( s 1 , … , s p , s 1 ′ , … , s q ′ ) w w ′ ww' w w ′ s s s w w w s ′ s' s ′ w ′ w' w ′
假设、l ( w ) ≤ p l(w) \leq p l ( w ) ≤ p l ( w ′ ) ≤ q l(w') \leq q l ( w ′ ) ≤ q l ( w w ′ ) = p + q l(ww') = p + q l ( w w ′ ) = p + q l ( w ) = p l(w) = p l ( w ) = p l ( w ′ ) = q l(w') = q l ( w ′ ) = q
2. 二面角群
令 M M M { 1 , − 1 } \{1, -1\} { 1 , − 1 } m m m ≥ 2 \geq 2 ≥ 2 m = ∞ m = \infty m = ∞ M M M ( − 1 ) . x = − x (-1).x = -x ( − 1 ) . x = − x Z / m Z \mathbf{Z}/m\mathbf{Z} Z / m Z Z \mathbf{Z} Z Z / m Z \mathbf{Z}/m\mathbf{Z} Z / m Z M M M M M M Z \mathbf{Z} Z D m D_m D m D m D_m D m ( ε , x ) (\varepsilon, x) ( ε , x ) ε = ± 1 \varepsilon = \pm 1 ε = ± 1 x ∈ Z / m Z x \in \mathbf{Z}/m\mathbf{Z} x ∈ Z / m Z x ∈ Z x \in \mathbf{Z} x ∈ Z D m D_m D m
( ε , x ) . ( ε ′ , x ′ ) = ( ε ε ′ , ε ′ x + x ′ ) .
(\varepsilon, x).(\varepsilon', x') = (\varepsilon \varepsilon', \varepsilon' x + x').
( ε , x ) . ( ε ′ , x ′ ) = ( ε ε ′ , ε ′ x + x ′ ) . 我们用 1 模
m m m （分别为
ι = 1 \iota = 1 ι = 1 ）和集合表示
ι \iota ι 类
ρ = ( − 1 , 0 ) , ρ ′ = ( − 1 , ι ) , π = ( 1 , ι ) .
\rho = (-1, 0), \quad \rho' = (-1, \iota), \quad \pi = (1, \iota).
ρ = ( − 1 , 0 ) , ρ ′ = ( − 1 , ι ) , π = ( 1 , ι ) . 然后是
ρ 2 = ρ ′ 2 = 1 \rho^2 = {\rho'}^2 = 1 ρ 2 = ρ ′ 2 = 1 和
π = ρ ρ ′ \pi = \rho \rho' π = ρ ρ ′ 。 公式
π n = ( 1 , n ι ) , ρ π n = ( − 1 , n ι )
\pi^n = (1, n\iota), \quad \rho \pi^n = (-1, n\iota)
π n = ( 1 , n ι ) , ρ π n = ( − 1 , n ι ) 证明
D m D_m D m 是由
{ ρ , ρ ′ } \{\rho, \rho'\} { ρ , ρ ′ } 生成的二面角群。
(i) p = s s ′ p = ss' p = s s ′ W W W P P P W W W T = { 1 , s } T = \{1, s\} T = { 1 , s } P P P ( W : P ) = 2 (W : P) = 2 ( W : P ) = 2
(ii) 令 m m m p p p m ≥ 2 m \geq 2 m ≥ 2 W W W 2 m 2m 2 m D m D_m D m W W W φ \varphi φ φ ( ρ ) = s \varphi(\rho) = s φ ( ρ ) = s φ ( ρ ′ ) = s ′ \varphi(\rho') = s' φ ( ρ ′ ) = s ′
(i) 我们有 s p s − 1 = s s s ′ s = s ′ s = p − 1 sps^{-1} = sss's = s's = p^{-1} s p s − 1 = ss s ′ s = s ′ s = p − 1
s p n s − 1 = p − n
sp^n s^{-1} = p^{-n}
s p n s − 1 = p − n 每整数n n n W W W { s , s ′ } \{ s, s' \} { s , s ′ } { s , p } \{ s, p \} { s , p } P P P T P TP T P W W W T P TP T P s s s s ′ = s p s' = sp s ′ = s p W = T P = P ∪ s P W = TP = P \cup sP W = T P = P ∪ s P W ≠ P W \neq P W = P W = P W = P W = P W W W p 2 = s 2 s ′ 2 = 1 p^2 = s^2 {s'}^2 = 1 p 2 = s 2 s ′ 2 = 1 W = P W = P W = P p p p W W W s s s s ′ s' s ′
(ii) 从s ≠ s ′ s \neq s' s = s ′ p ≠ 1 p \neq 1 p = 1 m ≥ 2 m \geq 2 m ≥ 2 P P P m m m ( W : P ) = 2 (W : P) = 2 ( W : P ) = 2 W W W 2 m 2m 2 m m m m Z / m Z \mathbf{Z}/m\mathbf{Z} Z / m Z Z \mathbf{Z} Z P P P π \pi π p p p φ ′ \varphi' φ ′ M = { 1 , − 1 } M = \{ 1, -1 \} M = { 1 , − 1 } T T T − 1 -1 − 1 s s s φ ′ ′ \varphi'' φ ′′ W W W T T T P P P ρ π n ρ − 1 = π − n \rho \pi^n \rho^{-1} = \pi^{-n} ρ π n ρ − 1 = π − n φ ′ \varphi' φ ′ φ ′ ′ \varphi'' φ ′′ φ \varphi φ D m D_m D m W W W φ ( ρ ) = s \varphi(\rho) = s φ ( ρ ) = s φ ( π ) = p \varphi(\pi) = p φ ( π ) = p φ ( ρ ′ ) = s ′ \varphi(\rho') = s' φ ( ρ ′ ) = s ′ φ \varphi φ D m D_m D m { ρ , ρ ′ } \{ \rho, \rho' \} { ρ , ρ ′ }
3. Coxeter 的第一个特性群
回想一下，从现在开始，我们假设 S 的元素是阶 2。
如果 (W, S) 满足以下条件，则称其为 Coxeter 系统：
(C) 对于 S 中的 s, s'，令 m(s, s') 为 ss' 的阶，并令 I 为对 (s, s') 的集合，使得 m(s, s') 是有限的。 生成的集合 S 和关系 (ss')^{m(s,s')} = 1，对于 I 形式群 W 的演示文稿 1 中的 (s, s')。
当 (W, S) 是 Coxeter 系统时，人们还滥用语言说 W 是 Coxeter 群。
令m为整数≥ 2 \geq 2 ≥ 2 ∞ \infty ∞ m = ∞ m = \infty m = ∞ s 2 = s ′ 2 = 1 s^2 = {s'}^2 = 1 s 2 = s ′ 2 = 1 s 2 = s ′ 2 = ( s s ′ ) m = 1 s^2 = {s'}^2 = (ss')^m = 1 s 2 = s ′ 2 = ( s s ′ ) m = 1 ρ \rho ρ ρ ′ \rho' ρ ′ ρ 2 = ( ρ ′ ) 2 = 1 \rho^2 = (\rho')^2 = 1 ρ 2 = ( ρ ′ ) 2 = 1 ( ρ ρ ′ ) m = 1 (\rho\rho')^m = 1 ( ρ ρ ′ ) m = 1 f ( s ) = ρ f(s) = \rho f ( s ) = ρ f ( s ′ ) = ρ ′ f(s') = \rho' f ( s ′ ) = ρ ′ ρ ρ ′ \rho\rho' ρ ρ ′ 2 m 2m 2 m
根据结构的迁移，得出每个二面角群都是 Coxeter 群。
引用了 1 引用了 命题 2
令 S n \mathfrak{S}_n S n n ≥ 2 n \geq 2 n ≥ 2 s i s_i s i i + 1 i+1 i + 1 1 ≤ i < n 1 \leq i < n 1 ≤ i < n ( S n , S ) (\mathfrak{S}_n, S) ( S n , S )
引用了 2 引用了 习题 4, § 1 , IV § 2. 。奶系统
对于有限 Coxeter 群、cf. 的分类4、§ 4。
假设 (W, S) 是 Coxeter 系统。那么，当且仅当满足以下条件时，S的两个元素 s和s'在W中是共轭2 ^2 2
被以下引用 2 被以下引用 习题 9, § 1 , 习题 12, § 4
(I) S 的元素存在有限序列( s 1 , … , s q ) (s_1, \ldots, s_q) ( s 1 , … , s q ) s 1 = s , s q = s ′ s_1 = s, s_q = s' s 1 = s , s q = s ′ s j s j + 1 s_j s_{j+1} s j s j + 1 1 ≤ j < q 1 \leq j < q 1 ≤ j < q
令 S 中的 s 和 s' 使得 p = s s ′ p = ss' p = s s ′ 2 n + 1 2n + 1 2 n + 1 s p − n = p n s sp^{-n} = p^n s s p − n = p n s
p n s p − n = p n p n s − p − 1 s = s ′ s s = s ′ ,
p^n sp^{-n} = p^n p^n s - p^{-1} s = s' ss = s',
p n s p − n = p n p n s − p − 1 s = s ′ ss = s ′ , 并且 s ′ s' s ′
对于 S 中的所有 s，令 A s A_s A s s ′ ∈ S s' \in S s ′ ∈ S s j s_j s j s j + 1 s_{j+1} s j + 1 1 ≤ j < q 1 \leq j < q 1 ≤ j < q A s A_s A s s ′ s' s ′
令 f f f M = { 1 , − 1 } M = \{1, -1\} M = { 1 , − 1 } A s A_s A s S − A s S - A_s S − A s s ′ s' s ′ s ′ ′ s'' s ′′ s ′ s ′ ′ s's'' s ′ s ′′ s ′ s' s ′ s ′ ′ s'' s ′′ A s A_s A s S − A s S - A_s S − A s f ( s ′ ) f ( s ′ ′ ) = 1 f(s') f(s'') = 1 f ( s ′ ) f ( s ′′ ) = 1 f ( s ′ ) f ( s ′ ′ ) = − 1 f(s') f(s'') = -1 f ( s ′ ) f ( s ′′ ) = − 1 ( f ( s ′ ) f ( s ′ ′ ) ) m = 1 (f(s') f(s''))^m = 1 ( f ( s ′ ) f ( s ′′ ) ) m = 1 s ′ s' s ′ s ′ s' s ′ f ( s ′ ) = g ( s ′ ) = 1 f(s') = g(s') = 1 f ( s ′ ) = g ( s ′ ) = 1 s ′ ∈ A s s' \in A_s s ′ ∈ A s
4. 在 COXETER 中减少分解群
假设 (W, S) 是 Coxeter 系统。令T为S的元素W中的共轭集合。对于S的元素任意有限序列s = ( s 1 , … , s q ) s = (s_1, \ldots, s_q) s = ( s 1 , … , s q ) Φ ( s ) \Phi(s) Φ ( s ) ( t 1 , … , t q ) (t_1, \ldots, t_q) ( t 1 , … , t q )
t j = ( s 1 … s j − 1 ) s j ( s 1 … s j − 1 ) − 1 for 1 ≤ j ≤ q .
t_j = (s_1 \ldots s_{j-1}) s_j (s_1 \ldots s_{j-1})^{-1} \quad \text{for } 1 \leq j \leq q.
t j = ( s 1 … s j − 1 ) s j ( s 1 … s j − 1 ) − 1 for 1 ≤ j ≤ q . 然后是t 1 = s 1 t_1 = s_1 t 1 = s 1 s 1 … s q = t q t q − 1 … t 1 s_1 \ldots s_q = t_q t_{q-1} \ldots t_1 s 1 … s q = t q t q − 1 … t 1 t ∈ T t \in T t ∈ T j j j n ( s , t ) n(s, t) n ( s , t ) 1 ≤ j ≤ q 1 \leq j \leq q 1 ≤ j ≤ q t j = t t_j = t t j = t
R = { 1 , − 1 } × T .
R = \{1, -1\} \times T.
R = { 1 , − 1 } × T .
(i) 设w ∈ W w \in W w ∈ W t ∈ T t \in T t ∈ T s = ( s 1 , … , s q ) s = (s_1, \ldots, s_q) s = ( s 1 , … , s q ) ( − 1 ) n ( s , t ) (-1)^{n(s, t)} ( − 1 ) n ( s , t ) η ( w , t ) \eta(w, t) η ( w , t ) w = s 1 … s q w = s_1 \ldots s_q w = s 1 … s q
被以下引用 3 被以下引用 习题 16, § 1 , 引理 3 , 推论, VI § 1
(ii) 对于 w ∈ W w \in W w ∈ W U w U_w U w
U w ( ε , t ) = ( ε ⋅ η ( w − 1 , t ) , w t w − 1 ) ( ε = ± 1 , t ∈ T ) .
U_w(\varepsilon, t) = (\varepsilon \cdot \eta(w^{-1}, t), wtw^{-1}) \quad (\varepsilon = \pm 1, t \in T).
U w ( ε , t ) = ( ε ⋅ η ( w − 1 , t ) , w t w − 1 ) ( ε = ± 1 , t ∈ T ) . 从W到R的置换的群，映射w ↦ U w w \mapsto U_w w ↦ U w
(2) 回想一下，如果存在 W 的内自同构且变换彼此相交，则群 W 的两个元素（分别为两个子集）被称为共轭。
对于 s ∈ S s \in S s ∈ S U s U_s U s R R R
U s ( ε , t ) = ( ε . ( − 1 ) δ s , t , s t s − 1 ) ( ε = ± 1 , t ∈ T ) ,
U_s(\varepsilon, t) = (\varepsilon.(-1)^{\delta_{s,t}}, st s^{-1}) \quad (\varepsilon = \pm 1, t \in T),
U s ( ε , t ) = ( ε . ( − 1 ) δ s , t , s t s − 1 ) ( ε = ± 1 , t ∈ T ) , 其中
δ s , t \delta_{s,t} δ s , t 是克罗内克符号。
U s 2 = I d R U_s^2 = \mathrm{Id}_R U s 2 = Id R 是直接的，这表明
U s U_s U s 是
R R R 的置换。
令s = ( s 1 , … , s q ) s = (s_1, \ldots, s_q) s = ( s 1 , … , s q ) S S S w = s q … s 1 w = s_q \ldots s_1 w = s q … s 1 U s = U s q … U s 1 U_s = U_{s_q} \ldots U_{s_1} U s = U s q … U s 1 q q q
U s ( ε , t ) = ( ε . ( − 1 ) n ( s , t ) , w t w − 1 ) .
U_s(\varepsilon, t) = (\varepsilon.(-1)^{n(s,t)}, wt w^{-1}).
U s ( ε , t ) = ( ε . ( − 1 ) n ( s , t ) , w t w − 1 ) . 对于
q = 0 , 1 q = 0, 1 q = 0 , 1 来说这是很清楚的。如果
q > 1 q > 1 q > 1 ，则输入
s ′ = ( s 1 , … , s q − 1 ) s' = (s_1, \ldots, s_{q-1}) s ′ = ( s 1 , … , s q − 1 ) 并
w ′ = s q − 1 … s 1 .
w' = s_{q-1} \ldots s_1.
w ′ = s q − 1 … s 1 . 使用归纳假设，我们得到
U s ( ε , t ) = U s q ( U s q ( ε . ( − 1 ) n ( s ′ , t ) , w ′ t w ′ − 1 ) ) = ( ε . ( − 1 ) n ( s ′ , t ) + δ s q , w ′ t w ′ − 1 , w t w − 1 ) .
\begin{align*}
U_s(\varepsilon, t) &= U_{s_q}(U_{s_q}(\varepsilon.(-1)^{n(s',t)}, w't {w'}^{-1})) \\
&= (\varepsilon.(-1)^{n(s',t)+\delta_{s_q,w't {w'}^{-1}}}, wt w^{-1}).
\end{align*}
U s ( ε , t ) = U s q ( U s q ( ε . ( − 1 ) n ( s ′ , t ) , w ′ t w ′ − 1 )) = ( ε . ( − 1 ) n ( s ′ , t ) + δ s q , w ′ t w ′ − 1 , w t w − 1 ) . 但是
Φ ( s ) = ( Φ ( s ′ ) , w ′ − 1 s q w ′ ) \Phi(s) = (\Phi(s'), {w'}^{-1} s_q w') Φ ( s ) = ( Φ ( s ′ ) , w ′ − 1 s q w ′ ) 和
n ( s , t ) = n ( s ′ , t ) + δ w ′ , − 1 s q w ′ , t n(s, t) = n(s', t) + \delta_{w', -1} s_q w', t n ( s , t ) = n ( s ′ , t ) + δ w ′ , − 1 s q w ′ , t ，证明公式（14）。
现在让s , s ′ ∈ S s, s' \in S s , s ′ ∈ S p = s s ′ p = ss' p = s s ′ m m m s = ( s 1 , … , s 2 m ) s = (s_1, \ldots, s_{2m}) s = ( s 1 , … , s 2 m ) S S S j j j s j = s s_j = s s j = s j j j s j = s ′ s_j = s' s j = s ′ s 2 m … s 1 = p − m = 1 s_{2m} \ldots s_1 = p^{-m} = 1 s 2 m … s 1 = p − m = 1
t j = p j − 1 s for 1 ≤ j ≤ 2 m .
t_j = p^{j-1} s \quad \text{for } 1 \leq j \leq 2m.
t j = p j − 1 s for 1 ≤ j ≤ 2 m . 由于
p p p 是阶
m m m ，元素
t 1 , … , t m t_1, \ldots, t_m t 1 , … , t m 是不同的，
t j + m = t j t_{j+m} = t_j t j + m = t j 对于
1 ≤ j ≤ m 1 \leq j \leq m 1 ≤ j ≤ m 是不同的。 对于所有
t ∈ T t \in T t ∈ T ，整数
n ( s , t ) n(s, t) n ( s , t ) 因此等于0或2，并且（14）表明
U s = I d R U_s = \mathrm{Id}_R U s = Id R 。 其他字，
( U s U s ′ ) m = I d R (U_s U_{s'})^m = \mathrm{Id}_R ( U s U s ′ ) m = Id R 。 Thus, by the definition of Coxeter systems, there exists a homomorphism
w ↦ U w w \mapsto U_w w ↦ U w from
W W W to the group of permutations of
R R R such that
U s U_s U s is given by the right-hand side of (13). 然后，对于每个序列
s = ( s 1 , … , s q ) s = (s_1, \ldots, s_q) s = ( s 1 , … , s q ) ，
U w = U s U_w = U_s U w = U s 使得
w = s q … s 1 w = s_q \ldots s_1 w = s q … s 1 和引理1紧随(14)而来。
设s = ( s 1 , … , s q ) s = (s_1, \ldots, s_q) s = ( s 1 , … , s q ) Φ ( s ) = ( t 1 , … , t q ) \Phi(s) = (t_1, \ldots, t_q) Φ ( s ) = ( t 1 , … , t q ) w = s 1 … s q w = s_1 \ldots s_q w = s 1 … s q T w T_w T w T T T η ( w , t ) = − 1 \eta(w, t) = -1 η ( w , t ) = − 1 t i t_i t i s s s w w w T w = { t 1 , … , t q } T_w = \{ t_1, \ldots, t_q \} T w = { t 1 , … , t q } C a r d ( T w ) = l ( w ) \mathrm{Card}(T_w) = l(w) Card ( T w ) = l ( w )
被以下引用 4 被以下引用 引理 3 , 注 , 推论 2, IV § 2 , 推论 3, IV § 2
显然T w ⊂ { t 1 , … , t q } T_w \subset \{ t_1, \ldots, t_q \} T w ⊂ { t 1 , … , t q } s s s C a r d ( T w ) ≤ l ( w ) \mathrm{Card}(T_w) \leq l(w) Card ( T w ) ≤ l ( w ) t i t_i t i t t t { t 1 , … , t q } \{ t_1, \ldots, t_q \} { t 1 , … , t q } n ( s , t ) n(s, t) n ( s , t ) T w = { t 1 , … , t q } T_w = \{ t_1, \ldots, t_q \} T w = { t 1 , … , t q } q = C a r d ( T w ) ≤ l ( w ) q = \mathrm{Card}(T_w) \leq l(w) q = Card ( T w ) ≤ l ( w ) s s s t i = t j t_i = t_j t i = t j i < j i < j i < j s i = u s j u − 1 s_i = us_j u^{-1} s i = u s j u − 1 u = s i + 1 … s j − 1 u = s_{i+1} \ldots s_{j-1} u = s i + 1 … s j − 1
w = s 1 … s i − 1 s i + 1 … s j − 1 s j + 1 … s q .
w = s_1 \ldots s_{i-1} s_{i+1} \ldots s_{j-1} s_{j+1} \ldots s_q.
w = s 1 … s i − 1 s i + 1 … s j − 1 s j + 1 … s q . 这表明s s s w w w
设w ∈ W w \in W w ∈ W s ∈ S s \in S s ∈ S l ( s w ) ≤ l ( w ) l(sw) \leq l(w) l ( s w ) ≤ l ( w ) s = ( s 1 , … , s q ) s = (s_1, \ldots, s_q) s = ( s 1 , … , s q ) S S S w = s 1 … s q w = s_1 \ldots s_q w = s 1 … s q j j j 1 ≤ j ≤ q 1 \leq j \leq q 1 ≤ j ≤ q
引用了 2 · 被以下引用 2 引用了 引理 1 , 引理 2
被以下引用 § 1. 。考克塞特群 , 定理 1
s s 1 … s j − 1 = s 1 … s j − 1 s j .
ss_1 \ldots s_{j-1} = s_1 \ldots s_{j-1} s_j.
s s 1 … s j − 1 = s 1 … s j − 1 s j . 设p p p w w w w ′ = s w w' = sw w ′ = s w l ( w ′ ) ≡ l ( w ) + 1 l(w') \equiv l(w) + 1 l ( w ′ ) ≡ l ( w ) + 1 注 得出。 2.假设l ( w ′ ) ≤ l ( w ) l(w') \leq l(w) l ( w ′ ) ≤ l ( w )
∣ l ( w ) − l ( w ′ ) ∣ ≤ l ( w w ′ − 1 ) = l ( s ) = 1
|l(w) - l(w')| \leq l(w{w'}^{-1}) = l(s) = 1
∣ l ( w ) − l ( w ′ ) ∣ ≤ l ( w w ′ − 1 ) = l ( s ) = 1 从而导致l ( w ′ ) = p − 1 l(w') = p - 1 l ( w ′ ) = p − 1
( s 1 ′ , … , s p − 1 ′ )
(s'_1, \ldots, s'_{p-1})
( s 1 ′ , … , s p − 1 ′ ) w ′ w' w ′ s = ( s , s 1 ′ , … , s p − 1 ′ ) s = (s, s'_1, \ldots, s'_{p-1}) s = ( s , s 1 ′ , … , s p − 1 ′ ) Φ ( s ′ ) = ( t 1 ′ , … , t p ′ ) \Phi(s') = (t'_1, \ldots, t'_p) Φ ( s ′ ) = ( t 1 ′ , … , t p ′ ) s ′ s' s ′ w w w t 1 ′ = s t'_1 = s t 1 ′ = s t 1 ′ , … , t p ′ t'_1, \ldots, t'_p t 1 ′ , … , t p ′ n ( s ′ , s ) = 1 n(s', s) = 1 n ( s ′ , s ) = 1 w w w s s s n ( s , s ) ≡ n ( s ′ , s ) n(s, s) \equiv n(s', s) n ( s , s ) ≡ n ( s ′ , s ) n ( s , s ) ≠ 0 n(s, s) \neq 0 n ( s , s ) = 0 s s s t j t_j t j Φ ( s ) \Phi(s) Φ ( s )
5. 交换条件
“交换条件”是关于( W , S ) (W, S) ( W , S )
(E) 令w ∈ W w \in W w ∈ W s ∈ S s \in S s ∈ S l ( s w ) ≤ l ( w ) l(sw) \leq l(w) l ( s w ) ≤ l ( w ) w w w ( s 1 , … , s q ) (s_1, \ldots, s_q) ( s 1 , … , s q ) j j j 1 ≤ j ≤ q 1 \leq j \leq q 1 ≤ j ≤ q
s s 1 … s j − 1 = s 1 … s j − 1 s j .
ss_1 \ldots s_{j-1} = s_1 \ldots s_{j-1} s_j.
s s 1 … s j − 1 = s 1 … s j − 1 s j . 我们假设在这个数中( W , S ) (W, S) ( W , S ) ( W , S ) (W, S) ( W , S )
令 s ∈ S , w ∈ W s \in S, w \in W s ∈ S , w ∈ W s = ( s 1 , … , s q ) s = (s_1, \ldots, s_q) s = ( s 1 , … , s q ) w w w
被以下引用 4 被以下引用 § 1. 。考克塞特群 , 引理 4 , 推论 1 , 定理 1
a) l ( s w ) = l ( w ) + 1 l(sw) = l(w) + 1 l ( s w ) = l ( w ) + 1 ( s , s 1 , … , s q ) (s, s_1, \ldots, s_q) ( s , s 1 , … , s q ) s w sw s w
b) l ( s w ) = l ( w ) − 1 l(sw) = l(w) - 1 l ( s w ) = l ( w ) − 1 j j j 1 ≤ j ≤ q 1 \leq j \leq q 1 ≤ j ≤ q ( s 1 , … , s j − 1 , s j + 1 , … , s q ) (s_1, \ldots, s_{j-1}, s_{j+1}, \ldots, s_q) ( s 1 , … , s j − 1 , s j + 1 , … , s q ) s w sw s w ( s , s 1 , … , s j − 1 , s j + 1 , … , s q ) (s, s_1, \ldots, s_{j-1}, s_{j+1}, \ldots, s_q) ( s , s 1 , … , s j − 1 , s j + 1 , … , s q ) w w w
把w ′ = s w w' = sw w ′ = s w
∣ l ( w ) − l ( w ′ ) ∣ ≤ l ( s ) = 1.
|l(w) - l(w')| \leq l(s) = 1.
∣ l ( w ) − l ( w ′ ) ∣ ≤ l ( s ) = 1. 我们区分两种情况：
a) l ( w ′ ) > l ( w ) l(w') > l(w) l ( w ′ ) > l ( w ) l ( w ′ ) = q + 1 l(w') = q + 1 l ( w ′ ) = q + 1 w ′ = s s 1 … s q w' = ss_1 \ldots s_q w ′ = s s 1 … s q
( s , s 1 , … , s q )
(s, s_1, \ldots, s_q)
( s , s 1 , … , s q ) 分解减少为
w ′ w' w ′ 。
b) l ( w ′ ) ≤ l ( w ) l(w') \leq l(w) l ( w ′ ) ≤ l ( w ) j j j 1 ≤ j ≤ q 1 \leq j \leq q 1 ≤ j ≤ q w = s s 1 … s j − 1 s j + 1 … s q w = ss_1 \ldots s_{j-1} s_{j+1} \ldots s_q w = s s 1 … s j − 1 s j + 1 … s q
w ′ = s 1 … s j − 1 s j + 1 … s q .
w' = s_1 \ldots s_{j-1} s_{j+1} \ldots s_q.
w ′ = s 1 … s j − 1 s j + 1 … s q . 从
q − 1 ≤ l ( w ′ ) ≤ q q - 1 \leq l(w') \leq q q − 1 ≤ l ( w ′ ) ≤ q 开始，
l ( w ′ ) = q − 1 l(w') = q - 1 l ( w ′ ) = q − 1 和
( s 1 , … , s j − 1 , s j + 1 , … , s q ) (s_1, \ldots, s_{j-1}, s_{j+1}, \ldots, s_q) ( s 1 , … , s j − 1 , s j + 1 , … , s q ) 是
w ′ w' w ′ 的减少的分解。
令w ∈ W w \in W w ∈ W q ≥ 1 q \geq 1 q ≥ 1 w w w F ( s ) = F ( s ′ ) F(s) = F(s') F ( s ) = F ( s ′ ) s = ( s 1 , … , s q ) ; s ′ = ( s 1 ′ , … , s q ′ ) s = (s_1, \ldots, s_q); s' = (s'_1, \ldots, s'_q) s = ( s 1 , … , s q ) ; s ′ = ( s 1 ′ , … , s q ′ )
引用了 1 · 被以下引用 1 引用了 命题 4
被以下引用 命题 5
a) s 1 = s 1 ′ s_1 = s'_1 s 1 = s 1 ′ s q = s q ′ s_q = s'_q s q = s q ′ s ′ s' s ′ j j j k k k s j = s k ′ = s s_j = s'_k = s s j = s k ′ = s s k = s j ′ = s ′ s_k = s'_j = s' s k = s j ′ = s ′
那么F是常数。
A) 设 s , s ′ ∈ D s, s' \in D s , s ′ ∈ D t = ( s 1 ′ , s 1 , … , s q − 1 ) t = (s'_1, s_1, \ldots, s_{q-1}) t = ( s 1 ′ , s 1 , … , s q − 1 ) F ( s ) ≠ F ( s ′ ) F(s) \neq F(s') F ( s ) = F ( s ′ ) t ∈ D t \in D t ∈ D F ( t ) ≠ F ( s ) F(t) \neq F(s) F ( t ) = F ( s ) w = s 1 ′ … s q ′ w = s'_1 \ldots s'_q w = s 1 ′ … s q ′ s 1 ′ w = s 2 ′ … s q ′ s'_1 w = s'_2 \ldots s'_q s 1 ′ w = s 2 ′ … s q ′ < q < q < q j j j 1 ≤ j ≤ q 1 \leq j \leq q 1 ≤ j ≤ q u = ( s 1 ′ , s 1 , … , s j − 1 , s j + 1 , … , s q ) u = (s'_1, s_1, \ldots, s_{j-1}, s_{j+1}, \ldots, s_q) u = ( s 1 ′ , s 1 , … , s j − 1 , s j + 1 , … , s q ) F ( u ) = F ( s ′ ) F(u) = F(s') F ( u ) = F ( s ′ ) j ≠ q j \neq q j = q F ( s ) = F ( u ) F(s) = F(u) F ( s ) = F ( u ) F ( s ) = F ( s ′ ) F(s) = F(s') F ( s ) = F ( s ′ ) j = q j = q j = q t = u ∈ D t = u \in D t = u ∈ D F ( t ) = F ( s ′ ) ≠ F ( s ) F(t) = F(s') \neq F(s) F ( t ) = F ( s ′ ) = F ( s )
B) 让 s 和 s ′ s' s ′ j j j 0 ≤ j ≤ q + 1 0 \leq j \leq q + 1 0 ≤ j ≤ q + 1 s j s_j s j
s 0 = ( s 1 ′ , … , s q ′ )
s_0 = (s'_1, \ldots, s'_q)
s 0 = ( s 1 ′ , … , s q ′ ) s 1 = ( s 1 , … , s q )
s_1 = (s_1, \ldots, s_q)
s 1 = ( s 1 , … , s q ) s q + 1 − k = ( s 1 , s 1 ′ , … , s 1 , s 1 ′ , s 1 , s 2 , … , s k )
s_{q+1-k} = (s_1, s'_1, \ldots, s_1, s'_1, s_1, s_2, \ldots, s_k)
s q + 1 − k = ( s 1 , s 1 ′ , … , s 1 , s 1 ′ , s 1 , s 2 , … , s k ) 对于
q − k q - k q − k 偶数和
0 ≤ k ≤ q 0 \leq k \leq q 0 ≤ k ≤ q s q + 1 − k = ( s 1 ′ , s 1 , … , s 1 , s 1 ′ , s 1 , s 2 , … , s k )
s_{q+1-k} = (s'_1, s_1, \ldots, s_1, s'_1, s_1, s_2, \ldots, s_k)
s q + 1 − k = ( s 1 ′ , s 1 , … , s 1 , s 1 ′ , s 1 , s 2 , … , s k ) 对于
q − k q - k q − k 奇数和
0 ≤ k ≤ q 0 \leq k \leq q 0 ≤ k ≤ q 。
用 (H_j) 断言 “s_j \in D, s_{j+1} \in D \text{ 和 } F(s_j) \neq F(s_{j+1})”表示。根据 A)，(H_j) \ 意味着 0 ≤ j < q 0 \leq j < q 0 ≤ j < q s 0 = s ′ s_0 = s' s 0 = s ′ s 1 = s s_1 = s s 1 = s F ( s ) = F ( s ′ ) F(s) = F(s') F ( s ) = F ( s ′ )
设 M 为幺半群（其中单位元 1）且 f 映射从 S 到 M。对于 s, s' \in S，令 m(s, s') 为 ss' 的阶并放入
引用了 1 · 被以下引用 3 引用了 引理 4
被以下引用 习题 13, § 1 , 命题 7 , 定理 1
a ( s , s ′ ) = { ( f ( s ) f ( s ′ ) ) l if m ( s , s ′ ) = 2 l , l finite ( f ( s ) f ( s ′ ) ) l f ( s ) if m ( s , s ′ ) = 2 l + 1 , l finite 1 if m ( s , s ′ ) = ∞ .
a(s, s') = \begin{cases}
(f(s)f(s'))^l & \text{if } m(s, s') = 2l, l \text{ finite} \\
(f(s)f(s'))^lf(s) & \text{if } m(s, s') = 2l + 1, l \text{ finite} \\
1 & \text{if } m(s, s') = \infty.
\end{cases}
a ( s , s ′ ) = ⎩ ⎨ ⎧ ( f ( s ) f ( s ′ ) ) l ( f ( s ) f ( s ′ ) ) l f ( s ) 1 if m ( s , s ′ ) = 2 l , l finite if m ( s , s ′ ) = 2 l + 1 , l finite if m ( s , s ′ ) = ∞. 如果当s ≠ s ′ s \neq s' s = s ′ a ( s , s ′ ) = a ( s ′ , s ) a(s, s') = a(s', s) a ( s , s ′ ) = a ( s ′ , s )
g ( w ) = f ( s 1 ) … f ( s q )
g(w) = f(s_1) \ldots f(s_q)
g ( w ) = f ( s 1 ) … f ( s q ) 适用于所有 w ∈ W w \in W w ∈ W ( s 1 , … , s q ) (s_1, \ldots, s_q) ( s 1 , … , s q )
对于任何w ∈ W w \in W w ∈ W D w D_w D w F w F_w F w D w D_w D w
F w ( s 1 , … , s q ) = f ( s 1 ) … f ( s q ) .
F_w(s_1, \ldots, s_q) = f(s_1) \ldots f(s_q).
F w ( s 1 , … , s q ) = f ( s 1 ) … f ( s q ) . 我们将通过 w 的长度上的归纳证明
F w F_w F w 是常数，这将建立命题 5。
l ( w ) = 0 , 1 l(w) = 0, 1 l ( w ) = 0 , 1 的情况是微不足道的，我们假设
q ≥ 2 q \geq 2 q ≥ 2 并且我们的断言被证明对于元素 w 和
l ( w ) < q l(w) < q l ( w ) < q 。设 w 为长度 q 和
s , s ′ ∈ D w s, s' \in D_w s , s ′ ∈ D w ；通过引理 4 足以证明引理的情况 a) 和 b) 中的
F w ( s ) = F w ( s ′ ) F_w(s) = F_w(s') F w ( s ) = F w ( s ′ ) 。
a) 公式
F w ( s 1 , … , s q ) = f ( s 1 ) F w ′ ′ ( s 2 , … , s q ) = F w ′ ( s 1 , … , s q − 1 ) f ( s q )
F_w(s_1, \ldots, s_q) = f(s_1)F_{w''}(s_2, \ldots, s_q) = F_{w'}(s_1, \ldots, s_{q-1})f(s_q)
F w ( s 1 , … , s q ) = f ( s 1 ) F w ′′ ( s 2 , … , s q ) = F w ′ ( s 1 , … , s q − 1 ) f ( s q ) 对于
w ′ = s 1 … s q − 1 w' = s_1 \ldots s_{q-1} w ′ = s 1 … s q − 1 、
w ′ ′ = s 2 … s q w'' = s_2 \ldots s_q w ′′ = s 2 … s q 和归纳假设，如果
s 1 = s 1 ′ s_1 = s'_1 s 1 = s 1 ′ 或
s q = s q ′ s_q = s'_q s q = s q ′ ，则显示
F w ( s ) = F w ( s ′ ) F_w(s) = F_w(s') F w ( s ) = F w ( s ′ ) 。
b) 假设 S 存在两个元素和 s'，使得对于 j 奇数和 k 偶数，s j = s k ′ = s s_j = s'_k = s s j = s k ′ = s s k = s j ′ = s ′ s_k = s'_j = s' s k = s j ′ = s ′ s ≠ s ′ s \neq s' s = s ′ s = s m s = s_m s = s m s ′ = s m ′ s' = s'_m s ′ = s m ′ F w ( s ) = a ( s , s ′ ) F_w(s) = a(s, s') F w ( s ) = a ( s , s ′ ) F w ( s ′ ) = a ( s ′ , s ) F_w(s') = a(s', s) F w ( s ′ ) = a ( s ′ , s )
F w ( s ) = F w ( s ′ ) .
F_w(s) = F_w(s').
F w ( s ) = F w ( s ′ ) .
6. 考克斯特的特征群
no. 4 中的引理 3 表明任何 Coxeter 系统满足 (E)。
相反，假设满足(E)。令G为从S到G的群和f映射，使得每当s和s ′ s' s ′ s s ′ ss' s s ′ ( f ( s ) f ( s ′ ) ) m = 1 (f(s)f(s'))^m = 1 ( f ( s ) f ( s ′ ) ) m = 1
g ( w ) = f ( s 1 ) … f ( s q )
g(w) = f(s_1) \ldots f(s_q)
g ( w ) = f ( s 1 ) … f ( s q ) 每当
w = s 1 … s q w = s_1 \ldots s_q w = s 1 … s q 为长度 q 时。要证明 (W, S) 是 Coxeter 系统，只需证明 g 是同态，即公式的推论
g ( s w ) = f ( s ) g ( w ) for s ∈ S , w ∈ W
g(sw) = f(s)g(w) \quad \text{for } s \in S, w \in W
g ( s w ) = f ( s ) g ( w ) for s ∈ S , w ∈ W 由于 S 生成 W。根据 no. 5 的 Prop. 4，只有两种情况是可能的：
a) l ( s w ) = l ( w ) + 1 l(sw) = l(w) + 1 l ( s w ) = l ( w ) + 1 ( s 1 , … , s q ) (s_1, \ldots, s_q) ( s 1 , … , s q ) w w w ( s , s 1 , … , s q ) (s, s_1, \ldots, s_q) ( s , s 1 , … , s q ) s w sw s w
b) l ( s w ) = l ( w ) − 1 l(sw) = l(w) - 1 l ( s w ) = l ( w ) − 1 w ′ = s w w' = sw w ′ = s w w = s w ′ w = sw' w = s w ′ l ( s w ′ ) = l ( w ′ ) + 1 l(sw') = l(w') + 1 l ( s w ′ ) = l ( w ′ ) + 1 g ( w ) = f ( s ) g ( s w ) g(w) = f(s)g(sw) g ( w ) = f ( s ) g ( s w ) ( f ( s ) ) 2 = 1 (f(s))^2 = 1 ( f ( s ) ) 2 = 1 f ( s ) g ( w ) = g ( s w ) f(s)g(w) = g(sw) f ( s ) g ( w ) = g ( s w )
7. 划分的家庭
假设 (W, S) 是 Coxeter 系统。对于任何s ∈ S s \in S s ∈ S P s P_s P s w w w l ( s w ) > l ( w ) l(sw) > l(w) l ( s w ) > l ( w )
（一）⋂ s ∈ S P s = { 1 } \bigcap_{s \in S} P_s = \{1\} ⋂ s ∈ S P s = { 1 }
事实上，让w ≠ 1 w \neq 1 w = 1 ( s 1 , … , s q ) (s_1, \ldots, s_q) ( s 1 , … , s q ) w w w q ≥ 1 q \geq 1 q ≥ 1 ( s 2 , … , s q ) (s_2, \ldots, s_q) ( s 2 , … , s q ) s 2 w s_2w s 2 w l ( w ) = q l(w) = q l ( w ) = q l ( s 1 w ) = q − 1 l(s_1w) = q - 1 l ( s 1 w ) = q − 1 w ∉ P s 1 w \notin P_{s_1} w ∈ / P s 1
(B) 对于 S 中的任何 s，W 的集合 P s P_s P s s P s sP_s s P s
设w ∈ W w \in W w ∈ W s ∈ S s \in S s ∈ S
a) l ( s w ) = l ( w ) + 1 l(sw) = l(w) + 1 l ( s w ) = l ( w ) + 1 w ∈ P s w \in P_s w ∈ P s l ( s w ) = l ( w ) − 1 l(sw) = l(w) - 1 l ( s w ) = l ( w ) − 1 w ′ = s w w' = sw w ′ = s w w = s w ′ w = sw' w = s w ′
l ( w ′ ) < l ( s w ′ )
l(w') < l(sw')
l ( w ′ ) < l ( s w ′ ) w ′ ∈ P s w' \in P_s w ′ ∈ P s w ∈ s P s w \in sP_s w ∈ s P s (C) 设 s、s' 在 S 中，w 在 W 中。如果 w ∈ P s w \in P_s w ∈ P s w s ′ ∈ P s ws' \in P_s w s ′ ∈ P s s w = w s ′ sw = ws' s w = w s ′
设q q q w w w w ∈ P s w \in P_s w ∈ P s l ( s w ) = q + 1 l(sw) = q + 1 l ( s w ) = q + 1 w s ′ ∈ P s ws' \in P_s w s ′ ∈ P s l ( s w s ′ ) = l ( w s ′ ) − 1 ≤ q l(sws') = l(ws') - 1 \leq q l ( s w s ′ ) = l ( w s ′ ) − 1 ≤ q l ( s w s ′ ) = l ( s w ) ± 1 l(sws') = l(sw) \pm 1 l ( s w s ′ ) = l ( s w ) ± 1 l ( w s ′ ) = q + 1 l(ws') = q + 1 l ( w s ′ ) = q + 1 l ( s w s ′ ) = q l(sws') = q l ( s w s ′ ) = q
令( s 1 , … , s q ) (s_1, \ldots, s_q) ( s 1 , … , s q ) w w w s q + 1 = s ′ s_{q+1} = s' s q + 1 = s ′ ( s 1 , … , s q , s q + 1 ) (s_1, \ldots, s_q, s_{q+1}) ( s 1 , … , s q , s q + 1 ) w s ′ ws' w s ′ q + 1 q + 1 q + 1 j j j 1 ≤ j ≤ q + 1 1 \leq j \leq q + 1 1 ≤ j ≤ q + 1
s s 1 … s j − 1 = s 1 … s j .
ss_1 \ldots s_{j-1} = s_1 \ldots s_j.
s s 1 … s j − 1 = s 1 … s j . 1 ≤ j ≤ q 1 \leq j \leq q 1 ≤ j ≤ q s w = s 1 … s j − 1 s j + 1 … s q sw = s_1 \ldots s_{j-1} s_{j+1} \ldots s_q s w = s 1 … s j − 1 s j + 1 … s q l ( s w ) = q + 1 l(sw) = q + 1 l ( s w ) = q + 1 j = q + 1 j = q + 1 j = q + 1 s w = w s ′ sw = ws' s w = w s ′ 相反，我们得到以下结果：
(A') 1 ∈ P s 1 \in P_s 1 ∈ P s 对于所有人 s ∈ S s \in S s ∈ S 集合 P s P_s P s 和 s P s sP_s s P s 对于所有 s ∈ S s \in S s ∈ S
那么 ，( W , S ) (W, S) ( W , S ) P s P_s P s w w w W W W l ( s w ) > l ( w ) l(sw) > l(w) l ( s w ) > l ( w )
令s ∈ S s \in S s ∈ S w ∈ W w \in W w ∈ W
a) w ∉ P s w \notin P_s w ∈ / P s ( s 1 , … , s q ) (s_1, \ldots, s_q) ( s 1 , … , s q ) w w w
w j = s 1 … s j
w_j = s_1 \ldots s_j
w j = s 1 … s j 1 ≤ j ≤ q 1 \leq j \leq q 1 ≤ j ≤ q ；还放了
w 0 = 1 w_0 = 1 w 0 = 1 。由于
w 0 ∈ P s w_0 \in P_s w 0 ∈ P s 由 (A') 且由于
w = w q w = w_q w = w q 不在
P s P_s P s 中，因此存在整数
j j j 与
1 ≤ j ≤ q 1 \leq j \leq q 1 ≤ j ≤ q ，使得
w j − 1 ∈ P s w_{j-1} \in P_s w j − 1 ∈ P s 和
w j = w j − 1 s j w_j = w_{j-1} s_j w j = w j − 1 s j 不会属于到
P s P_s P s 。通过（C），
s w j − 1 = w j − 1 s j .
sw_{j-1} = w_{j-1} s_j.
s w j − 1 = w j − 1 s j . 由此我们证明了公式
s s 1 … s j − 1 = s 1 … s j − 1 s j
ss_1 \ldots s_{j-1} = s_1 \ldots s_{j-1} s_j
s s 1 … s j − 1 = s 1 … s j − 1 s j 这意味着
s w = s 1 … s j − 1 s j + 1 … s q sw = s_1 \ldots s_{j-1} s_{j+1} \ldots s_q s w = s 1 … s j − 1 s j + 1 … s q 和
l ( s w ) < l ( w ) l(sw) < l(w) l ( s w ) < l ( w ) 。
b) w ∈ P s w \in P_s w ∈ P s w ′ = s w w' = sw w ′ = s w w ′ ∉ P s w' \notin P_s w ′ ∈ / P s l ( s w ′ ) < l ( w ′ ) l(sw') < l(w') l ( s w ′ ) < l ( w ′ ) l ( w ) < l ( s w ) l(w) < l(sw) l ( w ) < l ( s w )
a) 和 b) 的比较证明 P s P_s P s w ∈ W w \in W w ∈ W l ( s w ) > l ( w ) l(sw) > l(w) l ( s w ) > l ( w ) ( W , S ) (W, S) ( W , S )
8. 考克塞特的子群群
在这个数中，我们假设( W , S ) (W, S) ( W , S ) S S S X X X X X X W W W W X W_X W X
让w w w W W W S S S S w S_w S w { s 1 , … , s q } = S w \{s_1, \ldots, s_q\} = S_w { s 1 , … , s q } = S w w w w ( s 1 , … , s q ) (s_1, \ldots, s_q) ( s 1 , … , s q )
引用了 1 引用了 命题 5
用M M M S S S ( A , B ) ↦ A ∪ B (A, B) \mapsto A \cup B ( A , B ) ↦ A ∪ B M M M ∅ \varnothing ∅ f ( s ) = { s } f(s) = \{s\} f ( s ) = { s } s ∈ S s \in S s ∈ S M M M f f f m ( s , s ′ ) m(s, s') m ( s , s ′ ) S S S s , s ′ s, s' s , s ′ a ( s , s ′ ) = { s , s ′ } a(s, s') = \{s, s'\} a ( s , s ′ ) = { s , s ′ } W W W M M M g : w ↦ S w g : w \mapsto S_w g : w ↦ S w g ( w ) = f ( s 1 ) ∪ ⋯ ∪ f ( s q ) g(w) = f(s_1) \cup \cdots \cup f(s_q) g ( w ) = f ( s 1 ) ∪ ⋯ ∪ f ( s q ) S w = { s 1 , … , s q } S_w = \{s_1, \ldots, s_q\} S w = { s 1 , … , s q } w ∈ W w \in W w ∈ W w w w ( s 1 , … , s q ) (s_1, \ldots, s_q) ( s 1 , … , s q )
若w = s 1 … s q w = s_1 \ldots s_q w = s 1 … s q s 1 , … , s q s_1, \ldots, s_q s 1 , … , s q S S S w − 1 = s q … s 1 w^{-1} = s_q \ldots s_1 w − 1 = s q … s 1
S w − 1 = S w .
S_{w^{-1}} = S_w.
S w − 1 = S w . no. 5的第4条命题表明S s w ′ ⊂ { s } ∪ S w ′ S_{sw'} \subset \{s\} \cup S_{w'} S s w ′ ⊂ { s } ∪ S w ′ s ∈ S s \in S s ∈ S w ′ ∈ W w' \in W w ′ ∈ W
S w w ′ ⊂ S w ∪ S w ′
S_{ww'} \subset S_w \cup S_{w'}
S w w ′ ⊂ S w ∪ S w ′ 由 w w w w ∈ W w \in W w ∈ W U U U S w ⊂ X S_w \subset X S w ⊂ X W W W X ⊂ U ⊂ W X X \subset U \subset W_X X ⊂ U ⊂ W X U = W X U = W_X U = W X
这是从 Cor 得出的。 1 和 S S S s s s S s = { s } S_s = \{s\} S s = { s }
集合 S S S W W W
如果X ⊂ S X \subset S X ⊂ S W W W W = W X W = W_X W = W X X = S ∩ W X = S X = S \cap W_X = S X = S ∩ W X = S
引用了 1 引用了 推论 2
对于S S S X X X W X W_X W X w w w w w w W X W_X W X X X X l S ( w ) l_S(w) l S ( w )
令 ( s 1 , … , s q ) (s_1, \ldots, s_q) ( s 1 , … , s q ) w w w W W W 1 ≤ j ≤ q 1 \leq j \leq q 1 ≤ j ≤ q w = s 1 … s q w = s_1 \ldots s_q w = s 1 … s q s j ∈ X s_j \in X s j ∈ X w w w q = l S ( w ) q = l_S(w) q = l S ( w ) X ⊂ S X \subset S X ⊂ S q ′ < q q' < q q ′ < q
引用了 1 · 被以下引用 1 引用了 推论 1
被以下引用 定理 2
(ii) 设 ( X i ) i ∈ I (X_i)_{i \in I} ( X i ) i ∈ I S S S X = ⋂ i ∈ I X i X = \bigcap_{i \in I} X_i X = ⋂ i ∈ I X i W X = ⋂ i ∈ I W X i W_X = \bigcap_{i \in I} W_{X_i} W X = ⋂ i ∈ I W X i
(iii) 令X X X X ′ X' X ′ S S S W X ⊂ W X ′ W_X \subset W_{X'} W X ⊂ W X ′ W X = W X ′ W_X = W_{X'} W X = W X ′ X ⊂ X ′ X \subset X' X ⊂ X ′ X = X ′ X = X' X = X ′
X X X X X X W X W_X W X x ∈ X x \in X x ∈ X w ∈ W X w \in W_X w ∈ W X l X ( x w ) ≤ l X ( w ) = q l_X(xw) \leq l_X(w) = q l X ( x w ) ≤ l X ( w ) = q
l S ( x w ) ≤ l S ( w ) = q .
l_S(xw) \leq l_S(w) = q.
l S ( x w ) ≤ l S ( w ) = q . 令
x 1 , … , x q x_1, \ldots, x_q x 1 , … , x q 为
X X X 的元素，使得
w = x 1 … x q w = x_1 \ldots x_q w = x 1 … x q 。由于
( W , S ) (W, S) ( W , S ) 满足交换条件（no. 6的Th.1），因此存在整数
j j j 使得
1 ≤ j ≤ q 1 \leq j \leq q 1 ≤ j ≤ q 和
x x 1 … x j − 1 = x 1 … x j − 1 x j xx_1 \ldots x_{j-1} = x_1 \ldots x_{j-1} x_j x x 1 … x j − 1 = x 1 … x j − 1 x j 。因此，
( W X , X ) (W_X, X) ( W X , X ) 满足交换条件，因此是Coxeter系统（no. 6的Th.1）。这就证明了(i)。
断言 (ii) 和 (iii) 紧随 Cor.提案 7 中的 1 条。
9. 考克塞特矩阵和考克塞特图
令 I I I I I I M = ( m i j ) i , j ∈ I M = (m_{ij})_{i,j \in I} M = ( m ij ) i , j ∈ I + ∞ +\infty + ∞
m i i = 1 for all i ∈ I ; (25)
m_{ii} = 1 \quad \text{for all } i \in I; \tag{25}
m ii = 1 for all i ∈ I ; ( 25 ) m i j ≥ 2 for i , j ∈ I with i ≠ j . (26)
m_{ij} \geq 2 \quad \text{for } i, j \in I \text{ with } i \neq j. \tag{26}
m ij ≥ 2 for i , j ∈ I with i = j . ( 26 ) 引用了 2 引用了 V § 4. 。 COXETER 群的几何表示 , 定理 2
型 I I I Γ 3 \Gamma^3 Γ 3 I I I f f f + ∞ +\infty + ∞ ≥ 3 \geq 3 ≥ 3 Γ \Gamma Γ ( Γ , f ) (\Gamma, f) ( Γ , f )
Coxeter 图 ( Γ , f ) (\Gamma, f) ( Γ , f ) I I I M M M
图Γ \Gamma Γ I I I { i , j } \{i, j\} { i , j } I I I m i j ≥ 3 m_{ij} \geq 3 m ij ≥ 3 f f f M M M { i , j } \{i, j\} { i , j } m i j m_{ij} m ij
显然，这在型I I I I I I
为了帮助读者理解我们的论点，我们经常用图表来表示型 I I I { i , j } \{i, j\} { i , j } f ( { i , j } ) f(\{i, j\}) f ({ i , j })
3 请参阅附录了解定义以及此处使用的图的属性。
如果 (W, S) 是一个 Coxeter 系统、矩阵 M = ( m ( s , s ′ ) ) s , s ′ ∈ S M = (m(s, s'))_{s, s' \in S} M = ( m ( s , s ′ ) ) s , s ′ ∈ S m ( s , s ′ ) m(s, s') m ( s , s ′ ) s s ′ ss' s s ′ s ∈ S s \in S s ∈ S s 2 = 1 s^2 = 1 s 2 = 1 m ( s , s ) = 1 m(s, s) = 1 m ( s , s ) = 1 s s ′ = ( s ′ s ) − 1 ss' = (s's)^{-1} s s ′ = ( s ′ s ) − 1 s ≠ s ′ s \neq s' s = s ′ m ( s , s ′ ) = m ( s ′ , s ) ⩾ 2 m(s, s') = m(s', s) \geqslant 2 m ( s , s ′ ) = m ( s ′ , s ) ⩾ 2 ≠ 1 \neq 1 = 1 ( Γ , f ) (\Gamma, f) ( Γ , f ) Γ \Gamma Γ s s s s ′ s' s ′ s s s s ′ s' s ′ 2 m 2m 2 m ( 1 m m 1 ) \begin{pmatrix} 1 & m \\ m & 1 \end{pmatrix} ( 1 m m 1 )
∘ — m ∘
\circ \overset{m}{\text{---}} \circ
∘ — m ∘ 当m ⩾ 3 m \geqslant 3 m ⩾ 3
∘ — ∘
\circ \text{---} \circ
∘ — ∘ 如果 m = 3 m = 3 m = 3
当m = 2 m = 2 m = 2 S n \mathfrak{S}_n S n
∘ — ∘ — ∘ ⋯ ∘ — ∘
\circ \text{---} \circ \text{---} \circ \cdots \circ \text{---} \circ
∘ — ∘ — ∘ ⋯ ∘ — ∘ （n − 1 n - 1 n − 1
我们稍后（第 V, § 4 章）证明，相反，任何 Coxeter 矩阵都是 Coxeter 系统的矩阵。
如果 Coxeter 图的底层图是*连通*（附录，no. 2）且*非空*，则 Coxeter 系统 (W, S) 被称为*不可约*。等价地，S 是非空的，并且不存在 S 的划分分成 S 的两个不同的子集 S' 和 S''，使得 S' 的每个元素与 S'' 的每个元素交换。更一般地，令( Γ i ) i ∈ I (\Gamma_i)_{i \in I} ( Γ i ) i ∈ I Γ \Gamma Γ S i S_i S i Γ i \Gamma_i Γ i W i = W S i W_i = W_{S_i} W i = W S i S i S_i S i ( W i , S i ) (W_i, S_i) ( W i , S i ) i ∈ I i \in I i ∈ I W i W_i W i
令( S i ) i ∈ I (S_i)_{i \in I} ( S i ) i ∈ I S S S S i S_i S i S j S_j S j i ≠ j i \neq j i = j i ∈ I i \in I i ∈ I W i W_i W i S i S_i S i W W W ( W i ) i ∈ I (W_i)_{i \in I} ( W i ) i ∈ I
引用了 1 · 被以下引用 1 引用了 定理 2
被以下引用 V § 3. 。反射产生的位移群
很明显，对于j ≠ i j \neq i j = i W j W_j W j i ∈ I i \in I i ∈ I W i ′ W'_i W i ′ S i ′ = ⋃ j ≠ i S j S'_i = \bigcup_{j \neq i} S_j S i ′ = ⋃ j = i S j
W i ∩ W i ′ = W ∅ = { 1 }
W_i \cap W'_i = W_\varnothing = \{1\}
W i ∩ W i ′ = W ∅ = { 1 } 通过 Th。 no. 8 中的 2 个。由于
W W W 是由
W i W_i W i 的并生成的，这就证明了命题。
习题
参见习题为 § 1 。
[^1]: 群 G 是子群族( G i ) i ∈ I (G_i)_{i \in I} ( G i ) i ∈ I i ∈ J i \in J i ∈ J G i G_i G i G J G_J G J i ∈ J i \in J i ∈ J G i G_i G i G J G_J G J i ≠ j i \neq j i = j G i G_i G i G j G_j G j ∏ i ∈ I g i \prod_{i \in I} g_i ∏ i ∈ I g i g i ∈ G i g_i \in G_i g i ∈ G i g i = 1 g_i = 1 g i = 1 i i i G i G_i G i i ∈ I i \in I i ∈ I G i ∩ G J = { 1 } G_i \cap G_J = \{1\} G i ∩ G J = { 1 } i ∉ J i \notin J i ∈ / J