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§ 4。 根系的分类
1. 有限考克塞特群
在这个截面中，我们将要确定，直到同构，所有根系，因此所有晶体群（§ 2, no. 5）。更一般地，我们首先确定有限维实数向量空间中的反射生成的所有有限群：这相当于（第 V, § 4, no. 8 章）确定所有有限 Coxeter 群，或（第 V, § 4, no. 8，第 2 章）确定有限维的所有 Coxeter 矩阵阶使得相关的双线性形式是正的且非简并的。
令 M = ( m i j ) i , j ∈ I M = (m_{ij})_{i,j \in I} M = ( m ij ) i , j ∈ I l l l
q i j = − cos ( π / m i j ) .
q_{ij} = -\cos(\pi / m_{ij}).
q ij = − cos ( π / m ij ) . 回想一下，q i i = 1 q_{ii} = 1 q ii = 1 q i j = q j i q_{ij} = q_{ji} q ij = q j i ≤ − 1 / 2 \leq -1/2 ≤ − 1/2 i ≠ j i \neq j i = j E = R l E = \mathbf{R}^l E = R l ( e i ) i ∈ I (e_i)_{i \in I} ( e i ) i ∈ I E E E M M M E E E ( x ∣ y ) (x|y) ( x ∣ y ) q q q E E E x ↦ ( x ∣ x ) x \mapsto (x|x) x ↦ ( x ∣ x ) x = ∑ i ∈ I ξ i x i x = \sum_{i \in I} \xi_i x_i x = ∑ i ∈ I ξ i x i
∥ x ∥ 2 = q ( x ) = ∑ i , j ∈ I q i j ξ i ξ j .
\| x \| ^2 = q(x) = \sum_{i,j \in I} q_{ij} \xi_i \xi_j.
∥ x ∥ 2 = q ( x ) = i , j ∈ I ∑ q ij ξ i ξ j . 用 ( X , f ) (X, f) ( X , f ) M M M a a a X X X f ( a ) f(a) f ( a ) a a a
在这个数的其余部分中，由M M M W ( M ) W(M) W ( M ) q q q X X X X X X W ( M ) W(M) W ( M ) X X X
从q q q m i j m_{ij} m ij W ( M ) W(M) W ( M )
对于所有 i i i ∑ j ≠ i q i j 2 < 1 \sum_{j \neq i} q_{ij}^2 < 1 ∑ j = i q ij 2 < 1
被以下引用 4 被以下引用 引理 2 , 引理 3 , 引理 4 , 引理 5
令J J J j ∈ I j \in I j ∈ I q i j ≠ 0 q_{ij} \neq 0 q ij = 0 { i , j } \{i, j\} { i , j } X X X j , j ′ ∈ J j, j' \in J j , j ′ ∈ J j ≠ j ′ j \neq j' j = j ′ { j , j ′ } \{j, j'\} { j , j ′ } i , j , j ′ i, j, j' i , j , j ′ ( e j ∣ e j ′ ) = 0 (e_j|e_{j'}) = 0 ( e j ∣ e j ′ ) = 0 F = ∑ j ∈ J R e j F = \sum_{j \in J} \mathbf{R} e_j F = ∑ j ∈ J R e j ( e j ) j ∈ J (e_j)_{j \in J} ( e j ) j ∈ J F F F e i e_i e i F F F d d d d 2 = 1 − ∑ j ∈ J ( e i ∣ e j ) 2 = 1 − ∑ j ∈ J q i j 2 = 1 − ∑ j ≠ i q i j 2 d^2 = 1 - \sum_{j \in J} (e_i|e_j)^2 = 1 - \sum_{j \in J} q_{ij}^2 = 1 - \sum_{j \neq i} q_{ij}^2 d 2 = 1 − ∑ j ∈ J ( e i ∣ e j ) 2 = 1 − ∑ j ∈ J q ij 2 = 1 − ∑ j = i q ij 2
事实上，如果i i i h h h q i j 2 ≥ 1 4 q_{ij}^2 \geq \frac{1}{4} q ij 2 ≥ 4 1 h 4 < 1 \frac{h}{4} < 1 4 h < 1 h ≤ 3 h \leq 3 h ≤ 3
如果 i i i
引用了 1 引用了 引理 1
如果没有，鉴于关系 cos π 4 = 2 2 \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} cos 4 π = 2 2
∑ j ≠ i q i j 2 ≥ 1 4 + 1 4 + ( 2 2 ) 2 = 1
\sum_{j \neq i} q_{ij}^2 \geq \frac{1}{4} + \frac{1}{4} + (\frac{\sqrt{2}}{2})^2 = 1
j = i ∑ q ij 2 ≥ 4 1 + 4 1 + ( 2 2 ) 2 = 1 这是不可能的（引理1）。
如果存在阶≥ 6 \geq 6 ≥ 6 l = 2 l = 2 l = 2
引用了 1 · 被以下引用 2 引用了 引理 1
被以下引用 引理 9 , 定理 1
事实上，让 { i , j } \{i, j\} { i , j } l > 2 l > 2 l > 2 i , j i, j i , j i i i j ′ j' j ′ X X X cos π 6 = 3 2 \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} cos 6 π = 2 3
∑ k ≠ i q i k 2 ≥ 1 4 + ( 3 2 ) 2 = 1
\sum_{k \neq i} q_{ik}^2 \geq \frac{1}{4} + (\frac{\sqrt{3}}{2})^2 = 1
k = i ∑ q ik 2 ≥ 4 1 + ( 2 3 ) 2 = 1 这是不可能的（引理1）。
顶点不能属于到阶≥ 4 \geq 4 ≥ 4
引用了 1 · 被以下引用 1 引用了 引理 1
被以下引用 引理 7
令i i i ∑ j ≠ i q i j 2 ≥ ( 2 2 ) 2 + ( 2 2 ) 2 = 1 \sum_{j \neq i} q_{ij}^2 \geq (\frac{\sqrt{2}}{2})^2 + (\frac{\sqrt{2}}{2})^2 = 1 ∑ j = i q ij 2 ≥ ( 2 2 ) 2 + ( 2 2 ) 2 = 1
令{ i , j } \{i, j\} { i , j } X X X M M M 通过识别 i i i 和 j j j I ′ I' I ′ i i i j j j I I I p = { i , j } p = \{i, j\} p = { i , j } I ′ I' I ′ I I I i i i j j j I ′ I' I ′ k , k ′ k, k' k , k ′ I ′ I' I ′ { k , k ′ } \{k, k'\} { k , k ′ }
1) k k k k ′ k' k ′ p p p { k , k ′ } \{k, k'\} { k , k ′ } X X X m k k ′ m_{kk'} m k k ′
2）k = p k = p k = p { i , k ′ } , { j , k ′ } \{i, k'\}, \{j, k'\} { i , k ′ } , { j , k ′ } X X X { i , k ′ } \{i, k'\} { i , k ′ } X X X { p , k ′ } \{p, k'\} { p , k ′ } m i k ′ m_{ik'} m i k ′ { j , k ′ } \{j, k'\} { j , k ′ } X X X m j k ′ m_{jk'} m j k ′ X X X
令M ′ = ( m i j ′ ) i , j ∈ I ′ M' = (m'_{ij})_{i,j \in I'} M ′ = ( m ij ′ ) i , j ∈ I ′ q i j ′ = − cos π m i j ′ q'_{ij} = -\cos \frac{\pi}{m'_{ij}} q ij ′ = − cos m ij ′ π k ≠ p k \neq p k = p t p k ′ = q i k + q j k t'_{pk} = q_{ik} + q_{jk} t p k ′ = q ik + q j k ( ξ i ) ∈ R I ′ (\xi_i) \in \mathbf{R}^{I'} ( ξ i ) ∈ R I ′
∑ k , k ′ ∈ I ′ q k k ′ ′ ξ k ξ k ′ = ∑ k , k ′ ∈ I q k k ′ ξ k ξ k ′ + ξ p 2 − ξ i 2 − ξ j 2 − 2 q i j ξ i ξ j (1)
\sum_{k, k' \in I'} q'_{kk'} \xi_k \xi_{k'} = \sum_{k, k' \in I} q_{kk'} \xi_k \xi_{k'} + \xi_p^2 - \xi_i^2 - \xi_j^2 - 2q_{ij} \xi_i \xi_j \tag{1}
k , k ′ ∈ I ′ ∑ q k k ′ ′ ξ k ξ k ′ = k , k ′ ∈ I ∑ q k k ′ ξ k ξ k ′ + ξ p 2 − ξ i 2 − ξ j 2 − 2 q ij ξ i ξ j ( 1 ) = ∑ k , k ′ ∈ I q k k ′ ξ k ξ k ′ − ( 1 + 2 q i j ) ξ p 2 .
= \sum_{k, k' \in I} q_{kk'} \xi_k \xi_{k'} - (1 + 2q_{ij}) \xi_p^2.
= k , k ′ ∈ I ∑ q k k ′ ξ k ξ k ′ − ( 1 + 2 q ij ) ξ p 2 .
如果 { i , j } \{i, j\} { i , j } W ( M ′ ) W(M') W ( M ′ )
引用了 2 · 被以下引用 1 引用了 § 4. 。 根系的分类 , 定理 2
被以下引用 引理 7
事实上，q i j = − 1 2 q_{ij} = -\frac{1}{2} q ij = − 2 1
∑ k , k ′ ∈ I ′ q k k ′ ′ ξ k ξ k ′ = ∑ k , k ′ ∈ I q k k ′ ξ k ξ k ′
\sum_{k,k' \in I'} q'_{kk'} \xi_k \xi_{k'} = \sum_{k,k' \in I} q_{kk'} \xi_k \xi_{k'}
k , k ′ ∈ I ′ ∑ q k k ′ ′ ξ k ξ k ′ = k , k ′ ∈ I ∑ q k k ′ ξ k ξ k ′ ( ξ k ) k ∈ I ′ ↦ ∑ k , k ′ ∈ I ′ q k k ′ ′ ξ k ξ k ′ (\xi_k)_{k \in I'} \mapsto \sum_{k,k' \in I'} q'_{kk'} \xi_k \xi_{k'} ( ξ k ) k ∈ I ′ ↦ ∑ k , k ′ ∈ I ′ q k k ′ ′ ξ k ξ k ′
a) X 具有独特的分支点（第 IV、附录、no. 1 章）。并且 X 的所有边都是阶 3。 b) X 是一条链，最多有阶 ≥ 4 \geq 4 ≥ 4
我们在l上通过归纳争论。
a) 假设 X 具有分支点 i。然后我属于到阶3的3条边。{ i , k 1 } , { i , k 2 } , { i , k 3 } \{i, k_1\}, \{i, k_2\}, \{i, k_3\} { i , k 1 } , { i , k 2 } , { i , k 3 } l = 4 l = 4 l = 4 k 1 k_1 k 1 k 1 k_1 k 1 p p p X ′ X' X ′ X ′ X' X ′ k 1 k_1 k 1 k 1 k_1 k 1 p p p X ′ X' X ′
b) 假设X没有影响点。那么X是一条链（第IV，附录，no. 3，第3条命题）。令{ i , j } \{i, j\} { i , j } ≥ 4 \geq 4 ≥ 4 l = 2 l = 2 l = 2 k ≠ j k \neq j k = j { i , k } \{i, k\} { i , k } p p p X ′ X' X ′ X ′ X' X ′ { p , j } \{p, j\} { p , j } ≥ 4 \geq 4 ≥ 4 X ′ X' X ′ ≥ 4 \geq 4 ≥ 4 { i , j } \{i, j\} { i , j } ≥ 4 \geq 4 ≥ 4
令i 1 , i 2 , … , i p i_1, i_2, \ldots, i_p i 1 , i 2 , … , i p { i 1 , i 2 } , { i 2 , i 3 } , … , { i p − 1 , i p } \{i_1, i_2\}, \{i_2, i_3\}, \ldots, \{i_{p-1}, i_p\} { i 1 , i 2 } , { i 2 , i 3 } , … , { i p − 1 , i p } q ( ∑ r = 1 p r e i r ) = 1 2 p ( p + 1 ) q(\sum_{r=1}^p r e_{i_r}) = \frac{1}{2} p(p+1) q ( ∑ r = 1 p r e i r ) = 2 1 p ( p + 1 )
被以下引用 2 被以下引用 引理 10 , 引理 9
如果s > r + 1 s > r + 1 s > r + 1 ( e i r ∣ e i r ) = 1 , ( e i r ∣ e i r + 1 ) = − 1 2 , ( e i r ∣ e i s ) = 0 (e_{i_r}|e_{i_r}) = 1, (e_{i_r}|e_{i_{r+1}}) = -\frac{1}{2}, (e_{i_r}|e_{i_s}) = 0 ( e i r ∣ e i r ) = 1 , ( e i r ∣ e i r + 1 ) = − 2 1 , ( e i r ∣ e i s ) = 0
q ( ∑ r = 1 p r e i r ) = ∑ r = 1 p r 2 − 2 ∑ r = 1 p − 1 1 2 r ( r + 1 ) = p 2 − ∑ r = 1 p − 1 r .
q(\sum_{r=1}^p r e_{i_r}) = \sum_{r=1}^p r^2 - 2 \sum_{r=1}^{p-1} \frac{1}{2} r(r+1) = p^2 - \sum_{r=1}^{p-1} r.
q ( r = 1 ∑ p r e i r ) = r = 1 ∑ p r 2 − 2 r = 1 ∑ p − 1 2 1 r ( r + 1 ) = p 2 − r = 1 ∑ p − 1 r . 集合论，第 1 章III。 § 5, no. 8。科。根据提案 14，这等于
p 2 − 1 2 p ( p − 1 ) = 1 2 p ( p + 1 ) .
p^2 - \frac{1}{2} p(p-1) = \frac{1}{2} p(p+1).
p 2 − 2 1 p ( p − 1 ) = 2 1 p ( p + 1 ) .
假设 X 是一条具有顶点 1, 2, ..., l 和边 {1, 2}, {2, 3}, ..., {l - 1, l} 的链。
引用了 2 · 被以下引用 1 引用了 引理 4 , 引理 8
被以下引用 定理 1
(i) 如果边 {2, 3}, {3, 4}, ..., {l - 2, l - 1} 之一的大小为阶 ≥ 4 \geq 4 ≥ 4
!具有 4 个节点连接成链的图
(ii) 如果边 {1, 2} 的大小为阶 5，则该图为以下之一：
!具有 5 个节点以链连接的图
我们可以假设 l > 2 l > 2 l > 2 { i , i + 1 } \{i, i + 1\} { i , i + 1 } ≥ 4 \geq 4 ≥ 4 1 ≤ i ≤ l − 1 1 \leq i \leq l - 1 1 ≤ i ≤ l − 1
x = e 1 + 2 e 2 + ⋯ + i e i , y = e i + 2 e l − 1 + ⋯ + ( l − i ) e i + 1 , and j = l − i .
x = e_1 + 2e_2 + \cdots + ie_i,\ y = e_i + 2e_{l-1} + \cdots + (l - i)e_{i+1}, \text{ and } j = l - i.
x = e 1 + 2 e 2 + ⋯ + i e i , y = e i + 2 e l − 1 + ⋯ + ( l − i ) e i + 1 , and j = l − i . 引理 8、∥ x ∥ 2 = 1 2 i ( i + 1 ) \| x \| ^2 = \frac{1}{2} i(i + 1) ∥ x ∥ 2 = 2 1 i ( i + 1 ) ∥ y ∥ 2 = 1 2 j ( j + 1 ) \| y \| ^2 = \frac{1}{2} j(j + 1) ∥ y ∥ 2 = 2 1 j ( j + 1 ) ( x ∣ y ) = i j ( e i ∣ e i + 1 ) = − i j cos π m (x|y) = ij(e_i|e_{i+1}) = -ij \cos \frac{\pi}{m} ( x ∣ y ) = ij ( e i ∣ e i + 1 ) = − ij cos m π m = 4 m = 4 m = 4
( x ∣ y ) 2 < ∥ x ∥ 2 ∥ y ∥ 2 .
(x|y)^2 < \| x \| ^2 \| y \| ^2.
( x ∣ y ) 2 < ∥ x ∥ 2 ∥ y ∥ 2 . 这给出了
1 4 i j ( i + 1 ) ( j + 1 ) > i 2 j 2 cos 2 π m
\frac{1}{4} ij(i + 1)(j + 1) > i^2 j^2 \cos^2 \frac{\pi}{m}
4 1 ij ( i + 1 ) ( j + 1 ) > i 2 j 2 cos 2 m π 所以
( i + 1 ) ( j + 1 ) > 4 i j cos 2 π m ≥ 2 i j .
(i + 1)(j + 1) > 4ij \cos^2 \frac{\pi}{m} \geq 2ij.
( i + 1 ) ( j + 1 ) > 4 ij cos 2 m π ≥ 2 ij . 首先，给出i j − i − j − 1 < 0 ij - i - j - 1 < 0 ij − i − j − 1 < 0 ( i − 1 ) ( j − 1 ) < 2 (i - 1)(j - 1) < 2 ( i − 1 ) ( j − 1 ) < 2
1 < i < l − 1 ,
1 < i < l - 1,
1 < i < l − 1 , 然后1 < j < l − 1 1 < j < l - 1 1 < j < l − 1 i = j = 2 i = j = 2 i = j = 2 1 ^1 1
9 > 16 cos 2 π m , thus cos 2 π m < cos 2 π 5
9 > 16 \cos^2 \frac{\pi}{m}, \text{ thus } \cos^2 \frac{\pi}{m} < \cos^2 \frac{\pi}{5}
9 > 16 cos 2 m π , thus cos 2 m π < cos 2 5 π 因此m = 4 m = 4 m = 4 i = 1 i = 1 i = 1 m = 5 m = 5 m = 5 2 j + 2 > 4 j 3 + 5 8 2j + 2 > 4j \frac{3+\sqrt{5}}{8} 2 j + 2 > 4 j 8 3 + 5 j 5 − 1 2 < 2 , j < 5 + 1 < 4 j \frac{\sqrt{5}-1}{2} < 2, j < \sqrt{5} + 1 < 4 j 2 5 − 1 < 2 , j < 5 + 1 < 4 l = j + 1 ≤ 4 l = j + 1 \leq 4 l = j + 1 ≤ 4
1 ^1 1 z 4 + z 3 + z 2 + z + 1 = 0 z^4 + z^3 + z^2 + z + 1 = 0 z 4 + z 3 + z 2 + z + 1 = 0 x = 1 2 ( z + 1 z ) x = \frac{1}{2}(z + \frac{1}{z}) x = 2 1 ( z + z 1 ) ( 2 x ) 2 − 2 + 2 x + 1 = 0 (2x)^2 - 2 + 2x + 1 = 0 ( 2 x ) 2 − 2 + 2 x + 1 = 0 4 x 2 + 2 x − 1 = 0 4x^2 + 2x - 1 = 0 4 x 2 + 2 x − 1 = 0 x = − 1 ± 5 4 x = \frac{-1 \pm \sqrt{5}}{4} x = 4 − 1 ± 5
2 cos 2 π 5 − 1 = cos 2 π 5 = 5 − 1 4 , cos 4 π 5 = 3 + 5 8 > 5 8 > 9 16 , cos π 5 = 1 + 5 4 .
2 \cos^2 \frac{\pi}{5} - 1 = \cos \frac{2\pi}{5} = \frac{\sqrt{5} - 1}{4}, \quad \cos^4 \frac{\pi}{5} = \frac{3 + \sqrt{5}}{8} > \frac{5}{8} > \frac{9}{16}, \quad \cos \frac{\pi}{5} = \frac{1 + \sqrt{5}}{4}.
2 cos 2 5 π − 1 = cos 5 2 π = 4 5 − 1 , cos 4 5 π = 8 3 + 5 > 8 5 > 16 9 , cos 5 π = 4 1 + 5 .
如果 X 具有分支点 i，则完整子图 X - {i} 是三个链的并，并且如果 p - 1、q - 1、r - 1 是这些链的长度，则三元组 {p, q, r} 等于三元组之一，直到置换 {1, 2, 2}，{1, 2, 3}，{1, 2, 4}，{1, 1, m}（对于某些 m ≥ 1 \geq 1 ≥ 1
引用了 3 · 被以下引用 1 引用了 引理 2 , 引理 7 , 引理 8
被以下引用 定理 1
顶点 i 属于到 3 条边 (引理 2)，并且没有其他分支点 (引理 7)，因此完整子图 X - {i} 是 3 个链 X_1、X_2、X_3 的和，每个链都有一个链接到 i 的终端顶点X。令 {i_1, i_2}, {i_2, i_3}, ..., {i_{p-1}, i_p} 为 X_1 的边，{j_1, j_2}, {j_2, j_3}, ..., {j_{q-1}, j_q} 为 X_2 的边，{k_1, k_2}, {k_2, k_3}, ..., {k_{r-1}， k_r} X_3 的那些，其中 i_1、j_1、k_1 链接到 X 中的 i。我们可以假设p ≥ q ≥ r ≥ 1 p \geq q \geq r \geq 1 p ≥ q ≥ r ≥ 1
x = e i p + 2 e i p − 1 + ⋯ + p e i 1
x = e_{i_p} + 2e_{i_{p-1}} + \cdots + pe_{i_1}
x = e i p + 2 e i p − 1 + ⋯ + p e i 1 y = e j q + 2 e j q − 1 + ⋯ + q e j 1
y = e_{j_q} + 2e_{j_{q-1}} + \cdots + qe_{j_1}
y = e j q + 2 e j q − 1 + ⋯ + q e j 1 z = e k r + 2 e k r − 1 + ⋯ + r e k 1 .
z = e_{k_r} + 2e_{k_{r-1}} + \cdots + re_{k_1}.
z = e k r + 2 e k r − 1 + ⋯ + r e k 1 . 由于 X 的所有边都是阶 3 (引理 7)，引理 8 给出 ∥ x ∥ 2 = 1 2 p ( p + 1 ) \| x \| ^2 = \frac{1}{2} p(p+1) ∥ x ∥ 2 = 2 1 p ( p + 1 ) ∥ y ∥ 2 = 1 2 q ( q + 1 ) \| y \| ^2 = \frac{1}{2} q(q+1) ∥ y ∥ 2 = 2 1 q ( q + 1 ) ∥ z ∥ 2 = 1 2 r ( r + 1 ) \| z \| ^2 = \frac{1}{2} r(r+1) ∥ z ∥ 2 = 2 1 r ( r + 1 ) e i e_i e i e i 2 , e i 3 , … , e i p e_{i_2}, e_{i_3}, \ldots, e_{i_p} e i 2 , e i 3 , … , e i p ( e i ∣ x ) = p ( e i ∣ e i 1 ) = − 1 2 p (e_i | x) = p(e_i | e_{i_1}) = -\frac{1}{2} p ( e i ∣ x ) = p ( e i ∣ e i 1 ) = − 2 1 p ( e i ∣ y ) = − 1 2 q (e_i | y) = -\frac{1}{2} q ( e i ∣ y ) = − 2 1 q ( e i ∣ z ) = − 1 2 r (e_i | z) = -\frac{1}{2} r ( e i ∣ z ) = − 2 1 r ∥ x ∥ − 1 x , ∥ y ∥ − 1 y , ∥ z ∥ − 1 z \| x \|^{-1} x, \| y \|^{-1} y, \| z \|^{-1} z ∥ x ∥ − 1 x , ∥ y ∥ − 1 y , ∥ z ∥ − 1 z e i e_i e i e i e_i e i
1 − ( e i ∣ x ∥ x ∥ ) 2 − ( e i ∣ y ∥ y ∥ ) 2 − ( e i ∣ z ∥ z ∥ ) 2
1 - (e_i | \frac{x}{\| x \|})^2 - (e_i | \frac{y}{\| y \|})^2 - (e_i | \frac{z}{\| z \|})^2
1 − ( e i ∣ ∥ x ∥ x ) 2 − ( e i ∣ ∥ y ∥ y ) 2 − ( e i ∣ ∥ z ∥ z ) 2 = 1 − 1 2 p p + 1 − 1 2 q q + 1 − 1 2 r r + 1
= 1 - \frac{1}{2} \frac{p}{p+1} - \frac{1}{2} \frac{q}{q+1} - \frac{1}{2} \frac{r}{r+1}
= 1 − 2 1 p + 1 p − 2 1 q + 1 q − 2 1 r + 1 r = 1 − 1 2 + 1 2 1 p + 1 − 1 2 + 1 2 1 q + 1 − 1 2 + 1 2 1 r + 1 .
= 1 - \frac{1}{2} + \frac{1}{2} \frac{1}{p+1} - \frac{1}{2} + \frac{1}{2} \frac{1}{q+1} - \frac{1}{2} + \frac{1}{2} \frac{1}{r+1}.
= 1 − 2 1 + 2 1 p + 1 1 − 2 1 + 2 1 q + 1 1 − 2 1 + 2 1 r + 1 1 . 由于这个数量 > 0，我们有
( p + 1 ) − 1 + ( q + 1 ) − 1 + ( r + 1 ) − 1 > 1.
(p+1)^{-1} + (q+1)^{-1} + (r+1)^{-1} > 1.
( p + 1 ) − 1 + ( q + 1 ) − 1 + ( r + 1 ) − 1 > 1. (4)
因此3 ( r + 1 ) − 1 > 1 3(r+1)^{-1} > 1 3 ( r + 1 ) − 1 > 1 r < 2 r < 2 r < 2 r = 1 r = 1 r = 1
( p + 1 ) − 1 + ( q + 1 ) − 1 > 1 2
(p+1)^{-1} + (q+1)^{-1} > \frac{1}{2}
( p + 1 ) − 1 + ( q + 1 ) − 1 > 2 1 (5)
因此2 ( q + 1 ) − 1 > 1 2 2(q+1)^{-1} > \frac{1}{2} 2 ( q + 1 ) − 1 > 2 1 q ≤ 2 q \leq 2 q ≤ 2 q = 2 q = 2 q = 2
( p + 1 ) − 1 > 1 6 . hence p ≤ 4.
(p+1)^{-1} > \frac{1}{6}. \quad \text{hence } p \leq 4.
( p + 1 ) − 1 > 6 1 . hence p ≤ 4.
任何不可约有限 Coxeter 系统 (W, S) 的图同构于以下之一：
引用了 4 · 被以下引用 6 引用了 引理 4 , 引理 7 , 引理 9 , 引理 10
被以下引用 习题 10, § 4 , 习题 11, § 4 , 习题 12, § 4 , 推论 , 定理 2 , 定理 3
A_l (l \geq 1\ 顶点) B_l (l \geq 2\ 顶点) D_l (l \geq 4\ 顶点) E_6 E_7 E_8 F_4 4 G_2 6 H_3 5 H_4 5 I_2(p) p (p = 5\ 或\ p \geq 7)。
这些图中没有两个是同构的。
事实上，令 W = (m_{ij}) 为 (W, S) 的 Coxeter 矩阵，并令 l = Card(S)。如果其中一个 m_{ij} 为 \geq 6，则 l = 2 (引理 4)，并且 (W, S) 的 Coxeter 图为型 G_2 或 I_2(p)，p \geq 7。现在假设所有 m_{ij} \leq 5。
a) 如果 m_{ij} 不全部等于 3，则 (W, S) 的图 X 是一个链，并且 m_{ij} 中恰好有一个等于 4 或 5 (引理 7)。如果 m_{ij} 之一等于 5，则引理 9 表明我们有型 H_3、H_4 或 I_2(5) 之一。如果 m_{ij} 之一等于 4，则引理 9 表明我们有型 B_1 或 F_4 之一。
b) 现在假设所有 m_{ij} 都等于 3。如果 X 是一条链，则 Coxeter 图为型 A_l。如果不是，则引理10表明它是型E_6、E_7、E_8或D_1。
显然，列出的 Coxeter 图中没有两个是同构的。
反之：
这对于 I_2(p) 来说是清楚的，相应的群是阶 2p 的二面角群（第 IV, § 1, no. 9 章）。对于 H_4 相应的二次形式是
ξ 1 2 + ξ 2 2 + ξ 3 2 + ξ 4 2 − ξ 1 ξ 2 − ξ 2 ξ 3 − 2 ( cos π 5 ) ξ 3 ξ 4
\xi_1^2 + \xi_2^2 + \xi_3^2 + \xi_4^2 - \xi_1 \xi_2 - \xi_2 \xi_3 - 2(\cos \frac{\pi}{5}) \xi_3 \xi_4
ξ 1 2 + ξ 2 2 + ξ 3 2 + ξ 4 2 − ξ 1 ξ 2 − ξ 2 ξ 3 − 2 ( cos 5 π ) ξ 3 ξ 4 = ξ 1 2 + ξ 2 2 + ξ 3 2 + ξ 4 2 − ξ 1 ξ 2 − ξ 2 ξ 3 − 1 + 5 2 ξ 3 ξ 4
= \xi_1^2 + \xi_2^2 + \xi_3^2 + \xi_4^2 - \xi_1 \xi_2 - \xi_2 \xi_3 - \frac{1 + \sqrt{5}}{2} \xi_3 \xi_4
= ξ 1 2 + ξ 2 2 + ξ 3 2 + ξ 4 2 − ξ 1 ξ 2 − ξ 2 ξ 3 − 2 1 + 5 ξ 3 ξ 4 = ( ξ 2 − ξ 1 + ξ 3 2 ) 2 + ( ξ 4 − 1 + 5 4 ξ 3 ) 2 + 3 4 ( ξ 1 − 1 3 ξ 3 ) 2 + 7 − 3 5 24 ξ 3 2 .
= (\xi_2 - \frac{\xi_1 + \xi_3}{2})^2 + (\xi_4 - \frac{1 + \sqrt{5}}{4} \xi_3)^2 + \frac{3}{4} (\xi_1 - \frac{1}{3} \xi_3)^2 + \frac{7 - 3 \sqrt{5}}{24} \xi_3^2.
= ( ξ 2 − 2 ξ 1 + ξ 3 ) 2 + ( ξ 4 − 4 1 + 5 ξ 3 ) 2 + 4 3 ( ξ 1 − 3 1 ξ 3 ) 2 + 24 7 − 3 5 ξ 3 2 . 由于7 − 3 5 7 - 3 \sqrt{5} 7 − 3 5 > 0 > 0 > 0 H 3 H_3 H 3
对于型A l , B l , … , G 2 A_l, B_l, \ldots, G_2 A l , B l , … , G 2
2. 丹金图
通过滥用语言，我们将调用具有以下属性的*规范图* 对 ( Γ , f ) (\Gamma, f) ( Γ , f )
1) Γ \Gamma Γ ( Γ , f ) (\Gamma, f) ( Γ , f )
2) 如果E E E ( i , j ) (i, j) ( i , j ) { i , j } \{i, j\} { i , j } Γ \Gamma Γ f f f E E E R \mathbf{R} R f ( i , j ) f ( j , i ) = 1 f(i, j) f(j, i) = 1 f ( i , j ) f ( j , i ) = 1 ( i , j ) ∈ E (i, j) \in E ( i , j ) ∈ E
规范图存在明显的概念和同构。
令 R R R V V V ( X , f ) (X, f) ( X , f ) R R R Dynkin 图 。 X X X { B } × B \{B\} \times B { B } × B W ( R ) W(R) W ( R ) I I I B B B R R R N = ( n i j ) i , j ∈ I N = (n_{ij})_{i,j \in I} N = ( n ij ) i , j ∈ I M = ( m i j ) i , j ∈ I M = (m_{ij})_{i,j \in I} M = ( m ij ) i , j ∈ I R R R i i i j j j X X X n i j ≠ 0 n_{ij} \neq 0 n ij = 0
f ( i , j ) = n i j n j i .
f(i, j) = \frac{n_{ij}}{n_{ji}}.
f ( i , j ) = n j i n ij . n i j = 0 n_{ij} = 0 n ij = 0 n j i = 0 n_{ji} = 0 n j i = 0 ( X , f ) (X, f) ( X , f ) 令( x ∣ y ) (x|y) ( x ∣ y ) V V V W ( R ) W(R) W ( R ) B = ( α i ) i ∈ I B = (\alpha_i)_{i \in I} B = ( α i ) i ∈ I R R R X X X i i i j j j
( α i ∣ α j ) ≠ 0
(\alpha_i|\alpha_j) \neq 0
( α i ∣ α j ) = 0 f ( i , j ) = ( α i ∣ α i ) ( α j ∣ α j ) .
f(i, j) = \frac{(\alpha_i|\alpha_i)}{(\alpha_j|\alpha_j)}.
f ( i , j ) = ( α j ∣ α j ) ( α i ∣ α i ) . 鉴于§ 1、第3和第5个结果，唯一的可能性如下，直到交换i i i j j j
1) i i i j j j n i j = n j i = 0 ; m i j = 2 ; n_{ij} = n_{ji} = 0; m_{ij} = 2; n ij = n j i = 0 ; m ij = 2 ; f ( i , j ) = f ( j , i ) = 1 ; n i j = n j i = − 1 ; m i j = 3 ; f(i, j) = f(j, i) = 1; n_{ij} = n_{ji} = -1; m_{ij} = 3; f ( i , j ) = f ( j , i ) = 1 ; n ij = n j i = − 1 ; m ij = 3 ; f ( i , j ) = 2 , f ( j , i ) = 1 / 2 ; n i j = − 2 , n j i = − 1 ; m i j = 4 ; f(i, j) = 2, f(j, i) = 1/2; n_{ij} = -2, n_{ji} = -1; m_{ij} = 4; f ( i , j ) = 2 , f ( j , i ) = 1/2 ; n ij = − 2 , n j i = − 1 ; m ij = 4 ; f ( i , j ) = 3 , f ( j , i ) = 1 / 3 ; n i j = − 3 , n j i = − 1 ; m i j = 6. f(i, j) = 3, f(j, i) = 1/3; n_{ij} = -3, n_{ji} = -1; m_{ij} = 6. f ( i , j ) = 3 , f ( j , i ) = 1/3 ; n ij = − 3 , n j i = − 1 ; m ij = 6.
由此我们可以看出，Dynkin 图 R R R R R R R R R
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 = ( α i ) i ∈ I 1 B_1 = (\alpha_i)_{i \in I_1} B 1 = ( α i ) i ∈ I 1 B 2 = ( α i ) i ∈ I 2 B_2 = (\alpha_i)_{i \in I_2} B 2 = ( α i ) i ∈ I 2 R 1 R_1 R 1 R 2 R_2 R 2 λ \lambda λ R 1 R_1 R 1 R 2 R_2 R 2 i ∈ I 1 i \in I_1 i ∈ I 1 V 1 V_1 V 1 V 2 V_2 V 2 R 1 R_1 R 1 R 2 R_2 R 2 α i \alpha_i α i α λ ( i ) \alpha_{\lambda(i)} α λ ( i )
被以下引用 2 被以下引用 推论 , 定理 2, VIII § 4
很明显，R R R R R R φ \varphi φ A ( R ) A(R) A ( R ) R R R
显然，φ ( g ) = I d \varphi(g) = \mathrm{Id} φ ( g ) = Id g ∈ W ( R ) g \in W(R) g ∈ W ( R ) R R R A ( R ) A(R) A ( R ) E E E ψ \psi ψ R R R B B B φ ∘ ψ = I d \varphi \circ \psi = \mathrm{Id} φ ∘ ψ = Id A ( R ) A(R) A ( R ) E E E W ( R ) W(R) W ( R )
在实践中，Dynkin图( X , f ) (X, f) ( X , f ) X X X i i i j j j i i i j j j f ( i , j ) > 1 f(i, j) > 1 f ( i , j ) > 1 α i \alpha_i α i α j \alpha_j α j > > > i i i j j j j j j
∘ ⟹ ∘ (for f ( i , j ) = 2 ). ∘ ⇛ ∘ (for f ( i , j ) = 3 ).
\begin{array}{cc}
\circ \Longrightarrow \circ & \text{(for } f(i, j) = 2 \text{).} \\
\circ \Rrightarrow \circ & \text{(for } f(i, j) = 3 \text{).}
\end{array}
∘ ⟹ ∘ ∘ ⇛ ∘ (for f ( i , j ) = 2 ). (for f ( i , j ) = 3 ). 很明显，可以从此图表恢复Dynkin图( X , f ) (X, f) ( X , f )
We remark that the diagram associated to the Coxeter graph of W ( R ) W(R) W ( R ) R R R
A l ∘ − ⋯ − ∘ ( l ≥ 1 vertices ) B l ∘ − ⋯ − ∘ − ∘ ( l ≥ 2 vertices ) C l ∘ − ⋯ − ∘ − ∘ ( l ≥ 3 vertices ) D l ∘ − ⋯ − ∘ − ∘ − ∘ ( l ≥ 4 vertices ) E 6 ∘ − ⋯ − ∘ E 7 ∘ − ⋯ − ∘ − ∘ − ∘ E 8 ∘ − ⋯ − ∘ − ∘ − ∘ − ∘ F 4 ∘ − ∘ − ∘ − ∘ G 2 ∘ − ∘ − ∘
\begin{array}{ll}
A_l & \circ-\cdots-\circ \quad (l \geq 1 \text{ vertices }) \\
B_l & \circ-\cdots-\circ-\circ \quad (l \geq 2 \text{ vertices }) \\
C_l & \circ-\cdots-\circ-\circ \quad (l \geq 3 \text{ vertices }) \\
D_l & \circ-\cdots-\circ-\circ-\circ \quad (l \geq 4 \text{ vertices }) \\
E_6 & \circ-\cdots-\circ \\
E_7 & \circ-\cdots-\circ-\circ-\circ \\
E_8 & \circ-\cdots-\circ-\circ-\circ-\circ \\
F_4 & \circ-\circ-\circ-\circ \\
G_2 & \circ-\circ-\circ
\end{array}
A l B l C l D l E 6 E 7 E 8 F 4 G 2 ∘ − ⋯ − ∘ ( l ≥ 1 vertices ) ∘ − ⋯ − ∘ − ∘ ( l ≥ 2 vertices ) ∘ − ⋯ − ∘ − ∘ ( l ≥ 3 vertices ) ∘ − ⋯ − ∘ − ∘ − ∘ ( l ≥ 4 vertices ) ∘ − ⋯ − ∘ ∘ − ⋯ − ∘ − ∘ − ∘ ∘ − ⋯ − ∘ − ∘ − ∘ − ∘ ∘ − ∘ − ∘ − ∘ ∘ − ∘ − ∘ 这些图中没有两个是同构的，并且存在不可约的约化根系将它们中的每一个作为其 Dynkin 图（最多同构）。
第一个断言紧随 Th 之后。 1、鉴于前面的注，图 H_3、H_4 和 I_2(p)（对于 p = 5 和 p \geq 7）的 Coxeter 群不是晶体学的事实，以及与 Coxeter 图 F_4 相关的 Dynkin 图的双（或三重）键的两个可能的不等式(resp. G_2) 给出同构 Dynkin 图。第二个断言是明确的，第三个将根据第5至13号中执行的每个型、构造的不可约约化根系的显式构造得出。
3. AFFINE WEYL 群和完整的 DYNKIN 图
令 R 为不可约约化根系，令 (X, f) 为其 Dynkin 图。我们将得到定义另一个范数图 (\tilde{X}, \tilde{f})，我们将其称为 R 的完整 Dynkin 图。 \tilde{X} 顶点的集合 \tilde{I} 由集合 I 组成X 的顶点和用 0 表示的顶点，而不是属于到 I。到 R 的定义 \tilde{f}、选择基 B = (\alpha_i)_{i \in I} 和标量乘积 \langle x | W(R) 下 y \rangle 不变。设 \alpha_0 为 B 定义的阶的最高根的负数。 \in \tilde{I} 中的两个不同顶点 i, j 相连当且仅当 \langle \alpha_i | \alpha_j \rangle \neq 0 然后我们把
f ~ ( i , j ) = ⟨ α i ∣ α i ⟩ ⟨ α j ∣ α j ⟩ .
\tilde{f}(i, j) = \frac{\langle \alpha_i | \alpha_i \rangle}{\langle \alpha_j | \alpha_j \rangle}.
f ~ ( i , j ) = ⟨ α j ∣ α j ⟩ ⟨ α i ∣ α i ⟩ . 正是直接的这样定义的图\tilde{X}和映射\tilde{f}不依赖于B的选择或标量乘积。
若R的秩l等于1，则I=\{i\}且\alpha_0=-\alpha_1；因此 \tilde{f}(0, i) = 1。如果 l \geq 2，\alpha_0 不与 \alpha_i 和 \langle \alpha_0 | 中的任何一个成比例。 \alpha_i \rangle 为 \leq 0 （§ 1, no. 8，命题 25）。对于 \tilde{I} 的不同元素的任何对 (i, j)，唯一的可能性是前面数中由 1), 2), 3), 4) 表示的那些（对于例，n_{0i} = n(\alpha_0, \alpha_i) 和 m_{0i} = 阶的 s_{\alpha_0} s_{\alpha_i} 对于所有 i \in I)。
在 l \geq 2 的情况下，完整的 Dynkin 图由图表表示，其约定与前面的数、有时相同，用虚线表示连接顶点 0 和其他顶点的键。我们注在这样的键上的不等式符号>，如果存在的话，总是指向不同于0的顶点，因为\alpha_0是最长的根（§ 1, no. 8，命题25）。图 (X, f) 是 (\tilde{X}, \tilde{f}) 删除顶点 0 得到的子图。
A(R) 的作用在 (X, f) 扩展到作用在 (\tilde{X}, \tilde{f}) 上，保持 0 固定，而 W(R) 在 (\tilde{X}, \tilde{f}) 上作用很简单。
我们保留§ 2的记号、§ 2, no. 2的第5号提案以及Th。第 1 章V, § 3, no. 2，表明仿射 Weyl 群 W_a(R) 的 Coxeter 图 \Sigma 可以通过与从 (X, f) 获得 W(R) 的 Coxeter 图相同的规则从 (\tilde{X}, \tilde{f}) 获得。 另一方面，令 G 为 W_a(R) (§ 2, no. 3) 的归一化器。 对于 \Sigma 的任何 g \in G 对应自同构 \varphi(g) 且 \varphi(g) = \mathrm{Id} 如果 g \in W_a(R)。 相反，根据第 11 章的 Prop. 11，给定自同构 \lambda of \Sigma，则有。 V, § 4, no. 9, 唯一元素 g = \psi(\lambda) 保留给定的凹室C，并且使得 \varphi(g) = \lambda。 由于G是元素的子群G_C的半直接乘积，保留C和W_a(R) (§ 2, no. 3)，我们通过从G/W_a（或从G_C）到Aut(\Sigma)到商同构，推导出\varphi定义。 直接的同构与从 A(R)/W(R) 到 G/W_a 的规范映射重合与同态从 A(R)/W(R) 到 Aut(\Sigma) 由同态从 A(R)/W(R) 诱导到Aut(\tilde{X}, \tilde{f}) 定义如上。 由 § 2, no. 3，群 Aut(\Sigma) 同构于 的半直接乘积
A(R)/W(R) 通过 P(R^*)/Q(R^*)。 P(R^*)/Q(R^*) 与群 I C = G C ∩ W a ′ I_C = G_C \cap W'_a I C = G C ∩ W a ′ Σ \Sigma Σ I C I_C I C γ i \gamma_i γ i Σ \Sigma Σ i i i
令 (W, S) 为不可约 Coxeter 系统，且 S 有限。当且仅当 (W, S) 的 Coxeter 图与以下之一同构时，相关的二次形式（第 V, § 4, no. 1 章）为正且简并：
A ~ 1 ∘ ⟶ ∞ A ~ l ( l ≥ 2 ) ∘ ⟶ ⋯ ⟶ ∘ (circuit with l + 1 vertices) B ~ 2 ∘ ⟶ 4 ⟶ ∘ B ~ l ( l ≥ 3 ) ∘ ⟶ ⋯ ⟶ ∘ ⟶ 4 (l + 1 vertices) C ~ l ( l ≥ 3 ) ∘ ⟶ 4 ⟶ ⋯ ⟶ 4 (l + 1 vertices) D ~ l ( l ≥ 4 ) ∘ ⟶ ⋯ ⟶ ∘ (l + 1 vertices) E ~ 6 ∘ ⟶ ∘ ⟶ ∘ ⟶ ∘ E ~ 7 ∘ ⟶ ∘ ⟶ ∘ ⟶ ∘ ⟶ ∘ E ~ 8 ∘ ⟶ ∘ ⟶ ∘ ⟶ ∘ ⟶ ∘ ⟶ ∘ F ~ 4 ∘ ⟶ ∘ ⟶ 4 ⟶ ∘ G ~ 2 ∘ ⟶ 6 ⟶ ∘
\begin{array}{ll}
\tilde{A}_1 & \circ \longrightarrow \infty \\
\tilde{A}_l \quad (l \geq 2) & \circ \longrightarrow \cdots \longrightarrow \circ \quad \text{(circuit with } l + 1 \text{ vertices)} \\
\tilde{B}_2 & \circ \longrightarrow 4 \longrightarrow \circ \\
\tilde{B}_l \quad (l \geq 3) & \circ \longrightarrow \cdots \longrightarrow \circ \longrightarrow 4 \quad \text{(l + 1 vertices)} \\
\tilde{C}_l \quad (l \geq 3) & \circ \longrightarrow 4 \longrightarrow \cdots \longrightarrow 4 \quad \text{(l + 1 vertices)} \\
\tilde{D}_l \quad (l \geq 4) & \circ \longrightarrow \cdots \longrightarrow \circ \quad \text{(l + 1 vertices)} \\
\tilde{E}_6 & \circ \longrightarrow \circ \longrightarrow \circ \longrightarrow \circ \\
\tilde{E}_7 & \circ \longrightarrow \circ \longrightarrow \circ \longrightarrow \circ \longrightarrow \circ \\
\tilde{E}_8 & \circ \longrightarrow \circ \longrightarrow \circ \longrightarrow \circ \longrightarrow \circ \longrightarrow \circ \\
\tilde{F}_4 & \circ \longrightarrow \circ \longrightarrow 4 \longrightarrow \circ \\
\tilde{G}_2 & \circ \longrightarrow 6 \longrightarrow \circ
\end{array}
A ~ 1 A ~ l ( l ≥ 2 ) B ~ 2 B ~ l ( l ≥ 3 ) C ~ l ( l ≥ 3 ) D ~ l ( l ≥ 4 ) E ~ 6 E ~ 7 E ~ 8 F ~ 4 G ~ 2 ∘ ⟶ ∞ ∘ ⟶ ⋯ ⟶ ∘ (circuit with l + 1 vertices) ∘ ⟶ 4 ⟶ ∘ ∘ ⟶ ⋯ ⟶ ∘ ⟶ 4 (l + 1 vertices) ∘ ⟶ 4 ⟶ ⋯ ⟶ 4 (l + 1 vertices) ∘ ⟶ ⋯ ⟶ ∘ (l + 1 vertices) ∘ ⟶ ∘ ⟶ ∘ ⟶ ∘ ∘ ⟶ ∘ ⟶ ∘ ⟶ ∘ ⟶ ∘ ∘ ⟶ ∘ ⟶ ∘ ⟶ ∘ ⟶ ∘ ⟶ ∘ ∘ ⟶ ∘ ⟶ 4 ⟶ ∘ ∘ ⟶ 6 ⟶ ∘ 引用了 1 引用了 V § 4. 。 COXETER 群的几何表示
这些 Coxeter 图中没有两个是同构的。
由章。 V, § 4, no. 9 和 § 2, no. 5 的第 8 条，二次形式为正且退化的 Coxeter 系统是对应到不可约约化根系的仿射 Weyl 群的那些。因此，定理是根据第 7 号中完成的 Dynkin 图的确定得出的。下面5至13。
4. 根系构造的预备课程
令V为维数l ≥ 1 l \geq 1 l ≥ 1 ( x ∣ y ) (x|y) ( x ∣ y ) Λ \Lambda Λ α ∈ L \alpha \in L α ∈ L ( α ∣ α ) ∈ Λ (\alpha|\alpha) \in \Lambda ( α ∣ α ) ∈ Λ ( α , β ) (\alpha, \beta) ( α , β ) 2 ( α ∣ β ) ( α ∣ α ) 2 \frac{(\alpha|\beta)}{(\alpha|\alpha)} 2 ( α ∣ α ) ( α ∣ β ) α ∈ R \alpha \in R α ∈ R s α s_{\alpha} s α x ↦ x − 2 ( x ∣ α ) ( α ∣ α ) α x \mapsto x - 2 \frac{(x|\alpha)}{(\alpha|\alpha)} \alpha x ↦ x − 2 ( α ∣ α ) ( x ∣ α ) α β ∈ R \beta \in R β ∈ R 2 ( β ∣ α ) ( α ∣ α ) ∈ Z 2 \frac{(\beta|\alpha)}{(\alpha|\alpha)} \in \mathbf{Z} 2 ( α ∣ α ) ( β ∣ α ) ∈ Z s α ( β ) ∈ L s_{\alpha}(\beta) \in L s α ( β ) ∈ L ∥ s α ( β ) ∥ = ∥ β ∥ \| s_{\alpha}(\beta) \| = \| \beta \| ∥ s α ( β ) ∥ = ∥ β ∥ s α ( β ) ∈ R s_{\alpha}(\beta) \in R s α ( β ) ∈ R Λ \Lambda Λ λ \lambda λ 4 λ 4\lambda 4 λ
现在令 V 为 E = R n E = \mathbf{R}^n E = R n ( ε 1 , … , ε n ) (\varepsilon_1, \ldots, \varepsilon_n) ( ε 1 , … , ε n ) E ∗ E^* E ∗ V ∗ V^* V ∗ L 0 , L 1 , L 2 , L 3 L_0, L_1, L_2, L_3 L 0 , L 1 , L 2 , L 3
1) L 0 L_0 L 0 ( ε i ) (\varepsilon_i) ( ε i ) Z \mathbf{Z} Z α , β ∈ L 0 \alpha, \beta \in L_0 α , β ∈ L 0 ( α ∣ β ) ∈ Z (\alpha|\beta) \in \mathbf{Z} ( α ∣ β ) ∈ Z α ∈ L 0 \alpha \in L_0 α ∈ L 0 ( α ∣ α ) = 1 (\alpha|\alpha) = 1 ( α ∣ α ) = 1 ± ε i \pm \varepsilon_i ± ε i 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n ( α ∣ α ) = 2 (\alpha|\alpha) = 2 ( α ∣ α ) = 2 i < j i < j i < j ± ε i ± ε j \pm \varepsilon_i \pm \varepsilon_j ± ε i ± ε j ± ε i ± ε j \pm \varepsilon_i \pm \varepsilon_j ± ε i ± ε j ± \pm ±
2) L 1 L_1 L 1 x = ∑ i = 1 n ξ i ε i ∈ L 0 x = \sum_{i=1}^n \xi_i \varepsilon_i \in L_0 x = ∑ i = 1 n ξ i ε i ∈ L 0 L 0 L_0 L 0 Z \mathbf{Z} Z ∑ i = 1 n ξ i \sum_{i=1}^n \xi_i ∑ i = 1 n ξ i ξ i \xi_i ξ i ξ i 2 \xi_i^2 ξ i 2 ( x ∣ x ) (x|x) ( x ∣ x ) L 1 ′ L_1' L 1 ′ ε i ± ε j \varepsilon_i \pm \varepsilon_j ε i ± ε j L 1 L_1 L 1 ∑ i = 1 n ξ i ε i ≡ ( ∑ i = 1 n ξ i ) ε n ( m o d L 1 ′ ) \sum_{i=1}^n \xi_i \varepsilon_i \equiv (\sum_{i=1}^n \xi_i) \varepsilon_n \pmod{L_1'} ∑ i = 1 n ξ i ε i ≡ ( ∑ i = 1 n ξ i ) ε n ( mod L 1 ′ ) 2 ε n = ( ε 1 + ε n ) − ( ε 1 − ε n ) ∈ L 1 ′ 2\varepsilon_n = (\varepsilon_1 + \varepsilon_n) - (\varepsilon_1 - \varepsilon_n) \in L_1' 2 ε n = ( ε 1 + ε n ) − ( ε 1 − ε n ) ∈ L 1 ′ L 1 ′ = L 1 L_1' = L_1 L 1 ′ = L 1 L 0 L_0 L 0 L 1 L_1 L 1 ε 1 \varepsilon_1 ε 1 L 0 / L 1 L_0/L_1 L 0 / L 1 Z / 2 Z \mathbf{Z}/2\mathbf{Z} Z /2 Z
3）L 2 = L 0 + Z ( 1 2 ∑ i = 1 n ε i ) L_2 = L_0 + \mathbf{Z}(\frac{1}{2} \sum_{i=1}^n \varepsilon_i) L 2 = L 0 + Z ( 2 1 ∑ i = 1 n ε i ) x = ∑ i = 1 n ξ i ε i x = \sum_{i=1}^n \xi_i \varepsilon_i x = ∑ i = 1 n ξ i ε i L 2 L_2 L 2
2 ξ i ∈ Z , ξ i − ξ j ∈ Z for all i and j .
2\xi_i \in \mathbf{Z}, \quad \xi_i - \xi_j \in \mathbf{Z} \text{ for all } i \text{ and } j.
2 ξ i ∈ Z , ξ i − ξ j ∈ Z for all i and j . ( ε k ∣ 1 2 ∑ i = 1 n ε i ) = 1 2 (\varepsilon_k | \frac{1}{2} \sum_{i=1}^n \varepsilon_i ) = \frac{1}{2} ( ε k ∣ 2 1 ∑ i = 1 n ε i ) = 2 1 k k k ∥ 1 2 ∑ i = 1 n ε i ∥ 2 = n 4 \| \frac{1}{2} \sum_{i=1}^n \varepsilon_i \| ^2 = \frac{n}{4} ∥ 2 1 ∑ i = 1 n ε i ∥ 2 = 4 n n n n α , β ∈ L 2 \alpha, \beta \in L_2 α , β ∈ L 2 ( α ∣ β ) ∈ 1 2 Z (\alpha|\beta) \in \frac{1}{2}\mathbf{Z} ( α ∣ β ) ∈ 2 1 Z L 2 / L 0 L_2/L_0 L 2 / L 0 Z / 2 Z \mathbf{Z}/2\mathbf{Z} Z /2 Z 4）L 3 = L 1 + Z ( 1 2 ∑ i = 1 n ε i ) L_3 = L_1 + \mathbf{Z} (\frac{1}{2} \sum_{i=1}^n \varepsilon_i ) L 3 = L 1 + Z ( 2 1 ∑ i = 1 n ε i ) n n n L 3 L_3 L 3 ∑ i = 1 n ξ i ε i \sum_{i=1}^n \xi_i \varepsilon_i ∑ i = 1 n ξ i ε i ∑ i = 1 n ξ i ∈ 2 Z \sum_{i=1}^n \xi_i \in 2\mathbf{Z} ∑ i = 1 n ξ i ∈ 2 Z ( α ∣ β ) ∈ Z (\alpha|\beta) \in \mathbf{Z} ( α ∣ β ) ∈ Z α , β ∈ L 3 \alpha, \beta \in L_3 α , β ∈ L 3
与L 0 L_0 L 0 L 1 , L 2 L_1, L_2 L 1 , L 2 L 0 L_0 L 0 L 2 , L 1 L_2, L_1 L 2 , L 1 L 3 L_3 L 3
x : = ∑ i = 1 n ξ i ε i ∈ L 2
x := \sum_{i=1}^n \xi_i \varepsilon_i \in L_2
x := i = 1 ∑ n ξ i ε i ∈ L 2 使得( x ∣ 1 2 ∑ i = 1 n ε i ) ∈ Z (x| \frac{1}{2} \sum_{i=1}^n \varepsilon_i ) \in \mathbf{Z} ( x ∣ 2 1 ∑ i = 1 n ε i ) ∈ Z ∑ i = 1 n ξ i ∈ 2 Z \sum_{i=1}^n \xi_i \in 2\mathbf{Z} ∑ i = 1 n ξ i ∈ 2 Z n ≡ 0 ( m o d 4 ) n \equiv 0 \pmod{4} n ≡ 0 ( mod 4 ) L 3 L_3 L 3
阿贝尔的群 L 2 / L 1 L_2/L_1 L 2 / L 1 Z / 4 Z \mathbf{Z}/4\mathbf{Z} Z /4 Z Z / 2 Z × Z / 2 Z \mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/2\mathbf{Z} Z /2 Z × Z /2 Z n n n
p ( 1 2 ∑ i = 1 n ε i ) ∈ L 1 ⇔ p ≡ 0 ( m o d 4 )
p(\frac{1}{2} \sum_{i=1}^n \varepsilon_i ) \in L_1 \Leftrightarrow p \equiv 0 \pmod{4}
p ( 2 1 i = 1 ∑ n ε i ) ∈ L 1 ⇔ p ≡ 0 ( mod 4 ) 所以L 2 / L 1 L_2/L_1 L 2 / L 1 n n n
p ( 1 2 ∑ i = 1 n ε i ) ∈ L 1 ⇔ p ≡ 0 ( m o d 2 )
p(\frac{1}{2} \sum_{i=1}^n \varepsilon_i ) \in L_1 \Leftrightarrow p \equiv 0 \pmod{2}
p ( 2 1 i = 1 ∑ n ε i ) ∈ L 1 ⇔ p ≡ 0 ( mod 2 ) 所以L 2 / L 1 L_2/L_1 L 2 / L 1 Z / 2 Z × Z / 2 Z \mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/2\mathbf{Z} Z /2 Z × Z /2 Z
我们将在接下来的九个项目中使用这个记号。和在盘子里。对于 Th 中 Dynkin 图的每个型。 3、我们要明确描述：
(I) 根系 R 和根的数。 R 的 (II) 基 B，且相应的正根。 基 B 将由整数 1, ..., l 索引。 （III）考克斯特数小时（§ 1, no. 11）。 (IV) 最高根 α ~ \tilde{\alpha} α ~ R − R^- R − γ ( R ) \gamma(R) γ ( R )
(XI) 群 A ( R ) / W ( R ) A(R)/W(R) A ( R ) / W ( R ) W ( R ) W(R) W ( R ) w 0 w_0 w 0 B B B − B -B − B
完成的 Dynkin 图上的 P ( R − ) / Q ( R − ) P(R^-)/Q(R^-) P ( R − ) / Q ( R − ) P ( R − ) / Q ( R − ) P(R^-)/Q(R^-) P ( R − ) / Q ( R − ) A ( R ) / W ( R ) A(R)/W(R) A ( R ) / W ( R )
对于 Th 中的每个 Dynkin 图。 3、这些数据将被收集在板块I至IX中，并按照上述统一方式排序。我们还给予：
(XIII) 嘉当矩阵，从中导出 Dynkin 图，如no. 2 中所述。
5. 系统型 B l B_l B l l ≥ 2 l \geq 2 l ≥ 2
(I) 我们在V = R l V = \mathbf{R}^l V = R l L 0 L_0 L 0 R R R α ∈ L 0 \alpha \in L_0 α ∈ L 0 ( α ∣ α ) = 1 (\alpha|\alpha) = 1 ( α ∣ α ) = 1 ( α ∣ α ) = 2 (\alpha|\alpha) = 2 ( α ∣ α ) = 2 ± ε i \pm \varepsilon_i ± ε i 1 ≤ i ≤ l 1 \leq i \leq l 1 ≤ i ≤ l ± ε i ± ε j \pm \varepsilon_i \pm \varepsilon_j ± ε i ± ε j 1 ≤ i < j ≤ l 1 \leq i < j \leq l 1 ≤ i < j ≤ l α , β ∈ R \alpha, \beta \in R α , β ∈ R R R R V V V 2 ( α ∣ β ) / ( α ∣ α ) ∈ Z 2(\alpha|\beta)/(\alpha|\alpha) \in \mathbf{Z} 2 ( α ∣ β ) / ( α ∣ α ) ∈ Z R R R V V V n = 2 l + 4 l ( l − 1 ) / 2 = 2 l 2 n = 2l + 4^{l(l-1)/2} = 2l^2 n = 2 l + 4 l ( l − 1 ) /2 = 2 l 2
(II) 放
α 1 = ε 1 − ε 2 , α 2 = ε 2 − ε 3 , … , α l − 1 = ε l − 1 − ε 1 , α l = ε l .
\alpha_1 = \varepsilon_1 - \varepsilon_2,\ \alpha_2 = \varepsilon_2 - \varepsilon_3,\ldots,\alpha_{l-1} = \varepsilon_{l-1} - \varepsilon_1,\ \alpha_l = \varepsilon_l.
α 1 = ε 1 − ε 2 , α 2 = ε 2 − ε 3 , … , α l − 1 = ε l − 1 − ε 1 , α l = ε l . 然后
ε i = α i + α i + 1 + ⋯ + α l ( 1 ≤ i ≤ l )
\varepsilon_i = \alpha_i + \alpha_{i+1} + \cdots + \alpha_l \quad (1 \leq i \leq l)
ε i = α i + α i + 1 + ⋯ + α l ( 1 ≤ i ≤ l ) ε i + ε j = ( α i + α i + 1 + ⋯ + α l ) + ( α j + α j + 1 + ⋯ + α l ) ( 1 ≤ i < j ≤ l )
\varepsilon_i + \varepsilon_j = (\alpha_i + \alpha_{i+1} + \cdots + \alpha_l) + (\alpha_j + \alpha_{j+1} + \cdots + \alpha_l) \quad (1 \leq i < j \leq l)
ε i + ε j = ( α i + α i + 1 + ⋯ + α l ) + ( α j + α j + 1 + ⋯ + α l ) ( 1 ≤ i < j ≤ l ) ε i − ε j = α i + α i + 1 + ⋯ + α j − 1 ( 1 ≤ i < j ≤ l ) .
\varepsilon_i - \varepsilon_j = \alpha_i + \alpha_{i+1} + \cdots + \alpha_{j-1} \quad (1 \leq i < j \leq l).
ε i − ε j = α i + α i + 1 + ⋯ + α j − 1 ( 1 ≤ i < j ≤ l ) . 因此( α 1 , α 2 , … , α l ) (\alpha_1, \alpha_2, \ldots, \alpha_l) ( α 1 , α 2 , … , α l ) R R R i < l i < l i < l ∥ α i ∥ 2 = 2 \| \alpha_i \|^2 = 2 ∥ α i ∥ 2 = 2 ∥ α l ∥ 2 = 1 \| \alpha_l \|^2 = 1 ∥ α l ∥ 2 = 1 ( α i ∣ α i + 1 ) = − 1 (\alpha_i|\alpha_{i+1}) = -1 ( α i ∣ α i + 1 ) = − 1 1 ≤ i ≤ l − 1 1 \leq i \leq l-1 1 ≤ i ≤ l − 1 ( α i ∣ α j ) = 0 (\alpha_i|\alpha_j) = 0 ( α i ∣ α j ) = 0 j > i + 1 j > i + 1 j > i + 1 R R R B l B_l B l R R R ε i \varepsilon_i ε i ε i ± ε j \varepsilon_i \pm \varepsilon_j ε i ± ε j i < j i < j i < j
(III) 作者：Th.第 1 章第 1(ii) 节V, § 6, no. 2,
h = n / l = 2 l .
h = n/l = 2l.
h = n / l = 2 l . (IV)设α ~ = ε 1 + ε 2 = α 1 + 2 α 2 + 2 α 3 + ⋯ + 2 α l \tilde{\alpha} = \varepsilon_1 + \varepsilon_2 = \alpha_1 + 2\alpha_2 + 2\alpha_3 + \cdots + 2\alpha_l α ~ = ε 1 + ε 2 = α 1 + 2 α 2 + 2 α 3 + ⋯ + 2 α l ( α i ) (\alpha_i) ( α i ) 2 l − 1 = h − 1 2l - 1 = h - 1 2 l − 1 = h − 1 α ~ \tilde{\alpha} α ~ R R R ( α ~ ∣ α i ) = 0 (\tilde{\alpha}|\alpha_i) = 0 ( α ~ ∣ α i ) = 0 i ≠ 2 i \neq 2 i = 2 ( α ~ ∣ α 2 ) = 1 (\tilde{\alpha}|\alpha_2) = 1 ( α ~ ∣ α 2 ) = 1 l = 2 l = 2 l = 2 l ≥ 3 l \geq 3 l ≥ 3 α 2 \alpha_2 α 2 2 \sqrt{2} 2 R R R
l = 2 l = 2 l = 2 α 2 α 1
\begin{array}{c}
\alpha_2 \\
\alpha_1
\end{array}
α 2 α 1 l ≥ 3 l \geq 3 l ≥ 3 α 1 α 2 α 3 ⋯ α l − 1 α l
\begin{array}{cccccc}
\alpha_1 & & & & & \\
& \alpha_2 & \alpha_3 & \cdots & \alpha_{l-1} & \alpha_l
\end{array}
α 1 α 2 α 3 ⋯ α l − 1 α l (V)公式α − = 2 α ( α ∣ α ) \alpha^- = \frac{2\alpha}{(\alpha|\alpha)} α − = ( α ∣ α ) 2 α ± 2 ε i \pm 2\varepsilon_i ± 2 ε i 1 ≤ i ≤ l 1 \leq i \leq l 1 ≤ i ≤ l ± ε i ± ε j \pm \varepsilon_i \pm \varepsilon_j ± ε i ± ε j 1 ≤ i < j ≤ l 1 \leq i < j \leq l 1 ≤ i < j ≤ l R − R^- R − R − R^- R − R R R R − R^- R − C l C_l C l
与β = ε 1 \beta = \varepsilon_1 β = ε 1 4 l − 2 4l - 2 4 l − 2 ± ε 1 \pm \varepsilon_1 ± ε 1 ± ε 1 ± ε j \pm \varepsilon_1 \pm \varepsilon_j ± ε 1 ± ε j 2 ≤ j ≤ l 2 \leq j \leq l 2 ≤ j ≤ l α \alpha α n ( α , β ) = ± 2 n(\alpha, \beta) = \pm 2 n ( α , β ) = ± 2 Φ R \Phi_R Φ R β \beta β ( 4 l − 2 ) − 1 (4l - 2)^{-1} ( 4 l − 2 ) − 1 Φ R ( x , y ) = ( x ∣ y ) / ( 4 l − 2 ) \Phi_R(x, y) = (x|y)/(4l - 2) Φ R ( x , y ) = ( x ∣ y ) / ( 4 l − 2 ) x = y = β x = y = \beta x = y = β
2 + 1 4 ( 4 l − 4 ) = γ ( R ) 1 4 l − 2
2 + \frac{1}{4}(4l - 4) = \gamma(R) \frac{1}{4l - 2}
2 + 4 1 ( 4 l − 4 ) = γ ( R ) 4 l − 2 1 γ ( R ) = ( l + 1 ) ( 4 l − 2 ) \gamma(R) = (l + 1)(4l - 2) γ ( R ) = ( l + 1 ) ( 4 l − 2 ) (VI) 权 ω i \omega_i ω i 1 ≤ i ≤ l 1 \leq i \leq l 1 ≤ i ≤ l ( ω i ∣ α j ) = δ i j (\omega_i|\alpha_j) = \delta_{ij} ( ω i ∣ α j ) = δ ij
ω i = ε 1 + ε 2 + ⋯ + ε i = α 1 + 2 α 2 + ⋯ + ( i − 1 ) α i − 1 + i ( α 1 + α i + 1 + ⋯ + α l ) ( i < l ) ω l = 1 2 ( ε 1 + ε 2 + ⋯ + ε l ) = 1 2 ( α 1 + 2 α 2 + ⋯ + l α l ) .
\begin{align*}
\omega_i &= \varepsilon_1 + \varepsilon_2 + \cdots + \varepsilon_i \\
&= \alpha_1 + 2\alpha_2 + \cdots + (i - 1)\alpha_{i-1} + i(\alpha_1 + \alpha_{i+1} + \cdots + \alpha_l) \quad (i < l) \\
\omega_l &= \frac{1}{2}(\varepsilon_1 + \varepsilon_2 + \cdots + \varepsilon_l) = \frac{1}{2}(\alpha_1 + 2\alpha_2 + \cdots + l\alpha_l).
\end{align*}
ω i ω l = ε 1 + ε 2 + ⋯ + ε i = α 1 + 2 α 2 + ⋯ + ( i − 1 ) α i − 1 + i ( α 1 + α i + 1 + ⋯ + α l ) ( i < l ) = 2 1 ( ε 1 + ε 2 + ⋯ + ε l ) = 2 1 ( α 1 + 2 α 2 + ⋯ + l α l ) . (VII) 和的正根是
2 ρ = ( 2 l − 1 ) ε 1 + ( 2 l − 3 ) ε 2 + ⋯ + 3 ε l − 1 + ε l = ( 2 l − 1 ) α 1 + 2 ( 2 l − 2 ) α 2 + ⋯ + i ( 2 l − i ) α i + ⋯ + l 2 α l .
\begin{align*}
2\rho &= (2l - 1)\varepsilon_1 + (2l - 3)\varepsilon_2 + \cdots + 3\varepsilon_{l-1} + \varepsilon_l \\
&= (2l - 1)\alpha_1 + 2(2l - 2)\alpha_2 + \cdots + i(2l - i)\alpha_i + \cdots + l^2\alpha_l.
\end{align*}
2 ρ = ( 2 l − 1 ) ε 1 + ( 2 l − 3 ) ε 2 + ⋯ + 3 ε l − 1 + ε l = ( 2 l − 1 ) α 1 + 2 ( 2 l − 2 ) α 2 + ⋯ + i ( 2 l − i ) α i + ⋯ + l 2 α l . (VIII) 我们有Q ( R ) = L 0 Q(R) = L_0 Q ( R ) = L 0 P ( R ) P(R) P ( R ) Q ( R ) Q(R) Q ( R ) ω l \omega_l ω l L 2 L_2 L 2 P ( R ) / Q ( R ) P(R)/Q(R) P ( R ) / Q ( R ) Z / 2 Z \mathbf{Z}/2\mathbf{Z} Z /2 Z
(IX) 和 (X) 在 R l \mathbf{R}^l R l s ε i , − ε j s_{\varepsilon_i, -\varepsilon_j} s ε i , − ε j i ≠ j i \neq j i = j ε i \varepsilon_i ε i ε j \varepsilon_j ε j k k k ε k \varepsilon_k ε k i i i j j j s ε i , − ε i s_{\varepsilon_i, -\varepsilon_i} s ε i , − ε i G 1 G_1 G 1 S l S_l S l k k k i i i s ε i s_{\varepsilon_i} s ε i ε i \varepsilon_i ε i − ε i -\varepsilon_i − ε i ε k \varepsilon_k ε k s ε i s_{\varepsilon_i} s ε i G 2 G_2 G 2 ( Z / 2 Z ) l (\mathbf{Z}/2\mathbf{Z})^l ( Z /2 Z ) l W ( R ) W(R) W ( R ) G 1 G_1 G 1 G 2 G_2 G 2 G 2 G_2 G 2 W ( R ) W(R) W ( R ) W ( R ) W(R) W ( R ) G l G_l G l ( Z / 2 Z ) l (\mathbf{Z}/2\mathbf{Z})^l ( Z /2 Z ) l 2 l . l ! 2^l.l! 2 l . l !
对称的代数S ( R l ) S(\mathbf{R}^l) S ( R l ) R l \mathbf{R}^l R l P ( ξ 1 , … , ξ l ) P(\xi_1, \ldots, \xi_l) P ( ξ 1 , … , ξ l ) W ( R ) W(R) W ( R )
P ( ξ 1 , ξ 2 , … , ξ l ) = P ( ± ξ 1 , ± ξ 2 , … , ± ξ l )
P(\xi_1, \xi_2, \ldots, \xi_l) = P(\pm \xi_1, \pm \xi_2, \ldots, \pm \xi_l)
P ( ξ 1 , ξ 2 , … , ξ l ) = P ( ± ξ 1 , ± ξ 2 , … , ± ξ l ) P ( ξ 1 , … , ξ l ) = Q ( ξ 1 2 , … , ξ l 2 )
P(\xi_1, \ldots, \xi_l) = Q(\xi_1^2, \ldots, \xi_l^2)
P ( ξ 1 , … , ξ l ) = Q ( ξ 1 2 , … , ξ l 2 ) 其中Q是多项式，并且Q是对称多项式；这些条件就足够了。因而（代数。章节V，附录I），S ( R l ) W ( R ) S(\mathbf{R}^l)^{W(R)} S ( R l ) W ( R )
t i = ∑ τ ∈ S l ξ τ ( 1 ) 2 ξ τ ( 2 ) 2 ⋯ ξ τ ( l ) 2 ( 1 ≤ i ≤ l ) .
t_i = \sum_{\tau \in S_l} \xi_{\tau(1)}^2 \xi_{\tau(2)}^2 \cdots \xi_{\tau(l)}^2 \quad (1 \leq i \leq l).
t i = τ ∈ S l ∑ ξ τ ( 1 ) 2 ξ τ ( 2 ) 2 ⋯ ξ τ ( l ) 2 ( 1 ≤ i ≤ l ) . 而且，S ( R l ) W ( R ) S(\mathbf{R}^l)^{W(R)} S ( R l ) W ( R ) R \mathbf{R} R t i t_i t i t i t_i t i
1 , 3 , 5 , … , 2 l − 1. 1, 3, 5, \ldots, 2l - 1. 1 , 3 , 5 , … , 2 l − 1. (XI) Dynkin 图中唯一的自同构是单位元素。因而，A ( R ) = W ( R ) A(R) = W(R) A ( R ) = W ( R ) − 1 ∈ W ( R ) -1 \in W(R) − 1 ∈ W ( R ) − 1 -1 − 1 − B -B − B w 0 = − 1 w_0 = -1 w 0 = − 1
(XII) 群 P ( R − ) / Q ( R − ) P(R^-)/Q(R^-) P ( R − ) / Q ( R − ) P ( R ) / Q ( R ) P(R)/Q(R) P ( R ) / Q ( R ) Z / 2 Z \mathbf{Z}/2\mathbf{Z} Z /2 Z α 0 \alpha_0 α 0 α 1 \alpha_1 α 1
6. 系统型 C l C_l C l l ≥ 2 l \geq 2 l ≥ 2
(I)型C l C_l C l B l B_l B l C l C_l C l R ′ \mathbf{R}' R ′ ± 2 ε i \pm 2\varepsilon_i ± 2 ε i 1 ≤ i ≤ l 1 \leq i \leq l 1 ≤ i ≤ l ± ε i ± ε j \pm \varepsilon_i \pm \varepsilon_j ± ε i ± ε j 1 ≤ i < j ≤ l 1 \leq i < j \leq l 1 ≤ i < j ≤ l C l C_l C l 2 l 2 2l^2 2 l 2
R的(II)基可以通过在no. 5中考虑系统的基的映射α ↦ 2 α ( α ∣ α ) \alpha \mapsto \frac{2\alpha}{(\alpha|\alpha)} α ↦ ( α ∣ α ) 2 α
α 1 = ε 1 − ε 2 , α 2 = ε 2 − ε 3 , … , α l − 1 = ε l − 1 − ε l , α l = 2 ε l .
\alpha_1 = \varepsilon_1 - \varepsilon_2, \quad \alpha_2 = \varepsilon_2 - \varepsilon_3, \ldots, \quad \alpha_{l-1} = \varepsilon_{l-1} - \varepsilon_l, \quad \alpha_l = 2\varepsilon_l.
α 1 = ε 1 − ε 2 , α 2 = ε 2 − ε 3 , … , α l − 1 = ε l − 1 − ε l , α l = 2 ε l . 正根是 2 ε i 2\varepsilon_i 2 ε i ε i ± ε j \varepsilon_i \pm \varepsilon_j ε i ± ε j i < j i < j i < j
(III) Coxeter 数与逆系相同：h = 2 l h = 2l h = 2 l
(IV)设α ~ = 2 ε 1 = 2 α 1 + 2 α 2 + ⋯ + 2 α l − 1 + α l \tilde{\alpha} = 2\varepsilon_1 = 2\alpha_1 + 2\alpha_2 + \cdots + 2\alpha_{l-1} + \alpha_l α ~ = 2 ε 1 = 2 α 1 + 2 α 2 + ⋯ + 2 α l − 1 + α l ( α i ) (\alpha_i) ( α i ) 2 l − 1 = h − 1 2l - 1 = h - 1 2 l − 1 = h − 1 α ~ \tilde{\alpha} α ~ ( α ~ ∣ α i ) = 0 (\tilde{\alpha}|\alpha_i) = 0 ( α ~ ∣ α i ) = 0 i ≠ l i \neq l i = l ( α ~ ∣ α l ) = 2 (\tilde{\alpha}|\alpha_l) = 2 ( α ~ ∣ α l ) = 2
!C_l 的 Dynkin 图
(V) 我们已经确定了R − R^- R − B l B_l B l Φ R \Phi_R Φ R 2 ε i 2\varepsilon_i 2 ε i
( ( l + 1 ) ( 4 l − 2 ) ) − 1 ( ( 4 l − 2 ) − 1 ) − 1 = ( l + 1 ) − 1 ;
((l+1)(4l-2))^{-1} ((4l-2)^{-1})^{-1} = (l+1)^{-1};
(( l + 1 ) ( 4 l − 2 ) ) − 1 (( 4 l − 2 ) − 1 ) − 1 = ( l + 1 ) − 1 ; 因此，Φ R ( x , y ) = ( x ∣ y ) / 4 ( l + 1 ) \Phi_R(x, y) = (x|y)/4(l+1) Φ R ( x , y ) = ( x ∣ y ) /4 ( l + 1 )
我们有γ ( R ) = γ ( R − ) = ( l + 1 ) ( 4 l − 2 ) \gamma(R) = \gamma(R^-) = (l+1)(4l-2) γ ( R ) = γ ( R − ) = ( l + 1 ) ( 4 l − 2 )
(VI) 基本权很容易找到：
ω i = ε 1 + ε 2 + ⋯ + ε i
\omega_i = \varepsilon_1 + \varepsilon_2 + \cdots + \varepsilon_i
ω i = ε 1 + ε 2 + ⋯ + ε i = α 1 + 2 α 2 + ⋯ + ( i − 1 ) α i − 1 + i ( α i + α i + 1 + ⋯ + 1 2 α l ) ( i ≤ l ) .
= \alpha_1 + 2\alpha_2 + \cdots + (i-1)\alpha_{i-1} + i(\alpha_i + \alpha_{i+1} + \cdots + \frac{1}{2}\alpha_l) \quad (i \leq l).
= α 1 + 2 α 2 + ⋯ + ( i − 1 ) α i − 1 + i ( α i + α i + 1 + ⋯ + 2 1 α l ) ( i ≤ l ) . (VII) 和的正根为
2 ρ = 2 l ε 1 + ( 2 l − 2 ) ε 2 + ⋯ + 4 ε l − 1 + 2 ε l
2\rho = 2l\varepsilon_1 + (2l-2)\varepsilon_2 + \cdots + 4\varepsilon_{l-1} + 2\varepsilon_l
2 ρ = 2 l ε 1 + ( 2 l − 2 ) ε 2 + ⋯ + 4 ε l − 1 + 2 ε l = 2 l α 1 + 2 ( 2 l − 1 ) α 3 + ⋯ + i ( 2 l − i + 1 ) α i + ⋯
= 2l\alpha_1 + 2(2l-1)\alpha_3 + \cdots + i(2l-i+1)\alpha_i + \cdots
= 2 l α 1 + 2 ( 2 l − 1 ) α 3 + ⋯ + i ( 2 l − i + 1 ) α i + ⋯ ⋯ + ( l − 1 ) ( l + 2 ) α l − 1 + 1 2 l ( l + 1 ) α l .
\cdots + (l-1)(l+2)\alpha_{l-1} + \frac{1}{2}l(l+1)\alpha_l.
⋯ + ( l − 1 ) ( l + 2 ) α l − 1 + 2 1 l ( l + 1 ) α l . (VIII) 通过no. 4和no. 5 (VIII)，Q ( R ) = L 1 , P ( R ) = L 0 ; P ( R ) / Q ( R ) Q(R) = L_1, P(R) = L_0; P(R)/Q(R) Q ( R ) = L 1 , P ( R ) = L 0 ; P ( R ) / Q ( R ) Z / 2 Z \mathbf{Z}/2\mathbf{Z} Z /2 Z
(IX) 和 (X) 这些数据仅取决于W ( R ) W(R) W ( R ) B l B_l B l
(XI) 与no. 5相同的论证表明A ( R ) = W ( R ) A(R) = W(R) A ( R ) = W ( R ) w 0 = − 1 w_0 = -1 w 0 = − 1
(XII) P ( R − ) / Q ( R − ) P(R^-)/Q(R^-) P ( R − ) / Q ( R − ) α j \alpha_j α j α l − j \alpha_{l-j} α l − j 0 ≤ j ≤ l 0 \leq j \leq l 0 ≤ j ≤ l
7. 系统型 A l A_l A l l ≥ 1 l \geq 1 l ≥ 1
(I) 和 (II) 令E = R l + 1 E = \mathbf{R}^{l+1} E = R l + 1 V V V ∑ i = 1 l + 1 ξ i = 0 \sum_{i=1}^{l+1} \xi_i = 0 ∑ i = 1 l + 1 ξ i = 0 l + 1 l+1 l + 1 l l l E E E B l + 1 B_{l+1} B l + 1 R ′ R' R ′
α 1 = ε 1 − ε 2 , α 2 = ε 2 − ε 3 , … , α l = ε l − ε l + 1 , α l + 1 = ε l + 1 .
\alpha_1 = \varepsilon_1 - \varepsilon_2,\ \alpha_2 = \varepsilon_2 - \varepsilon_3, \ldots, \alpha_l = \varepsilon_l - \varepsilon_{l+1},\ \alpha_{l+1} = \varepsilon_{l+1}.
α 1 = ε 1 − ε 2 , α 2 = ε 2 − ε 3 , … , α l = ε l − ε l + 1 , α l + 1 = ε l + 1 . α 1 , … , α l \alpha_1, \ldots, \alpha_l α 1 , … , α l V V V R = R ′ ∩ V R = R' \cap V R = R ′ ∩ V V V V ( α 1 , … , α l ) (\alpha_1, \ldots, \alpha_l) ( α 1 , … , α l ) R R R A l A_l A l R R R ε i − ε j \varepsilon_i - \varepsilon_j ε i − ε j i ≠ j , 1 ≤ i ≤ l + 1 , 1 ≤ j ≤ l + 1 i \neq j, 1 \leq i \leq l+1, 1 \leq j \leq l+1 i = j , 1 ≤ i ≤ l + 1 , 1 ≤ j ≤ l + 1 n = l ( l + 1 ) n = l(l+1) n = l ( l + 1 ) ε i − ε j \varepsilon_i - \varepsilon_j ε i − ε j i < j i < j i < j (III) 我们有h = n / l = l + 1 h = n/l = l+1 h = n / l = l + 1
(IV)设α ~ = ε 1 − ε l + 1 = α 1 + α 2 + ⋯ + α l \tilde{\alpha} = \varepsilon_1 - \varepsilon_{l+1} = \alpha_1 + \alpha_2 + \cdots + \alpha_l α ~ = ε 1 − ε l + 1 = α 1 + α 2 + ⋯ + α l ( α i ) (\alpha_i) ( α i ) l = h − 1 l = h-1 l = h − 1 α ~ \tilde{\alpha} α ~ l = 1 l = 1 l = 1 α ~ = α 1 \tilde{\alpha} = \alpha_1 α ~ = α 1 ( α ~ ∣ α 1 ) = 2 (\tilde{\alpha}|\alpha_1) = 2 ( α ~ ∣ α 1 ) = 2 W a ( R ) W_a(R) W a ( R )
∞ α 1
\begin{array}{c}
\infty \\
\alpha_1
\end{array}
∞ α 1 l ≥ 2 l \geq 2 l ≥ 2 ( α ~ ∣ α i ) = 0 (\tilde{\alpha}|\alpha_i) = 0 ( α ~ ∣ α i ) = 0 0 < i < l 0 < i < l 0 < i < l ( α ~ ∣ α 1 ) = ( α ~ ∣ α l ) = 1 (\tilde{\alpha}|\alpha_1) = (\tilde{\alpha}|\alpha_l) = 1 ( α ~ ∣ α 1 ) = ( α ~ ∣ α l ) = 1
!丁金图
(V) 使用标量乘积识别V及其对偶，我们对所有α ∈ R \alpha \in \mathbf{R} α ∈ R α ~ = 2 α ( α ∣ α ) = α \tilde{\alpha} = \frac{2\alpha}{(\alpha|\alpha)} = \alpha α ~ = ( α ∣ α ) 2 α = α R ∗ = R R^* = R R ∗ = R
对于形式Φ R \Phi_R Φ R h − 1 / 2 = ( l + 1 ) − 1 / 2 h^{-1/2} = (l+1)^{-1/2} h − 1/2 = ( l + 1 ) − 1/2 Φ r ( x , y ) = ( x ∣ y ) / 2 ( l + 1 ) \Phi_r(x, y) = (x|y)/2(l+1) Φ r ( x , y ) = ( x ∣ y ) /2 ( l + 1 )
我们有 γ ( R ) = ( l + 1 ) 2 \gamma(R) = (l+1)^2 γ ( R ) = ( l + 1 ) 2
(VI) 设( ω i ) 1 ≤ i ≤ l (\omega_i)_{1 \leq i \leq l} ( ω i ) 1 ≤ i ≤ l
ω i = ∑ j = 1 l + 1 ξ i j ε j . with ξ i j ∈ R .
\omega_i = \sum_{j=1}^{l+1} \xi_{ij} \varepsilon_j. \quad \text{with } \xi_{ij} \in \mathbf{R}.
ω i = j = 1 ∑ l + 1 ξ ij ε j . with ξ ij ∈ R . 条件 ( ω i ∣ α j ) = δ i j (\omega_i|\alpha_j) = \delta_{ij} ( ω i ∣ α j ) = δ ij ω i ∈ V \omega_i \in V ω i ∈ V
ξ i i − ξ i , i + 1 = 1 , ξ i j − ξ i , j + 1 = 0 for j ≠ i . ∑ j = 1 l + 1 ξ i j = 0 ,
\xi_{ii} - \xi_{i,i+1} = 1, \quad \xi_{ij} - \xi_{i,j+1} = 0 \text{ for } j \neq i. \quad \sum_{j=1}^{l+1} \xi_{ij} = 0,
ξ ii − ξ i , i + 1 = 1 , ξ ij − ξ i , j + 1 = 0 for j = i . j = 1 ∑ l + 1 ξ ij = 0 , 从而容易导致
ω i = ε 1 + ⋯ + ε i − i l + 1 ( ε 1 + ⋯ + ε l + 1 ) = 1 l + 1 ( ( l − i + 1 ) ( α 1 + 2 α 2 + ⋯ + ( i − 1 ) α i − 1 ) + i ( ( l − i + 1 ) α i + ( l − i ) α i + 1 + ⋯ + α l ) ) .
\begin{align*}
\omega_i &= \varepsilon_1 + \cdots + \varepsilon_i - \frac{i}{l+1}(\varepsilon_1 + \cdots + \varepsilon_{l+1}) \\
&= \frac{1}{l+1}((l-i+1)(\alpha_1 + 2\alpha_2 + \cdots + (i-1)\alpha_{i-1}) \\
&\qquad + i((l-i+1)\alpha_i + (l-i)\alpha_{i+1} + \cdots + \alpha_l)).
\end{align*}
ω i = ε 1 + ⋯ + ε i − l + 1 i ( ε 1 + ⋯ + ε l + 1 ) = l + 1 1 (( l − i + 1 ) ( α 1 + 2 α 2 + ⋯ + ( i − 1 ) α i − 1 ) + i (( l − i + 1 ) α i + ( l − i ) α i + 1 + ⋯ + α l )) . (VII) 和的正根是
2 ρ = l ε 1 + ( l − 2 ) ε 2 + ( l − 4 ) ε 3 + ⋯ − ( l − 2 ) ε l − l ε l + 1 = l α 1 + 2 ( l − 1 ) α 2 + ⋯ + i ( l − i + 1 ) α i + ⋯ + l α l .
\begin{align*}
2\rho &= l\varepsilon_1 + (l-2)\varepsilon_2 + (l-4)\varepsilon_3 + \cdots - (l-2)\varepsilon_l - l\varepsilon_{l+1} \\
&= l\alpha_1 + 2(l-1)\alpha_2 + \cdots + i(l-i+1)\alpha_i + \cdots + l\alpha_l.
\end{align*}
2 ρ = l ε 1 + ( l − 2 ) ε 2 + ( l − 4 ) ε 3 + ⋯ − ( l − 2 ) ε l − l ε l + 1 = l α 1 + 2 ( l − 1 ) α 2 + ⋯ + i ( l − i + 1 ) α i + ⋯ + l α l . (VIII) 引入no. 4的E = R l + 1 E = \mathbf{R}^{l+1} E = R l + 1 L 0 L_0 L 0 p p p E E E V V V
Q ( R ) = Q ( R ′ ) ∩ V ⊄ L 0 ∩ V , and P ( R ) = p ( P ( R ′ ) ) ;
Q(R) = Q(R') \cap V \not\subset L_0 \cap V, \quad \text{and} \quad P(R) = p(P(R'));
Q ( R ) = Q ( R ′ ) ∩ V ⊂ L 0 ∩ V , and P ( R ) = p ( P ( R ′ )) ; 鉴于R ′ R' R ′ V V V P ( R ) = p ( Q ( R ′ ) ) = p ( L 0 ) P(R) = p(Q(R')) = p(L_0) P ( R ) = p ( Q ( R ′ )) = p ( L 0 ) P ( R ) P(R) P ( R ) ε i − ε j \varepsilon_i - \varepsilon_j ε i − ε j p ( ε 1 ) = ε 1 − ( l + 1 ) − 1 ∑ i = 1 l + 1 ε i p(\varepsilon_1) = \varepsilon_1 - (l+1)^{-1} \sum_{i=1}^{l+1} \varepsilon_i p ( ε 1 ) = ε 1 − ( l + 1 ) − 1 ∑ i = 1 l + 1 ε i
P ( R ) = Q ( R ) + Z ( ε 1 − ( l + 1 ) − 1 ∑ i = 1 l + 1 ε i ) .
P(R) = Q(R) + \mathbf{Z}(\varepsilon_1 - (l+1)^{-1} \sum_{i=1}^{l+1} \varepsilon_i).
P ( R ) = Q ( R ) + Z ( ε 1 − ( l + 1 ) − 1 i = 1 ∑ l + 1 ε i ) . 现在l + 1 l + 1 l + 1 m > 0 m > 0 m > 0 m p ( ε 1 ) ∈ Q ( R ) mp(\varepsilon_1) \in Q(R) m p ( ε 1 ) ∈ Q ( R ) P ( R ) / Q ( R ) P(R)/Q(R) P ( R ) / Q ( R ) Z / ( l + 1 ) Z \mathbf{Z}/(l+1)\mathbf{Z} Z / ( l + 1 ) Z l + 1 l + 1 l + 1
(IX) 和 (X) 对于V V V g g g φ ( g ) \varphi(g) φ ( g ) E E E g g g ε 1 + ε 2 + ⋯ + ε l \varepsilon_1 + \varepsilon_2 + \cdots + \varepsilon_l ε 1 + ε 2 + ⋯ + ε l g g g s ε i − ε j ∣ V s_{\varepsilon_i - \varepsilon_j}|V s ε i − ε j ∣ V φ ( g ) \varphi(g) φ ( g ) s ε i − ε j s_{\varepsilon_i - \varepsilon_j} s ε i − ε j ε i \varepsilon_i ε i ε j \varepsilon_j ε j ε k \varepsilon_k ε k k k k i i i j j j
X = { ε 1 , ε 2 , … , ε l + 1 } .
X = \{\varepsilon_1, \varepsilon_2, \ldots, \varepsilon_{l+1}\}.
X = { ε 1 , ε 2 , … , ε l + 1 } . 那么从W ( R ) W(R) W ( R ) X X X g ↦ φ ( g ) ∣ X g \mapsto \varphi(g)|X g ↦ φ ( g ) ∣ X W ( R ) W(R) W ( R ) S l + 1 S_{l+1} S l + 1 ( l + 1 ) ! (l+1)! ( l + 1 )!
对称的代数S ( E ) S(E) S ( E ) E E E P ( ξ 1 , ξ 2 , … , ξ l + 1 ) P(\xi_1, \xi_2, \ldots, \xi_{l+1}) P ( ξ 1 , ξ 2 , … , ξ l + 1 ) G = φ ( W ( R ) ) G = \varphi(W(R)) G = φ ( W ( R )) G G G S ( E ) S(E) S ( E ) S ( E ) G S(E)^G S ( E ) G
s i ′ = ∑ τ ∈ S l + 1 ξ τ ( 1 ) ξ τ ( 2 ) ⋯ ξ τ ( i ) ( 1 ≤ i ≤ l + 1 ) .
s'_i = \sum_{\tau \in S_{l+1}} \xi_{\tau(1)} \xi_{\tau(2)} \cdots \xi_{\tau(i)} \quad (1 \leq i \leq l+1).
s i ′ = τ ∈ S l + 1 ∑ ξ τ ( 1 ) ξ τ ( 2 ) ⋯ ξ τ ( i ) ( 1 ≤ i ≤ l + 1 ) . 代数S ( V ) S(V) S ( V ) E E E V V V P ∈ S ( E ) G P \in S(E)^G P ∈ S ( E ) G P P P V V V W ( R ) W(R) W ( R ) Q ∈ S ( V ) W ( R ) Q \in S(V)^{W(R)} Q ∈ S ( V ) W ( R ) P ∈ S ( E ) P \in S(E) P ∈ S ( E ) Q Q Q P P P ( ( l + 1 ) ! ) − 1 ∑ g ∈ G g ( P ) ((l+1)!)^{-1} \sum_{g \in G} g(P) (( l + 1 )! ) − 1 ∑ g ∈ G g ( P ) P P P V V V P ∈ S ( E ) G P \in S(E)^G P ∈ S ( E ) G S ( V ) W ( R ) S(V)^{W(R)} S ( V ) W ( R ) s i = s i ′ ∣ V s_i = s'_i|V s i = s i ′ ∣ V s 1 = 0 s_1 = 0 s 1 = 0 S ( V ) W ( R ) S(V)^{W(R)} S ( V ) W ( R ) R \mathbf{R} R l l l s i s_i s i 2 ≤ i ≤ l + 1 2 \leq i \leq l+1 2 ≤ i ≤ l + 1 s i s_i s i 2 , 3 , … , l + 1 2, 3, \ldots, l+1 2 , 3 , … , l + 1 W ( R ) W(R) W ( R )
1 , 2 , 3 , … , l .
1, 2, 3, \ldots, l.
1 , 2 , 3 , … , l . (XI) 适用于 l = 1 l = 1 l = 1 A ( R ) = W ( R ) = Z / 2 Z A(R) = W(R) = \mathbf{Z}/2\mathbf{Z} A ( R ) = W ( R ) = Z /2 Z w 0 = − 1 w_0 = -1 w 0 = − 1
对于l ≥ 2 l \geq 2 l ≥ 2 ε ∈ A ( R ) \varepsilon \in A(R) ε ∈ A ( R ) α i \alpha_i α i α l + 1 − i \alpha_{l+1-i} α l + 1 − i ε \varepsilon ε A ( R ) / W ( R ) A(R)/W(R) A ( R ) / W ( R ) Z / 2 Z \mathbf{Z}/2\mathbf{Z} Z /2 Z − 1 -1 − 1 A ( R ) A(R) A ( R ) W ( R ) W(R) W ( R ) A ( R ) A(R) A ( R ) W ( R ) × Z / 2 Z W(R) \times \mathbf{Z}/2\mathbf{Z} W ( R ) × Z /2 Z w 0 = − ε w_0 = -\varepsilon w 0 = − ε
(XII)群P ( R − ) / Q ( R − ) P(R^-)/Q(R^-) P ( R − ) / Q ( R − ) l + 1 l+1 l + 1 l ≥ 2 l \geq 2 l ≥ 2 A ( R ) / W ( R ) A(R)/W(R) A ( R ) / W ( R ) x ↦ − x x \mapsto -x x ↦ − x P ( R − ) / Q ( R − ) P(R^-)/Q(R^-) P ( R − ) / Q ( R − )
8. 系统型 D l D_l D l l ≥ 3 l \geq 3 l ≥ 3
（I）考虑V = R ′ V = \mathbf{R}' V = R ′ L 0 L_0 L 0 α ∈ L 0 \alpha \in L_0 α ∈ L 0 R R R ( α ∣ α ) = 2 (\alpha|\alpha) = 2 ( α ∣ α ) = 2 ± ε i ± ε j \pm \varepsilon_i \pm \varepsilon_j ± ε i ± ε j 1 ≤ i < j ≤ l 1 \leq i < j \leq l 1 ≤ i < j ≤ l α , β ∈ R \alpha, \beta \in R α , β ∈ R R R R V V V 2 ( α ∣ β ) / ( α ∣ α ) ∈ Z 2(\alpha|\beta)/(\alpha|\alpha) \in \mathbf{Z} 2 ( α ∣ β ) / ( α ∣ α ) ∈ Z R R R V V V n = 2 l ( l − 1 ) n = 2l(l-1) n = 2 l ( l − 1 )
(II) 放
α 1 = ε 1 − ε 2 , α 2 = ε 2 − ε 3 … … α l − 1 = ε l − 1 − ε l , α l = ε l − 1 + ε l .
\alpha_1 = \varepsilon_1 - \varepsilon_2,\ \alpha_2 = \varepsilon_2 - \varepsilon_3 \ldots \ldots \alpha_{l-1} = \varepsilon_{l-1} - \varepsilon_l,\ \alpha_l = \varepsilon_{l-1} + \varepsilon_l.
α 1 = ε 1 − ε 2 , α 2 = ε 2 − ε 3 …… α l − 1 = ε l − 1 − ε l , α l = ε l − 1 + ε l . 以下公式为直接的：
ε i − ε j = α i + α i + 1 + ⋯ + α j − 1 ( i < j )
\varepsilon_i - \varepsilon_j = \alpha_i + \alpha_{i+1} + \cdots + \alpha_{j-1}\quad (i < j)
ε i − ε j = α i + α i + 1 + ⋯ + α j − 1 ( i < j ) ε i + ε j = α i + α i + 1 + ⋯ + α j − 1 + 2 α j + 2 α j + 1 + ⋯
\varepsilon_i + \varepsilon_j = \alpha_i + \alpha_{i+1} + \cdots + \alpha_{j-1} + 2\alpha_j + 2\alpha_{j+1} + \cdots
ε i + ε j = α i + α i + 1 + ⋯ + α j − 1 + 2 α j + 2 α j + 1 + ⋯ ⋯ + 2 α l − 2 + α l − 1 + α l ( i < j ≤ l − 2 )
\cdots + 2\alpha_{l-2} + \alpha_{l-1} + \alpha_l\quad (i < j \leq l-2)
⋯ + 2 α l − 2 + α l − 1 + α l ( i < j ≤ l − 2 ) ε i + ε l − 1 = α i + α i + 1 + ⋯ + α l ( i < l − 1 )
\varepsilon_i + \varepsilon_{l-1} = \alpha_i + \alpha_{i+1} + \cdots + \alpha_l\quad (i < l-1)
ε i + ε l − 1 = α i + α i + 1 + ⋯ + α l ( i < l − 1 ) ε i + ε l = α i + α i + 1 + ⋯ + α l − 2 + α l ( i < l − 1 )
\varepsilon_i + \varepsilon_l = \alpha_i + \alpha_{i+1} + \cdots + \alpha_{l-2} + \alpha_l\quad (i < l-1)
ε i + ε l = α i + α i + 1 + ⋯ + α l − 2 + α l ( i < l − 1 ) ε l − 1 + ε l = α l .
\varepsilon_{l-1} + \varepsilon_l = \alpha_l.
ε l − 1 + ε l = α l . 所以如果 R R R ( α 1 , … , α l ) (\alpha_1, \ldots, \alpha_l) ( α 1 , … , α l ) ∥ α i ∥ 2 = 2 \| \alpha_i \| ^2 = 2 ∥ α i ∥ 2 = 2 i i i ( α i ∣ α j ) = 0 (\alpha_i|\alpha_j) = 0 ( α i ∣ α j ) = 0 i + 1 < j i + 1 < j i + 1 < j i = l − 2 , j = l i = l-2, j = l i = l − 2 , j = l ( α l − 2 ∣ α l ) = − 1 (\alpha_{l-2}|\alpha_l) = -1 ( α l − 2 ∣ α l ) = − 1 ( α i ∣ α i + 1 ) = − 1 (\alpha_i|\alpha_{i+1}) = -1 ( α i ∣ α i + 1 ) = − 1 i ≤ l − 2 i \leq l-2 i ≤ l − 2 ( α l − 1 ∣ α l ) = − 1 (\alpha_{l-1}|\alpha_l) = -1 ( α l − 1 ∣ α l ) = − 1 R R R D l D_l D l i < j i < j i < j ε i ± ε j \varepsilon_i \pm \varepsilon_j ε i ± ε j
(III) 我们有h = n / l = 2 ( l − 1 ) h = n/l = 2(l-1) h = n / l = 2 ( l − 1 )
(IV)设α ~ = ε 1 + ε 2 = α 1 + 2 α 2 + ⋯ + 2 α l − 2 + α l − 1 + α l \tilde{\alpha} = \varepsilon_1 + \varepsilon_2 = \alpha_1 + 2\alpha_2 + \cdots + 2\alpha_{l-2} + \alpha_{l-1} + \alpha_l α ~ = ε 1 + ε 2 = α 1 + 2 α 2 + ⋯ + 2 α l − 2 + α l − 1 + α l ( α i ) (\alpha_i) ( α i )
2 l − 3 = h − 1.
2l-3 = h-1.
2 l − 3 = h − 1. 因此α ~ \tilde{\alpha} α ~
如果l = 3 l = 3 l = 3
( α ~ ∣ α 1 ) = 0 , ( α ~ ∣ α 2 ) = ( α ~ ∣ α 3 ) = 1.
(\tilde{\alpha}|\alpha_1) = 0,\quad (\tilde{\alpha}|\alpha_2) = (\tilde{\alpha}|\alpha_3) = 1.
( α ~ ∣ α 1 ) = 0 , ( α ~ ∣ α 2 ) = ( α ~ ∣ α 3 ) = 1. 如果l ≥ 4 l \geq 4 l ≥ 4 ( α ~ ∣ α i ) = 0 (\tilde{\alpha}|\alpha_i) = 0 ( α ~ ∣ α i ) = 0 i ≠ 2 i \neq 2 i = 2 ( α ~ ∣ α 2 ) = 1 (\tilde{\alpha}|\alpha_2) = 1 ( α ~ ∣ α 2 ) = 1
!D_l 的 Dynkin 图
(V) 从( α ∣ α ) = 2 (\alpha|\alpha) = 2 ( α ∣ α ) = 2 α ∉ R \alpha \notin R α ∈ / R R − = R R^- = R R − = R Φ L ′ \Phi_{L'} Φ L ′ h − 1 / 2 = ( 2 l − 2 ) − 1 / 2 h^{-1/2} = (2l-2)^{-1/2} h − 1/2 = ( 2 l − 2 ) − 1/2
Φ R ( x , y ) = ( x ∣ y ) / ( 4 l − 4 ) and γ ( R ) = 4 ( l − 1 ) 2 .
\Phi_R(x, y) = (x|y)/(4l-4)\quad \text{and}\quad \gamma(R) = 4(l-1)^2.
Φ R ( x , y ) = ( x ∣ y ) / ( 4 l − 4 ) and γ ( R ) = 4 ( l − 1 ) 2 . (VI) 类似于 no. 7 中的计算给出基本权：
ω i = ε 1 + ε 2 + ⋯ + ε i = α 1 + 2 α 2 + ⋯ + ( i − 1 ) α i − 1 + i ( α i + α i + 1 + ⋯ + α l − 2 ) + 1 2 i ( α l − 1 + α l )
\omega_i = \varepsilon_1 + \varepsilon_2 + \cdots + \varepsilon_i \\
= \alpha_1 + 2\alpha_2 + \cdots + (i-1)\alpha_{i-1} + i(\alpha_i + \alpha_{i+1} + \cdots + \alpha_{l-2}) \\
\quad + \frac{1}{2}i(\alpha_{l-1} + \alpha_l)
ω i = ε 1 + ε 2 + ⋯ + ε i = α 1 + 2 α 2 + ⋯ + ( i − 1 ) α i − 1 + i ( α i + α i + 1 + ⋯ + α l − 2 ) + 2 1 i ( α l − 1 + α l ) 对于i < l − 1 i < l - 1 i < l − 1
ω l − 1 = 1 2 ( ε 1 + ε 2 + ⋯ + ε l − 2 + ε l − 1 − ε l ) = 1 2 ( α 1 + 2 α 2 + ⋯ + ( l − 2 ) α l − 2 + 1 2 l α l − 1 + 1 2 ( l − 2 ) α l ) ,
\omega_{l-1} = \frac{1}{2}(\varepsilon_1 + \varepsilon_2 + \cdots + \varepsilon_{l-2} + \varepsilon_{l-1} - \varepsilon_l) \\
= \frac{1}{2}(\alpha_1 + 2\alpha_2 + \cdots + (l-2)\alpha_{l-2} + \frac{1}{2}l\alpha_{l-1} + \frac{1}{2}(l-2)\alpha_l),
ω l − 1 = 2 1 ( ε 1 + ε 2 + ⋯ + ε l − 2 + ε l − 1 − ε l ) = 2 1 ( α 1 + 2 α 2 + ⋯ + ( l − 2 ) α l − 2 + 2 1 l α l − 1 + 2 1 ( l − 2 ) α l ) , ω l = 1 2 ( ε 1 + ε 2 + ⋯ + ε l − 2 + ε l − 1 + ε l ) = 1 2 ( α 1 + 2 α 2 + ⋯ + ( l − 2 ) α l − 2 + 1 2 ( l − 2 ) α l − 1 + 1 2 l α l ) .
\omega_l = \frac{1}{2}(\varepsilon_1 + \varepsilon_2 + \cdots + \varepsilon_{l-2} + \varepsilon_{l-1} + \varepsilon_l) \\
= \frac{1}{2}(\alpha_1 + 2\alpha_2 + \cdots + (l-2)\alpha_{l-2} + \frac{1}{2}(l-2)\alpha_{l-1} + \frac{1}{2}l\alpha_l).
ω l = 2 1 ( ε 1 + ε 2 + ⋯ + ε l − 2 + ε l − 1 + ε l ) = 2 1 ( α 1 + 2 α 2 + ⋯ + ( l − 2 ) α l − 2 + 2 1 ( l − 2 ) α l − 1 + 2 1 l α l ) . (VII) 和的正根是
2 ρ = 2 ( l − 1 ) ε 1 + 2 ( l − 2 ) ε 2 + ⋯ + 2 ε l − 1 = ∑ i = 1 l − 2 2 ( i ( l − i ( i + 1 ) 2 ) ) α i + l ( l − 1 ) 2 ( α l − 1 + α l ) .
2\rho = 2(l-1)\varepsilon_1 + 2(l-2)\varepsilon_2 + \cdots + 2\varepsilon_{l-1} \\
= \sum_{i=1}^{l-2} 2(i(l-\frac{i(i+1)}{2}))\alpha_i + \frac{l(l-1)}{2}(\alpha_{l-1} + \alpha_l).
2 ρ = 2 ( l − 1 ) ε 1 + 2 ( l − 2 ) ε 2 + ⋯ + 2 ε l − 1 = i = 1 ∑ l − 2 2 ( i ( l − 2 i ( i + 1 ) )) α i + 2 l ( l − 1 ) ( α l − 1 + α l ) . (VIII) ± ε i ± ε j \pm \varepsilon_i \pm \varepsilon_j ± ε i ± ε j L 1 L_1 L 1 Q ( R ) = L 1 Q(R) = L_1 Q ( R ) = L 1 Q ( R ∗ ) = L 1 Q(R^*) = L_1 Q ( R ∗ ) = L 1 P ( R ) = L 2 P(R) = L_2 P ( R ) = L 2 l l l P ( R ) / Q ( R ) P(R)/Q(R) P ( R ) / Q ( R ) Z / 4 Z \mathbf{Z}/4\mathbf{Z} Z /4 Z l l l Z / 2 Z × Z / 2 Z \mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/2\mathbf{Z} Z /2 Z × Z /2 Z P ( R ) / Q ( R ) P(R)/Q(R) P ( R ) / Q ( R ) ω l \omega_l ω l ω l − 1 \omega_{l-1} ω l − 1 P ( R ) / Q ( R ) P(R)/Q(R) P ( R ) / Q ( R ) ω l − 1 \omega_{l-1} ω l − 1 ω l \omega_l ω l
(IX) 和 (X) 在 R l \mathbf{R}^l R l s ε i , − ε j s_{\varepsilon_i, -\varepsilon_j} s ε i , − ε j i ≠ j i \neq j i = j ε i \varepsilon_i ε i ε j \varepsilon_j ε j ε k \varepsilon_k ε k k k k i i i j j j s ε i , − ε j s_{\varepsilon_i, -\varepsilon_j} s ε i , − ε j G 1 G_1 G 1 S l S_l S l s i j = s ε i , − ε j s ε i , + ε j s_{ij} = s_{\varepsilon_i, -\varepsilon_j} s_{\varepsilon_i, +\varepsilon_j} s ij = s ε i , − ε j s ε i , + ε j ε i \varepsilon_i ε i − ε i -\varepsilon_i − ε i ε j \varepsilon_j ε j − ε j -\varepsilon_j − ε j ε k \varepsilon_k ε k k k k i i i j j j R l \mathbf{R}^l R l u u u s i j s_{ij} s ij G 2 G_2 G 2 u ( ε i ) = ( − 1 ) ν i ε i u(\varepsilon_i) = (-1)^{\nu_i} \varepsilon_i u ( ε i ) = ( − 1 ) ν i ε i ∏ i = 1 l ( − 1 ) ν i = 1 \prod_{i=1}^l (-1)^{\nu_i} = 1 ∏ i = 1 l ( − 1 ) ν i = 1 G 2 G_2 G 2 ( Z / 2 Z ) l − 1 (\mathbf{Z}/2\mathbf{Z})^{l-1} ( Z /2 Z ) l − 1 G 2 G_2 G 2 W ( R ) W(R) W ( R ) W ( R ) W(R) W ( R ) S l S_l S l ( Z / 2 Z ) l − 1 (\mathbf{Z}/2\mathbf{Z})^{l-1} ( Z /2 Z ) l − 1 2 l − 1 l ! 2^{l-1} l! 2 l − 1 l !
no. 5的多项式函数t i t_i t i W ( R ) W(R) W ( R ) t = ξ 1 ξ 2 … ξ l t = \xi_1 \xi_2 \ldots \xi_l t = ξ 1 ξ 2 … ξ l t l = t 2 t_l = t^2 t l = t 2 P ( ξ 1 , … , ξ l ) P(\xi_1, \ldots, \xi_l) P ( ξ 1 , … , ξ l ) W ( R ) W(R) W ( R ) ξ 1 ν 1 ξ 2 ν 2 … ξ l ν l \xi_1^{\nu_1} \xi_2^{\nu_2} \ldots \xi_l^{\nu_l} ξ 1 ν 1 ξ 2 ν 2 … ξ l ν l P P P ν i \nu_i ν i ν j \nu_j ν j j j j ( − 1 ) ν i + ν j ξ 1 ν 1 ξ 2 ν 2 … ξ l ν l (-1)^{\nu_i + \nu_j} \xi_1^{\nu_1} \xi_2^{\nu_2} \ldots \xi_l^{\nu_l} ( − 1 ) ν i + ν j ξ 1 ν 1 ξ 2 ν 2 … ξ l ν l s i j ( P ) s_{ij}(P) s ij ( P ) ν i + ν j ≡ 0 ( m o d 2 ) \nu_i + \nu_j \equiv 0 \pmod{2} ν i + ν j ≡ 0 ( mod 2 ) ν j ≡ 1 ( m o d 2 ) \nu_j \equiv 1 \pmod{2} ν j ≡ 1 ( mod 2 )
P = P_1 + tP_2，其中 P_1 和 P_2 中的所有单项式都只有偶数指数。由于 P 在 ξ i \xi_i ξ i t 1 , t 2 , … , t l t_1, t_2, \ldots, t_l t 1 , t 2 , … , t l S ( R l ) W ( R ) S(\mathbf{R}^l)^{W(R)} S ( R l ) W ( R ) t 1 , t 2 , … , t l − 1 , t t_1, t_2, \ldots, t_{l-1}, t t 1 , t 2 , … , t l − 1 , t S ( R l ) W ( R ) S(\mathbf{R}^l)^{W(R)} S ( R l ) W ( R ) t 1 , t 2 , … , t l − 1 , t t_1, t_2, \ldots, t_{l-1}, t t 1 , t 2 , … , t l − 1 , t
1 , 3 , 5 , … , 2 l − 5 , 2 l − 3 , l − 1.
1, 3, 5, \ldots, 2l - 5, 2l - 3, l - 1.
1 , 3 , 5 , … , 2 l − 5 , 2 l − 3 , l − 1. Note that l − 1 l - 1 l − 1 l l l l l l
Dynkin 图的 (XI) 自同构是底层图的那些。因此： 1) 若l = 3 l = 3 l = 3 A ( R ) / W ( R ) A(R)/W(R) A ( R ) / W ( R ) Z / 2 Z \mathbf{Z}/2\mathbf{Z} Z /2 Z l = 4 l = 4 l = 4 A ( R ) / W ( R ) A(R)/W(R) A ( R ) / W ( R ) S 3 \mathfrak{S}_3 S 3 l ≥ 5 l \geq 5 l ≥ 5 l − 3 ≥ 2 l - 3 \geq 2 l − 3 ≥ 2 ε ∈ A ( R ) \varepsilon \in A(R) ε ∈ A ( R ) α l − 1 \alpha_{l-1} α l − 1 α l \alpha_l α l α i \alpha_i α i 1 ≤ i ≤ l − 2 1 \leq i \leq l - 2 1 ≤ i ≤ l − 2 A ( R ) / W ( R ) A(R)/W(R) A ( R ) / W ( R ) Z / 2 Z \mathbf{Z}/2\mathbf{Z} Z /2 Z A ( R ) A(R) A ( R ) R l \mathbf{R}^l R l u u u G 3 G_3 G 3 G 1 ≅ S l G_1 \cong \mathfrak{S}_l G 1 ≅ S l u ( ε i ) = ± ε i u(\varepsilon_i) = \pm \varepsilon_i u ( ε i ) = ± ε i i i i
如果l l l − 1 ∈ W ( R ) -1 \in W(R) − 1 ∈ W ( R ) w 0 = − 1 w_0 = -1 w 0 = − 1 l l l − 1 ∉ W ( R ) -1 \notin W(R) − 1 ∈ / W ( R ) A ( R ) = W ( R ) × { 1 , − 1 } A(R) = W(R) \times \{1, -1\} A ( R ) = W ( R ) × { 1 , − 1 } w 0 = − ε w_0 = -\varepsilon w 0 = − ε
(XII) 对于l l l P ( R − ) / Q ( R − ) P(R^-)/Q(R^-) P ( R − ) / Q ( R − ) ω 1 , ω l − 1 \omega_1, \omega_{l-1} ω 1 , ω l − 1 ω l \omega_l ω l ω l \omega_l ω l ω l − 1 \omega_{l-1} ω l − 1 α 0 \alpha_0 α 0 α l \alpha_l α l α l − 1 \alpha_{l-1} α l − 1 α 1 \alpha_1 α 1 α l − 1 \alpha_{l-1} α l − 1 α l \alpha_l α l α j \alpha_j α j α l − j \alpha_{l-j} α l − j
2 ≤ j ≤ l − 2.
2 \leq j \leq l - 2.
2 ≤ j ≤ l − 2. 我们有ω 1 = ω l ω l − 1 \omega_1 = \omega_l \omega_{l-1} ω 1 = ω l ω l − 1
对于l l l P ( R − ) / Q ( R − ) P(R^-)/Q(R^-) P ( R − ) / Q ( R − ) ω l − 1 \omega_{l-1} ω l − 1 ω l \omega_l ω l ω 1 \omega_1 ω 1 ω 1 \omega_1 ω 1 α 0 \alpha_0 α 0 α 1 \alpha_1 α 1 2 ≤ j ≤ l − 2 2 \leq j \leq l - 2 2 ≤ j ≤ l − 2 α j \alpha_j α j ω l \omega_l ω l α 0 \alpha_0 α 0 α l \alpha_l α l α 1 \alpha_1 α 1 α l − 1 \alpha_{l-1} α l − 1 α l \alpha_l α l α 1 \alpha_1 α 1 α l − 1 \alpha_{l-1} α l − 1 α 0 \alpha_0 α 0 α j \alpha_j α j α l − j \alpha_{l-j} α l − j 2 ≤ j ≤ l − 2 2 \leq j \leq l - 2 2 ≤ j ≤ l − 2 ω 1 = ω l 2 \omega_1 = \omega_l^2 ω 1 = ω l 2 ω l − 1 = ω l 3 \omega_{l-1} = \omega_l^3 ω l − 1 = ω l 3
对于l ≠ 4 l \neq 4 l = 4 A ( R ) / W ( R ) A(R)/W(R) A ( R ) / W ( R ) α l − 1 \alpha_{l-1} α l − 1 α l \alpha_l α l P ( R − ) / Q ( R − ) P(R^-)/Q(R^-) P ( R − ) / Q ( R − ) ω l − 1 \omega_{l-1} ω l − 1 ω l \omega_l ω l l l l P ( R − ) / Q ( R − ) P(R^-)/Q(R^-) P ( R − ) / Q ( R − ) x ↦ − x x \mapsto -x x ↦ − x
对于l = 4 l = 4 l = 4 A ( R ) / W ( R ) A(R)/W(R) A ( R ) / W ( R ) { 1 , 3 , 4 } \{1, 3, 4\} { 1 , 3 , 4 } { ω 1 , ω 3 , ω 4 } \{\omega_1, \omega_3, \omega_4\} { ω 1 , ω 3 , ω 4 }
9. 系统型 F 4 F_4 F 4
(I) 考虑R 4 \mathbf{R}^4 R 4 L 2 L_2 L 2 R R R α ∈ L 2 \alpha \in L_2 α ∈ L 2 ( α ∣ α ) = 1 (\alpha|\alpha) = 1 ( α ∣ α ) = 1 ( α ∣ α ) = 2 (\alpha|\alpha) = 2 ( α ∣ α ) = 2
± ε i , ± ε i ± ε j ( i < j ) , 1 2 ( ± ε 1 ± ε 2 ± ε 3 ± ε 4 ) .
\pm \varepsilon_i,\quad \pm \varepsilon_i \pm \varepsilon_j\ (i < j),\quad \frac{1}{2}(\pm \varepsilon_1 \pm \varepsilon_2 \pm \varepsilon_3 \pm \varepsilon_4).
± ε i , ± ε i ± ε j ( i < j ) , 2 1 ( ± ε 1 ± ε 2 ± ε 3 ± ε 4 ) . α ∈ R \alpha \in R α ∈ R α \alpha α 0 , ± 1 2 , ± 1 0, \pm \frac{1}{2}, \pm 1 0 , ± 2 1 , ± 1 ( 3 2 ) 2 > 2 (\frac{3}{2})^2 > 2 ( 2 3 ) 2 > 2 ± ε i , ± ε i ± ε j \pm \varepsilon_i, \pm \varepsilon_i \pm \varepsilon_j ± ε i , ± ε i ± ε j ± 1 2 \pm \frac{1}{2} ± 2 1 1 2 ( ± ε 1 ± ε 2 ± ε 3 ± ε 4 ) \frac{1}{2}(\pm \varepsilon_1 \pm \varepsilon_2 \pm \varepsilon_3 \pm \varepsilon_4) 2 1 ( ± ε 1 ± ε 2 ± ε 3 ± ε 4 ) 我们证明，对于 α , β ∈ R \alpha, \beta \in R α , β ∈ R 2 ( α ∣ β ) / ( ∣ α ∣ α ) ∈ Z 2(\alpha|\beta)/(|\alpha|\alpha) \in \mathbf{Z} 2 ( α ∣ β ) / ( ∣ α ∣ α ) ∈ Z α = ± ε i \alpha = \pm \varepsilon_i α = ± ε i α = 1 2 ( ± ε 1 ± ε 2 ± ε 3 ± ε 4 ) \alpha = \frac{1}{2}(\pm \varepsilon_1 \pm \varepsilon_2 \pm \varepsilon_3 \pm \varepsilon_4) α = 2 1 ( ± ε 1 ± ε 2 ± ε 3 ± ε 4 ) ( α ∣ α ) = 1 (\alpha|\alpha) = 1 ( α ∣ α ) = 1 α , β ∈ L 2 \alpha, \beta \in L_2 α , β ∈ L 2 ( α ∣ β ) ∈ 1 2 Z (\alpha|\beta) \in \frac{1}{2}\mathbf{Z} ( α ∣ β ) ∈ 2 1 Z α = ± ε i ± ε j \alpha = \pm \varepsilon_i \pm \varepsilon_j α = ± ε i ± ε j ( α ∣ α ) = 2 (\alpha|\alpha) = 2 ( α ∣ α ) = 2 α ∈ L 1 \alpha \in L_1 α ∈ L 1 β ∈ L 2 \beta \in L_2 β ∈ L 2 ( α ∣ β ) ∈ Z (\alpha|\beta) \in \mathbf{Z} ( α ∣ β ) ∈ Z R R R R 4 \mathbf{R}^4 R 4 n = 8 + ( 4 2 ) 4 + 2 4 = 48 n = 8 + \binom{4}{2} 4 + 2^4 = 48 n = 8 + ( 2 4 ) 4 + 2 4 = 48
(II) 给出R 4 \mathbf{R}^4 R 4 ( ε 1 , ε 2 , ε 3 , ε 4 ) (\varepsilon_1, \varepsilon_2, \varepsilon_3, \varepsilon_4) ( ε 1 , ε 2 , ε 3 , ε 4 ) ε 1 > ε 2 > ε 3 > ε 4 \varepsilon_1 > \varepsilon_2 > \varepsilon_3 > \varepsilon_4 ε 1 > ε 2 > ε 3 > ε 4
ε i , ε i ± ε j ( i < j ) , 1 2 ( ε 1 ± ε 2 ± ε 3 ± ε 4 ) .
\varepsilon_i,\quad \varepsilon_i \pm \varepsilon_j\ (i < j),\quad \frac{1}{2}(\varepsilon_1 \pm \varepsilon_2 \pm \varepsilon_3 \pm \varepsilon_4).
ε i , ε i ± ε j ( i < j ) , 2 1 ( ε 1 ± ε 2 ± ε 3 ± ε 4 ) . α 3 = ε 4 \alpha_3 = \varepsilon_4 α 3 = ε 4 R ε 3 + R ε 4 \mathbf{R}\varepsilon_3 + \mathbf{R}\varepsilon_4 R ε 3 + R ε 4 R ε 4 \mathbf{R}\varepsilon_4 R ε 4 α 2 = ε 3 − ε 4 \alpha_2 = \varepsilon_3 - \varepsilon_4 α 2 = ε 3 − ε 4 R ε 2 + R ε 3 + R ε 4 \mathbf{R}\varepsilon_2 + \mathbf{R}\varepsilon_3 + \mathbf{R}\varepsilon_4 R ε 2 + R ε 3 + R ε 4 R ε 3 + R ε 4 \mathbf{R}\varepsilon_3 + \mathbf{R}\varepsilon_4 R ε 3 + R ε 4 α 1 = ε 2 − ε 3 \alpha_1 = \varepsilon_2 - \varepsilon_3 α 1 = ε 2 − ε 3 R ε 2 + R ε 3 + R ε 4 \mathbf{R}\varepsilon_2 + \mathbf{R}\varepsilon_3 + \mathbf{R}\varepsilon_4 R ε 2 + R ε 3 + R ε 4 α 4 = 1 2 ( ε 1 − ε 2 − ε 3 − ε 4 ) \alpha_4 = \frac{1}{2}(\varepsilon_1 - \varepsilon_2 - \varepsilon_3 - \varepsilon_4) α 4 = 2 1 ( ε 1 − ε 2 − ε 3 − ε 4 ) α i \alpha_i α i ( α 1 , α 2 , α 3 , α 4 ) (\alpha_1, \alpha_2, \alpha_3, \alpha_4) ( α 1 , α 2 , α 3 , α 4 ) R R R ∥ α 1 ∥ = ∥ α 2 ∥ = 2 , ∥ α 3 ∥ = ∥ α 4 ∥ = 1 , ( α 1 ∣ α 2 ) = ( α 2 ∣ α 3 ) = − 1 , ( α 3 ∣ α 4 ) = − 1 2 , ( α 1 ∣ α 3 ) = ( α 1 ∣ α 4 ) = ( α 2 ∣ α 4 ) = 0 \| \alpha_1 \| = \| \alpha_2 \| = 2, \| \alpha_3 \| = \| \alpha_4 \| = 1, (\alpha_1|\alpha_2) = (\alpha_2|\alpha_3) = -1, (\alpha_3|\alpha_4) = -\frac{1}{2}, (\alpha_1|\alpha_3) = (\alpha_1|\alpha_4) = (\alpha_2|\alpha_4) = 0 ∥ α 1 ∥ = ∥ α 2 ∥ = 2 , ∥ α 3 ∥ = ∥ α 4 ∥ = 1 , ( α 1 ∣ α 2 ) = ( α 2 ∣ α 3 ) = − 1 , ( α 3 ∣ α 4 ) = − 2 1 , ( α 1 ∣ α 3 ) = ( α 1 ∣ α 4 ) = ( α 2 ∣ α 4 ) = 0 R R R F 4 F_4 F 4 (III) 我们有h = n l = 12 h = \frac{n}{l} = 12 h = l n = 12
(IV) 让α ~ = ε 1 + ε 2 = 2 α 1 + 3 α 2 + 4 α 3 + 2 α 4 \tilde{\alpha} = \varepsilon_1 + \varepsilon_2 = 2\alpha_1 + 3\alpha_2 + 4\alpha_3 + 2\alpha_4 α ~ = ε 1 + ε 2 = 2 α 1 + 3 α 2 + 4 α 3 + 2 α 4 α ~ \tilde{\alpha} α ~ ( α i ) (\alpha_i) ( α i ) 11 = h − 1 11 = h - 1 11 = h − 1 α ~ \tilde{\alpha} α ~ ( α ~ ∣ α 1 ) = 1 , ( α ~ ∣ α 2 ) = ( α ~ ∣ α 3 ) = ( α ~ ∣ α 4 ) = 0 (\tilde{\alpha}|\alpha_1) = 1, (\tilde{\alpha}|\alpha_2) = (\tilde{\alpha}|\alpha_3) = (\tilde{\alpha}|\alpha_4) = 0 ( α ~ ∣ α 1 ) = 1 , ( α ~ ∣ α 2 ) = ( α ~ ∣ α 3 ) = ( α ~ ∣ α 4 ) = 0
完整的 Dynkin 图为
!F_4 的 Dynkin 图
(V)公式α ∗ = 2 α ( α ∣ α ) \alpha^* = \frac{2\alpha}{(\alpha|\alpha)} α ∗ = ( α ∣ α ) 2 α R − R^- R − ± 2 ε i , ± ε i ± ε j , ± ε 1 ± ε 2 ± ε 3 ± ε 4 \pm 2\varepsilon_i, \pm \varepsilon_i \pm \varepsilon_j, \pm \varepsilon_1 \pm \varepsilon_2 \pm \varepsilon_3 \pm \varepsilon_4 ± 2 ε i , ± ε i ± ε j , ± ε 1 ± ε 2 ± ε 3 ± ε 4 R − R^- R − R R R R − R^- R − F 4 F_4 F 4
The roots not orthogonal to β = ε 1 \beta = \varepsilon_1 β = ε 1 ± ε 1 , ± ε 1 ± ε j ( j ≥ 2 ) \pm \varepsilon_1, \pm \varepsilon_1 \pm \varepsilon_j \ (j \geq 2) ± ε 1 , ± ε 1 ± ε j ( j ≥ 2 ) 1 2 ( ± ε 1 ± ε 2 ± ε 3 ± ε 4 ) \frac{1}{2}(\pm \varepsilon_1 \pm \varepsilon_2 \pm \varepsilon_3 \pm \varepsilon_4) 2 1 ( ± ε 1 ± ε 2 ± ε 3 ± ε 4 ) n ( α ∣ β ) = 2 ( α ∣ β ) n(\alpha|\beta) = 2(\alpha|\beta) n ( α ∣ β ) = 2 ( α ∣ β ) ± 2 \pm 2 ± 2 ± 1 \pm 1 ± 1 Φ R \Phi_R Φ R β \beta β 4 ( 14.4 + 16.1 ) − 1 = 1 18 4(14.4 + 16.1)^{-1} = \frac{1}{18} 4 ( 14.4 + 16.1 ) − 1 = 18 1
Φ R ( x , y ) = ( x ∣ y ) 18 .
\Phi_R(x, y) = \frac{(x|y)}{18}.
Φ R ( x , y ) = 18 ( x ∣ y ) . 我们现在应用§ 1, no. 12的公式(18)，以及x = y = β x = y = \beta x = y = β
2 + 12. 1 4 + 16. 1 / 4 1 = γ ( R ) . 1 18
2 + 12.\frac{1}{4} + 16.\frac{1/4}{1} = \gamma(R).\frac{1}{18}
2 + 12. 4 1 + 16. 1 1/4 = γ ( R ) . 18 1 所以
γ ( R ) = 2.3 4 .
\gamma(R) = 2.3^4.
γ ( R ) = 2. 3 4 . (VI) 计算基本权给出
ω 1 = ε 1 + ε 2 = 2 α 1 + 3 α 2 + 4 α 3 + 2 α 4 = α ~ ω 2 = 2 ε 1 + ε 2 + ε 3 = 3 α 1 + 6 α 2 + 8 α 3 + 4 α 4 ω 3 = 1 2 ( 3 ε 1 + ε 2 + ε 3 + ε 4 ) = 2 α 1 + 4 α 2 + 6 α 3 + 3 α 4 ω 4 = ε 1 = α 1 + 2 α 2 + 3 α 3 + 2 α 4 .
\begin{align*}
\omega_1 &= \varepsilon_1 + \varepsilon_2 = 2\alpha_1 + 3\alpha_2 + 4\alpha_3 + 2\alpha_4 = \tilde{\alpha} \\
\omega_2 &= 2\varepsilon_1 + \varepsilon_2 + \varepsilon_3 = 3\alpha_1 + 6\alpha_2 + 8\alpha_3 + 4\alpha_4 \\
\omega_3 &= \frac{1}{2}(3\varepsilon_1 + \varepsilon_2 + \varepsilon_3 + \varepsilon_4) = 2\alpha_1 + 4\alpha_2 + 6\alpha_3 + 3\alpha_4 \\
\omega_4 &= \varepsilon_1 = \alpha_1 + 2\alpha_2 + 3\alpha_3 + 2\alpha_4.
\end{align*}
ω 1 ω 2 ω 3 ω 4 = ε 1 + ε 2 = 2 α 1 + 3 α 2 + 4 α 3 + 2 α 4 = α ~ = 2 ε 1 + ε 2 + ε 3 = 3 α 1 + 6 α 2 + 8 α 3 + 4 α 4 = 2 1 ( 3 ε 1 + ε 2 + ε 3 + ε 4 ) = 2 α 1 + 4 α 2 + 6 α 3 + 3 α 4 = ε 1 = α 1 + 2 α 2 + 3 α 3 + 2 α 4 . 正根的 (VII) 和为
2 ρ = 11 ε 1 + 5 ε 2 + 3 ε 3 + ε 4 = 16 α 1 + 30 α 2 + 42 α 3 + 22 α 4 .
2\rho = 11\varepsilon_1 + 5\varepsilon_2 + 3\varepsilon_3 + \varepsilon_4 = 16\alpha_1 + 30\alpha_2 + 42\alpha_3 + 22\alpha_4.
2 ρ = 11 ε 1 + 5 ε 2 + 3 ε 3 + ε 4 = 16 α 1 + 30 α 2 + 42 α 3 + 22 α 4 . (VIII) 我们有 Q ( R ) = L 2 Q(R) = L_2 Q ( R ) = L 2 P ( R ) = Q ( R ) P(R) = Q(R) P ( R ) = Q ( R )
(IX) 指数族有 4 项，从 h = 12 h = 12 h = 12 W ( R ) W(R) W ( R )
(X) 和 (XI) Dynkin 图中唯一的自同构是恒等式，因而A ( R ) = W ( R ) A(R) = W(R) A ( R ) = W ( R ) w 0 = − 1 w_0 = -1 w 0 = − 1 R ′ R' R ′ R R R ± ε i ± ε j \pm \varepsilon_i \pm \varepsilon_j ± ε i ± ε j R ′ R' R ′ D 4 D_4 D 4 A ( R ) A(R) A ( R ) A ( R ′ ) A(R') A ( R ′ ) A ( R ′ ) A(R') A ( R ′ ) L 1 L_1 L 1 R ′ R' R ′ L 2 L_2 L 2 R R R W ( R ) = A ( R ) = A ( R ′ ) W(R) = A(R) = A(R') W ( R ) = A ( R ) = A ( R ′ ) W ( R ) W(R) W ( R ) S 3 S_3 S 3 W ( R ′ ) W(R') W ( R ′ ) W ( R ′ ) W(R') W ( R ′ ) S 4 S_4 S 4 ( Z / 2 Z ) 3 (\mathbf{Z}/2\mathbf{Z})^3 ( Z /2 Z ) 3 W ( R ) W(R) W ( R ) 3 ! 4 ! 2 3 = 2 7 .3 2 3!4!2^3 = 2^7.3^2 3 ! 4 ! 2 3 = 2 7 . 3 2
10. 系统型 E_8
(I) 考虑R 8 \mathbf{R}^8 R 8 α ∈ L 3 \alpha \in L_3 α ∈ L 3 ( α ∣ α ) = 2 (\alpha|\alpha) = 2 ( α ∣ α ) = 2
± ε i ± ε j ( i < j ) , 1 2 ∑ i = 1 8 ( − 1 ) ν ( i ) ε i ( ∑ i = 1 8 ν ( i ) even ) .
\pm \varepsilon_i \pm \varepsilon_j \quad (i < j), \quad \frac{1}{2} \sum_{i=1}^8 (-1)^{\nu(i)} \varepsilon_i \quad (\sum_{i=1}^8 \nu(i) \text{ even}).
± ε i ± ε j ( i < j ) , 2 1 i = 1 ∑ 8 ( − 1 ) ν ( i ) ε i ( i = 1 ∑ 8 ν ( i ) even ) . 相反，如果元素 α ∈ L 3 \alpha \in L_3 α ∈ L 3 ( α ∣ α ) = 2 (\alpha|\alpha) = 2 ( α ∣ α ) = 2 0 , ± 1 2 , ± 1 0, \pm \frac{1}{2}, \pm 1 0 , ± 2 1 , ± 1 ± ε i ± ε j \pm \varepsilon_i \pm \varepsilon_j ± ε i ± ε j ± 1 2 \pm \frac{1}{2} ± 2 1
1 2 ∑ i = 1 8 ( − 1 ) ν ( i ) ε i
\frac{1}{2} \sum_{i=1}^8 (-1)^{\nu(i)} \varepsilon_i
2 1 i = 1 ∑ 8 ( − 1 ) ν ( i ) ε i 与 ∑ i = 1 8 ν ( i ) \sum_{i=1}^8 \nu(i) ∑ i = 1 8 ν ( i )
我们已经看到 (no. 4) 对于所有 α , β ∈ L 3 \alpha, \beta \in L_3 α , β ∈ L 3 ( α ∣ β ) ∈ Z (\alpha|\beta) \in \mathbf{Z} ( α ∣ β ) ∈ Z n = ( 8 2 ) .4 + 2 7 = 240 n = \binom{8}{2} . 4 + 2^7 = 240 n = ( 2 8 ) .4 + 2 7 = 240
(II) 设ρ \rho ρ ( 0 , 1 , 2 , 3 , 4 , 5 , 6 , 23 ) (0, 1, 2, 3, 4, 5, 6, 23) ( 0 , 1 , 2 , 3 , 4 , 5 , 6 , 23 ) L 3 L_3 L 3 ρ \rho ρ ± ε i ± ε j \pm \varepsilon_i \pm \varepsilon_j ± ε i ± ε j 1 2 ∑ i = 1 8 ( − 1 ) ν ( i ) ε i \frac{1}{2} \sum_{i=1}^8 (-1)^{\nu(i)} \varepsilon_i 2 1 ∑ i = 1 8 ( − 1 ) ν ( i ) ε i ρ \rho ρ ∑ i = 1 6 i ( − 1 ) ν ( i + 1 ) + 23 ( − 1 ) ν ( 8 ) = 0 \sum_{i=1}^6 i(-1)^{\nu(i+1)} + 23(-1)^{\nu(8)} = 0 ∑ i = 1 6 i ( − 1 ) ν ( i + 1 ) + 23 ( − 1 ) ν ( 8 ) = 0 ∑ i = 1 6 i < 23 \sum_{i=1}^6 i < 23 ∑ i = 1 6 i < 23 α ∈ R \alpha \in R α ∈ R ( α ∣ ρ ) > 0 (\alpha|\rho) > 0 ( α ∣ ρ ) > 0 ± ε i + ε j ( i < j ) \pm \varepsilon_i + \varepsilon_j \ (i < j) ± ε i + ε j ( i < j )
1 2 ( ε 8 + ∑ i = 1 7 ( − 1 ) ν ( i ) ε i )
\frac{1}{2} (\varepsilon_8 + \sum_{i=1}^7 (-1)^{\nu(i)} \varepsilon_i)
2 1 ( ε 8 + i = 1 ∑ 7 ( − 1 ) ν ( i ) ε i ) 与 ∑ i = 1 7 ν ( i ) \sum_{i=1}^7 \nu(i) ∑ i = 1 7 ν ( i ) α ∈ R \alpha \in R α ∈ R ( α ∣ ρ ) ∈ Z (\alpha|\rho) \in \mathbf{Z} ( α ∣ ρ ) ∈ Z ( α ∣ ρ ) (\alpha|\rho) ( α ∣ ρ )
α 1 = 1 2 ( ε 1 + ε 8 ) − 1 2 ( ε 2 + ε 3 + ε 4 + ε 5 + ε 6 + ε 7 ) , α 2 = ε 1 + ε 2 , α 3 = ε 2 − ε 1 , α 4 = ε 3 − ε 2 , α 5 = ε 4 − ε 3 , α 6 = ε 5 − ε 4 , α 7 = ε 6 − ε 5 , α 8 = ε 7 − ε 6 .
\begin{align*}
\alpha_1 &= \frac{1}{2} (\varepsilon_1 + \varepsilon_8) - \frac{1}{2} (\varepsilon_2 + \varepsilon_3 + \varepsilon_4 + \varepsilon_5 + \varepsilon_6 + \varepsilon_7), \\
\alpha_2 &= \varepsilon_1 + \varepsilon_2, \ \alpha_3 = \varepsilon_2 - \varepsilon_1, \ \alpha_4 = \varepsilon_3 - \varepsilon_2, \ \alpha_5 = \varepsilon_4 - \varepsilon_3, \\
\alpha_6 &= \varepsilon_5 - \varepsilon_4, \ \alpha_7 = \varepsilon_6 - \varepsilon_5, \ \alpha_8 = \varepsilon_7 - \varepsilon_6.
\end{align*}
α 1 α 2 α 6 = 2 1 ( ε 1 + ε 8 ) − 2 1 ( ε 2 + ε 3 + ε 4 + ε 5 + ε 6 + ε 7 ) , = ε 1 + ε 2 , α 3 = ε 2 − ε 1 , α 4 = ε 3 − ε 2 , α 5 = ε 4 − ε 3 , = ε 5 − ε 4 , α 7 = ε 6 − ε 5 , α 8 = ε 7 − ε 6 . 还有R 8 \mathbf{R}^8 R 8 ( α 1 , α 2 , … , α 8 ) (\alpha_1, \alpha_2, \ldots, \alpha_8) ( α 1 , α 2 , … , α 8 )
( α 4 ∣ α 5 ) = ( α 5 ∣ α 6 ) = ( α 6 ∣ α 7 ) = ( α 7 ∣ α 8 ) = ( α 4 ∣ α 2 ) = ( α 4 ∣ α 3 ) = ( α 3 ∣ α 1 ) = − 1 ,
(\alpha_4|\alpha_5) = (\alpha_5|\alpha_6) = (\alpha_6|\alpha_7) = (\alpha_7|\alpha_8) = (\alpha_4|\alpha_2) = (\alpha_4|\alpha_3) = (\alpha_3|\alpha_1) = -1,
( α 4 ∣ α 5 ) = ( α 5 ∣ α 6 ) = ( α 6 ∣ α 7 ) = ( α 7 ∣ α 8 ) = ( α 4 ∣ α 2 ) = ( α 4 ∣ α 3 ) = ( α 3 ∣ α 1 ) = − 1 , ( α i ∣ α j ) = 0 (\alpha_i|\alpha_j) = 0 ( α i ∣ α j ) = 0
(III) 我们有h = n 8 = 30 h = \frac{n}{8} = 30 h = 8 n = 30
(IV) 让
α ~ = ε 7 + ε 8 = 2 α 1 + 3 α 2 + 4 α 3 + 6 α 4 + 5 α 5 + 4 α 6 + 3 α 7 + 2 α 8 ,
\tilde{\alpha} = \varepsilon_7 + \varepsilon_8 = 2\alpha_1 + 3\alpha_2 + 4\alpha_3 + 6\alpha_4 + 5\alpha_5 + 4\alpha_6 + 3\alpha_7 + 2\alpha_8,
α ~ = ε 7 + ε 8 = 2 α 1 + 3 α 2 + 4 α 3 + 6 α 4 + 5 α 5 + 4 α 6 + 3 α 7 + 2 α 8 , ( α i ) (\alpha_i) ( α i ) 29 = h − 1 29 = h - 1 29 = h − 1 α ~ \tilde{\alpha} α ~ α 8 \alpha_8 α 8 ( α ~ ∣ α 8 ) = 1 (\tilde{\alpha}|\alpha_8) = 1 ( α ~ ∣ α 8 ) = 1 α i \alpha_i α i !E_8 的 Dynkin 图
(V) 由于( α ∣ α ) = 2 (\alpha|\alpha) = 2 ( α ∣ α ) = 2 α ∈ R \alpha \in R α ∈ R R − = R R^- = R R − = R
对于 Φ R \Phi_R Φ R 1 30 \frac{1}{30} 30 1 Φ R ( x , y ) = ( x ∣ y ) / 60 \Phi_R(x, y) = (x|y)/60 Φ R ( x , y ) = ( x ∣ y ) /60 γ ( R ) = 900 \gamma(R) = 900 γ ( R ) = 900
(VI) 计算基本权给出
ω 1 = 2 ε 8 = 4 α 1 + 5 α 2 + 7 α 3 + 10 α 4 + 8 α 5 + 6 α 6 + 4 α 7 + 2 α 8 ω 2 = 1 2 ( ε 1 + ε 2 + ε 3 + ε 4 + ε 5 + ε 6 + ε 7 + 5 ε 8 ) = 5 α 1 + 8 α 2 + 10 α 3 + 15 α 4 + 12 α 5 + 9 α 6 + 6 α 7 + 3 α 8 ω 3 = 1 2 ( − ε 1 + ε 2 + ε 3 + ε 4 + ε 5 + ε 6 + ε 7 + 7 ε 8 ) = 7 α 1 + 10 α 2 + 14 α 3 + 20 α 4 + 16 α 5 + 12 α 6 + 8 α 7 + 4 α 8 ω 4 = ε 3 + ε 4 + ε 5 + ε 6 + ε 7 + 5 ε 8 = 10 α 1 + 15 α 2 + 20 α 3 + 30 α 4 + 24 α 5 + 18 α 6 + 12 α 7 + 6 α 8 ω 5 = ε 4 + ε 5 + ε 6 + ε 7 + 4 ε 8 = 8 α 1 + 12 α 2 + 16 α 3 + 24 α 4 + 20 α 5 + 15 α 6 + 10 α 7 + 5 α 8 ω 6 = ε 5 + ε 6 + ε 7 + 3 ε 8 = 6 α 1 + 9 α 2 + 12 α 3 + 18 α 4 + 15 α 5 + 12 α 6 + 8 α 7 + 4 α 8 ω 7 = ε 6 + ε 7 + 2 ε 8 = 4 α 1 + 6 α 2 + 8 α 3 + 2 α 4 + 10 α 5 + 8 α 6 + 6 α 7 + 3 α 8 ω 8 = ε 7 + ε 8 = 5 α 1 + 8 α 2 + 10 α 3 + 15 α 4 + 12 α 5 + 9 α 6 + 6 α 7 + 3 α 8 = α ~ .
\begin{align*}
\omega_1 &= 2\varepsilon_8 = 4\alpha_1 + 5\alpha_2 + 7\alpha_3 + 10\alpha_4 + 8\alpha_5 + 6\alpha_6 + 4\alpha_7 + 2\alpha_8 \\
\omega_2 &= \frac{1}{2}(\varepsilon_1 + \varepsilon_2 + \varepsilon_3 + \varepsilon_4 + \varepsilon_5 + \varepsilon_6 + \varepsilon_7 + 5\varepsilon_8) \\
&= 5\alpha_1 + 8\alpha_2 + 10\alpha_3 + 15\alpha_4 + 12\alpha_5 + 9\alpha_6 + 6\alpha_7 + 3\alpha_8 \\
\omega_3 &= \frac{1}{2}(-\varepsilon_1 + \varepsilon_2 + \varepsilon_3 + \varepsilon_4 + \varepsilon_5 + \varepsilon_6 + \varepsilon_7 + 7\varepsilon_8) \\
&= 7\alpha_1 + 10\alpha_2 + 14\alpha_3 + 20\alpha_4 + 16\alpha_5 + 12\alpha_6 + 8\alpha_7 + 4\alpha_8 \\
\omega_4 &= \varepsilon_3 + \varepsilon_4 + \varepsilon_5 + \varepsilon_6 + \varepsilon_7 + 5\varepsilon_8 \\
&= 10\alpha_1 + 15\alpha_2 + 20\alpha_3 + 30\alpha_4 + 24\alpha_5 + 18\alpha_6 + 12\alpha_7 + 6\alpha_8 \\
\omega_5 &= \varepsilon_4 + \varepsilon_5 + \varepsilon_6 + \varepsilon_7 + 4\varepsilon_8 \\
&= 8\alpha_1 + 12\alpha_2 + 16\alpha_3 + 24\alpha_4 + 20\alpha_5 + 15\alpha_6 + 10\alpha_7 + 5\alpha_8 \\
\omega_6 &= \varepsilon_5 + \varepsilon_6 + \varepsilon_7 + 3\varepsilon_8 \\
&= 6\alpha_1 + 9\alpha_2 + 12\alpha_3 + 18\alpha_4 + 15\alpha_5 + 12\alpha_6 + 8\alpha_7 + 4\alpha_8 \\
\omega_7 &= \varepsilon_6 + \varepsilon_7 + 2\varepsilon_8 \\
&= 4\alpha_1 + 6\alpha_2 + 8\alpha_3 + 2\alpha_4 + 10\alpha_5 + 8\alpha_6 + 6\alpha_7 + 3\alpha_8 \\
\omega_8 &= \varepsilon_7 + \varepsilon_8 \\
&= 5\alpha_1 + 8\alpha_2 + 10\alpha_3 + 15\alpha_4 + 12\alpha_5 + 9\alpha_6 + 6\alpha_7 + 3\alpha_8 = \tilde{\alpha}.
\end{align*}
ω 1 ω 2 ω 3 ω 4 ω 5 ω 6 ω 7 ω 8 = 2 ε 8 = 4 α 1 + 5 α 2 + 7 α 3 + 10 α 4 + 8 α 5 + 6 α 6 + 4 α 7 + 2 α 8 = 2 1 ( ε 1 + ε 2 + ε 3 + ε 4 + ε 5 + ε 6 + ε 7 + 5 ε 8 ) = 5 α 1 + 8 α 2 + 10 α 3 + 15 α 4 + 12 α 5 + 9 α 6 + 6 α 7 + 3 α 8 = 2 1 ( − ε 1 + ε 2 + ε 3 + ε 4 + ε 5 + ε 6 + ε 7 + 7 ε 8 ) = 7 α 1 + 10 α 2 + 14 α 3 + 20 α 4 + 16 α 5 + 12 α 6 + 8 α 7 + 4 α 8 = ε 3 + ε 4 + ε 5 + ε 6 + ε 7 + 5 ε 8 = 10 α 1 + 15 α 2 + 20 α 3 + 30 α 4 + 24 α 5 + 18 α 6 + 12 α 7 + 6 α 8 = ε 4 + ε 5 + ε 6 + ε 7 + 4 ε 8 = 8 α 1 + 12 α 2 + 16 α 3 + 24 α 4 + 20 α 5 + 15 α 6 + 10 α 7 + 5 α 8 = ε 5 + ε 6 + ε 7 + 3 ε 8 = 6 α 1 + 9 α 2 + 12 α 3 + 18 α 4 + 15 α 5 + 12 α 6 + 8 α 7 + 4 α 8 = ε 6 + ε 7 + 2 ε 8 = 4 α 1 + 6 α 2 + 8 α 3 + 2 α 4 + 10 α 5 + 8 α 6 + 6 α 7 + 3 α 8 = ε 7 + ε 8 = 5 α 1 + 8 α 2 + 10 α 3 + 15 α 4 + 12 α 5 + 9 α 6 + 6 α 7 + 3 α 8 = α ~ . (VII) 正根的一半和是基本权的和（§ 1, no. 10，命题29）；这给出了
ρ = ε 2 + 2 ε 3 + 3 ε 4 + 4 ε 5 + 5 ε 6 + 6 ε 7 + 23 ε 8 = 46 α 1 + 68 α 2 + 91 α 3 + 135 α 4 + 110 α 5 + 84 α 6 + 57 α 7 + 29 α 8 .
\begin{align*}
\rho &= \varepsilon_2 + 2\varepsilon_3 + 3\varepsilon_4 + 4\varepsilon_5 + 5\varepsilon_6 + 6\varepsilon_7 + 23\varepsilon_8 \\
&= 46\alpha_1 + 68\alpha_2 + 91\alpha_3 + 135\alpha_4 + 110\alpha_5 + 84\alpha_6 + 57\alpha_7 + 29\alpha_8.
\end{align*}
ρ = ε 2 + 2 ε 3 + 3 ε 4 + 4 ε 5 + 5 ε 6 + 6 ε 7 + 23 ε 8 = 46 α 1 + 68 α 2 + 91 α 3 + 135 α 4 + 110 α 5 + 84 α 6 + 57 α 7 + 29 α 8 . (VIII)群Q(R)由ε i ± ε j \varepsilon_i \pm \varepsilon_j ε i ± ε j 1 2 ∑ i = 1 8 ε i \frac{1}{2} \sum_{i=1}^8 \varepsilon_i 2 1 ∑ i = 1 8 ε i
(IX) 指数族有 8 个项，从 h = 30 h = 30 h = 30
(X) 来自 (IX) 和第 1 章。 V。 § 6, no. 2，科。由命题 3 的 1 可知，W(R) 的阶为
2.8.12.14.18.20.24.30 = 2 14 .3 5 .5 2 .7.
2.8.12.14.18.20.24.30 = 2^{14}.3^5.5^2.7.
2.8.12.14.18.20.24.30 = 2 14 . 3 5 . 5 2 .7. (XI) Dynkin 图的唯一自同构是恒等式，因为从分支点发出的三个链具有不同的长度。因而，A ( R ) = W ( R ) A(R) = W(R) A ( R ) = W ( R ) w 0 = − 1 w_0 = -1 w 0 = − 1
11. 系统型 E_7
(I)和(II)令E = R 8 E = \mathbf{R}^8 E = R 8 α 1 , … , α 7 \alpha_1, \ldots, \alpha_7 α 1 , … , α 7 ω = ε 7 + ε 8 \omega = \varepsilon_7 + \varepsilon_8 ω = ε 7 + ε 8
让R = R 8 ∩ V R = R_8 \cap V R = R 8 ∩ V ( α 1 , … , α 7 ) (\alpha_1, \ldots, \alpha_7) ( α 1 , … , α 7 )
± ε i ± ε j ( 1 ≤ i ≤ j ≤ 6 ) , ± ( ε 7 − ε 8 ) ,
\pm \varepsilon_i \pm \varepsilon_j \quad (1 \leq i \leq j \leq 6), \quad \pm (\varepsilon_7 - \varepsilon_8),
± ε i ± ε j ( 1 ≤ i ≤ j ≤ 6 ) , ± ( ε 7 − ε 8 ) , ± 1 2 ( ε 7 − ε 8 + ∑ i = 1 8 ( − 1 ) ν ( i ) ε i ) with ∑ i = 1 8 ν ( i ) odd .
\pm \frac{1}{2} (\varepsilon_7 - \varepsilon_8 + \sum_{i=1}^8 (-1)^{\nu(i)} \varepsilon_i) \quad \text{with } \sum_{i=1}^8 \nu(i) \text{ odd}.
± 2 1 ( ε 7 − ε 8 + i = 1 ∑ 8 ( − 1 ) ν ( i ) ε i ) with i = 1 ∑ 8 ν ( i ) odd . 根中的数是n = 2 + ( 6 2 ) .4 + 2 6 = 126 n = 2 + \binom{6}{2}.4 + 2^6 = 126 n = 2 + ( 2 6 ) .4 + 2 6 = 126
± ε i + ε j ( 1 ≤ i < j ≤ 6 ) , − ε 7 + ε 8 ,
\pm \varepsilon_i + \varepsilon_j \quad (1 \leq i < j \leq 6), \quad -\varepsilon_7 + \varepsilon_8,
± ε i + ε j ( 1 ≤ i < j ≤ 6 ) , − ε 7 + ε 8 , 1 2 ( − ε 7 + ε 8 + ∑ i = 1 6 ( − 1 ) ν ( i ) ε i ) with ∑ i = 1 6 ν ( i ) odd .
\frac{1}{2} (-\varepsilon_7 + \varepsilon_8 + \sum_{i=1}^6 (-1)^{\nu(i)} \varepsilon_i) \quad \text{with } \sum_{i=1}^6 \nu(i) \text{ odd}.
2 1 ( − ε 7 + ε 8 + i = 1 ∑ 6 ( − 1 ) ν ( i ) ε i ) with i = 1 ∑ 6 ν ( i ) odd . (III) 我们有h = n l = 18 h = \frac{n}{l} = 18 h = l n = 18
(IV)设α ~ = ε 8 − ε 7 = 2 α 1 + 2 α 2 + 3 α 3 + 4 α 4 + 3 α 5 + 2 α 6 + α 7 \tilde{\alpha} = \varepsilon_8 - \varepsilon_7 = 2\alpha_1 + 2\alpha_2 + 3\alpha_3 + 4\alpha_4 + 3\alpha_5 + 2\alpha_6 + \alpha_7 α ~ = ε 8 − ε 7 = 2 α 1 + 2 α 2 + 3 α 3 + 4 α 4 + 3 α 5 + 2 α 6 + α 7 ( α i ) (\alpha_i) ( α i ) h − 1 h - 1 h − 1 2 ≤ i ≤ 7 2 \leq i \leq 7 2 ≤ i ≤ 7 ( α ~ ∣ α 1 ) = 1 (\tilde{\alpha}|\alpha_1) = 1 ( α ~ ∣ α 1 ) = 1 α i \alpha_i α i
!E_7 的 Dynkin 图
(V) 从( α ∣ α ) = 2 (\alpha|\alpha) = 2 ( α ∣ α ) = 2 α ∈ R \alpha \in R α ∈ R R − = R R^- = R R − = R Φ R \Phi_R Φ R 1 18 \frac{1}{18} 18 1
Φ R ( x , y ) = ( x ∣ y ) / 36 , and γ ( R ) = 2 2 ⋅ 3 4
\Phi_R(x, y) = (x|y)/36, \quad \text{and} \quad \gamma(R) = 2^2 \cdot 3^4
Φ R ( x , y ) = ( x ∣ y ) /36 , and γ ( R ) = 2 2 ⋅ 3 4 (VI) 计算基本权给出
ω 1 = ε 8 − ε 7 = 2 α 1 + 2 α 2 + 3 α 3 + 4 α 4 + 3 α 5 + 2 α 6 + α 7 ω 2 = 1 2 ( ε 1 + ε 2 + ε 3 + ε 4 + ε 5 + ε 6 − 2 ε 7 + 2 ε 8 ) = 1 2 ( 4 α 1 + 7 α 2 + 8 α 3 + 12 α 4 + 9 α 5 + 8 α 6 + 3 α 7 ) ω 3 = 1 2 ( − ε 1 + ε 2 + ε 3 + ε 4 + ε 5 + ε 6 − 3 ε 7 + 3 ε 8 ) = 3 α 1 + 4 α 2 + 6 α 3 + 8 α 4 + 6 α 5 + 4 α 6 + 2 α 7 ω 4 = ε 3 + ε 4 + ε 5 + ε 6 + 2 ( ε 8 − ε 7 ) = 4 α 1 + 6 α 2 + 8 α 3 + 12 α 4 + 9 α 5 + 6 α 6 + 3 α 7 ω 5 = ε 4 + ε 5 + ε 6 + 3 2 ( ε 8 − ε 7 ) = 1 2 ( 6 α 1 + 9 α 2 + 12 α 3 + 18 α 4 + 15 α 5 + 10 α 6 + 5 α 7 ) ω 6 = ε 5 + ε 6 − ε 7 + ε 8 = 2 α 1 + 3 α 2 + 4 α 3 + 6 α 4 + 5 α 5 + 4 α 6 + 2 α 7 ω 7 = ε 6 + 1 2 ( ε 8 − ε 7 ) = 1 2 ( 2 α 1 + 3 α 2 + 4 α 3 + 6 α 4 + 5 α 5 + 4 α 6 + 3 α 7 ) .
\begin{align*}
\omega_1 &= \varepsilon_8 - \varepsilon_7 = 2\alpha_1 + 2\alpha_2 + 3\alpha_3 + 4\alpha_4 + 3\alpha_5 + 2\alpha_6 + \alpha_7 \\
\omega_2 &= \frac{1}{2}(\varepsilon_1 + \varepsilon_2 + \varepsilon_3 + \varepsilon_4 + \varepsilon_5 + \varepsilon_6 - 2\varepsilon_7 + 2\varepsilon_8) \\
&= \frac{1}{2}(4\alpha_1 + 7\alpha_2 + 8\alpha_3 + 12\alpha_4 + 9\alpha_5 + 8\alpha_6 + 3\alpha_7) \\
\omega_3 &= \frac{1}{2}(-\varepsilon_1 + \varepsilon_2 + \varepsilon_3 + \varepsilon_4 + \varepsilon_5 + \varepsilon_6 - 3\varepsilon_7 + 3\varepsilon_8) \\
&= 3\alpha_1 + 4\alpha_2 + 6\alpha_3 + 8\alpha_4 + 6\alpha_5 + 4\alpha_6 + 2\alpha_7 \\
\omega_4 &= \varepsilon_3 + \varepsilon_4 + \varepsilon_5 + \varepsilon_6 + 2(\varepsilon_8 - \varepsilon_7) \\
&= 4\alpha_1 + 6\alpha_2 + 8\alpha_3 + 12\alpha_4 + 9\alpha_5 + 6\alpha_6 + 3\alpha_7 \\
\omega_5 &= \varepsilon_4 + \varepsilon_5 + \varepsilon_6 + \frac{3}{2}(\varepsilon_8 - \varepsilon_7) \\
&= \frac{1}{2}(6\alpha_1 + 9\alpha_2 + 12\alpha_3 + 18\alpha_4 + 15\alpha_5 + 10\alpha_6 + 5\alpha_7) \\
\omega_6 &= \varepsilon_5 + \varepsilon_6 - \varepsilon_7 + \varepsilon_8 \\
&= 2\alpha_1 + 3\alpha_2 + 4\alpha_3 + 6\alpha_4 + 5\alpha_5 + 4\alpha_6 + 2\alpha_7 \\
\omega_7 &= \varepsilon_6 + \frac{1}{2}(\varepsilon_8 - \varepsilon_7) \\
&= \frac{1}{2}(2\alpha_1 + 3\alpha_2 + 4\alpha_3 + 6\alpha_4 + 5\alpha_5 + 4\alpha_6 + 3\alpha_7).
\end{align*}
ω 1 ω 2 ω 3 ω 4 ω 5 ω 6 ω 7 = ε 8 − ε 7 = 2 α 1 + 2 α 2 + 3 α 3 + 4 α 4 + 3 α 5 + 2 α 6 + α 7 = 2 1 ( ε 1 + ε 2 + ε 3 + ε 4 + ε 5 + ε 6 − 2 ε 7 + 2 ε 8 ) = 2 1 ( 4 α 1 + 7 α 2 + 8 α 3 + 12 α 4 + 9 α 5 + 8 α 6 + 3 α 7 ) = 2 1 ( − ε 1 + ε 2 + ε 3 + ε 4 + ε 5 + ε 6 − 3 ε 7 + 3 ε 8 ) = 3 α 1 + 4 α 2 + 6 α 3 + 8 α 4 + 6 α 5 + 4 α 6 + 2 α 7 = ε 3 + ε 4 + ε 5 + ε 6 + 2 ( ε 8 − ε 7 ) = 4 α 1 + 6 α 2 + 8 α 3 + 12 α 4 + 9 α 5 + 6 α 6 + 3 α 7 = ε 4 + ε 5 + ε 6 + 2 3 ( ε 8 − ε 7 ) = 2 1 ( 6 α 1 + 9 α 2 + 12 α 3 + 18 α 4 + 15 α 5 + 10 α 6 + 5 α 7 ) = ε 5 + ε 6 − ε 7 + ε 8 = 2 α 1 + 3 α 2 + 4 α 3 + 6 α 4 + 5 α 5 + 4 α 6 + 2 α 7 = ε 6 + 2 1 ( ε 8 − ε 7 ) = 2 1 ( 2 α 1 + 3 α 2 + 4 α 3 + 6 α 4 + 5 α 5 + 4 α 6 + 3 α 7 ) . 正根的 (VII) 和 2 ρ 2\rho 2 ρ 2 ∑ i = 1 7 ω i 2 \sum_{i=1}^7 \omega_i 2 ∑ i = 1 7 ω i
2 ρ = 2 ε 2 + 4 ε 3 + 6 ε 4 + 8 ε 5 + 10 ε 6 − 17 ε 7 + 17 ε 8 = 34 α 1 + 49 α 2 + 66 α 3 + 96 α 4 + 75 α 5 + 52 α 6 + 27 α 7 .
\begin{align*}
2\rho &= 2\varepsilon_2 + 4\varepsilon_3 + 6\varepsilon_4 + 8\varepsilon_5 + 10\varepsilon_6 - 17\varepsilon_7 + 17\varepsilon_8 \\
&= 34\alpha_1 + 49\alpha_2 + 66\alpha_3 + 96\alpha_4 + 75\alpha_5 + 52\alpha_6 + 27\alpha_7.
\end{align*}
2 ρ = 2 ε 2 + 4 ε 3 + 6 ε 4 + 8 ε 5 + 10 ε 6 − 17 ε 7 + 17 ε 8 = 34 α 1 + 49 α 2 + 66 α 3 + 96 α 4 + 75 α 5 + 52 α 6 + 27 α 7 . (VIII) 由no. 10 (VIII) 和§ 1, no. 10。提案 28，我们有
Q ( R ) = Q ( R 8 ) ∩ V = L 3 ∩ V and P ( R ) = p ( P ( R 8 ) ) = p ( L 3 ) ,
Q(R) = Q(R_8) \cap V = L_3 \cap V \quad \text{and} \quad P(R) = p(P(R_8)) = p(L_3),
Q ( R ) = Q ( R 8 ) ∩ V = L 3 ∩ V and P ( R ) = p ( P ( R 8 )) = p ( L 3 ) , p p p E E E V V V Q ( R ) Q(R) Q ( R ) ( α 1 , … , α 7 ) (\alpha_1, \ldots, \alpha_7) ( α 1 , … , α 7 ) P ( R ) P(R) P ( R ) Q ( R ) Q(R) Q ( R ) p ( α 8 ) = α 8 − 1 2 ω .
p(\alpha_8) = \alpha_8 - \frac{1}{2}\omega.
p ( α 8 ) = α 8 − 2 1 ω . 我们有ω ∈ P ( R 8 ) \omega \in P(R_8) ω ∈ P ( R 8 ) 1 2 ω ∉ P ( R 8 ) \frac{1}{2}\omega \notin P(R_8) 2 1 ω ∈ / P ( R 8 ) 2 p ( α 8 ) ∈ Q ( R ) 2p(\alpha_8) \in Q(R) 2 p ( α 8 ) ∈ Q ( R ) p ( α 8 ) ∉ Q ( R ) p(\alpha_8) \notin Q(R) p ( α 8 ) ∈ / Q ( R ) P ( R ) / Q ( R ) P(R)/Q(R) P ( R ) / Q ( R ) Z / 2 Z \mathbf{Z}/2\mathbf{Z} Z /2 Z ω 7 \omega_7 ω 7
连接指标为 2。
(IX) 序列 W ( R ) W(R) W ( R ) h = 18 h = 18 h = 18 m m m m + m = 18 m + m = 18 m + m = 18
1 , 5 , 7 , 9 , 11 , 13 , 17. 1, 5, 7, 9, 11, 13, 17. 1 , 5 , 7 , 9 , 11 , 13 , 17. (X) 来自 (IX) 和章节。 V, § 6, no. 2，科。根据命题 3 的 1，W ( R ) W(R) W ( R )
2.6.8.10.12.14.18 = 2 10 .3 4 .5.7. 2.6.8.10.12.14.18 = 2^{10}.3^4.5.7. 2.6.8.10.12.14.18 = 2 10 . 3 4 .5.7. (XI) Dynkin 图中唯一的自同构是恒等式，因此A ( R ) = W ( R ) A(R) = W(R) A ( R ) = W ( R ) w 0 = − 1 w_0 = -1 w 0 = − 1
(XII) P ( R − ) / Q ( R − ) P(R^-)/Q(R^-) P ( R − ) / Q ( R − ) α 0 \alpha_0 α 0 α 7 \alpha_7 α 7 α 1 \alpha_1 α 1 α 6 \alpha_6 α 6 α 3 \alpha_3 α 3 α 5 \alpha_5 α 5 α 2 \alpha_2 α 2 α 4 \alpha_4 α 4
12. 系统型 E_6
(I)和(II)令E = R 8 E = \mathbf{R}^8 E = R 8 R 8 R_8 R 8 E E E V V V R 8 R_8 R 8 α 1 , … , α 6 \alpha_1, \ldots, \alpha_6 α 1 , … , α 6 E E E R 8 R_8 R 8 ω = ε 7 + ε 8 \omega = \varepsilon_7 + \varepsilon_8 ω = ε 7 + ε 8 π = ε 6 + ε 7 + 2 ε 8 \pi = \varepsilon_6 + \varepsilon_7 + 2\varepsilon_8 π = ε 6 + ε 7 + 2 ε 8
让R = R 8 ∩ V R = R_8 \cap V R = R 8 ∩ V ( α 1 , … , α 6 ) (\alpha_1, \ldots, \alpha_6) ( α 1 , … , α 6 ) E 6 E_6 E 6
± ε i ± ε j ( 1 ≤ i < j ≤ 5 ) ,
\pm \varepsilon_i \pm \varepsilon_j \quad (1 \leq i < j \leq 5),
± ε i ± ε j ( 1 ≤ i < j ≤ 5 ) , ± 1 2 ( ε 8 − ε 7 − ε 6 + ∑ i = 1 5 ( − 1 ) ν ( i ) ε i ) with ∑ i = 1 5 ν ( i ) even .
\pm \frac{1}{2} (\varepsilon_8 - \varepsilon_7 - \varepsilon_6 + \sum_{i=1}^5 (-1)^{\nu(i)} \varepsilon_i) \text{ with } \sum_{i=1}^5 \nu(i) \text{ even}.
± 2 1 ( ε 8 − ε 7 − ε 6 + i = 1 ∑ 5 ( − 1 ) ν ( i ) ε i ) with i = 1 ∑ 5 ν ( i ) even . 根中的数是n = ( 5 2 ) .4 + 2 5 = 72 n = \binom{5}{2}.4 + 2^5 = 72 n = ( 2 5 ) .4 + 2 5 = 72
± ε i + ε j ( 1 ≤ i < j ≤ 5 ) ,
\pm \varepsilon_i + \varepsilon_j \quad (1 \leq i < j \leq 5),
± ε i + ε j ( 1 ≤ i < j ≤ 5 ) , 1 2 ( ε 8 − ε 7 − ε 6 + ∑ i = 1 5 ( − 1 ) ν ( i ) ε i ) with ∑ i = 1 5 ν ( i ) even .
\frac{1}{2} (\varepsilon_8 - \varepsilon_7 - \varepsilon_6 + \sum_{i=1}^5 (-1)^{\nu(i)} \varepsilon_i) \text{ with } \sum_{i=1}^5 \nu(i) \text{ even}.
2 1 ( ε 8 − ε 7 − ε 6 + i = 1 ∑ 5 ( − 1 ) ν ( i ) ε i ) with i = 1 ∑ 5 ν ( i ) even . (III) 我们有h = n 6 = 12 h = \frac{n}{6} = 12 h = 6 n = 12
(IV) 让
α ~ = 1 2 ( ε 1 + ε 2 + ε 3 + ε 4 + ε 5 − ε 6 − ε 7 + ε 8 )
\tilde{\alpha} = \frac{1}{2} (\varepsilon_1 + \varepsilon_2 + \varepsilon_3 + \varepsilon_4 + \varepsilon_5 - \varepsilon_6 - \varepsilon_7 + \varepsilon_8)
α ~ = 2 1 ( ε 1 + ε 2 + ε 3 + ε 4 + ε 5 − ε 6 − ε 7 + ε 8 ) = α 1 + 2 α 2 + 2 α 3 + 3 α 4 + 2 α 5 + α 6 ,
= \alpha_1 + 2\alpha_2 + 2\alpha_3 + 3\alpha_4 + 2\alpha_5 + \alpha_6,
= α 1 + 2 α 2 + 2 α 3 + 3 α 4 + 2 α 5 + α 6 , 即根。它相对于( α i ) (\alpha_i) ( α i ) 11 = h − 1 11 = h - 1 11 = h − 1 α ~ \tilde{\alpha} α ~ α 1 , α 3 , α 4 , α 5 , α 6 \alpha_1, \alpha_3, \alpha_4, \alpha_5, \alpha_6 α 1 , α 3 , α 4 , α 5 , α 6 ( α ~ ∣ α 2 ) = 1 (\tilde{\alpha}|\alpha_2) = 1 ( α ~ ∣ α 2 ) = 1
!丁金图
(V) 从( α ∣ α ) = 2 (\alpha|\alpha) = 2 ( α ∣ α ) = 2 α ∈ R \alpha \in R α ∈ R R ′ = R R' = R R ′ = R Φ R \Phi_R Φ R 1 12 \frac{1}{12} 12 1
Φ R ( x , y ) = ( x ∣ y ) / 24 , and γ ( R ) = 144.
\Phi_R(x, y) = (x|y)/24, \quad \text{and} \quad \gamma(R) = 144.
Φ R ( x , y ) = ( x ∣ y ) /24 , and γ ( R ) = 144. (VI) 计算基本权得出：
ω 1 = 2 3 ( ε 8 − ε 7 − ε 6 ) = 1 3 ( 4 α 1 + 3 α 2 + 5 α 3 + 6 α 4 + 4 α 5 + 2 α 6 ) ω 2 = 1 2 ( ε 1 + ε 2 + ε 3 + ε 4 + ε 5 − ε 6 − ε 7 + ε 8 ) = α 1 + 2 α 2 + 2 α 3 + 3 α 4 + 2 α 5 + α 6 = α ~ ω 3 = 5 6 ( ε 8 − ε 7 − ε 6 ) + 1 2 ( − ε 1 + ε 2 + ε 3 + ε 4 + ε 5 ) = 1 3 ( 5 α 1 + 6 α 2 + 10 α 3 + 12 α 4 + 8 α 5 + 4 α 6 ) ω 4 = ε 3 + ε 4 + ε 5 − ε 6 − ε 7 + ε 8 = 2 α 1 + 3 α 2 + 4 α 3 + 6 α 4 + 4 α 5 + 2 α 6 ω 5 = 2 3 ( ε 8 − ε 7 − ε 6 ) + ε 4 + ε 5 = 1 3 ( 4 α 1 + 6 α 2 + 8 α 3 + 12 α 4 + 10 α 5 + 5 α 6 ) ω 6 = 1 3 ( ε 8 − ε 7 − ε 6 ) + ε 5 = 1 3 ( 2 α 1 + 3 α 2 + 4 α 3 + 6 α 4 + 5 α 5 + 4 α 6 ) .
\begin{align*}
\omega_1 &= \frac{2}{3}(\varepsilon_8 - \varepsilon_7 - \varepsilon_6) = \frac{1}{3}(4\alpha_1 + 3\alpha_2 + 5\alpha_3 + 6\alpha_4 + 4\alpha_5 + 2\alpha_6) \\
\omega_2 &= \frac{1}{2}(\varepsilon_1 + \varepsilon_2 + \varepsilon_3 + \varepsilon_4 + \varepsilon_5 - \varepsilon_6 - \varepsilon_7 + \varepsilon_8) \\
&= \alpha_1 + 2\alpha_2 + 2\alpha_3 + 3\alpha_4 + 2\alpha_5 + \alpha_6 = \tilde{\alpha} \\
\omega_3 &= \frac{5}{6}(\varepsilon_8 - \varepsilon_7 - \varepsilon_6) + \frac{1}{2}(-\varepsilon_1 + \varepsilon_2 + \varepsilon_3 + \varepsilon_4 + \varepsilon_5) \\
&= \frac{1}{3}(5\alpha_1 + 6\alpha_2 + 10\alpha_3 + 12\alpha_4 + 8\alpha_5 + 4\alpha_6) \\
\omega_4 &= \varepsilon_3 + \varepsilon_4 + \varepsilon_5 - \varepsilon_6 - \varepsilon_7 + \varepsilon_8 \\
&= 2\alpha_1 + 3\alpha_2 + 4\alpha_3 + 6\alpha_4 + 4\alpha_5 + 2\alpha_6 \\
\omega_5 &= \frac{2}{3}(\varepsilon_8 - \varepsilon_7 - \varepsilon_6) + \varepsilon_4 + \varepsilon_5 \\
&= \frac{1}{3}(4\alpha_1 + 6\alpha_2 + 8\alpha_3 + 12\alpha_4 + 10\alpha_5 + 5\alpha_6) \\
\omega_6 &= \frac{1}{3}(\varepsilon_8 - \varepsilon_7 - \varepsilon_6) + \varepsilon_5 \\
&= \frac{1}{3}(2\alpha_1 + 3\alpha_2 + 4\alpha_3 + 6\alpha_4 + 5\alpha_5 + 4\alpha_6).
\end{align*}
ω 1 ω 2 ω 3 ω 4 ω 5 ω 6 = 3 2 ( ε 8 − ε 7 − ε 6 ) = 3 1 ( 4 α 1 + 3 α 2 + 5 α 3 + 6 α 4 + 4 α 5 + 2 α 6 ) = 2 1 ( ε 1 + ε 2 + ε 3 + ε 4 + ε 5 − ε 6 − ε 7 + ε 8 ) = α 1 + 2 α 2 + 2 α 3 + 3 α 4 + 2 α 5 + α 6 = α ~ = 6 5 ( ε 8 − ε 7 − ε 6 ) + 2 1 ( − ε 1 + ε 2 + ε 3 + ε 4 + ε 5 ) = 3 1 ( 5 α 1 + 6 α 2 + 10 α 3 + 12 α 4 + 8 α 5 + 4 α 6 ) = ε 3 + ε 4 + ε 5 − ε 6 − ε 7 + ε 8 = 2 α 1 + 3 α 2 + 4 α 3 + 6 α 4 + 4 α 5 + 2 α 6 = 3 2 ( ε 8 − ε 7 − ε 6 ) + ε 4 + ε 5 = 3 1 ( 4 α 1 + 6 α 2 + 8 α 3 + 12 α 4 + 10 α 5 + 5 α 6 ) = 3 1 ( ε 8 − ε 7 − ε 6 ) + ε 5 = 3 1 ( 2 α 1 + 3 α 2 + 4 α 3 + 6 α 4 + 5 α 5 + 4 α 6 ) . (VII) 正根的一半和是∑ i = 1 6 ω i \sum_{i=1}^6 \omega_i ∑ i = 1 6 ω i
ρ = ε 2 + 2 ε 3 + 3 ε 4 + 4 ε 5 + 4 ( ε 8 − ε 7 − ε 6 ) = 8 α 1 + 11 α 2 + 15 α 3 + 21 α 4 + 15 α 5 + 8 α 6 .
\begin{align*}
\rho &= \varepsilon_2 + 2\varepsilon_3 + 3\varepsilon_4 + 4\varepsilon_5 + 4(\varepsilon_8 - \varepsilon_7 - \varepsilon_6) \\
&= 8\alpha_1 + 11\alpha_2 + 15\alpha_3 + 21\alpha_4 + 15\alpha_5 + 8\alpha_6.
\end{align*}
ρ = ε 2 + 2 ε 3 + 3 ε 4 + 4 ε 5 + 4 ( ε 8 − ε 7 − ε 6 ) = 8 α 1 + 11 α 2 + 15 α 3 + 21 α 4 + 15 α 5 + 8 α 6 . (VIII) 根据no. 10 (VIII) 和§ 4, no. 10，第 28 号提案，
Q ( R ) = Q ( R 8 ) ∩ V = L 3 ∩ V and P ( R ) = p ( P ( R 8 ) ) = p ( L 3 ) ,
Q(R) = Q(R_8) \cap V = L_3 \cap V \quad \text{and} \quad P(R) = p(P(R_8)) = p(L_3),
Q ( R ) = Q ( R 8 ) ∩ V = L 3 ∩ V and P ( R ) = p ( P ( R 8 )) = p ( L 3 ) , 其中 p p p E E E V V V
p ( α 7 ) = α 7 − 2 3 π + ω , p ( α 8 ) = α 8 + π − 2 ω .
p(\alpha_7) = \alpha_7 - \frac{2}{3}\pi + \omega, \quad p(\alpha_8) = \alpha_8 + \pi - 2\omega.
p ( α 7 ) = α 7 − 3 2 π + ω , p ( α 8 ) = α 8 + π − 2 ω . Q ( R ) Q(R) Q ( R ) ( α 1 , … , α 6 ) (\alpha_1, \ldots, \alpha_6) ( α 1 , … , α 6 ) p ( α 8 ) ∈ P ( R 8 ) ∩ V = Q ( R 8 ) ∩ V = Q ( R ) p(\alpha_8) \in P(R_8) \cap V = Q(R_8) \cap V = Q(R) p ( α 8 ) ∈ P ( R 8 ) ∩ V = Q ( R 8 ) ∩ V = Q ( R ) P ( R ) P(R) P ( R ) Q ( R ) Q(R) Q ( R ) p ( α 7 ) p(\alpha_7) p ( α 7 ) 3 p ( α 7 ) ∈ Q ( R ) 3p(\alpha_7) \in Q(R) 3 p ( α 7 ) ∈ Q ( R ) p ( α 7 ) ∉ Q ( R ) p(\alpha_7) \notin Q(R) p ( α 7 ) ∈ / Q ( R ) P ( R ) / Q ( R ) P(R)/Q(R) P ( R ) / Q ( R ) Z / 3 Z \mathbf{Z}/3\mathbf{Z} Z /3 Z ω 6 \omega_6 ω 6 连接指标为 3。
(IX) 和 (X) 根据§ 2, no. 4，第 7 号提案，Weyl 群的阶为 6 ! 1.2.2.3.2.1.3 = 2 7 .3 4 .5 6!1.2.2.3.2.1.3 = 2^7.3^4.5 6 ! 1.2.2.3.2.1.3 = 2 7 . 3 4 .5 m , m ′ m, m' m , m ′
m + m ′ = 12 ,
m + m' = 12,
m + m ′ = 12 , ( m + 1 ) ( m ′ + 1 ) ( 1 + 1 ) ( 5 + 1 ) ( 7 + 1 ) ( 11 + 1 ) = 2 7 .3 4 .5 ,
(m + 1)(m' + 1)(1 + 1)(5 + 1)(7 + 1)(11 + 1) = 2^7.3^4.5,
( m + 1 ) ( m ′ + 1 ) ( 1 + 1 ) ( 5 + 1 ) ( 7 + 1 ) ( 11 + 1 ) = 2 7 . 3 4 .5 , ( m + 1 ) ( m ′ + 1 ) = 45 (m + 1)(m' + 1) = 45 ( m + 1 ) ( m ′ + 1 ) = 45 m + m ′ + 2 = 14 m + m' + 2 = 14 m + m ′ + 2 = 14 m = 4 , m ′ = 8 m = 4, m' = 8 m = 4 , m ′ = 8 1 , 4 , 5 , 7 , 8 , 11.
1, 4, 5, 7, 8, 11.
1 , 4 , 5 , 7 , 8 , 11. (XI) 和 (XII) 由于根都具有相同的长度，因此 Dynkin 图的自同构是基础图的自同构。除了身份之外，只有自同构ε \varepsilon ε α 1 , α 3 , α 4 , α 5 , α 6 , α 2 \alpha_1, \alpha_3, \alpha_4, \alpha_5, \alpha_6, \alpha_2 α 1 , α 3 , α 4 , α 5 , α 6 , α 2 α 6 , α 5 , α 4 , α 3 , α 1 , α 2 \alpha_6, \alpha_5, \alpha_4, \alpha_3, \alpha_1, \alpha_2 α 6 , α 5 , α 4 , α 3 , α 1 , α 2 A ( R ) / W ( R ) A(R)/W(R) A ( R ) / W ( R ) Z / 2 Z \mathbf{Z}/2\mathbf{Z} Z /2 Z − 1 ∈ W ( R ) -1 \in W(R) − 1 ∈ W ( R ) A ( R ) A(R) A ( R ) W ( R ) × { 1 , − 1 } W(R) \times \{1, -1\} W ( R ) × { 1 , − 1 } w 0 w_0 w 0 − ε -\varepsilon − ε A ( R ) / W ( R ) A(R)/W(R) A ( R ) / W ( R ) x ↦ − x x \mapsto -x x ↦ − x P ( R − ) / Q ( R − ) P(R^-)/Q(R^-) P ( R − ) / Q ( R − )
此外，P ( R − ) / Q ( R − ) P(R^-)/Q(R^-) P ( R − ) / Q ( R − )
13. 系统型 G_2
(I) 令 R 3 \mathbf{R}^3 R 3 E E E
ξ 1 + ξ 2 + ξ 3 = 0.
\xi_1 + \xi_2 + \xi_3 = 0.
ξ 1 + ξ 2 + ξ 3 = 0. R R R α ∈ L 0 ∩ V \alpha \in L_0 \cap V α ∈ L 0 ∩ V ( α ∣ α ) = 2 (\alpha|\alpha) = 2 ( α ∣ α ) = 2 ( α ∣ α ) = 6 (\alpha|\alpha) = 6 ( α ∣ α ) = 6 R R R ± ( ε 1 − ε 2 ) , ± ( ε 1 − ε 3 ) , ± ( ε 2 − ε 3 ) , ± ( 2 ε 1 − ε 2 − ε 3 ) ,
\pm (\varepsilon_1 - \varepsilon_2), \quad \pm (\varepsilon_1 - \varepsilon_3), \quad \pm (\varepsilon_2 - \varepsilon_3), \quad \pm (2\varepsilon_1 - \varepsilon_2 - \varepsilon_3),
± ( ε 1 − ε 2 ) , ± ( ε 1 − ε 3 ) , ± ( ε 2 − ε 3 ) , ± ( 2 ε 1 − ε 2 − ε 3 ) , ± ( 2 ε 2 − ε 1 − ε 3 ) , ± ( 2 ε 3 − ε 1 − ε 2 ) .
\pm (2\varepsilon_2 - \varepsilon_1 - \varepsilon_3), \quad \pm (2\varepsilon_3 - \varepsilon_1 - \varepsilon_2).
± ( 2 ε 2 − ε 1 − ε 3 ) , ± ( 2 ε 3 − ε 1 − ε 2 ) . 然后，R 生成 V，以及所有α , β ∈ R \alpha, \beta \in R α , β ∈ R 2 ( α ∣ β ) ( β ∣ β ) ∈ Z \frac{2(\alpha|\beta)}{(\beta|\beta)} \in \mathbf{Z} ( β ∣ β ) 2 ( α ∣ β ) ∈ Z β = ± ( ε i − ε j ) \beta = \pm (\varepsilon_i - \varepsilon_j) β = ± ( ε i − ε j ) i ≠ j i \neq j i = j β = 2 ε 1 − ε 2 − ε 3 \beta = 2\varepsilon_1 - \varepsilon_2 - \varepsilon_3 β = 2 ε 1 − ε 2 − ε 3 ( α ∣ β ) ∈ 3 Z (\alpha|\beta) \in 3\mathbf{Z} ( α ∣ β ) ∈ 3 Z n = 12 n = 12 n = 12
（II）放入α 1 = ε 1 − ε 2 , α 2 = − 2 ε 1 + ε 2 + ε 3 \alpha_1 = \varepsilon_1 - \varepsilon_2, \alpha_2 = -2\varepsilon_1 + \varepsilon_2 + \varepsilon_3 α 1 = ε 1 − ε 2 , α 2 = − 2 ε 1 + ε 2 + ε 3
± α 1 , ± ( α 1 + α 2 ) , ± ( 2 α 1 + α 2 ) , ± α 2 ,
\pm \alpha_1, \quad \pm (\alpha_1 + \alpha_2), \quad \pm (2\alpha_1 + \alpha_2), \quad \pm \alpha_2,
± α 1 , ± ( α 1 + α 2 ) , ± ( 2 α 1 + α 2 ) , ± α 2 , ± ( 3 α 1 + α 2 ) , ± ( 3 α 1 + 2 α 2 ) .
\pm (3\alpha_1 + \alpha_2), \quad \pm (3\alpha_1 + 2\alpha_2).
± ( 3 α 1 + α 2 ) , ± ( 3 α 1 + 2 α 2 ) . 因此，( α 1 , α 2 ) (\alpha_1, \alpha_2) ( α 1 , α 2 ) ∥ α 1 ∥ 2 = 2 , ∥ α 2 ∥ 2 = 6 , ( α 1 ∣ α 2 ) = − 3 \| \alpha_1 \| ^2 = 2, \| \alpha_2 \| ^2 = 6, (\alpha_1|\alpha_2) = -3 ∥ α 1 ∥ 2 = 2 , ∥ α 2 ∥ 2 = 6 , ( α 1 ∣ α 2 ) = − 3 α 1 , α 1 + α 2 , 2 α 1 + α 2 , 3 α 1 + α 2 , 3 α 1 + 2 α 2 \alpha_1, \alpha_1 + \alpha_2, 2\alpha_1 + \alpha_2, 3\alpha_1 + \alpha_2, 3\alpha_1 + 2\alpha_2 α 1 , α 1 + α 2 , 2 α 1 + α 2 , 3 α 1 + α 2 , 3 α 1 + 2 α 2
(III) 我们有h = n 2 = 6 h = \frac{n}{2} = 6 h = 2 n = 6
(IV)最高根为α ~ = 3 α 1 + 2 α 2 = − ε 1 − ε 2 + 2 ε 3 \tilde{\alpha} = 3\alpha_1 + 2\alpha_2 = -\varepsilon_1 - \varepsilon_2 + 2\varepsilon_3 α ~ = 3 α 1 + 2 α 2 = − ε 1 − ε 2 + 2 ε 3 ( α ~ ∣ α 1 ) = 0 , ( α ~ ∣ α 2 ) = 3 (\tilde{\alpha}|\alpha_1) = 0, (\tilde{\alpha}|\alpha_2) = 3 ( α ~ ∣ α 1 ) = 0 , ( α ~ ∣ α 2 ) = 3
!丁金图
(V) 逆系是向量的以下集合：
± α 1 , ± ( α 1 + α 2 ) , ± ( 2 α 1 + α 2 ) , ± 1 3 α 2 ,
\pm \alpha_1, \quad \pm (\alpha_1 + \alpha_2), \quad \pm (2\alpha_1 + \alpha_2), \quad \pm \frac{1}{3}\alpha_2,
± α 1 , ± ( α 1 + α 2 ) , ± ( 2 α 1 + α 2 ) , ± 3 1 α 2 , ± 1 3 ( 3 α 1 + α 2 ) , ± 1 3 ( 3 α 1 + 2 α 2 ) .
\pm \frac{1}{3}(3\alpha_1 + \alpha_2), \quad \pm \frac{1}{3}(3\alpha_1 + 2\alpha_2).
± 3 1 ( 3 α 1 + α 2 ) , ± 3 1 ( 3 α 1 + 2 α 2 ) . 有10个根不与α 1 \alpha_1 α 1 n ( β , α 1 ) = ± 1 n(\beta, \alpha_1) = \pm 1 n ( β , α 1 ) = ± 1 n ( β , α 1 ) = ± 3 n(\beta, \alpha_1) = \pm 3 n ( β , α 1 ) = ± 3 n ( β , α 1 ) = ± 2 n(\beta, \alpha_1) = \pm 2 n ( β , α 1 ) = ± 2 β = ± α 1 \beta = \pm \alpha_1 β = ± α 1 Φ R \Phi_R Φ R α 1 \alpha_1 α 1 4 ( 4.1 + 4.9 + 2.4 ) − 1 = 1 12 4(4.1 + 4.9 + 2.4)^{-1} = \frac{1}{12} 4 ( 4.1 + 4.9 + 2.4 ) − 1 = 12 1 Φ R ( x , y ) = ( x ∣ y ) / 24 \Phi_R(x, y) = (x|y)/24 Φ R ( x , y ) = ( x ∣ y ) /24 x = y = α 1 x = y = \alpha_1 x = y = α 1
2 + 4. 1 4 + 4. 1 4 = γ ( R ) . 1 12
2 + 4.\frac{1}{4} + 4.\frac{1}{4} = \gamma(R).\frac{1}{12}
2 + 4. 4 1 + 4. 4 1 = γ ( R ) . 12 1 所以γ ( R ) = 48 \gamma(R) = 48 γ ( R ) = 48
(VI) 和 (VII) 正根的一半和为
ρ = 5 α 1 + 3 α 2 .
\rho = 5\alpha_1 + 3\alpha_2.
ρ = 5 α 1 + 3 α 2 . 基本权ω 1 \omega_1 ω 1 ω 2 \omega_2 ω 2 α 2 \alpha_2 α 2 α 1 \alpha_1 α 1 2 α 1 + α 2 2\alpha_1 + \alpha_2 2 α 1 + α 2 3 α 1 + 2 α 2 3\alpha_1 + 2\alpha_2 3 α 1 + 2 α 2
ω 1 + ω 2 = ρ = 5 α 1 + 3 α 2 = ( 2 α 1 + α 2 ) + ( 3 α 1 + 2 α 2 ) .
\omega_1 + \omega_2 = \rho = 5\alpha_1 + 3\alpha_2 = (2\alpha_1 + \alpha_2) + (3\alpha_1 + 2\alpha_2).
ω 1 + ω 2 = ρ = 5 α 1 + 3 α 2 = ( 2 α 1 + α 2 ) + ( 3 α 1 + 2 α 2 ) . 因此，
ω 1 = 2 α 1 + α 2 , ω 2 = 3 α 1 + 2 α 2 = α ~ .
\omega_1 = 2\alpha_1 + \alpha_2, \quad \omega_2 = 3\alpha_1 + 2\alpha_2 = \tilde{\alpha}.
ω 1 = 2 α 1 + α 2 , ω 2 = 3 α 1 + 2 α 2 = α ~ . 对于例，(VIII) Q(R) 由ε 1 − ε 2 \varepsilon_1 - \varepsilon_2 ε 1 − ε 2 ε 1 − ε 3 \varepsilon_1 - \varepsilon_3 ε 1 − ε 3
(IX) 指数族有 2 个项；由于 1 和 h − 1 = 5 h - 1 = 5 h − 1 = 5
(X) 我们有 ( α 1 , α 2 ‾ ) = 5 π 6 (\overline{\alpha_1, \alpha_2}) = \frac{5\pi}{6} ( α 1 , α 2 ) = 6 5 π
(XI) Dynkin 图表唯一的自同构是恒等式，因此Λ ( R ) = W ( R ) \Lambda(R) = W(R) Λ ( R ) = W ( R ) w 0 = − 1 w_0 = -1 w 0 = − 1
14. 不可约非约根系
不可约的、非约化的根系统可以通过使用 Props 从不可约的、约化的系统中获得。 § 1 的 13 和 14。 no. 4。对于每个整数l ≥ 1 l \geq 1 l ≥ 1 l l l 2 l ( l + 1 ) 2l(l+1) 2 l ( l + 1 )
± ε i , ± 2 ε i , ε i ± ε j ( i < j ) .
\pm \varepsilon_i,\ \pm 2\varepsilon_i,\ \varepsilon_i \pm \varepsilon_j\ (i < j).
± ε i , ± 2 ε i , ε i ± ε j ( i < j ) . 习题
若R为根系，则用行列式1的W ( R ) W(R) W ( R ) W + ( R ) W^+(R) W + ( R )
参见习题为 § 4 。
