置换
置换是指将一组对象重新排列的一种方式
置换 p = ( 1 2 3 ⋯ n p 1 p 2 p 3 ⋯ p n ) : 1 2 3 n ↓ ↓ ↓ ⋯ ↓ p 1 p 2 p 3 p n \text{置换 }p=
\begin{pmatrix}
1 & 2 & 3 & \cdots & n \\
p_1&p_2&p_3 & \cdots& p_n
\end{pmatrix}:\quad
\begin{array}
{cccccc}1 & 2 & 3 & &n \\
\downarrow & \downarrow & \downarrow & \cdots & \downarrow \\
p_1 & p_2 & p_3 & &p_n
\end{array} 置换 p = ( 1 p 1 2 p 2 3 p 3 ⋯ ⋯ n p n ) : 1 ↓ p 1 2 ↓ p 2 3 ↓ p 3 ⋯ n ↓ p n
对于含有 n n n S n S_n S n S n S_n S n n ! n! n !
定义置换群中 S n S_n S n m m m
( a 1 a 2 ⋯ a m ) = ( 1 2 ⋯ p a 1 a 2 ⋯ a m a m + 1 ⋯ n 1 2 ⋯ p a 2 a 3 ⋯ a 1 a m + 1 ⋯ n ) (\textcolor{cyan}{a_{1}\,a_{2}\,\cdots\,a_{m}})=
\begin{pmatrix}
1 & 2 & \cdots & p & \textcolor{cyan}{a_{1}} & \textcolor{cyan}{a_{2}} & \cdots & \textcolor{cyan}{a_{m}} & a_{m+1} & \cdots & n \\
1 & 2 & \cdots & p & \textcolor{cyan}{a_{2}} & \textcolor{cyan}{a_{3}} & \cdots & \textcolor{cyan}{a_{1}} & a_{m+1} & \cdots & n
\end{pmatrix}
( a 1 a 2 ⋯ a m ) = ( 1 1 2 2 ⋯ ⋯ p p a 1 a 2 a 2 a 3 ⋯ ⋯ a m a 1 a m + 1 a m + 1 ⋯ ⋯ n n )
表示 a 1 → a 2 , a 2 → a 3 , ⋯ , a m − 1 → a m , a m → a 1 a_1\to a_2,a_2\to a_3,\cdots,a_{m-1}\to a_m,a_m\to a_1 a 1 → a 2 , a 2 → a 3 , ⋯ , a m − 1 → a m , a m → a 1 m m m 1 1 1 ( a i ) (a_i) ( a i ) 2 2 2 ( a i a j ) (a_i\,a_j) ( a i a j )
循环的性质
( a 1 a 2 ⋯ a m ) m = e (a_1a_2\cdots a_m)^m=e ( a 1 a 2 ⋯ a m ) m = e 没有共同循环符号的两个循环是独立的，独立的循环才可交换( 13 ) ( 245 ) = ( 245 ) ( 13 ) (13)(245)=(245)(13) \\[1em]
( 13 ) ( 245 ) = ( 245 ) ( 13 )
轮换不变性质( a 1 a 2 ⋯ a m ) = ( a 2 a 3 ⋯ a m a 1 ) = ( a m a 1 a 2 ⋯ a m − 1 ) (a_1a_2\cdots a_m)=(a_2a_3\cdots a_ma_1)=(a_ma_1a_2\cdots a_{m-1})
( a 1 a 2 ⋯ a m ) = ( a 2 a 3 ⋯ a m a 1 ) = ( a m a 1 a 2 ⋯ a m − 1 )
循环的合并：有且仅有一个共同符号的两个循环之积可合并为一个循环( a i ⋯ a j a k ) ( a k a p ⋯ a q ) = ( a i ⋯ a j a k a p ⋯ a q ) (a_i\cdots a_j\textcolor{red}{a_k})(\textcolor{red}{a_k}a_p\cdots a_q)
= (a_i\cdots a_j\textcolor{red}{a_k}a_p\cdots a_q)
( a i ⋯ a j a k ) ( a k a p ⋯ a q ) = ( a i ⋯ a j a k a p ⋯ a q )
循环的拆解：一个循环拆成两个较短的循环之积( a i ⋯ a j a k a p ⋯ a q ) = ( a i ⋯ a j a k ) ( a k a p ⋯ a q ) (a_i\cdots a_j\textcolor{red}{a_k}a_p\cdots a_q)
= (a_i\cdots a_j\textcolor{red}{a_k})(\textcolor{red}{a_k}a_p\cdots a_q)
( a i ⋯ a j a k a p ⋯ a q ) = ( a i ⋯ a j a k ) ( a k a p ⋯ a q )
符号逆序就是就是循环的逆 ( a 1 a 2 ⋯ a m ) − 1 = ( a m a m − 1 ⋯ a 1 ) (a_1a_2\cdots a_m)^{-1}=(a_ma_{m-1}\cdots a_1) ( a 1 a 2 ⋯ a m ) − 1 = ( a m a m − 1 ⋯ a 1 )
循环内的符号只是代表位置，如第 a i a_i a i
对换的性质
任何循环都可以写成一些对换的乘积
( a 1 a 2 ⋯ a m ) = ( a 1 a m ) ( a 1 a m − 1 ) ⋯ ( a 1 a 2 ) (a_1a_2\cdots a_m)=(a_1a_m)(a_1a_{m-1})\cdots(a_1a_2)
( a 1 a 2 ⋯ a m ) = ( a 1 a m ) ( a 1 a m − 1 ) ⋯ ( a 1 a 2 )
( a i a j ) = ( a j a i ) , ( a i a j a k ) ( a k a j ) = ( a i a j ) (a_ia_j)=(a_ja_i),(a_ia_ja_k)(a_ka_j)=(a_ia_j)
( a i a j ) = ( a j a i ) , ( a i a j a k ) ( a k a j ) = ( a i a j )
基本对换 ( 1 k ) (1k) ( 1 k ) S n S_n S n ( m ≠ 1 , n ≠ 1 ) \left(m\neq1,n\neq1\right) ( m = 1 , n = 1 )
( m n ) = ( 1 m ) ( 1 n ) ( 1 m ) = ( 1 n ) ( 1 m ) ( 1 n ) = ( j m ) ( j n ) ( j m ) = ( j n ) ( j m ) ( j n ) ( j ≠ m , j ≠ n ) \begin{aligned}
(mn) & =(1m)(1n)(1m)=(1n)(1m)(1n) \\
& =(jm)(jn)(jm)=(jn)(jm)(jn)(j\neq m,j\neq n)
\end{aligned} ( mn ) = ( 1 m ) ( 1 n ) ( 1 m ) = ( 1 n ) ( 1 m ) ( 1 n ) = ( jm ) ( jn ) ( jm ) = ( jn ) ( jm ) ( jn ) ( j = m , j = n )
偶（奇）数个符号的循环可以写成奇（偶）数个对换的乘积，称为奇（偶）循环
S n S_n S n S n S_n S n S n S_n S n n ! / 2 n!/2 n ! /2
所有奇置换不构成子群，因为不满足封闭性
共轭类、配分和 Young 图
长度为 k k k k k k ν 1 \nu_1 ν 1 1 1 1 ν 2 \nu_2 ν 2 2 2 2 ν n \nu_n ν n n n n
( 1 ν 1 2 ν 2 ⋯ n ν n ) (1^{\nu_1}2^{\nu_2}\cdots n^{\nu_n})
( 1 ν 1 2 ν 2 ⋯ n ν n )
且有 ν 1 + 2 ν 2 + ⋯ + n ν n = n \nu_1+2\nu_2+\cdots+n\nu_n=n ν 1 + 2 ν 2 + ⋯ + n ν n = n
定理：置换群中元素属于同一类的充要条件是具有相同的循环结构
也就是说，置换群 S n S_n S n ( l ) = ( 1 ν 1 2 ν 2 ⋯ n ν n ) (l)=(1^{\nu_1}2^{\nu_2}\cdots n^{\nu_n}) ( l ) = ( 1 ν 1 2 ν 2 ⋯ n ν n )
定理：S n S_n S n ( l ) = ( 1 ν 1 2 ν 2 ⋯ n ν n ) (l)=(1^{\nu_1}2^{\nu_2}\cdots n^{\nu_n}) ( l ) = ( 1 ν 1 2 ν 2 ⋯ n ν n )
n ( l ) = n ! 1 ν 1 2 ν 2 ⋯ n ν n ν 1 ! ν 2 ! ⋯ ν n ! n_{(l)}=\frac{n!}{1^{\nu_1}2^{\nu_2}\cdots n^{\nu_n}\nu_1!\nu_2!\cdots\nu_n!}
n ( l ) = 1 ν 1 2 ν 2 ⋯ n ν n ν 1 ! ν 2 ! ⋯ ν n ! n !
若令
ν 1 + ν 2 + ⋯ + ν n = λ 1 ν 2 + ⋯ + ν n = λ 2 ⋮ ⋮ ν n = λ n \begin{array}
{rcl}\nu_1+\nu_2+\cdots+\nu_n & = & \lambda_1 \\
\nu_2+\cdots+\nu_n & = & \lambda_2 \\
\vdots & & \vdots \\
\nu_n & = & \lambda_n
\end{array} ν 1 + ν 2 + ⋯ + ν n ν 2 + ⋯ + ν n ⋮ ν n = = = λ 1 λ 2 ⋮ λ n
则有
λ 1 + λ 2 + ⋯ + λ n = ν 1 + 2 ν 2 + ⋯ + n ν n = n λ 1 ≥ λ 2 ≥ ⋯ ≥ λ n ≥ 0 \begin{aligned}
& \lambda_1+\lambda_2+\cdots+\lambda_n= \nu_1+2\nu_2+\cdots+n\nu_n=n \\
& \lambda_1\geq\lambda_2\geq\cdots\geq\lambda_n\geq0
\end{aligned} λ 1 + λ 2 + ⋯ + λ n = ν 1 + 2 ν 2 + ⋯ + n ν n = n λ 1 ≥ λ 2 ≥ ⋯ ≥ λ n ≥ 0
满足上述条件的 [ λ ] = [ λ 1 , λ 2 , ⋯ , λ n ] \begin{bmatrix} \lambda \end{bmatrix}= \begin{bmatrix} \lambda_1,\lambda_2,\cdots,\lambda_n \end{bmatrix} [ λ ] = [ λ 1 , λ 2 , ⋯ , λ n ] n n n S 4 S_4 S 4
( l ) : ( 1 4 ) ( 1 2 2 ) ( 2 2 ) ( 13 ) ( 4 ) [ λ ] : [ 4 ] [ 3 , 1 ] [ 2 2 ] [ 2 , 1 2 ] [ 1 4 ] \begin{array}
{ccccccc}(l){:} & (1^{4}) & (1^{2}2) & (2^{2}) & (13) & (4) \\
\\
[\lambda]{:} & [4] & [3,1] & [2^{2}] & [2,1^{2}] & [1^{4}]
\end{array} ( l ) : [ λ ] : ( 1 4 ) [ 4 ] ( 1 2 2 ) [ 3 , 1 ] ( 2 2 ) [ 2 2 ] ( 13 ) [ 2 , 1 2 ] ( 4 ) [ 1 4 ]
可见 ν i {\nu_i} ν i λ i {\lambda_i} λ i λ i {\lambda_i} λ i ( l ) = ( 1 v 1 2 v 2 ⋯ n v n ) (l)=(1^{v_{1}}2^{v_{2}}\cdots n^{v_{n}}) ( l ) = ( 1 v 1 2 v 2 ⋯ n v n ) [ λ ] = [ λ 1 , λ 2 , ⋯ , λ n ] [\lambda]=[\lambda_{1},\lambda_{2},\cdots,\lambda_{n}] [ λ ] = [ λ 1 , λ 2 , ⋯ , λ n ]
注意：上面公式仅表示 { ν i } \{\nu_i\} { ν i } { λ i } \{\lambda_i\} { λ i }
n n n [ λ 1 , λ 2 , ⋯ , λ n ] [\lambda_{1},\lambda_{2},\cdots,\lambda_{n}] [ λ 1 , λ 2 , ⋯ , λ n ] S n S_n S n Young 图 \text{Young 图} Young 图 λ 1 \lambda_1 λ 1 λ 2 \lambda_2 λ 2 n n n λ n \lambda_n λ n
因此，配分 ⟺ \Longleftrightarrow ⟺ ⟺ \Longleftrightarrow ⟺ ⟺ \Longleftrightarrow ⟺
n n n Young \text{Young} Young S n S_n S n S n S_n S n
Young \text{Young} Young
杨盘与图形方法
不可约表示维数的计算
Young \text{Young} Young 1 , 2 , ⋯ , n 1,2,\cdots,n 1 , 2 , ⋯ , n Young \text{Young } Young Young \text{Young} Young Young \text{Young} Young
定理：S n S_n S n { λ } \{\lambda\} { λ } Young \text{Young} Young
对偶表示具有相同的维数
标准 Young \text{Young} Young [ λ ] \left[\lambda\right] [ λ ] m m m Young \text{Young} Young r i = λ i + m − i r_i=\lambda_i+m-i r i = λ i + m − i
d [ λ ] ( S n ) = n ! r 1 ! r 2 ! ⋯ r m ! ∏ i < j m ( r i − r j ) \boxed{ d_{[\lambda]}(S_n)=\frac{n!}{r_1!r_2!\cdots r_m!}\prod_{i<j}^m(r_i-r_j)}
d [ λ ] ( S n ) = r 1 ! r 2 ! ⋯ r m ! n ! i < j ∏ m ( r i − r j )
还可以用构形规则计算标准 Young \text{Young} Young Young \text{Young} Young h i j h_{ij} h ij
钩形规则: d [ λ ] ( S n ) = n ! ∏ i , j h i j \boxed{\text{钩形规则:}\quad d_{[\lambda]}(S_n)=\frac{n!}{\prod_{i,j}h_{ij}}}
钩形规则 : d [ λ ] ( S n ) = ∏ i , j h ij n !
不可约表示的特征标
通过Young \text{Young} Young
根据不可约表示的配分 λ \lambda λ Young \text{Young} Young
按照类 ( l ) = ( 1 ν 1 2 ν 2 ⋯ n ν n ) (l)=(1^{\nu_1}2^{\nu_2}\cdots n^{\nu_n}) ( l ) = ( 1 ν 1 2 ν 2 ⋯ n ν n ) k k k k k k ( k = 1 , 2 , ⋯ , n ) \left(k=1,2,\cdots,n\right) ( k = 1 , 2 , ⋯ , n ) ν k \nu_k ν k k − k- k − ν k \nu_k ν k ν 1 + ν 2 + ⋯ + ν n \nu_1+\nu_2+\cdots+\nu_n ν 1 + ν 2 + ⋯ + ν n
将这些数字填入 Young \text{Young} Young Young \text{Young} Young
对每个不同的数字，如果含有该数字的行数为偶数，则分别给出一个负号，多个负号则相乘。
所有可能的填充方式得到的 Young \text{Young} Young
结论一：任意 n ≥ 2 n\geq2 n ≥ 2 S n S_n S n [ n ] \left[n\right] [ n ] [ 1 n ] \left[1^{n}\right] [ 1 n ]
其中 χ ( l ) [ n ] = 1 \chi^{[n]}_{(l)}=1 χ ( l ) [ n ] = 1 χ ( l ) [ 1 n ] \chi^{[1^{n}]}_{(l)} χ ( l ) [ 1 n ] + 1 +1 + 1 − 1 -1 − 1
结论二：对偶图 [ λ ] [\lambda] [ λ ] [ λ ′ ] [\lambda'] [ λ ′ ] χ ( l ) [ λ ′ ] = χ ( l ) [ λ ] χ ( l ) [ 1 n ] \chi_{(l)}^{[\lambda^{\prime}]}=\chi_{(l)}^{[\lambda]}\chi_{(l)}^{[1^n]} χ ( l ) [ λ ′ ] = χ ( l ) [ λ ] χ ( l ) [ 1 n ]
推论 ：自对偶的不可约表示在奇置换下的特征标为 0 0 0
Young 算符
两个一维不可约表示的基
置换群不可约表示的基是什么？如何求？我们先看两个一维不可约表示 [ n ] \left[n\right] [ n ] [ 1 n ] \left[1^{n}\right] [ 1 n ] n n n f ( x 1 , x 2 , ⋯ , x n ) f(x_1,x_2,\cdots,x_n) f ( x 1 , x 2 , ⋯ , x n )
全对称函数 ψ S = S ^ f ( 1 , 2 , ⋯ , n ) = ∑ P ∈ S n P f ( 1 , 2 , ⋯ , n ) 全反对称函数 ψ A = A ^ f ( 1 , 2 , ⋯ , n ) = ∑ P ∈ S n ( − 1 ) P P f ( 1 , 2 , ⋯ , n ) \begin{aligned}
\text{全对称函数} \qquad & \psi^{S}=\hat{S}f(1,2,\cdots,n)=\sum_{P\in S_{n}}Pf(1,2,\cdots,n) \\[1em]
\text{全反对称函数} \qquad & \psi^{A}=\hat{A}f(1,2,\cdots,n)=\sum_{P\in S_{n}}(-1)^{P}Pf(1,2,\cdots,n)
\end{aligned} 全对称函数 全反对称函数 ψ S = S ^ f ( 1 , 2 , ⋯ , n ) = P ∈ S n ∑ P f ( 1 , 2 , ⋯ , n ) ψ A = A ^ f ( 1 , 2 , ⋯ , n ) = P ∈ S n ∑ ( − 1 ) P P f ( 1 , 2 , ⋯ , n )
其中 ( − 1 ) P \left(-1\right)^P ( − 1 ) P
( − 1 ) P ≡ { + 1 , P 为偶置换 − 1 , P 为奇置换 } = χ P [ 1 n ] (-1)^P\equiv
\begin{Bmatrix}
+1, & \quad P\text{ 为偶置换} \\[1em]
-1, & \quad P\text{ 为奇置换}
\end{Bmatrix}=\chi_P^{[1^n]} ( − 1 ) P ≡ ⎩ ⎨ ⎧ + 1 , − 1 , P 为偶置换 P 为奇置换 ⎭ ⎬ ⎫ = χ P [ 1 n ]
1 对 ∀ Q ∈ S n \forall Q\in S_{n} ∀ Q ∈ S n
Q ψ S = Q ∑ P P f = ∑ P ( Q P ) f = ψ S Q\psi^{S}=Q\sum_{P}Pf=\sum_{P}(QP)f=\psi^{S}
Q ψ S = Q P ∑ P f = P ∑ ( QP ) f = ψ S
可见 ψ S \psi^{S} ψ S [ n ] \left[n\right] [ n ] ψ [ n ] = ψ S \psi^{[n]}=\psi^S ψ [ n ] = ψ S
2 对 ∀ Q ∈ S n , Q ψ A = Q ∑ P ( − 1 ) P P f = ∑ P ( − 1 ) P ( Q P ) f \forall Q\in S_n,Q\psi^A=Q\sum_P(-1)^PPf=\sum_P(-1)^P(QP)f ∀ Q ∈ S n , Q ψ A = Q ∑ P ( − 1 ) P P f = ∑ P ( − 1 ) P ( QP ) f Q Q Q ( − 1 ) P = ( − 1 ) Q ( − 1 ) Q P (-1)^P=(-1)^Q(-1)^{QP} ( − 1 ) P = ( − 1 ) Q ( − 1 ) QP
Q ψ A = ∑ P ( − 1 ) P ( Q P ) f = ( − 1 ) Q ∑ P ( − 1 ) Q P ( Q P ) f = χ Q [ 1 n ] ψ A Q\psi^{A}=\sum_{P}(-1)^{P}(QP)f=(-1)^{Q}\sum_{P}(-1)^{QP}(QP)f=\chi_{Q}^{[1^{n}]}\psi^{A}
Q ψ A = P ∑ ( − 1 ) P ( QP ) f = ( − 1 ) Q P ∑ ( − 1 ) QP ( QP ) f = χ Q [ 1 n ] ψ A
可见 ψ A \psi^{A} ψ A [ 1 n ] \left[1^{n}\right] [ 1 n ] ψ [ 1 n ] = ψ A \psi^{[1^{n}]}=\psi^{A} ψ [ 1 n ] = ψ A
恒等表示 [ n ] \left[n\right] [ n ] [ 1 n ] \left[1^{n}\right] [ 1 n ]
Young 算符的构造
定义对 Young \text{Young} Young i i i H i ^ \hat{H_i} H i ^ i i i j j j V j ^ \hat{V_j} V j ^ j j j
用第 i i i H i ^ \hat{H_i} H i ^ S i ^ = ∑ H i ^ H i ^ \hat{S_i}=\sum_{\hat{H_i}}\hat{H_i} S i ^ = ∑ H i ^ H i ^ j j j V j ^ \hat{V_j} V j ^ A j ^ = ∑ V j ^ ( − 1 ) V j ^ V j ^ \hat{A_j}=\sum_{\hat{V_j}}(-1)^{\hat{V_j}}\hat{V_j} A j ^ = ∑ V j ^ ( − 1 ) V j ^ V j ^
定义水平置换算符为 H ^ = ∏ i S ^ i = ∏ i ( ∑ H ^ i H ^ i ) \widehat{H}=\prod_i\hat{S}_i=\prod_i\left(\sum_{\widehat{H}_i}\widehat{H}_i\right) H = ∏ i S ^ i = ∏ i ( ∑ H i H i ) V ^ = ∏ j A ^ j = ∏ j ( ∑ V ^ j ( − 1 ) V ^ j V ^ j ) \widehat{V}=\prod_j\hat{A}_j=\prod_j\left(\sum_{\widehat{V}_j}(-1)^{\widehat{V}_j}\widehat{V}_j\right) V = ∏ j A ^ j = ∏ j ( ∑ V j ( − 1 ) V j V j )
对每个标准 Young \text{Young} Young T k T_k T k Young \text{Young} Young
Y ^ k [ λ ] = V ^ H ^ = ∏ j ( ∑ V ^ j ( − 1 ) V ^ j V ^ j ) ∏ i ( ∑ H ^ i H ^ i ) \widehat{Y}_k^{[\lambda]}=\widehat{V}\widehat{H}=\prod_j\left(\sum_{\widehat{V}_j}(-1)^{\widehat{V}_j}\widehat{V}_j\right)\prod_i\left(\sum_{\widehat{H}_i}\widehat{H}_i\right)
Y k [ λ ] = V H = j ∏ V j ∑ ( − 1 ) V j V j i ∏ H i ∑ H i
Young \text{Young} Young Young \text{Young} Young [ λ ] [\lambda] [ λ ] d [ λ ] d_{\left[\lambda\right]} d [ λ ] Young \text{Young} Young d [ λ ] d_{\left[\lambda\right]} d [ λ ] { ψ k } \{\psi_k\} { ψ k } d [ λ ] d_{\left[\lambda\right]} d [ λ ] Young \text{Young} Young { T k } \{T_k\} { T k }
由任意 n n n ϕ ( 1 , 2 , ⋯ , n ) \phi\left(1,2,\cdots,n\right) ϕ ( 1 , 2 , ⋯ , n ) Young \text{Young} Young S n S_n S n [ λ ] [\lambda] [ λ ]
ψ k ( 1 , 2 , ⋯ , n ) = Y ^ k [ λ ] P ^ k ϕ ( 1 , 2 , ⋯ , n ) , k = 1 , 2 , ⋯ , d [ λ ] \psi_k(1,2,\cdots,n)=\hat{Y}_k^{[\lambda]}\hat{P}_k\phi(1,2,\cdots,n),\quad k=1,2,\cdots,d_{[\lambda]}
ψ k ( 1 , 2 , ⋯ , n ) = Y ^ k [ λ ] P ^ k ϕ ( 1 , 2 , ⋯ , n ) , k = 1 , 2 , ⋯ , d [ λ ]
其中 P ^ k \hat{P}_k P ^ k T 1 T_1 T 1 T k T_k T k
ψ k = Y ^ k [ λ ] P ^ k ϕ \psi_k=\widehat{Y}_k^{[\lambda]}\widehat{P}_k\phi
ψ k = Y k [ λ ] P k ϕ
对 Young \text{Young} Young Y ^ 1 [ λ ] \hat{Y}_1^{[\lambda]} Y ^ 1 [ λ ] P ^ k \hat{P}_k P ^ k Young \text{Young} Young Y ^ k [ λ ] \hat{Y}_k^{[\lambda]} Y ^ k [ λ ] P ^ k \hat{P}_k P ^ k Y ^ 1 [ λ ] \hat{Y}_1^{[\lambda]} Y ^ 1 [ λ ] □ \square □
P ^ k Y ^ 1 [ λ ] □ = Y ^ k [ λ ] P ^ k □ → 口的任意性 P ^ k Y ^ 1 [ λ ] = Y ^ k [ λ ] P ^ k \widehat{P}_k\widehat{Y}_1^{[\lambda]}\square=\widehat{Y}_k^{[\lambda]}\widehat{P}_k\square\quad\xrightarrow{\text{口的任意性}}\quad\widehat{P}_k\widehat{Y}_1^{[\lambda]}=\widehat{Y}_k^{[\lambda]}\widehat{P}_k
P k Y 1 [ λ ] □ = Y k [ λ ] P k □ 口的任意性 P k Y 1 [ λ ] = Y k [ λ ] P k
⟹ Y ^ k [ λ ] = P ^ k Y ^ 1 [ λ ] P ^ k − 1 \Longrightarrow\quad\widehat{Y}_k^{[\lambda]}=\widehat{P}_k\widehat{Y}_1^{[\lambda]}\widehat{P}_k^{-1}
⟹ Y k [ λ ] = P k Y 1 [ λ ] P k − 1
那么不可约表示 [ λ ] \left[\lambda\right] [ λ ] { ψ k } \{\psi_k\} { ψ k }
ψ k = Y ^ k [ λ ] P ^ k ϕ = P ^ k Y ^ 1 [ λ ] ϕ = P ^ k ψ 1 \boxed{\psi_k=\widehat{Y}_k^{[\lambda]}\widehat{P}_k\phi=\widehat{P}_k\widehat{Y}_1^{[\lambda]}\phi=\widehat{P}_k\psi_1}
ψ k = Y k [ λ ] P k ϕ = P k Y 1 [ λ ] ϕ = P k ψ 1
我们只需构造一个 Young \text{Young} Young [ λ ] \left[\lambda\right] [ λ ] ψ 1 \psi_1 ψ 1 ψ 1 \psi_1 ψ 1 P ^ k \hat{P}_k P ^ k
分支律与外积
分支律：分导表示的约化
以循环表示群元，当 n ′ < n n'<n n ′ < n S n ′ S_{n'} S n ′ S n S_n S n S n S_n S n S n ′ S_{n'} S n ′ S n S_{n} S n S n − 1 S_{n-1} S n − 1
分支律：将 S n S_n S n [ λ ] [\lambda] [ λ ] Young \text{Young} Young Young \text{Young} Young n − 1 n-1 n − 1 Young \text{Young} Young [ λ ] [\lambda] [ λ ] S n − 1 S_{n-1} S n − 1 S n − 1 S_{n-1} S n − 1
分支律只适用于 S n − 1 S_{n-1} S n − 1 S n S_n S n S n ′ S_{n'} S n ′ S n S_n S n ( n − n ′ ≥ 2 ) \left(n-n'\geq 2\right) ( n − n ′ ≥ 2 ) n − n ′ n-n' n − n ′
外积：诱导表示的约化
一个 n + m n+m n + m S n + m S_{n+m} S n + m ( n + m ) ! (n+m)! ( n + m )! n n n S n S_n S n n ! n! n ! m m m S m S_m S m m ! m! m ! S n ⊗ S m ⊂ S n + m S_n\otimes S_m\subset S_{n+m} S n ⊗ S m ⊂ S n + m n ! m ! n!m! n ! m ! S n + m S_{n+m} S n + m
如果已知 S n S_n S n [ μ ] [\mu] [ μ ] S m S_m S m [ ν ] [\nu] [ ν ] S n + m S_{n+m} S n + m
从子群 S n ⊗ S m S_n\otimes S_m S n ⊗ S m [ μ ] ⊗ [ ν ] [\mu]\otimes[\nu] [ μ ] ⊗ [ ν ] S n + m S_{n+m} S n + m S n S_n S n [ μ ] [\mu] [ μ ] S m S_m S m [ ν ] [\nu] [ ν ] [ μ ] × [ ν ] [\mu]\times[\nu] [ μ ] × [ ν ]
诱导表示 [ μ ] × [ ν ] [\mu]\times[\nu] [ μ ] × [ ν ] d [ μ ] d [ ν ] ( n + m ) ! n ! m ! d_{[\mu]}d_{[\nu]}\frac{(n+m)!}{n!m!} d [ μ ] d [ ν ] n ! m ! ( n + m )!
[ μ ] × [ ν ] = ∑ λ ⊕ a μ ν λ [ λ ] d [ μ ] d [ ν ] ( n + m ) ! n ! m ! = ∑ λ a μ ν λ d [ λ ] [\mu]\times[\nu]=\sum_\lambda\oplus a_{\mu\nu}^\lambda\left[\lambda\right] \\[1em]
d_{[\mu]}d_{[\nu]}\frac{(n+m)!}{n!m!}=\sum_\lambda a_{\mu\nu}^\lambda d_{[\lambda]} [ μ ] × [ ν ] = λ ∑ ⊕ a μν λ [ λ ] d [ μ ] d [ ν ] n ! m ! ( n + m )! = λ ∑ a μν λ d [ λ ]
我们来看如何约化外积表示 [ μ ] × [ ν ] [\mu]\times[\nu] [ μ ] × [ ν ] Littlewood-Richardson \text{Littlewood-Richardson} Littlewood-Richardson
[ μ ] × [ m ] \left[\mu\right]\times\left[m\right] [ μ ] × [ m ] [ m ] \left[m\right] [ m ] m m m [ μ ] \left[\mu\right] [ μ ] Young \text{Young} Young Young \text{Young} Young Young \text{Young} Young [ μ ] × [ 1 m ] \left[\mu\right]\times\left[1^{m}\right] [ μ ] × [ 1 m ] [ 1 m ] \left[1^{m}\right] [ 1 m ] m m m [ μ ] \left[\mu\right] [ μ ] Young \text{Young} Young Young \text{Young} Young Young \text{Young} Young [ μ ] × [ ν ] \left[\mu\right]\times\left[\nu\right] [ μ ] × [ ν ] a a a Young \text{Young} Young [ ν ] \left[\nu\right] [ ν ] 1 1 1 b b b 2 2 2 a a a Young \text{Young} Young [ μ ] \left[\mu\right] [ μ ] b b b c c c Young \text{Young} Young a a a ≥ \geq ≥ b b b ≥ \geq ≥ c c c ≥ ⋯ \geq\cdots ≥ ⋯
[ μ ] × [ ν ] \left[\mu\right]\times\left[\nu\right] [ μ ] × [ ν ] S n ⊗ S m S_{n}\otimes S_{m} S n ⊗ S m [ μ ] ⊗ [ ν ] [\mu]\otimes[\nu] [ μ ] ⊗ [ ν ] S n + m S_{n+m} S n + m S n S_{n} S n S m S_{m} S m S n ⊗ S m = S m ⊗ S n S_n\otimes S_m=S_m\otimes S_n S n ⊗ S m = S m ⊗ S n [ μ ] ⊗ [ ν ] ≠ [ ν ] ⊗ [ μ ] [\mu]\otimes[\nu]\neq[\nu]\otimes[\mu] [ μ ] ⊗ [ ν ] = [ ν ] ⊗ [ μ ] T r ( [ μ ] ⊗ [ ν ] ) = T r ( [ ν ] ⊗ [ μ ] ) = T r [ ν ] T r [ μ ] \mathrm{Tr}([\mu]\otimes[\nu])=\mathrm{Tr}([\nu]\otimes[\mu])=\mathrm{Tr}[\nu]Tr[\mu] Tr ([ μ ] ⊗ [ ν ]) = Tr ([ ν ] ⊗ [ μ ]) = Tr [ ν ] T r [ μ ] [ ν ] ⊗ [ μ ] 与 [ μ ] ⊗ [ ν ] [\nu]\otimes[\mu]\text{ 与 }[\mu]\otimes[\nu] [ ν ] ⊗ [ μ ] 与 [ μ ] ⊗ [ ν ] S n + m S_{n+m} S n + m [ ν ] × [ μ ] = [ μ ] × [ ν ] [\nu]\times[\mu]=[\mu]\times[\nu] [ ν ] × [ μ ] = [ μ ] × [ ν ]
在等价的意义下，外积具有交换律
[ μ ] × [ ν ] = [ ν ] × [ μ ] [\mu]\times[\nu]=[\nu]\times[\mu]
[ μ ] × [ ν ] = [ ν ] × [ μ ]
