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Schur マニュアルの問題を解く　その７

7.5 The Final Test

問３

3. Show that if under U_{n(n-3)\over2}\rightarrow S_n 　\{1\}\rightarrow< 2 > then for an arbitrary irreducible representation \lambda we have
{\lambda}\rightarrow<1>\otimes[\sum_{\rho,\mu,\nu}(-1)^{\omega_\rho}\{\lambda/\rho\circ\mu\}\cdot\{\mu\}\cdot(\{1^2\}\otimes\{\tilde{\rho}/\nu\})\cdot\{\nu/M\}] (7.11)

  • Plethysm の公式より

公式集
A\otimes(B\pm C)=A\otimes B\pm A\otimes C (3.34)
A\otimes(BC)=(A\otimes B)\cdot(A\otimes C) (3.35)
A\otimes(B\otimes C)=(A\otimes B)\otimes C (3.36)
(A+B)\otimes\{\mu\}=\sum_{\zeta}(A\otimes\{\mu/\zeta\})\cdot(B\otimes\{\zeta\}) (3.37)
(A-B)\otimes\{\mu\}=\sum_{\zeta}(-1)^{\omega_\zeta}(A\otimes\{\mu/\zeta\})\cdot(B\otimes\{\tilde\zeta\}) (3.38)
(AB)\otimes{\mu}=\sum_\rho(A\otimes\{\rho\})\cdot(B\otimes\{\mu\circ\rho\}) (3.39)



\{1\}\rightarrow<2> なので、\{\lambda\}={1}\otimes\{\lambda\} と合わせると、\{\lambda\}\rightarrow<2>\otimes\{\lambda\} を計算すればよい。

ここで問１で求めた <2>=<1>^2-<1^2>-<1>-<0> の関係式をつかうと求めるべき計算式は以下のようになる。

<2>\otimes\{\lambda\}=<1>\otimes(\{1\}^2-(\{1^2\}+\{1\}+\{0\}))\otimes\{\lambda\}
式 (3.36) と式 (3.38) より
=<1>\otimes\sum_\rho(-1)^{\omega_\rho}(\{1\}^2\otimes\{\lambda/\rho\})\cdot( (\{1^2\}+\{1\}+\{0\})\otimes\{\rho'\})

式 (3.39) と式 (3.37) より
=<1>\otimes\sum_\rho(-1)^{\omega_\rho}(\sum_\mu (\{1\}\otimes\{\mu\})\cdot(\{1\}\otimes\ \{(\lambda/\rho)\circ\mu\})\cdot(\sum_\nu(\{1^2\}\otimes\{\tilde\rho/\nu\})\cdot( (\{1\}+\{0\})\otimes\{\nu\}))
ここで {1}\otimes{\lambda}={\lambda} および U_n\rightarrow U_{n-1} のブランチングに用いられる
(\{1\}+\{0\})\otimes\nu= \{1/M\}\otimes\{\nu\}=\{\nu/M\}
の関係より、

=<1>\otimes\sum_\rho(-1)^{\omega_\rho}(\sum_\mu\{\mu\}\cdot\{(\lambda/\rho)\circ\mu\})\cdot(\sum_\nu(\{1^2\}\otimes\{\tilde\rho/\nu\}\cdot\{\nu/M\})

総和記号をまとめれば、問題の式が示される。


問４

4. Let x = f^{<21>}_n. Show that if under U_x\rightarrow S_n \{1\} \rightarrow< 21 > then for an arbitrary irreducible representation \lambda we have
\{\lambda\}\rightarrow<1>\otimes\sum_{\rho,\mu,\nu,\tau}(-1)^{\omega_\rho} (\{1^2\}\otimes\{\lambda/M\rho\circ\mu\}\cdot\{\mu\})\cdot(\{1^3\}\otimes\{\tilde\rho/\tau\}) \cdot\{\tau\circ\nu\}\cdot\{\nu\} (7.12)

  • 問３と同様に

問１の<21>=<1^2><1>-<1^3>-<1>^2+<0>より

\begin{align} <21> &\rightarrow <21>\otimes\{\lambda\}\\ &=[<1>\otimes (\{1^2\}\{1\}-\{1^3\}-\{1\}^2+\{0\}) ] \otimes\{\lambda\} \end{align}

\begin{align} &(\{1^2\}\{1\}-\{1^3\}-\{1\}^2+\{0\})\otimes\{\lambda\} \\ &= \sum_\rho(-1)^{\omega_\rho}(\{1^2\}\cdot\{1\}+\{0\})\otimes\{\lambda/\rho\})\cdot((\{1^3\}+\{1\}^2)\otimes\{\tilde\rho\}) \\ &= \sum_\rho(-1)^{\omega_\rho}(\sum_\zeta(\{1^2\}\{1\})\otimes\{\lambda/\rho\cdot\zeta\}) \cdot(\{0\}\otimes\{\zeta\})\cdot(\sum_\tau(\{1^3\}\otimes\{\tilde\rho/\tau\})\cdot(\{1\}^2\otimes\tau) ) \end{align}
ここで
\begin{align} &\sum_\zeta(\{1^2\}\{1\})\otimes\{\lambda/\rho\cdot\zeta\} \\ &=\sum_\zeta(\sum_\mu(\{1^2\}\otimes\{\lambda/\rho\cdot\zeta\circ\mu\})\cdot(\{1\}\otimes\{\mu\})\cdot(\{0\}\otimes\{\zeta\}) ) \\ &=\{1^2\}\otimes\{\lambda/\rho M\circ\mu\}\cdot\{\mu\} \end{align}
但し、
\{0\}\otimes\{\zeta\}=1\,\,\,\mathrm{if}\,\,\zeta=\{1\},\{2\},\{3\}\cdots \mathrm{else}\,\,\, 0.
M\equiv\sum_m\{m\}=\{1\}+\{2\}+\{3\}+\cdots

を用いた。
また、
\begin{align} &\sum_\tau(\{1^3\}\otimes\{\tilde\rho/\tau\})\cdot(\{1\}^2\otimes\tau) \\ &=\sum_\tau(\{1^3\}\otimes\{\tilde\rho/\tau\}) \cdot (\sum_\nu(\{1\}\otimes\nu) \cdot (\{1\}\otimes\{\tau\circ\nu\} ) \\ &=\sum_\tau(\{1^3\}\otimes\{\tilde\rho/\tau\})\cdot(\sum_\nu\{\nu\}\cdot\{\tau\circ\nu\}) \end{align}
と変形できる。
これらをまとめれば、問題の式が示される。

問５

5. Write a function to obtain the branching rule for Eq. 7.11 for n = 8 and obtain the
decomposition for the \{21\} irreducible representation for U_{20}\rightarrow S_8.

  • 模範解答が与えられているので実行結果のみ
SchurQ7-5.fn

[rv1*0*1^2*0*0] は各群の規約表現などの初期値。

gr u20
enter rv1
dim[rv1]
gr 5 u8 u8 u8 u8 u8
rule [rv1*0*1^2*0*0]  sum sk 1 conj 4
rule last  ch_phase 4
rule last  sum i 1 eq 2
cont 1, 2  o last
rule last  sum sk 3 eq 4
cont 2, 3  pl last
cont 1, 2  o  last
rule last  sk 2, with M
cont 1, 2  o last
gr s8 
rule last  i_pl 1
sup false
rule last  make 1, 8
dim last
stop
実行結果
DP> readfn 5 'ShurQ7-5.fn'
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-

DP> fn 5
Group is U(20)
enter rv1 21
Dimension = 2660
Groups are U(8) * U(8) * U(8) * U(8) * U(8)
Groups are U(8) * U(8) * U(8) * U(8)
Groups are U(8) * U(8) * U(8)
Groups are U(8) * U(8)
Group is U(8)
Group is S(8)
2{71} + 5{62} + 4{61^2} + 4{53} + 9{521} + 3{51^3} + 2{4^2} + 6{431} + 5{42^2} + 5{421^2} + {41^4} + 2{3^2 2} + 3{3^2 1^2}
 + 2{32^2 1} + {321^3}
Dimension = 2660
DP>

問６

6. Write a function to obtain the branching rule for Eq. 7.12 for n = 8 and obtain the
decomposition for the {21} irreducible representation for U_{64}\rightarrow S_8.

  • 模範解答が与えられているので実行結果のみ

contraction で、演算子を省略すると外積になるようで、マニュアルの解答例では省略してあるが、ここでは明示的に書いた。（cont 1, 2 last　＝>　cont 1, 2 o last）

'ShurQ7-6.fn'
gr u64
enter rv1
dim[rv1]
gr 7 u8 u8 u8 u8 u8 u8 u8
rule [1^2*rv1*0*1^3*0*0*0]  sk 2 with M
rule last  sum sk 2 conj 5
rule last  ch_phase 5
rule last  sum i 2 eq 3
cont 1, 2  pl last
cont 1, 2  o  last
rule last  sum sk 3 eq 4
cont 2, 3  pl last
cont 1, 2  o  last
rule last  sum i 2 eq 3
cont 1, 2  o last
cont 1, 2  o last
gr s8 
rule last  i_pl 1
sup false
rule last  make 1, 8
dim last
stop
実行結果
DP> readfn 6 'ShurQ7-6.fn'
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-
=-

DP> fn 6
Group is U(64)
enter rv1 21
Dimension = 87360
Groups are U(8) * U(8) * U(8) * U(8) * U(8) * U(8) * U(8)
Groups are U(8) * U(8) * U(8) * U(8) * U(8) * U(8)
Groups are U(8) * U(8) * U(8) * U(8) * U(8)
Groups are U(8) * U(8) * U(8) * U(8)
Groups are U(8) * U(8) * U(8)
Groups are U(8) * U(8)
Group is U(8)
Group is S(8)
2{8} + 19{71} + 53{62} + 54{61^2} + 69{53} + 156{521} + 79{51^3} + 33{4^2} + 160{431} + 126{42^2} + 195{421^2} + 70{41^4}
 + 91{3^2 2} + 118{3^2 1^2} + 142{32^2 1} + 124{321^3} + 37{31^5} + 25{2^4} + 51{2^3 1^2} + 35{2^2 1^4} + 10{21^6} + {1^8}
Dimension = 87360
DP>