Schur マニュアルの問題を解く　その１

Schur 付属のマニュアルの問題を解きます

6-3節

問１

1. The branching rule for the maximal embedding of $Sp_{2k}$ in $U_{2k}$ is given as $\{\lambda\}\downarrow\{\lambda/B\}$ . Inversely, for $Sp_{2k}\uparrow U_{2k}$ we have $<\lambda >\uparrow\{\lambda/A\}$ .
(a) Use SCHUR to show that for $2k = 8$
$< 21>\uparrow\{21\}-\{1\}$ and $\{21\}\downarrow< 21> + < 1 >$
(b) Use the knowledge gained in (a.) to write a single sequence of commands to evaluate
the $Sp_8$ Kronecker product $< 21>\times< 21>$ . Check that your result is dimensionally
correct and compare your result with that obtained by simply setting the group as $Sp_8$
and using the command “prod21,21”.

Schur function mode で {21/A}, {21/B} の確認

REP> sfn
Schur Function Mode    
SFN> sk 21, a
{21} - {1}
SFN> sk 21, b
{21} + {1}
SFN>

Representation mode と Branching Rule mode で解けという事でしょうか？

6-３節はまだ BRMode の説明をしていなので、違うかもしれませんｗ
まず branching のテーブルを表示させて欲しい branching の番号を知ります。

DP> ?tab
 TABleOfBranchingRules
      Format:-
        Modes:- DP
 Description:- Table A.2 : Table of branching rules

NOTES:
Rule No. Group         Subgroup             Rule and number(s) in BRM 
1 :-  U(n)            ->  O(n)                       1,n   
2 :-  U(n)            ->  Sp(n)                      2,n
3 :-  U(n)            ->  U(n-1)                     3,n
4 :-  U(m+n)          ->  U(m) x U(n)                4,m,n
5 :-  U(mn)           ->  U(m) x U(n)                5,m,n
6 :-  U(2k)           ->  U(k)                       6,2k
7 :-  U(n)            ->  SO(3)                      7,n
8 :-  SU(m+n)         ->  U(1) x SU(m) x SU(n)       8,m,n
9 :-  Sp(2k)          ->  SO(3)                      9,2k
10:-  Sp(2k)          ->  U(1) x SU(k)              10,2k
11:-  Sp(2k)          ->  SU(2) x SO(k)             11,2k
12:-  Sp(2k)          ->  U(2k)                     12,2k
13:-  Sp(2k)          ->  U(k)                      13,2k

DP> brm  
Branch Mode
enter branching & rule numbers> 12 8
Sp(8) to SU(8) 
BRM> 21
{21} - {1}
BRM> stop
enter branching & rule numbers> 2 8
U(8) to Sp(8) 
BRM> 21
<21> + <1>
BRM> exit

DPrep Mode (with function)
DP> rep 
REP mode       
REP> gr sp8
Group is Sp(8)
REP> prop 21
<dynkin label> (1100)
dimension = 160   20*2nd-casimir=27
2nd-dynkin = 60

REP> prod 21, 21                                                                                                                   
<42> + <41^2> + <4> + <3^2> + 2<321> + <31^3> + 3<31> + <2^3> + <2^2 1^2> + 2<2^2> + 3<21^2> + 2<2> + <1^4> + 2<1^2> + <0>
REP> dim last
dimension = 25600

REP> conv_s_to_rep sk o, sk 21,a, sk 21,a, b                                                                                       
<42> + <41^2> + <4> + <3^2> + 2<321> + <31^3> + 3<31> + <2^3> + <2^2 1^2> + 2<2^2> + 3<21^2> + 2<2> + <1^4> + 2<1^2> + <0>

確認：160x160 = 25600

問２

2. Read the information in Appendix A about the command au and use the command to show
that under the $SO_8$ automorphism $[1]\to[s;0]- \rightarrow[s;0]+$ . Use that observation to obtain
the content of the Kronecker products $[s;0]-\times[s;0]-$ and $[s;0]+\times[s;0]+$ from the result
obtained from the command “prod1,1”.

SO8 の三回対称なルート図に対応した自己同型

prod 1,1 → prod s0-,s0- → prod s0+,s0+ → prod 1,1

REP> rep
REP> gr so8
Group is SO(8)
REP> au so8, 1
Group is SO(8)
[s;0]-
REP> au so8, last
Group is SO(8)
[s;0]+
REP> au so8, last
Group is SO(8)
[1]
REP> prod 1,1
[2] + [1^2] + [0]
REP> prod s0+,s0+
[1^4]+ + [1^2] + [0]
REP> prod s0-,s0-
[1^4]- + [1^2] + [0]
REP> au so8, prod 1, 1                                                                                                             
Group is SO(8)
[1^4]- + [1^2] + [0]
REP> au so8, last
Group is SO(8)
[1^4]+ + [1^2] + [0]
REP> au so8, last
Group is SO(8)
[2] + [1^2] + [0]
REP>

B.R.Judd が Nieson & Koster の cfp の表に関して、SO(7) の対称性では説明のつかない 0 要素が多くあるから隠れた対称性がまだあるはず、と言っていた正体が、この SO(8) ⊃ SO(7),SO(7)',SO(7)'' の対称性に基づくものだったようです。ここで三つの SO(7) は線形従属で独立なのは２つだそうです。

問３

3. Set the group as $Sp_8$ and use SCHUR to calculate the Kronecker product $< 321>\times< 421>$ .
Now save the result by using the command “setr1 last” . Use the command ”zero” to eliminate
“last” and then issue the command “rv1” and you should once again see the output of the
Kronecker product. Now save the “rvar” to a diskfile using the command “save rvar ’filename’
” as described in Appendix A. If you now issue the command “setr1 zero” you will find on
saying “rvar 1” the data for the Kronecker product has been lost and “rvar 1” simply reports
zero. All is not lost as we can reload, even days later, the saved “rvar1” using the command
load as described in Appendix A. Thus issuing the command “load rvar’filename’ ” followed
by the command “rv1” will put to screen once again the Kronecker product. Note however,
we must set the group as appropriate to that of the saved rvar.

file への保存・読み出し

マニュアルには、"setr1 xxx" と書いてありますが、"setrv1 xxx" と「ｖ」を付け加えないとうまくゆかないです。

注：setrv 1 xxx はいいですが set rv1 xxx は許されません。

REP> gr sp8
Group is Sp(8)
REP> prod 321,421
<742> + <741^2> + <74> + <73^2> + 2<7321> + 3<731> + <72^3> + 2<72^2> + 3<721^2> + 2<72> + 2<71^2> + <7> + <652> + <651^2> + <65>
 + 2<643> + 4<6421> + 6<641> + 3<63^2 1> + 3<632^2> + 9<632> + 9<631^2> + 6<63> + 8<62^2 1> + 12<621> + 5<61^3> + 5<61> + <5^2 3>
 + 2<5^2 21> + 3<5^2 1> + <54^2> + 4<5431> + 3<542^2> + 10<542> + 9<541^2> + 6<54> + 3<53^2 2> + 7<53^2> + 18<5321> + 21<531>
 + 6<52^3> + 15<52^2> + 18<521^2> + 12<52> + 12<51^2> + 3<5> + <4^3 1> + 2<4^2 32> + 5<4^2 3> + 10<4^2 21> + 11<4^2 1> + <43^3>
 + 9<43^2 1> + 9<432^2> + 21<432> + 20<431^2> + 12<43> + 19<42^2 1> + 26<421> + 11<41^3> + 10<41> + 3<3^3 2> + 5<3^3> + 14<3^2 21>
 + 15<3^2 1> + 6<32^3> + 14<32^2> + 17<321^2> + 11<32> + 12<31^2> + 3<3> + 6<2^3 1> + 9<2^2 1> + 5<21^3> + 5<21> + 2<1^3> + <1>
REP> setrv1 last
REP> zero
zero
REP> last
zero
REP> rv1
<742> + <741^2> + <74> + <73^2> + 2<7321> + 3<731> + <72^3> + 2<72^2> + 3<721^2> + 2<72> + 2<71^2> + <7> + <652> + <651^2> + <65>
 + 2<643> + 4<6421> + 6<641> + 3<63^2 1> + 3<632^2> + 9<632> + 9<631^2> + 6<63> + 8<62^2 1> + 12<621> + 5<61^3> + 5<61> + <5^2 3>
 + 2<5^2 21> + 3<5^2 1> + <54^2> + 4<5431> + 3<542^2> + 10<542> + 9<541^2> + 6<54> + 3<53^2 2> + 7<53^2> + 18<5321> + 21<531>
 + 6<52^3> + 15<52^2> + 18<521^2> + 12<52> + 12<51^2> + 3<5> + <4^3 1> + 2<4^2 32> + 5<4^2 3> + 10<4^2 21> + 11<4^2 1> + <43^3>
 + 9<43^2 1> + 9<432^2> + 21<432> + 20<431^2> + 12<43> + 19<42^2 1> + 26<421> + 11<41^3> + 10<41> + 3<3^3 2> + 5<3^3> + 14<3^2 21>
 + 15<3^2 1> + 6<32^3> + 14<32^2> + 17<321^2> + 11<32> + 12<31^2> + 3<3> + 6<2^3 1> + 9<2^2 1> + 5<21^3> + 5<21> + 2<1^3> + <1>
REP> save rvar '6-3-3.rvar'
save[1]
REP> setrv1 zero
REP> rvar 1
zero
REP> rv1
zero
REP> load rvar '6-3-3.rvar'
load[1]
REP> rvar 1
<742> + <741^2> + <74> + <73^2> + 2<7321> + 3<731> + <72^3> + 2<72^2> + 3<721^2> + 2<72> + 2<71^2> + <7> + <652> + <651^2> + <65>
 + 2<643> + 4<6421> + 6<641> + 3<63^2 1> + 3<632^2> + 9<632> + 9<631^2> + 6<63> + 8<62^2 1> + 12<621> + 5<61^3> + 5<61> + <5^2 3>
 + 2<5^2 21> + 3<5^2 1> + <54^2> + 4<5431> + 3<542^2> + 10<542> + 9<541^2> + 6<54> + 3<53^2 2> + 7<53^2> + 18<5321> + 21<531>
 + 6<52^3> + 15<52^2> + 18<521^2> + 12<52> + 12<51^2> + 3<5> + <4^3 1> + 2<4^2 32> + 5<4^2 3> + 10<4^2 21> + 11<4^2 1> + <43^3>
 + 9<43^2 1> + 9<432^2> + 21<432> + 20<431^2> + 12<43> + 19<42^2 1> + 26<421> + 11<41^3> + 10<41> + 3<3^3 2> + 5<3^3> + 14<3^2 21>
 + 15<3^2 1> + 6<32^3> + 14<32^2> + 17<321^2> + 11<32> + 12<31^2> + 3<3> + 6<2^3 1> + 9<2^2 1> + 5<21^3> + 5<21> + 2<1^3> + <1>
REP>

fortran66のブログ

fortran について書きます。

Schur マニュアルの問題を解く　その１

Schur 付属のマニュアルの問題を解きます

6-3節

問１

問２

問３