App IV. Supplements to the Theory of Symmetric Groups

Lemma IV.1: xp = p x p–1

Let { h } & { v } be horizontal & vertical permutations of Young tableaux .

Let s, a & e be the associated symmetrizer, anti-symmetrizer, & irreducible symmetrizer, respectively.

Then the corresponding quantities for the Young tableau p are given by

1ph p h p 1pv p v p

1ps p s p 1pa p a p

1pe p e p

Proof: Let x be h, v, s, a, or e.

p p px x p p x px p p x 1px p x p

QED

Hence, only need be considered explicitly.

Lemma IV.2:

1. For a given tableau , { h } & { v } are each a subgroup of Sn.

2. The following identities hold:

h s s h s vv a a v a

vh e v e

s s s a a a

,h v

, Z

Proof:

1. { h } = Sm where m n. Ditto { v }.

2.h

h s h h

h

h

s (Rearrangement theorem)

vv

v a v v

v v

v

v

va

h e v h s a v vs a

s h

a v

ve

h

s s s h h

s !n s

vv

a a a v

v

a !n a

Lemma IV.3:

Given and p Sn. at least 2 numbers in one row of which appear in

the same column of p.

p h v

Proof of :

If p h v then 1p h v h h hv h 1hv h v h

Hence p p hv h h hv

p can be obtained from in 2 steps:

1. hh

2.

h phv

which permutes numbers in same row of

which permutes numbers in same column of p

p can't move 2 numbers in one row of to the same column of p.

Negation of this completes the proof.

Proof of :

Assume no 2 numbers are shared by a row of and a column of p

p can be obtained from as followings:

Starting from the 1st column of p

.

Since these numbers are in different rows of , they can be brought to the 1st column by a horizontal permutation.

Repeating the procedure for the other columns of p

, we geth , which differs from

p by a vertical permutation.

h hp v

1h v h h h v

i.e., p h v

Negation of this completes the proof.

Lemma IV.4:

Given and p Sn.

p h v ,h v p h p v ( ~ denotes transpositions )

Proof:

By Lemma IV.3,

p h v (a,b) in the same row of & the same column of p

Let t be the transposition of (a,b).

t h and pt v

Let pt h v 1p v p 1v p t p

1h p v t p p t p tt p p QED

Lemma IV.5:

Given and r . vh r v r

,h v r e

Proof: rG pp G

r p

qq

h r v h q v 1 1h p vp

p

p h q v

1 1q h p v

v r v pp G

p

where

1 1

v

ph p v

,h v

Lemma IV.4:

p h v p h p v

1 1 ph p v v p p

1 1h p v

0p

If p h v 1 1h p v e then ve p

,

v

eh v

r h v

e s a e eQED

Lemma IV.6:

Given 2 distinct diagrams > ,

1. 0q p p qa s s a

0q pe e 2. , np q S Proof:

Let r be the permutation that brings the numbers of the 2 tableaux to the same sequential order ( 1st row left to right, then 2nd row … )

Since the diagrams are distinct, r h v

By lemma IV.3, at 1 pair of numbers that appears simultaneously in one row of

p and one column of q.

Let t be the transposition of these 2 numbers. p qt h v

By lemma IV.2, p p pt s s t s tq q q qt a a t a a

p q p qs a s t a p qs a 0 p qs t a

q p q pa s a t s q pa s 0 q pa t s

q p q q p pe e s a s a q q p ps a s a 0

Lemma IV.7:

The linear group transformations on Vmn ,

k

k

j ji mi

k

D g g g G

spans the space K of all symmetry-preserving linear transformations.

Proof:A K

p

p

ii

j jA A

np S

mg G K Obviously,

A necessary & sufficient condition for { g Gm } to span K is that

0L g mg G L = 0

where L is a linear functional

j i

i jA L A L A: nmL V C by

Since A K

The symmetry-preserving version of L is

1

!p

pn

jj

i ip S

L Ln

L A L A We can assume L L

ij

i jL g L D g a b c r s t

r s t a b cL g g g

mg G

Let 0L g and consider g g where 1

L g L g a b c r r s s t tr s t a a b b c cL g g g

2a b c r s tr s t a b cn L g g O

Since 0L g mg G

0a b c r s tr s t a b cL g g

is arbitrary 0a b c s tr s t b cL g g

2 311 0

2a bd c r s u t

r s u t a b d cL g n n L g g O

0a b d c u tr s u t d cL g g

Repeating the argument gives 0a b d cr s u tL

QED