5. Representations of the Symmetric Groups Sn

5.1 One-Dimensional Representations 5.2 Partitions and Young Diagrams 5.3 Symmetrizers and Anti-Symmetrizers of Young Tableaux 5.4 Irreducible Representations of Sn

5.5 Symmetry Classes of Tensors

App III & IV

Importance of Sn:

• Cayley's theorem: Every group of order n is isomorphic to a subgroup of Sn

• Construction of IRs of classical groups, e.g., GL(n), U(n), SU(n), etc

• Identical particlesBasic tools:

Young diagrams, Young tableaux

Symmetrizers, anti-symmetrizers,

Irreducible symmetrizers ( Idempotents / Projections )

nth rank tensors on m-D space : Sn + GL(m)

5.1. 1-D Reps

Every Sn has a non-trivial invariant subgroup, the alternating group,

An = { All even permutations } Sn / An C2

Every Sn has two 1-D reps:

Identity rep 1 : D(p) = 1 p

2 : D(p) = (–) p

Theorem 5.1:

Symmetrizerp

s p & antisymmetrizer pp

a p

are essentially idempotent & primitive

Proof:

p

qs q pp

p

s sq nq S

q

ss qsq

s !n s

sqs ss !n s s is essentially idempotent & primitive

pp

qa q p 1q p

p

p

q a

qq

a a qa q

a !n a

qaqa aa !q n a a is essentially idempotent & primitive

0sqa sa nq S s & a generate inequivalent IRs of Sn on n

Basis vectors: : ns q s q S : na q a q S

Since qs s qqa a

1 1D q 2 qD q

5.2. Partitions and Young Diagrams

Definition 5.1: Partition of n , Young Diagram

A partition of integer n is an ordered sequence of integers 1 2, , , r

such that 1 2 r & 1 2 r n

2 partitions & are equal if j j j

if the 1st non-zero member in sequence j j positive

negativeis

A partition is represented graphically by a Young diagram of

n squares arranged in r rows, the jth of which contains j squares

Example 1: n = 3

3 distinct partitions:

{ 3 } : { 2, 1 } { 1, 1, 1 }

Example 2: n = 4

5 distinct partitions:

{ 4 } : { 3, 1 } { 2, 2 }

{ 1, 1, 1,1 }{ 2, 1, 1 }

Partition of n ~ Classes of Sn

Let there be j j-cycles in an element in a class of Sn

1

n

jj

n j

1 2

j j nj j

1 2 3 n

Then

k jj k

1k k 1

n

kk

n

1, ,k k n is a partition of n

( most k = 0 )

Theorem 5.2:

Number of distinct Young diagram for n

= Number of classes in Sn

= Number of inequivalent IRs in Sn

Example: S3 = { e, (123), (132) , (23), (13), (12) }

S3 1

2 3 1 2 3

e 3 0 0 3 0 0

(23), (13), (12) 1 1 0 2 1 0

(123), (132) 0 0 1 1 1 1

e = (1)(2)(3)

(12) = (12)(3)

Definition 5.2: Young Tableau, Normal Tableau, Standard Tableau

A tableau is a diagram filled with a distinct number (1,…,n) in each square.

(i) Young Tableau: Numbers filled with no particular order

(ii) Normal Tableau : Numbers filled consecutively from left to right & top to bottom

(iii) Standard Tableau p : Numbers ordered from left to right in each row & top to bottom in each column

Example: S4

1 2 34

1 23 4

1 2 43

1 32 4

Normal tableaux Standard tableaux p

p p p q pq

p = (3,4)

p = (2,3)

5.3. Symmetrizers & Anti-Symmetrizers of Young Tableaux

Definition 5.3: Horizontal and Vertical Permutations

Let p be any tableau.

A horizontal ( vertical ) permutation h p ( v p ) is a permutation that does not exchange numbers between different rows (columns).

Each cycle in h p ( v p ) must contain numbers that appear in the same row (column).

Definition 5.4: Symmetrizer, Anti-symmetrizer, Irreducible Symmetrizerp p

h

s h Symmetrizer:

vp p

v

a v

Anti-symmetrizer:

,

vp p p p p

h v

e s a h v

Irreducible symmetrizer:

Example: S3

Observations:

1) { h p } and { v p } are each a subgroup of S3.

2) s and a are total (anti-)symmetrizers of these subgroups. Also:

s and a are essentially idempotent.

h

s h h h

h

h

s ( Rearrangement theorem used on subgroup { h } )

vv

a v v v

v v

v

v

v a

h h

s s h h

,h h

h

n s h

s 1! !nn

,

v v

v v

a a v v

,

v

v v

v

v

a n a

3) e are primitive idempotents :

Cases e1 & e3 are obvious. For case e2, see Problem 5.3.

4) { e } generates a set of inequivalent IRs of S3.

Cases e1 & e3 are obvious.

Case e2 is proved by showing that

{ p e2 p S3 } spans a 2–D subspace (left ideal) of 3.

2 2e e e

212 12 12 13 12 321e e

2 12 13 321e e

12 132 13e 2e

223 23 23 12 23 13 23 321e

23 132 123 12 2r

231 31 31 12 31 321e e 31 123 23e 2 2e r

2123 123 123 12 123 31e e 123 13 32 e 2 2e r

2321 321 321 12 321 13 321 321e

321 23 12 123 2r QED

5) Similarly e2(23) also generate a 2-D IR but it is equivalent to that from e2.

The left ideal is however distinct from that of e2. It is spanned by 23

2 32 123 132 13r 232 13 12 231e e &

6) The group algebra is a direct sum of the 4 left ideals generated by the standard tableaux e1 , e2 , e2

(23) and e3 .

The identity can be decomposed as

231 2 2 3

1 1 1 16 3 3 6

e e e e e

DR is fully reduced by the e 's of the standard tableaux

Summary of the lemmas proved in Appendix IV:

Lemma IV.1: xp = p x p–1

Lemma IV.2: For a given tableau

{ h } & { v } are each a subgroup of Sn.

h s s h s vv a a v a

vh e v e

s s s a a a

,h v

, Z

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

Lemma IV.4: Given and p Sn

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

Lemma IV.5:

Given and r . vh r v r

,h v r e

Lemma IV.6:

Given 2 distinct diagrams > ,

0q p p qa s s a

0q pe e , np q S

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.

e pp

r pof

5.4. Irreducible Representations of Sn

( Superscripts p in p, s

p…, are omitted )Theorem 5.3:

s r a e nr S

e e e ( e is essentially idempotent )

For a given Young tableau :

and

Proof:

0

By lemma IV.2 :

h s r a v h s r a v vs r a

nr S

By lemma IV.5 :

s r a e

e where pp

s r a p of

e e s a s a s a s a e

where e of pp

s a s a p e e

Since e s and e a while (–)e = 1, we have αe 0 QED

Theorem 5.4: e IR

e is a primitive idempotent. It generates an IR of Sn on n.

Proof:

e r e s a r s a e ( Theorem 5.3 )nr SBy theorem III.3, e is a primitive idempotent. QED

Theorem 5.5: Equivalent IRs

IRs generated by e and ep , p Sn, are equivalent

Proof:1pe p e p e p p e

p e e p e ( Theorem 5.3 )0

Proof is completed by theorem III.4.

Theorem 5.6: Inequivalent IRs

e & e generate inequivalent IRs if ( different Young diagrams )

Proof:

Let > & p Sn .

1e p e e p e p p pe e p 0 Lemma IV.6

0pe r e r e p e nr S QED by theorem III.4

Example: e1, e2, e3 of S3 generate inequivalent IRs

Corollary: 0p qe e , np q S

Proof:

Case > is proved in lemma IV.6.

Case < is left as exercise.

Theorem 5.7: IRs of Sn

{ e } of all normal tableaux generate all inequivalent IRs of Sn.

Proof:

1. Number of inequivalent IRs = Number of Young diagrams ( theorem 5.2 )

2. Each normal diagram begets an e

3. Every e generates an inequivalent IR ( theorem 5.6 )Theorem 5.8: Decomposition of DR

1. Left ideals La generated by e a 's associated with distinct standard

tableaux are linearly independent.

2.,

n aaL

S

Proof:See W. Miller, "Symmetry Groups & Their Applications", Academic Press (1972)

5.5. Symmetry Classes of Tensors

Let Vm be an m-D vector space with basis { | i , i = 1, …, m }

{ g } be all invertible linear transformations on Vm

{ g } = GL(m, C) = General linear group = Gm

Natural m-D rep of Gm on Vm :jig i j g

Definition 5.5: Tensor Space Vmn

1

nnm m

jV V

| |n factors

V Vm m

g i j = ( i, j ) element of an invertible m m matrix

mg G

Natural basis of Vmn :

1 21

n

n jj

i i i i

1 2 ni i i 1 2 ni i i n

i

1 21 2

ni i inx i i i x n

mx V i

ni x

1 2 ni i ix = tensor components of x

j

in ng i j D g

Natural (nm)-D rep of Gm on Vmn :

1

k

k

nj j

iik

D g g

1 2

1 2

n

n

jj ji i ig g g

mg G

gg x x jgn

j x in

g i x

j iin

j D g x

ii jg jx D g x

Action of Sn on Vmn :

pp x x with1 2 1 2p p pn ni i ii i ipx x

np S

1 2 nnp i p i i i

1 1 11 2 np p pi i i 1p n

i

in

p x p i x 1 21 1 1

1 2

n

n

i i ip p pi i i x 1 2

1 2p p pni i i

ni i i x

1

ip ni x ip

ni x i

pni x ii p

px x

Natural n-D rep of Sn on Vmn :

j

in np i j D p 1p n

i

1p

j ji i

D p

1

pk

k

nji

k

11

k

pk

nji

k

pji

np S

Coming Attractions:

• D[Gm] & D[Sn] are in general reducible.

• D[Sn] can be decomposed using e 's of Sn.

• Since p & g commutes, Gm can be decomposed using e's of n from reduction of Vm

n.

Lemma 5.1: Gm & Sn are symmetry preserving on Vmn

q

q

jji i

D DFor both D[Gm] & D[Sn] :

where 1 2 nq q q qi i i i

1 2

1 2n

n

nq S

q q q

Proof: Follows directly from the explicit form of D[Gm] & D[Sn].

Theorem 5.9: p g = g p g Gm & p Sn

j

in np g i p j D g

1

j

ip nj D g

pj

inj D g

1n p ng p i g i

1p

j

inj D g

pj

inj D g

QED

Example 1: V22

Basis of V22 :

S2 = { e, (12) }

i kp g p i k g g i kk i g g g p

i kp g p i k g g i kk i g g g g p

Preview:

In the decomposition of Vmn using e

p, one gets

1. An irreducible invariant subspace wrt Sn: nT r e r S

2. An irreducible invariant subspace wrt Gm: p p nmT e V

r e L

3. The decomposed Vmn has basis | , , a , where

denotes a symmetry class / Young diagram,

labels the irreducible invariant subspaces under Sn, and

a labels the irreducible invariant subspaces under Gm

Definition 5.6: Tensors of Symmetry p & Tensors of Symmetry Class

Tensors of Symmetry p = p n

me V

Given Young diagram labelled :

Tensors of Symmetry Class = , nn mr e r V S

Given Young tableau p :

Theorem 5.10: Let nT r e r S ( α fixed )

1. T(α) is an irreducible invariant subspace wrt Sn.

2. T(α) realization of Sn on T(α) coincides

with IRs generated by e on n.Proof of 1:

| x T x r e For some nrS

p x p r e T np S QED

Proof of 2:

T 0e

Let { ri e } be a basis of L, then { ri e | } is a basis for T().

If i ip r e p r e jj ir e D p on n

then ji j ip r e r e D p on T() np S QED

s 1!s

p

e pn

= s = total symmetrizer

s sr e e nr S Ls is 1-D

s s nT r e r S se nmV

Ts( ) = totally symmetric tensors

! is n

p

e n p i 1

ip np

i pin

p

i

1!

pi is

p

en

Realization of Sn on Ts( ) is the identity representation D1(p) = 1 p

1!

qis n

q

p e p in

1

1!

qi

p nq

in

1!

p qi

nq

in

1!

qi

nq

in

se np S

Example 2: V23

1. se , 1, 1s

2. 13se , 2,1s

4. se , 4,1s

3. 13se , 3, 1s

s

Ts(1)

Ts(2)

Ts(3)

Ts(4)

4 distinct totally symmetric tensors ( = s ) can be generated:

( Symmetry class s )

Problem 5.6: Symmetry class a exists only in Vmn with m

n.

Each Ts() is invariant under 3

All 4 | s, j, 1 together span a subspace Ts invariant under G2

There is no symmetry class a for V23

Example 3: Totally anti-symmetric tensor

1 & only 1 totally anti-symmetric tensor in Vnn.

n = 2:

12 21 1 11 22 0

n = 3:

11

0

i j k

evenodd

otherwise

if ( i j k ) is permutations of (123)

Example 4: Vm2 , m 2

se i i i i, 1, ,i j m 12se i j i j j i

m ( m+1) / 2 distinct totally symmetric tensors:

m ( m–1) / 2 distinct totally anti-symmetric tensors:

0ae i i 12ae i j i j j i

Example 5: V23 mixed symmetry = m

1 23

m 1 12 314me e e

2 independent irreducible invariant subspaces of tensors with mixed symmetry can be generated.

Normal tableau

23 1 32

m 23 123 31 12

4m me e e e Standard tableau

1.

1 124me e

, 1, 1m1 24

23 23 ,1,1me m

1 24

, 1, 2m

| m,1,1 & | m,1,2 span the 2-D subspace Tm(1), invariant under 3

2.

1 124me e

, 2,1m1 24

23 23 , 2,1me m

1 24

, 2, 2m

| m,2,1 & | m,2,2 span the 2-D subspace Tm(2), invariant under 3

231 ,m m mT span e e

1 ,m m mT span e e 32me V

232 ,m m mT span e e

23 232 ,m m mT span e e

23 32me V

1m m mg e e g T 2g G

, 1, 1 , , 2,1span m m

Tm(1) is invariant under G2.

Ditto

, 1, 2 , , 2, 2span m m

Summary:

The 8-D V23 is decomposed into 4 1-D class s & 2 2-D class m

irreducible subspaces invariant under S3

32 1 2 3 4 1 2s s s s m mV T T T T T T

Basis | , , * for each T() is obtained by applying all standard tableaux

p to a single |

The 8-D V23 is decomposed into 1 4-D class s & 2 2-D class m

irreducible subspaces invariant under G2

32 1 2s m mV T T T

Basis | , *, p for each T (p) is obtained by applying each standard tableaux

p to all |

Theorem 5.11:

T T 1. Either or 0T T

2.

0T T if (different symmetries )

( disjoint )

Proof of 1:

0T T Either

or T() & T() has at least 1 non-zero element in common, i.e.,

s e s e , ns s S

r s e r s e nr S T T

Proof of 2:

T T is also invariant under Sn.

T() = span { ep | } is invariant under Sn

Since T() & T() are irreducible & 0T T

QED

Observations:

Theorem 5.11 implies nmV T

dim T() = Number of standard tableaux * of symmetry

It is permissible to the same D(Sn) for all 's :

, , , , b

ap a b D p np S

Each T() is invariant under Sn

a,b = 1, …, dim T()

Basis | , , * for each T() is obtained by applying ep of

all standard tableaux p to a single |

Theorem 5.12: , *,T a span a is invariant under Gm

IR of Gm on T (a) satisfies g a a D g

mg G

Proof:

Reminder: | , *, a is obtained by applying ea of standard

a to all |

Since r e T

theorem 5.9 g r e r e g T g

ag a g e

nr S

mg G

b

ab D g

ag p e a np S T g

cag p a g c D p b c

c ab D g D p

c

ap g a p c D g

b c

c ab D p D g

b c b c

c a c aD g D p D p D g

D g D p D p D g

np S

Schur's lemma: b baa

D g

QED D g E

Theorem 5.13: IRs of Gm

Reps of Gm on T(a) of Vmn are IRs.

Reminder: | , *, a is obtained by applying ea of standard

a to all |

Outline of Proof:For complete proof, see W. Miller, "Symmetry Groups & Their Applications", Academic Press (72)

Rationale: Since Gm is the largest group that commutes with Sn on T(a) of Vm

n, the operators D(g) should be complete & hence irreducible.

Let A be a linear operator on T(a) :

i i i j

jx y A x

Since x & y belong to the same symmetry class , A must be symmetry preserving, i.e.,

p

p

iij j

A A np S

Lemma 5.1 states that gGm on Vmn are symmetry preserving.

Lemma IV.7 A is a linear combination of D(g).

D(g) is irreducible