5. representations of the symmetric groups s n
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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
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
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