PHY 745 Group Theory
11-11:50 AM MWF Olin 102
Plan for Lecture 5:
Representations, characters, and
the “great” orthogonality theorem
Reading: Chapter 3 in DDJ
1. Finish proof of “Great Orthogonality
Theorem”
2. Character of a representation
3. Great orthogonality theorem for
charactersPHY 745 Spring 2017 -- Lecture 5
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1
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PHY 745 Spring 2017 -- Lecture 5
2
The great orthogonality theorem on unitary irreducible
representations
Notation: h  order of the group
R  element of the group
 ( R )  ith representation of R
i

denote matrix indices
li  dimension of the representation
   ( R)    ( R)
i
R
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*
j
h
  ij  
li
PHY 745 Spring 2017 -- Lecture 5
3
Proof of the great orthogonality theorem
• Prove that all representations can be unitary
matrices
• Prove Schur’s lemma part 1 – any matrix which
commutes with all matrices of an irreducible
representation must be a constant matrix
• Prove Schur’s lemma part 2
• Put all parts together
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PHY 745 Spring 2017 -- Lecture 5
4
Proof of the great orthogonality theorem
• Prove that all representations can be unitary
matrices
• Prove Schur’s lemma part 1 – any matrix which
commutes with all matrices of an irreducible
representation must be a constant matrix
• Prove Schur’s lemma part 2
• Put all parts together
   ( R)    ( R)
i
R
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*
j
h
  ij  
li
PHY 745 Spring 2017 -- Lecture 5
5
Construct a
2

1
matrix
M    2 ( R ) X 1 ( R 1 )
R
Note that:  2 ( S ) M  M 1 ( S )
For
2

1
where S is a member of the group
Schur's lemma shows that M  0
For X  0 except X   1:

2
1
1 
2
1
1

(
R
)
X

(
R
)


(
R
)

(
R
)

 




R
 R

    ( R) 
2
R
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
PHY 745 Spring 2017 -- Lecture 5
  ( R) 
1
*

0
6
Construct a
2

1
matrix
M    2 ( R ) X 1 ( R 1 )
R
Note that:  2 ( S ) M  M 1 ( S )
where S is a member of the group
For 2  1, Schur's lemma shows that M  sI I  ( 1  1 ) identity

1
1
1 
   ( R ) X  ( R )   s
 R

For X  0 except for X   1:
1
1
1

(
R
)

(
R
)  s 
 

R
Here s denotes the scalar constant for the particular choice of X .
Consider    and sum over all  :
1
1
1
1
1
1
1
1
1

(
R
)

(
R
)


(
R
)

(
R
)


(
R
R
)

h

 
 
 


 ( E )
R
R
R
  1 ( R )1 ( R 1 )  h   s   s
R
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
PHY 745 Spring 2017 -- Lecture 5
1
7
Proof continued:
1
1
1

(
R
)

(
R
)  s 
 

R
h   s
 s 
1
h
 
1
   ( R)  ( R
1
1
1
)     ( R )   ( R) 
1
R
1
R
   ( R)    ( R)
i
R
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*
j
*


h
  
1
h
  ij  
li
PHY 745 Spring 2017 -- Lecture 5
8
   ( R)    ( R)
i
*
j
R
h
  ij  
li
Geometric interpretation:
h dimensional vector space should be spanned by representations
i ( R ) each of which consists of li2 components.
  li2  h
i
For P (3) :
1+1+22  6
For P (4) :
1+1+2  3  3  24
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2
2
2
PHY 745 Spring 2017 -- Lecture 5
9
Characters
The character of a representation i for a group element R is defined:
 i ( R)  Tr  i ( R )     i ( R ) 


Note that the character is unique for each irreducible representation.
Suppose M '  U † MU
Tr(M ')  Tr( M )
Details:
†
†
M
'

U
M
U

U
U
 ii  ij jk ki  ki ij M jk   jk M jk  M jj
i
ijk
ijk
jk
j
Note that all members of a class has the same character
for any given representation.
Details:  i ( X 1RX )   ia ( X 1RX )

=  ia  ( X 1 ) i ( R)i a ( X )   i a ( X ) ia  ( X 1 ) i ( R)

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
i
=  
( E ) i ( R)   i ( R)  i ( R)

PHY 745 Spring
2017 -- Lecture 5

10
Great orthogonality theorem for characters
   ( R)    ( R)
*
i
j
R
Let    and   
   ( R)   ( R)
*
i
j
R
h
  ij  
li
and perform summations
h
  ij     
li 
j

(
R
)

 ( R)  h ij

*
i
R
In terms of classes C , each with N C elements :
j
N

(
C
)

 (C )  h ij
 C
i
C
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*
PHY 745 Spring 2017 -- Lecture 5
11
Example – P(3):
Classes:
C1  E
C2  A, B, C
C3  D, F
1 ( A)  1 ( B)  1 (C )  1 ( D)  1 ( E )  1 ( F )  1
 2 ( A)   2 ( B)   2 (C )  1
1 0
3
 E  

0
1


1


3  C    2
 3
 2
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 2 ( E )   2 ( D)   2 ( F )  1
1 0 
3
  A  

0

1



3
2
1
2
1


3  B    2
 3
 2
3 
1
 3
  12


2
   D  
 3  F    2

 3  1 
 3
2

 2
 2
PHY 745 Spring 2017 -- Lecture 5
3
2
1
2






1 
2 

3
2
12
Character table for P(3):
1 ( A)  1 ( B)  1 (C )  1 ( D)  1 ( E )  1 ( F )  1
 2 ( A)   2 ( B)   2 (C )  1
1 0
3
 E  

0
1


1


3  C    2
 3
 2
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1


3  B    2
 3
 2
1 0 
3
  A  

0

1



3
2
1
2
 3
  12
   D  

 3

 2
C1  E
Classes:
 2 ( E )   2 ( D)   2 ( F )  1
 3
  12
  F   
 3
 12 
 2
3
2
C2  A, B, C
1
2






 12 

3
2
C3  D, F
C1
3C2
2C3
1
1
1
1
2
1
-1
1
3
2
0
-1
PHY 745 Spring 2017 -- Lecture 5
3
2
13
Check orthogonality:
 N   (C )  
*
i
C
C
j
(C )  h ij
C1
3C2
2C3
1
1
1
1
2
1
-1
1
3
2
0
-1
2
N

(
C
)

 (C )  1  2 1  3  0  (1)  2  (1) 1  0
 C
3
*
C
3
N

(
C
)

 (C )  1  2  2  3  0  0  2  (1)  (1)  6
 C
3
*
C
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PHY 745 Spring 2017 -- Lecture 5
14
Some further conclusions:
j
N

(
C
)

 (C )  h ij
 C
*
i
C
The characters i behave as a vector space with the
dimension equal to the number of classes.
The number of characters=the number of classes
Second character identity:
h
i   (Ck )   (Cl )  N  kl
Ck
i
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*
i
PHY 745 Spring 2017 -- Lecture 5
15
The regular representation is composed of h x h
matrices constructed as follows, shown with the P(3)
example:
100000
reg
reg (E)= 0 1 0 0 0 0  (A)=
E-1
A-1
B-1
C-1
D-1
F-1
EABCDF
EABCDF
AEDFBC
BFEDCA
CDFEAB
FBCAED
DCABFE
reg
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(D)=
001000
000100
000010
000001

reg (B)=
000010
001000
000100
010000
000001
100000
001000
000010
100000
000001
010000
000100
reg
(F)=
reg
(C)=
000001
000100
010000
001000
100000
000010
PHY 745 Spring 2017 -- Lecture 5
010000
100000
000001
000010
000100
001000
000100
000001
000010
100000
001000
010000
16
Regular representation continued –
Note that the regular representation matrices satisfy the
multiplication table of the group and have the same class
structure as the group.
Since the characters of the irreducible representations form a spanning
vector space, we can decompose  reg ( R) into a linear combination:
*
1
i
where ai     ( R )   reg ( R )
h R
 ( R )   ai  ( R)
reg
i
i
By construction:
 reg ( R)  h RE
Also note:  i ( E ) =
i
  reg ( R )   i  i ( R )
i
 reg ( E )   i  i ( E )
i
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so that ai 
 h=
i
2
i
i
PHY 745 Spring 2017 -- Lecture 5
17
Example for P(3):
C1  E
Classes:
C2  A, B, C
C1
3C2
2C3
1
1
1
1
2
1
-1
1
3
2
0
-1
C3  D, F
 reg ( R )   i  i ( R)
i
 reg ( E )   i  i ( E )
 6=1  1  4
i
 reg (C2 )   i  i (C2 )
 0=1  1  0
i
 reg (C3 )   i  i (C3 )
 0=1+1  2
i
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PHY 745 Spring 2017 -- Lecture 5
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