COMPUTING CHARACTER TABLES OF FINITE GROUPS · COMPUTING CHARACTER TABLES OF FINITE GROUPS Jay...
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COMPUTING CHARACTER TABLES OF FINITE GROUPS
Jay Taylor(Università degli Studi di Padova)
Symmetry
Symmetry
Chemistry
Symmetry
Chemistry Biology
Symmetry
Chemistry Biology Physics
Symmetry
Chemistry Biology Physics
Groups
Symmetry
Chemistry Biology Physics
Groups
NumberTheory
Symmetry
Chemistry Biology Physics
Groups
NumberTheory Topology
Symmetry
Chemistry Biology Physics
Groups
NumberTheory Topology Algebraic
Geometry
DEFINITION
DEFINITIONA group is a pair with a set and a binary operation such that:
(G, ?) G ? : G⇥G ! G
DEFINITIONA group is a pair with a set and a binary operation such that:1. there exists an element such that for
all x 2 G
(G, ?) G ? : G⇥G ! G
e 2 Gx ? e = e ? x = x
DEFINITIONA group is a pair with a set and a binary operation such that:1. there exists an element such that for
all2. for every there exists an element such that
.
x 2 G
(G, ?) G ? : G⇥G ! G
e 2 Gx ? e = e ? x = x
g 2 G g-1 2 Gg ? g-1 = g-1 ? g = e
DEFINITIONA group is a pair with a set and a binary operation such that:1. there exists an element such that for
all2. for every there exists an element such that
.3. for all
x 2 G
(G, ?) G ? : G⇥G ! G
e 2 Gx ? e = e ? x = x
g 2 G g-1 2 Gg ? g-1 = g-1 ? g = e
a ? (b ? c) = (a ? b) ? c a, b, c 2 G.
DEFINITIONA group is a pair with a set and a binary operation such that:1. there exists an element such that for
all2. for every there exists an element such that
.3. for all
x 2 G
(G, ?) G ? : G⇥G ! G
e 2 Gx ? e = e ? x = x
g 2 G g-1 2 Gg ? g-1 = g-1 ? g = e
a ? (b ? c) = (a ? b) ? c a, b, c 2 G.
Examples
• with ,(Z,+) Z = {. . . ,-2,-1, 0, 1, 2, . . . }
DEFINITIONA group is a pair with a set and a binary operation such that:1. there exists an element such that for
all2. for every there exists an element such that
.3. for all
x 2 G
(G, ?) G ? : G⇥G ! G
e 2 Gx ? e = e ? x = x
g 2 G g-1 2 Gg ? g-1 = g-1 ? g = e
a ? (b ? c) = (a ? b) ? c a, b, c 2 G.
Examples
• with ,• with where denote the real numbers.
(Z,+) Z = {. . . ,-2,-1, 0, 1, 2, . . . }
(R⇥,⇥) R⇥ = R \ {0} R
DIHEDRAL GROUPS
DIHEDRAL GROUPS
DIHEDRAL GROUPS1 2
34
DIHEDRAL GROUPS
1
23
4
DIHEDRAL GROUPS1 2
34
DIHEDRAL GROUPS1 2
34
DIHEDRAL GROUPS1 2
34
DIHEDRAL GROUPS
1 2
34
DIHEDRAL GROUPS1 2
34
DIHEDRAL GROUPS1 2
34
DIHEDRAL GROUPS1 2
34
DIHEDRAL GROUPS1
23
4
DIHEDRAL GROUPS1
23
4
90o
DIHEDRAL GROUPS4 1
23
DIHEDRAL GROUPS4 1
23
90o
DIHEDRAL GROUPS
4
12
3
90o
DIHEDRAL GROUPS
4
12
3
90o
DIHEDRAL GROUPS2 1
43
90o
DIHEDRAL GROUPS2 1
43
90o
DIHEDRAL GROUPS
2 1
43
90o
DIHEDRAL GROUPS
2 1
43
180o
90o
DIHEDRAL GROUPS3 4
12
90o
DIHEDRAL GROUPS3 4
12
90o 180o
DIHEDRAL GROUPS
3
41
2
90o 180o
DIHEDRAL GROUPS
3
41
2
90o 180o
DIHEDRAL GROUPS1 4
32
90o 180o
DIHEDRAL GROUPS1 4
32
90o 180o
DIHEDRAL GROUPS
1 4
32
90o 180o
DIHEDRAL GROUPS
1 4
32
90o 180o
270o
DIHEDRAL GROUPS
90o 180o
DIHEDRAL GROUPS
90o 180o 270o
DIHEDRAL GROUPS
90o
180o 270o
DIHEDRAL GROUPS
90o
180o 270o
a b
DIHEDRAL GROUPS
90o
180o 270o
a b ab
DIHEDRAL GROUPS
90o
180o 270o
a b ab aba
DIHEDRAL GROUPS
90o
180o 270o
a b ab aba
abab
DIHEDRAL GROUPS
90o
180o 270o
a b ab aba
abab ababa
DIHEDRAL GROUPS
90o
180o 270o
a b ab aba
abab ababa ababab
DIHEDRAL GROUPS
90o
180o 270o
a b ab aba
abab ababa ababab e
DIHEDRAL GROUPS
90o
180o 270o
a b ab aba
abab ababa ababab e
a2 = e,
DIHEDRAL GROUPS
90o
180o 270o
a b ab aba
abab ababa ababab e
a2 = e, b2 = e,
DIHEDRAL GROUPS
90o
180o 270o
a b ab aba
abab ababa ababab e
a2 = e, b2 = e, (ab)4 = e
DIHEDRAL GROUPS
90o
180o 270o
a b ab aba
abab ababa ababab e
a2 = e, b2 = e, (ab)4 = eiI2(4) = ha, b |
DIHEDRAL GROUPS
DIHEDRAL GROUPS
72o 144o
216o 288o
a b ab aba
(ab)3
(ab)2
(ab)2a (ab)3a (ab)4 e
DIHEDRAL GROUPS
72o 144o
216o 288o
a b ab aba
(ab)3
(ab)2
(ab)2a (ab)3a (ab)4 e
I2(5) = ha, b | a2, b2, (ab)5i
DIHEDRAL GROUPS
72o 144o
216o 288o
a b ab aba
(ab)3
(ab)2
(ab)2a (ab)3a (ab)4 e
I2(m) = ha, b | a2, b2, (ab)mi
MATRIX REPRESENTATION
MATRIX REPRESENTATION
v1
v2
MATRIX REPRESENTATION
v1
v2
MATRIX REPRESENTATION
v1
v2 v1
v2
MATRIX REPRESENTATION
v1
v2 v1
v2
0 11 0
�
a
MATRIX REPRESENTATION
v1
v2
0 11 0
�
a
MATRIX REPRESENTATION
v1
v2
0 11 0
�
a
MATRIX REPRESENTATION
v1
v2
v1v2
0 11 0
�
a
MATRIX REPRESENTATION
v1
v2
v1v2
0 11 0
� 1 00 -1
�
a b
MATRIX REPRESENTATION
v1
v2
0 11 0
� 1 00 -1
�
a b
MATRIX REPRESENTATION
v1
v2
0 11 0
� 1 00 -1
�
a b
MATRIX REPRESENTATION
v1
v2
v1
v2
0 11 0
� 1 00 -1
�
a b
MATRIX REPRESENTATION
v1
v2
v1
v2
0 11 0
� 1 00 -1
� 0 -11 0
�
a b ab
MATRIX REPRESENTATION0 11 0
� 1 00 -1
� 0 -11 0
�
a b ab
-1 00 1
�
aba
-1 00 -1
� 0 -1-1 0
� 0 1-1 0
� 1 00 1
�
abab ababa ababab e
REPRESENTATIONS
REPRESENTATIONS
• an -dimensional -vector space.V n C
REPRESENTATIONS
• an -dimensional -vector space.
• the group of all invertible linear transformations .
V n C
GL(V) ⇠= GLn(C)f : V ! V
REPRESENTATIONS
• an -dimensional -vector space.
• the group of all invertible linear transformations .
• A representation of a group is a map such that for all we have
(G, ?) ⇢ : G ! GL(V)
a, b 2 G
V n C
GL(V) ⇠= GLn(C)f : V ! V
⇢(a ? b) = ⇢(a) � ⇢(b)
CHARACTERS• Let be a representation of a group . The
function defined by
is called the character of .
⇢ : G ! GLn(C) (G, ?)
�⇢ : G ! C
�⇢(a) = Tr(⇢(a))
⇢
CHARACTERS• Let be a representation of a group . The
function defined by
is called the character of .
e a b ab aba abab ababa ababab
0 11 0
�1 00 1
� 1 00 -1
� 0 -11 0
� -1 00 1
� -1 00 -1
�0 -1-1 0
� 0 1-1 0
�
2 0 0 0 0 0 0-2
⇢ : G ! GLn(C) (G, ?)
�⇢ : G ! C
�⇢(a) = Tr(⇢(a))
⇢
CONJUGACY
CONJUGACY• For any matrices recall thatA, B 2 GLn(C)
Tr(ABA-1) = Tr(B)
CONJUGACY• For any matrices recall that
• Hence for any two elements and any character we have
A, B 2 GLn(C)
Tr(ABA-1) = Tr(B)
a, b 2 G � : G ! C
�(a ? b ? a-1) = �(b)
CONJUGACY• For any matrices recall that
• Hence for any two elements and any character we have
• We say are conjugate if there exists an element such that .
A, B 2 GLn(C)
Tr(ABA-1) = Tr(B)
a, b 2 G � : G ! C
�(a ? b ? a-1) = �(b)
a, b 2 G x 2 G
x ? a ? x-1 = b
CONJUGACY• For any matrices recall that
• Hence for any two elements and any character we have
• We say are conjugate if there exists an element such that .
• This defines an equivalence relation on . The resulting equivalence classes are called conjugacy classes.
A, B 2 GLn(C)
Tr(ABA-1) = Tr(B)
a, b 2 G � : G ! C
�(a ? b ? a-1) = �(b)
a, b 2 G x 2 G
x ? a ? x-1 = b
G
CONJUGACY
90o 180o270o
a b
ab
aba
abab
ababa
ababab e
IRREDUCIBLE CHARACTERS
IRREDUCIBLE CHARACTERS• A representation is irreducible if there is no proper
subspace which is invariant under . By this we mean that for all we have .
⇢ : G ! GL(V)
W ✓ V G
⇢(g)W ✓ Wg 2 G
IRREDUCIBLE CHARACTERS• A representation is irreducible if there is no proper
subspace which is invariant under . By this we mean that for all we have .
• We have is irreducible if and only if
⇢ : G ! GL(V)
W ✓ V G
⇢(g)W ✓ W
1
|G|
X
g2G
�⇢(g)�⇢(g) = 1
g 2 G
⇢ : G ! GL(V)
IRREDUCIBLE CHARACTERS• A representation is irreducible if there is no proper
subspace which is invariant under . By this we mean that for all we have .
• We have is irreducible if and only if
• A character with this property is also called irreducible.
⇢ : G ! GL(V)
W ✓ V G
⇢(g)W ✓ W
1
|G|
X
g2G
�⇢(g)�⇢(g) = 1
g 2 G
⇢ : G ! GL(V)
IRREDUCIBLE CHARACTERS
a b
IRREDUCIBLE CHARACTERS
a b
1
8(22 + 0+ 0+ 0+ 0+ (-2)2 + 0+ 0) = 1
CHARACTER TABLES
Theorem
The number of distinct irreducible characters of a finite group is equal to the number of conjugacy classes.
CHARACTER TABLES
Theorem
The number of distinct irreducible characters of a finite group is equal to the number of conjugacy classes.
• Let be representatives for the conjugacy classes and let be the irreducible characters of . The square matrix
is called the character table of .
g1, . . . , gn 2 G
G�1, . . . ,�n
G
(�i(gj))16i,j6n
CHARACTER TABLES
�1
e a b ab (ab)2
�2
�3
�4
1 1 1 1 1
1 1 1
1 1 1
1 1 1
2 0 0 0 -2
-1 -1
-1
-1 -1
I2(4)
-1
CHARACTER TABLES
�1
e a b
�2
�3
�4
1 1 1 1 1
1 1
1 1
1 1 1
2 0 0
-1
-1
-1 -1
I2(2m) (ab)m(ab)r
(-1)r
(-1)r
(-1)m
(-1)m
j 2(-1)j"jr + "-jr
CHARACTER TABLES
�1
e a b
�2
�3
�4
1 1 1 1 1
1 1
1 1
1 1 1
2 0 0
-1
-1
-1 -1
I2(2m) (ab)m(ab)r
(-1)r
(-1)r
(-1)m
(-1)m
j 2(-1)j"jr + "-jr
1 6 j, r 6 m- 1 " = e⇡i/m
SYMMETRIC GROUPS
SYMMETRIC GROUPS
• is the group of all bijective functions .Sn f : {1, . . . , n} ! {1, . . . , n}
SYMMETRIC GROUPS
• is the group of all bijective functions .
• We call the symmetric group on points.
Sn f : {1, . . . , n} ! {1, . . . , n}
Sn n
SYMMETRIC GROUPS
• is the group of all bijective functions .
• We call the symmetric group on points.
• which can be very large even for small . For example
Sn f : {1, . . . , n} ! {1, . . . , n}
Sn n
|Sn| = n! n
|S20| = 2432902008176640000
SYMMETRIC GROUPS
SYMMETRIC GROUPS
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
Example ( )n = 3
SYMMETRIC GROUPS
�
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
Example ( )n = 3
SYMMETRIC GROUPS
� =
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
Example ( )n = 3
SYMMETRIC GROUPS
� = =
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
Example ( )n = 3
SYMMETRIC GROUPS• A function is called a cycle of length if there exists a
subset such that for any integer and acts on the elements of in the following wayi 62 X
x1
xk
xk-1
x4
x3
x2
f
f
f
f
f
f
f
f 2 Sn k
X = {x1, . . . , xk} ✓ {1, . . . , n} f(i) = i
f X
SYMMETRIC GROUPS
Lemma
Every element of is a product of disjoint cycles.Sn
SYMMETRIC GROUPS
Lemma
Every element of is a product of disjoint cycles.Sn
=
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
�
1
2
3
4
1
2
3
4
SYMMETRIC GROUPS
SYMMETRIC GROUPS• A partition of is a sequence of integers such
that and .n µ = (µ1, . . . , µk)
µ1 + · · ·+ µk = nµ1 > · · · > µk > 1
SYMMETRIC GROUPS• A partition of is a sequence of integers such
that and .
• For example the partitions of are
n µ = (µ1, . . . , µk)
µ1 + · · ·+ µk = nµ1 > · · · > µk > 1
(5), (4, 1), (3, 2), (3, 1, 1), (2, 2, 1), (2, 1, 1, 1), (1, 1, 1, 1, 1)
5
SYMMETRIC GROUPS• A partition of is a sequence of integers such
that and .
• For example the partitions of are
• Given let be a decomposition of into a product of disjoint cycles. If denotes the length of the cycle then the sequence is a partition of , after possibly reordering the entries. We call the cycle type of .
n µ = (µ1, . . . , µk)
µ1 + · · ·+ µk = nµ1 > · · · > µk > 1
(5), (4, 1), (3, 2), (3, 1, 1), (2, 2, 1), (2, 1, 1, 1), (1, 1, 1, 1, 1)
f 2 Sn f1 � · · · � fk f
µi fiµ(f) = (µ1, . . . , µk) n
µ(f) f
5
SYMMETRIC GROUPS
Theorem
Two elements of the symmetric group are conjugate if and only if they have the same cycle type.
SYMMETRIC GROUPS
Theorem
Two elements of the symmetric group are conjugate if and only if they have the same cycle type.
• We will write for the set of all partitions of .P(n) n
SYMMETRIC GROUPS
Theorem
Two elements of the symmetric group are conjugate if and only if they have the same cycle type.
• We will write for the set of all partitions of .
• For any partition we denote by an irreducible character of .
P(n) n
� 2 P(n) ��
Sn
SYMMETRIC GROUPS
Theorem
Two elements of the symmetric group are conjugate if and only if they have the same cycle type.
• We will write for the set of all partitions of .
• For any partition we denote by an irreducible character of .
P(n) n
|P(20)| = 627
� 2 P(n) ��
Sn
HOOKS OF PARTITIONS• Consider the partition .µ = (5, 5, 4, 3, 1, 1) 2 P(19)
HOOKS OF PARTITIONS• Consider the partition .µ = (5, 5, 4, 3, 1, 1) 2 P(19)
HOOKS OF PARTITIONS• Consider the partition .µ = (5, 5, 4, 3, 1, 1) 2 P(19)
HOOKS OF PARTITIONS• Consider the partition .µ = (5, 5, 4, 3, 1, 1) 2 P(19)
HOOKS OF PARTITIONS• Consider the partition .µ = (5, 5, 4, 3, 1, 1) 2 P(19)
H22
HOOKS OF PARTITIONS• Consider the partition .µ = (5, 5, 4, 3, 1, 1) 2 P(19)
h22 = 6
H22
HOOKS OF PARTITIONS• Consider the partition .µ = (5, 5, 4, 3, 1, 1) 2 P(19)
h22 = 6
l22 = 6
H22
HOOKS OF PARTITIONS• Consider the partition .µ = (5, 5, 4, 3, 1, 1) 2 P(19)
h22 = 6
l22 = 6
HOOKS OF PARTITIONS• Consider the partition .µ = (5, 5, 4, 3, 1, 1) 2 P(19)
h22 = 6
l22 = 6
R22
HOOKS OF PARTITIONS• Consider the partition .µ = (5, 5, 4, 3, 1, 1) 2 P(19)
h22 = 6
l22 = 6
R22
µ \ R22
CHARACTER TABLE
Theorem (Murnaghan–Nakayama Formula)
Write as a product of disjoint cycles. Assume is a cycle of length then the element
is contained in the symmetric group . For any partition we have
f 2 Sn
� 2 P(n)
f1 � · · · � fkfk m
g = f1 � · · · � fk-1
Sn-m
��(f) =X
hij=m
(-1)lij��\Rij(g)
CHARACTER TABLE11111 2111 221 311 32 41 5
5
41
32
311
221
2111
11111
1 1 1 1 1 1 1
1
1
1
1
1
1
1
1
1
1
2 1
1
1
0 0 0
0
0
0
0
0
0
4
4
5
5
6
S5
-1 -1
-1 -1
0 -2
-1 -1 -1
-2 -1
-1 -1 -1