App III. Group Algebra & Reduction of Regular Representations 1. Group Algebra 2. Left Ideals,...

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App III. Group Algebra & Reduction of Regular Representations 1. Group Algebra 2. Left Ideals, Projection Operators 3. Idempotents 4. Complete Reduction of the Regular Representation
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Transcript of App III. Group Algebra & Reduction of Regular Representations 1. Group Algebra 2. Left Ideals,...

App III. Group Algebra & Reduction of Regular Representations

1. Group Algebra

2. Left Ideals, Projection Operators

3. Idempotents

4. Complete Reduction of the Regular Representation

III.1. Group Algebra

Definition III.1: Group Algebra

The group algebra { ; • ,+, } of a finite group { G, • } is the set

,j jj jr g r g G r G C

Together with the algebraic rules:

j jjr q g r q , & ,r q G C

i ji jr q g g r q k i j

k i jg r q wherek

i j k i jg g g

Comments

• { ; • ,+ } is a ring with identity

• { ; +, } is a complex linear vector space spanned by { | gj }

• An inner product can be defined by (we won't be using it):i i

j jg g so that*j k

k jr q g g r q * jjr q

An element r of also serves as an operator on it via • as follows

kj k jrg r g g k m

k j mr g

or k mj m k jr g g r m k m

k jjD r r

so thatj

jr q r g q k m jm k jg r q m j

m jg D r q

m mg r q g rq m j

jD r qm k j

k j r q

Definition III.2: Representation of

Let be the space of linear operators on V.

A rep of on V is a homomorphism

U: r U(r)

that preserves the group algebra structure, i.e.,

Representation of

U r q U r U q

U rq U r U q, & ,r q G C

U() is an irreducible representation (IR) if V has no non-trivial invariant subspace wrt U()

Theorem III.1:

• U is rep of U is rep of G

• U is IR of U is IR of G

III.2. Left Ideals, Projection Operators

( V of DR of G ) =

Since 1

CnRD n D

where D are IRs

& nC = number of classes

( is decomposable)

1 1

C nn

aa

L

G

2

2

2

1

C

C

C

R

n

n

n

D O

D

D

D

O

D

n Blocks n Blocks

p r p r L

L is an invariant subspace:

r L

p

G L is a left ideal.If L doesn't contain a smaller ideal, it is minimal

~ irreducible invar subspace

Minimal left ideals can be found by means of projections (idempotents)

A projection Pa onto the minimal left ideal L

a must satisfy

1.

a aP r L r G i.e., a aP L G

2.

aq L aP q q

3.a aP r r P r G

4.a b ab aP P r P

1

n

L L

The projection onto is P

1

Cn

L

G

III.3. Idempotents

e has a unique decomposition e e

e L

Theorem III.2: P r r e

Proof:

1. P is linear: Proof left as exercise.

2.

r re r e

P r

r

3.

:P r r L

:P r r P P r q P rq rq e

r P q r qe rq e

4. :P P P e e e e e

e e e

P P P

Definition III.3:

{ e } are idempotents if e e e

{ e } are essentially idempotents if e e e c

All results remain valid if P & e are replaced by P & e

, resp.

Definition III.4:

A primitive idempotent generates a minimal left ideal.

Theorem III.3:

An idempotent e is primitive iff re r e c e r G

Proof ( ) :

e is primitive L re r G is a minimal left ideal

& realization of on L is irreducible

Define R by R q q e r e L q G

Rs q R sq sq e r e s q e r e sR q Rs sR

Schur's lemma re r e c e

s G

Proof ( ) :

Let re r e c e e e e r G If e is not primitive

ee e e e e' & e'' are idempotents eee e e e e e e e ce e e e 2c ee 2c e 2c c

0c e e 1c e e e is primitive

Theorem III.4:

Primitive idempotents e1 & e2 generate equivalent IRs iff

1 2 0e r e for some r

Proof () :

Let L1 & L2 be minimal left ideals generated by e1 & e2, resp.

Assume 1 2 0e r e s for some r

Let 1 2:S L L 1 2 1q q q sby

1 1S p q S p q 1p q s 1p q s 1p S q

1q G

S p = p S p

Schur's lemma L1 = L2 so that IRs on them are equivalent

1 1 2 2q e r e L

Proof () :

If the IRs D1 & D2 are equivalent, there exists S such that

1 2SD p D p S p G

or, equivalently, there exists mapping 1 2:S L L S p pS

Let 1 2s S e L

1 1 1S e S e e 1 1Se e 1 1e S e 1e s 1e s

1s e s

2s L 2s se

i.e.

1 2s e s se 1 2 0e se QED

Example: Reduction of DR of G = C3 = { e = a3, a, a2 = a–1 }

i) Idempotent e1 for the identity representation 1 :

1

1

gG

e gn

11

3e a a

Rearrangement theorem 1 1ge e g G

1 1 2,

1

g gG

e e g gn

2,

1

g gG

gn

1

1

gG

en

1e

1 1 1 1 1e g e e e e g G

r G1 1 1 1j

je r e e g r e 1j

j

e r 1r e

Theorem III.3 e1 is primitive

1 is irreducible

ii) Let 12e xe ya za Then

1 2 10 e e x y z e

12e xe ya za

1 12 2e e xe ya za xe ya za 2 2 2 12 2 2x yz e x y z a zx y a

2 2x yz x 22x y z y 22zx y z 0x y z

This can be solved using Mathematica. 4 sets of solutions are obtained:

0x y z

1 1 1, 1 3 , 1 33 6 6

x y i z i

1 1 1, 1 3 , 1 33 6 6

x y i z i

2 1 1, ,3 3 3

x y z

( Discarded )

2 /3 2 /31 1 1, ,3 3 3

i ix y e z e or

2 /3 2 /31 1 1, ,3 3 3

i ix y e z e or

2 1 1, ,3 3 3

x y z 112

3e e a a

e e e e e 112

3ae e a a a e 1 1 11

23

a e e a a e a

112

3e e e e a a 1 1 11 2 1 1

2 2 23 3 3 3

e a a a a e a e a

116 3 3

9e a a e e is indeed idempotent

112

3e ae e a a e 1 1 11 2 1 1

2 2 23 3 3 3

a a e a e a e a a

113 6 3

9e a a ae e is not primitive

2 /3 2 /31 1 1, ,3 3 3

i ix y e z e 1 11

3e e a a

2 /3ie

e e e e e 1 11

3ae e a a a e

1 1 1 11

3a e e a a e a

1 11

3e e e e a a

1 1 1 1 1 1 11 1 1 1

3 3 3 3e a a a a e a e a

1 113 3 3

9e a a

2 1

e e+ is indeed idempotent

e a e a e e 1 e

2 /3 2 /31 1 1, ,3 3 3

i ix y e z e 1 11

3e e a a

a e

1 e

e

1 1e a e a e e 1a e

e e+ is primitive

Changing –1 & e+ e– gives

e e e e a e e 1 1e a e e

e– is a primitive idempotent

1 11

3e e e e a a

a e e a e 1 1 1a e e a e

11

3e e e

0

e a e a e e 1 e e 0

1 1e a e a e e e e 0

e+ & e– generate inequivalent IRs.

1

11

3

1

1

1 1 1

1

1

C e a a basis

e s

e

e

Ex: Check the Orthogonality theorems

e e e Also:

III.4. Complete Reduction of the Regular Representation

Summary:

1.1

Cn

L

G

2. L r e r G

3.1

n

aa

L L

e e e e e

a b ab a re r e e primitive

Reduction of DR Finding all inequivalent ea's.

L is a 2-sided ideal, i.e., r L q r s L ,q s G

A 2-sided ideal is minimal if it doesn't contain another 2-sided ideal.

If a minimal 2-sided ideal L contains a minimal left-sided ideal La ,

then it is a direct sum of all minimal left-sided ideals of the same .

Proof:

Let La and L

b correspond to equivalent IRs ( belong to same ). Then

0a bs s e s e G s ( See proof of Theorem III.4 )

La and L

b are both in the 2-sided ideal L if either of them is.

a a b br L rs r e s e L sHence

Let La and L

b be both in the 2-sided ideal L . Then

b bs L s L G

they generate equivalent rep's. QED

Reduction of DR :

1. Decompose into minimal 2-sided ideals L.

2. Reduce each L into minimal left ideals L

a