example of induced representation

example of induced representation

To understand the definition of

induced representation

, let us work through a simple example in detail.

Let

G

be the group of

permutations

of three objects and let

H

be the

subgroup

of

even permutations

. We have

G

=

{

e

,

(

a

⁢

b

)

,

(

a

⁢

c

)

,

(

b

⁢

c

)

,

(

a

⁢

b

⁢

c

)

,

(

a

⁢

c

⁢

b

)

}

H

=

{

e

,

(

a

⁢

b

⁢

c

)

,

(

a

⁢

c

⁢

b

)

}

Let

V

be the one dimensional

representation

of

H

. Being one-dimensional,

V

is

spanned by

a single basis vector

v

. The action of

H

on

V

is given as

e

⁢

v

=

v

(

a

⁢

b

⁢

c

)

⁢

v

=

exp

⁡

(

2

⁢

π

⁢

i

/

3

)

⁢

v

(

a

⁢

c

⁢

b

)

⁢

v

=

exp

⁡

(

4

⁢

π

⁢

i

/

3

)

⁢

v

Since

H

has half as many

elements

as

G

, there are exactly two cosets,

σ

1

and

σ

2

in

G

/

H

where

σ

1

=

{

e

,

(

a

⁢

b

⁢

c

)

,

(

a

⁢

c

⁢

b

)

}

σ

2

=

{

(

a

⁢

b

)

,

(

a

⁢

c

)

,

(

b

⁢

c

)

}

Since there are two cosets, the

vector space

of the induced representation consists of the

direct sum

of two formal

translates

of

V

. A basis for this space is

{

σ

1

⁢

v

,

σ

2

⁢

v

}

.

We will now compute the action of

G

on this vector space. To do this, we need a choice of coset representatives. Let us choose

g

1

=

e

as a representative of

σ

1

and

g

2

=

(

a

⁢

b

)

as a representative of

σ

2

. As a preliminary step, we shall express the

product

of every element of

G

with a coset representative as the product of a coset representative and an element of

H

.

e

⋅

g

1

=

e

=

g

1

⋅

e

e

⋅

g

2

=

(

a

⁢

b

)

=

g

2

⋅

e

(

a

⁢

b

)

⋅

g

1

=

(

a

⁢

b

)

=

g

2

⋅

e

(

a

⁢

b

)

⋅

g

2

=

e

=

g

1

⋅

e

(

b

⁢

c

)

⋅

g

1

=

(

b

⁢

c

)

=

g

2

⋅

(

a

⁢

c

⁢

b

)

(

b

⁢

c

)

⋅

g

2

=

(

a

⁢

b

⁢

c

)

=

g

1

⋅

(

a

⁢

b

⁢

c

)

(

a

⁢

c

)

⋅

g

1

=

(

a

⁢

c

)

=

g

2

⋅

(

a

⁢

b

⁢

c

)

(

a

⁢

c

)

⋅

g

2

=

(

a

⁢

c

⁢

b

)

=

g

1

⋅

(

a

⁢

c

⁢

b

)

(

a

⁢

b

⁢

c

)

⋅

g

1

=

(

a

⁢

b

⁢

c

)

=

g

1

⋅

(

a

⁢

b

⁢

c

)

(

a

⁢

b

⁢

c

)

⋅

g

2

=

(

b

⁢

c

)

=

g

2

⋅

(

a

⁢

c

⁢

b

)

(

a

⁢

c

⁢

b

)

⋅

g

1

=

(

a

⁢

c

⁢

b

)

=

g

1

⋅

(

a

⁢

c

⁢

b

)

(

a

⁢

c

⁢

b

)

⋅

g

2

=

(

a

⁢

c

)

=

g

2

⋅

(

a

⁢

b

⁢

c

)

We will now compute of the action of

G

using the

formula

g

⁢

(

σ

⁢

v

)

=

τ

⁢

(

h

⁢

v

)

given in the definition.

e

⁢

(

σ

1

⁢

v

)

=

[

e

⋅

g

1

]

⁢

(

e

⁢

v

)

=

σ

1

⁢

v

e

⁢

(

σ

2

⁢

v

)

=

[

e

⋅

g

2

]

⁢

(

e

⁢

v

)

=

σ

2

⁢

v

(

a

⁢

b

)

⁢

(

σ

1

⁢

v

)

=

[

(

a

⁢

b

)

⋅

g

1

]

⁢

(

e

⁢

v

)

=

σ

2

⁢

v

(

a

⁢

b

)

⁢

(

σ

2

⁢

v

)

=

[

(

a

⁢

b

)

⋅

g

2

]

⁢

(

e

⁢

v

)

=

σ

1

⁢

v

(

b

⁢

c

)

⁢

(

σ

1

⁢

v

)

=

[

(

b

⁢

c

)

⋅

g

1

]

⁢

(

(

a

⁢

c

⁢

b

)

⁢

v

)

=

exp

⁡

(

4

⁢

π

⁢

i

/

3

)

⁢

σ

2

⁢

v

(

b

⁢

c

)

⁢

(

σ

2

⁢

v

)

=

[

(

b

⁢

c

)

⋅

g

2

]

⁢

(

(

a

⁢

b

⁢

c

)

⁢

v

)

=

exp

⁡

(

2

⁢

π

⁢

i

/

3

)

⁢

σ

1

⁢

v

(

a

⁢

c

)

⁢

(

σ

1

⁢

v

)

=

[

(

a

⁢

c

)

⋅

g

1

]

⁢

(

(

a

⁢

b

⁢

c

)

⁢

v

)

=

exp

⁡

(

2

⁢

π

⁢

i

/

3

)

⁢

σ

2

⁢

v

(

a

⁢

c

)

⁢

(

σ

2

⁢

v

)

=

[

(

a

⁢

c

)

⋅

g

2

]

⁢

(

(

a

⁢

c

⁢

b

)

⁢

v

)

=

exp

⁡

(

4

⁢

π

⁢

i

/

3

)

⁢

σ

1

⁢

v

(

a

⁢

b

⁢

c

)

⁢

(

σ

1

⁢

v

)

=

[

(

a

⁢

b

⁢

c

)

⋅

g

1

]

⁢

(

(

a

⁢

b

⁢

c

)

⁢

v

)

=

exp

⁡

(

2

⁢

π

⁢

i

/

3

)

⁢

(

σ

1

⁢

v

)

(

a

⁢

b

⁢

c

)

⁢

(

σ

2

⁢

v

)

=

[

(

a

⁢

b

⁢

c

)

⋅

g

2

]

⁢

(

(

a

⁢

c

⁢

b

)

⁢

v

)

=

exp

⁡

(

4

⁢

π

⁢

i

/

3

)

⁢

(

σ

2

⁢

v

)

(

a

⁢

c

⁢

b

)

⁢

(

σ

1

⁢

v

)

=

[

(

a

⁢

c

⁢

b

)

⋅

g

1

]

⁢

(

(

a

⁢

c

⁢

b

)

⁢

v

)

=

exp

⁡

(

4

⁢

π

⁢

i

/

3

)

⁢

(

σ

1

⁢

v

)

(

a

⁢

c

⁢

b

)

⁢

(

σ

2

⁢

v

)

=

[

(

a

⁢

c

⁢

b

)

⋅

g

2

]

⁢

(

(

a

⁢

b

⁢

c

)

⁢

v

)

=

exp

⁡

(

2

⁢

π

⁢

i

/

3

)

⁢

(

σ

2

⁢

v

)

Here the square brackets indicate the coset to which the group element inside the brackets belongs. For

instance

,

[

(

a

⁢

c

)

⋅

g

2

]

=

[

(

a

⁢

c

)

⋅

(

a

⁢

b

)

]

=

[

(

a

⁢

c

⁢

b

)

]

=

σ

1

since

(

a

⁢

c

⁢

b

)

∈

σ

1

.

The results of the calculation may be easier understood when expressed in matrix form

e

→

(

1

0

0

1

)

(

a

⁢

b

)

→

(

0

1

1

0

)

(

b

⁢

c

)

→

(

0

exp

⁡

(

2

⁢

π

⁢

i

/

3

)

exp

⁡

(

4

⁢

π

⁢

i

/

3

)

0

)

(

a

⁢

c

)

→

(

0

exp

⁡

(

4

⁢

π

⁢

i

/

3

)

exp

⁡

(

2

⁢

π

⁢

i

/

3

)

0

)

(

a

⁢

b

⁢

c

)

→

(

exp

⁡

(

2

⁢

π

⁢

i

/

3

)

0

0

exp

⁡

(

4

⁢

π

⁢

i

/

3

)

)

(

a

⁢

c

⁢

b

)

→

(

exp

⁡

(

4

⁢

π

⁢

i

/

3

)

0

0

exp

⁡

(

2

⁢

π

⁢

i

/

3

)

)

Having expressed the answer thus, it is not hard to verify that this is indeed a representation of

G

. For instance,

(

a

⁢

c

⁢

b

)

⋅

(

a

⁢

c

)

=

(

b

⁢

c

)

and

(

exp

⁡

(

4

⁢

π

⁢

i

/

3

)

0

0

exp

⁡

(

2

⁢

π

⁢

i

/

3

)

)

⁢

(

0

exp

⁡

(

4

⁢

π

⁢

i

/

3

)

exp

⁡

(

2

⁢

π

⁢

i

/

3

)

0

)

=

(

0

exp

⁡

(

2

⁢

π

⁢

i

/

3

)

exp

⁡

(

4

⁢

π

⁢

i

/

3

)

0

)

Title

example of induced representation

Canonical name

ExampleOfInducedRepresentation

Date of creation

2013-03-22 14:35:43

Last modified on

2013-03-22 14:35:43

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

8

Author

rspuzio (6075)

Entry type

Example

Classification

msc 20C99

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