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

$$G=\{e,(ab),(ac),(bc),(abc),(acb)\}$$

$$H=\{e,(abc),(acb)\}$$

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

$$ev=v$$

$$(abc)v=\exp (2\pi i/3)v$$

$$(acb)v=\exp (4\pi i/3)v$$

Since $H$ has half as many elements as $G$, there are exactly two cosets, ${\sigma }_1$ and ${\sigma }_2$ in $G/H$ where

$${\sigma }_1=\{e,(abc),(acb)\}$$

$${\sigma }_2=\{(ab),(ac),(bc)\}$$

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 $\{{\sigma }_{1}v,{\sigma }_{2}v\}$.

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 ${\sigma }_1$ and $g_2=(ab)$ as a representative of ${\sigma }_2$. 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 $H$.

$$e\cdot g_1=e=g_1\cdot e$$

$$e\cdot g_2=(ab)=g_2\cdot e$$

$$(ab)\cdot g_1=(ab)=g_2\cdot e$$

$$(ab)\cdot g_2=e=g_1\cdot e$$

$$(bc)\cdot g_1=(bc)=g_2\cdot (acb)$$

$$(bc)\cdot g_2=(abc)=g_1\cdot (abc)$$

$$(ac)\cdot g_1=(ac)=g_2\cdot (abc)$$

$$(ac)\cdot g_2=(acb)=g_1\cdot (acb)$$

$$(abc)\cdot g_1=(abc)=g_1\cdot (abc)$$

$$(abc)\cdot g_2=(bc)=g_2\cdot (acb)$$

$$(acb)\cdot g_1=(acb)=g_1\cdot (acb)$$

$$(acb)\cdot g_2=(ac)=g_2\cdot (abc)$$

We will now compute of the action of $G$ using the formula $g(\sigma v)=\tau (hv)$ given in the definition.

$$e({\sigma }_{1}v)=[e\cdot g_1](ev)={\sigma }_{1}v$$

$$e({\sigma }_{2}v)=[e\cdot g_2](ev)={\sigma }_{2}v$$

$$(ab)({\sigma }_{1}v)=[(ab)\cdot g_1](ev)={\sigma }_{2}v$$

$$(ab)({\sigma }_{2}v)=[(ab)\cdot g_2](ev)={\sigma }_{1}v$$

$$(bc)({\sigma }_{1}v)=[(bc)\cdot g_1]((acb)v)=\exp (4\pi i/3){\sigma }_{2}v$$

$$(bc)({\sigma }_{2}v)=[(bc)\cdot g_2]((abc)v)=\exp (2\pi i/3){\sigma }_{1}v$$

$$(ac)({\sigma }_{1}v)=[(ac)\cdot g_1]((abc)v)=\exp (2\pi i/3){\sigma }_{2}v$$

$$(ac)({\sigma }_{2}v)=[(ac)\cdot g_2]((acb)v)=\exp (4\pi i/3){\sigma }_{1}v$$

$$(abc)({\sigma }_{1}v)=[(abc)\cdot g_1]((abc)v)=\exp (2\pi i/3)({\sigma }_{1}v)$$

$$(abc)({\sigma }_{2}v)=[(abc)\cdot g_2]((acb)v)=\exp (4\pi i/3)({\sigma }_{2}v)$$

$$(acb)({\sigma }_{1}v)=[(acb)\cdot g_1]((acb)v)=\exp (4\pi i/3)({\sigma }_{1}v)$$

$$(acb)({\sigma }_{2}v)=[(acb)\cdot g_2]((abc)v)=\exp (2\pi i/3)({\sigma }_{2}v)$$

Here the square brackets indicate the coset to which the group element inside the brackets belongs. For instance, $[(ac)\cdot g_2]=[(ac)\cdot (ab)]=[(acb)]={\sigma }_1$ since $(acb)\in {\sigma }_1$.

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

Having expressed the answer thus, it is not hard to verify that this is indeed a representation of $G$. For instance, $(acb)\cdot (ac)=(bc)$ and

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