# 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

 $e\qquad\to\qquad\begin{pmatrix}1&0\cr 0&1\end{pmatrix}$
 $(ab)\qquad\to\qquad\begin{pmatrix}0&1\cr 1&0\end{pmatrix}$
 $(bc)\qquad\to\qquad\begin{pmatrix}0&\exp(2\pi i/3)\cr\exp(4\pi i/3)&0\end{pmatrix}$
 $(ac)\qquad\to\qquad\begin{pmatrix}0&\exp(4\pi i/3)\cr\exp(2\pi i/3)&0\end{pmatrix}$
 $(abc)\qquad\to\qquad\begin{pmatrix}\exp(2\pi i/3)&0\cr 0&\exp(4\pi i/3)\end{pmatrix}$
 $(acb)\qquad\to\qquad\begin{pmatrix}\exp(4\pi i/3)&0\cr 0&\exp(2\pi i/3)\end{pmatrix}$

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

 $\begin{pmatrix}\exp(4\pi i/3)&0\cr 0&\exp(2\pi i/3)\end{pmatrix}\begin{pmatrix% }0&\exp(4\pi i/3)\cr\exp(2\pi i/3)&0\end{pmatrix}=\begin{pmatrix}0&\exp(2\pi i% /3)\cr\exp(4\pi i/3)&0\end{pmatrix}$
Title example of induced representation ExampleOfInducedRepresentation 2013-03-22 14:35:43 2013-03-22 14:35:43 rspuzio (6075) rspuzio (6075) 8 rspuzio (6075) Example msc 20C99