Abstract Nonsense

Crushing one theorem at a time

Another Way of Looking at Induced Representations (Pt. I)

Point of Post: In this post we discuss another way to view induced representations and and show that it is actually equivalent to our previous definition since the two different constructed representations of the ambient group will be equivalent.

\text{ }


Of course, as is true for most things in mathematics, the notion of induced representation has several equivalent formulations. In this post we describe another one which may seem, at first, to be a simpler construction than our previous one. In fact, both are equally useful and come up in different aspects of the coming theory. Thus, it is wise to have both of these (among many more) equivalent formulations sitting in our back pocket.

\text{ }

Induced Representations Again

\text{ }

Let G be a finite group and H\leqslant G. Assume that there is a representation \psi:H\to\mathcal{U}\left(\mathscr{V}\right). We begin our construction by fixing a transversal \mathscr{T} for the set of cosets G/H. We then consider then the direct sum of (G:H) copies of \mathscr{V} indexed by the elements of \mathscr{T} where each copy is distinguished. Said more concretely for each t\in\mathscr{T} we define \mathscr{V}_t to be equal to \mathscr{V} and consider the space \displaystyle \bigoplus_{t\in\mathscr{T}}\mathscr{V}_t given the usual inner product structure on direct sum spaces with a normalization factor of |G|^{-1}–we will think of the elements of this space as \displaystyle \sum_{t\in\mathscr{T}}x_tv_t where v_t\in\mathscr{V}_t and x_t is a formal symbol ‘tagging’ v_t. We then note that since the set \left\{Ht\right\}_{t\in\mathscr{T}} forms a partition of G for each g\in G and t\in\mathscr{T} there exists unique h\in H and t'\in\mathscr{T} such that gt=t'h, we denote this t' by P_1(g,t) and this h as P_2(g,t). We then define, for each g\in G the map

\text{ }

\displaystyle \rho_g:\bigoplus_{t\in\mathscr{T}}\mathscr{V}_t\to\bigoplus_{t\in\mathscr{T}}\mathscr{V}_t:\sum_{t\in\mathscr{T}}x_t v_t\mapsto \sum_{t\in\mathscr{T}}x_{P_1(g,t)}\psi_{P_2(g,t)}(v_t)

\text{ }

We claim that in fact \rho_g is in \displaystyle \mathcal{U}\left(\bigoplus_{t\in\mathscr{T}}\mathscr{V}_t\right). Indeed, to see that \rho_g is a linear transformation we note that for any z,z'\in\mathbb{C} and \displaystyle \sum_{t\in\mathscr{T}}x_t v_t,\sum_{t\in\mathscr{t}}x_t v'_t

\text{ }

\displaystyle \begin{aligned}\rho_g\left(z\sum_{t\in\mathscr{T}}x_t v_t+z'\sum_{t\in\mathscr{T}}x_t v'_t\right) &= \rho_g\left(\sum_{t\in\mathscr{T}}x_t\left(xv_t+z'v'_t\right)\right)\\ &= \sum_{t\in\mathscr{T}}x_{P_1(g,t)}\psi_{P_2(g,t)}\left(zv_t+z'v'_t\right)\\ &= z\sum_{t\in\mathscr{T}}x_{P_1(g,t)}\psi_{P_1(g,t)}(v_t)+z'\sum_{t\in\in\mathscr{T}}x_{P_1(g,t)}\psi_{P_2(g,t)}(v'_t)\\ &= z\rho_g\left(\sum_{t\in\mathscr{T}}x_t v_t\right)+z'\rho_g\left(\sum_{t\in\mathscr{T}}x_t v'_t\right)\end{aligned}

\text{ }

To see that it’s unitary we merely note that


\displaystyle \begin{aligned}\left\langle\rho_g\left(\sum_{t\in\mathscr{T}}x_tv_t\right),\rho_g\left(\sum_{t\in\mathscr{T}}x_v w_t\right)\right\rangle &= \left\langle \sum_{t\in\mathscr{T}}x_{P_1(g,t)}\psi_{P_2(g,t)}(v_t),\sum_{t\in\mathscr{T}}x_{P_1(g,t)}\psi_{P_2(g,t)}(w_t)\right\rangle\\ &= \frac{1}{|G|}\sum_{t\in\mathscr{T}}\left\langle \psi_{P_2(g,t)}(v_t),\psi_{P_2(g,t)}(w_t)\right\rangle_\mathscr{V}\\ &= \frac{1}{|G|}\sum_{t\in\mathscr{V}}\left\langle v_t,w_t\right\rangle_\mathscr{V}\\ &= \left\langle \sum_{t\in\mathscr{T}}x_t v_t,\sum_{t\in\mathscr{T}}x_t w_t\right\rangle\end{aligned}

\text{ }

and thus \rho_g is in \displaystyle \mathcal{U}\left(\bigoplus_{t\in\mathscr{T}}\mathscr{V}_t\right) as claimed. Our last claim is that the map G\ni g\mapsto \rho_g is a homomorphism. To do this we first claim that for any g,g'\in G and any t\in\mathscr{T} one has that P_1(gg',t)=P_1\left(g,P_1(g',t)\right) and P_2(gg',t)=P_2\left(g',P_1(g',t)\right)P_2(g',t). To see this we merely note that if g'=P_1(g',t)P_2(g',t) then gg't=gP_1(g',t)P_2(g',t)=P_1\left(g,P_1(g',t)\right)P_2\left(g,P_1(g',t)\right)P_2(g',t). Then, for any g,g'\in G and any \displaystyle \sum_{t\in\mathscr{T}}x_tv_t\in\bigoplus_{t\in\mathscr{T}}\mathscr{V}_t one has that

\text{ }

\displaystyle \begin{aligned}\rho_{gg'}\left(\sum_{t\in\mathscr{T}}x_t v_t\right) &= \sum_{t\in\mathscr{T}}x_{P_1(gg',t)}\psi_{P_2(gg',t)}(v_t)\\ &= \sum_{t\in\mathscr{T}}x_{P_1(g,P_1(g',t))}\psi_{P_2(g,P_1(g',t))P_2(g',t)}(v_t)\\ &= \sum_{t\in\mathscr{T}}x_{P_1(g,P_1(g',t))}\psi_{P_2(g,P_1(g',t))}\left(\psi_{P_2(g',t)}(v_t)\right)\\ &= \rho_g\left(\sum_{t\in\mathscr{T}}x_{P_1(g',t)}\psi_{P_2(g',t)}(v_t)\right)\\ &= \rho_g\left(\rho_{g'}\left(\sum_{t\in\mathscr{T}}x_tv_t\right)\right)\end{aligned}

\text{ }

and since \displaystyle \sum_{t\in\mathscr{T}}x_t v_t was arbitrary we may conclude that \rho_{gg'}=\rho_g\circ\rho_{g'} and thus \rho is a homomorphism as claimed. Summing this all up we have that:

\text{ }

Theorem: Let G be a finite group, H\leqslant G,\mathscr{T} a transversal of G/h, and \psi:H\to\mathcal{U}\left(\mathscr{V}\right) a representation of H. Then, if the space \displaystyle \bigoplus_{t\in\mathscr{T}}\mathscr{V}_t (as described above) is given the usual inner product structure on direct sums with a normalization factor |G|^{-1} then the map 

\text{ }

\displaystyle \rho:G\to\mathcal{U}\left(\bigoplus_{t\in\mathscr{T}}\mathscr{V}_t\right),\;\; \rho_g:\sum_{t\in\mathscr{T}}x_t v_t\mapsto \sum_{t\in\mathscr{T}}x_{P_1(g,t)}\psi_{P_2(g,t)}(v_t)

\text{ }

(where P_1(g,t),P_2(g,t) are the unique elements of \mathscr{T},H respectively such that gt=P_1(g,t)P_2(g,t)) is a representation.

\text{ }

We shall call this representation \text{ind}^G_H(\psi).

\text{ }


1. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Math. Soc., 1996. Print.


April 24, 2011 - Posted by | Algebra, Representation Theory | , , , , , ,


  1. […] Point of Post: This post is a continuation of this one. […]

    Pingback by Another Way of Looking at Induced Representations (Pt. II) « Abstract Nonsense | April 24, 2011 | Reply

  2. […] a finite group and . Furthermore, suppose that is a representation and consider the associated induced representation . Let then be the character of . We say then that the character of is induced by the character […]

    Pingback by The Character of an Induced Representation « Abstract Nonsense | April 25, 2011 | Reply

  3. […] The interesting thing is that we derived this formula using the ‘type’ of induced representation we derived second. A natural question is that if we can use our first equivalent form of induced […]

    Pingback by Different Formula For the Character of an Induced Representation « Abstract Nonsense | April 26, 2011 | Reply

  4. […] the theory in this post can be traced back to our first derivation for the induced character of an induced representation when one noticed that by extending every (irreducible) character on to a character (thus a class […]

    Pingback by Induced Class Functions and the Space of Integral Class Functions (Pt. I) « Abstract Nonsense | April 27, 2011 | Reply

  5. […] Let be a finite group, and a representation. Then, if is the usual induced representation then for […]

    Pingback by Composition of the Restriction Map and the Induction Map « Abstract Nonsense | May 4, 2011 | Reply

  6. […] this post we put to rest probably one of the most niggling questions about induced representations. Namely, we have so far figured out how to take representations on and create a representation on […]

    Pingback by Mackey Irreducibility Criterion « Abstract Nonsense | May 6, 2011 | Reply

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: