# Abstract Nonsense

## Induced Representations

Point of Post: In this post we introduce the notion of induced representations and prove some of the basic properties about them (the fact that they are representations, etc.). We also define some preliminary notions

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Motivation

In the past we’ve encountered the situation where we had a representation $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ and we restricted $\rho$ to $G$ so that $\rho:H\to\mathcal{U}\left(\mathscr{V}\right)$ where $H\leqslant G$, this is usually denoted by $\text{Res}^G_H(\rho)$. In this post we pursue an essentially dual concept where given a representation $\rho:H\to\mathcal{U}\left(\mathscr{V}\right)$ we define a representation on $G$–in essence the representation on the subgroup ‘induces’ a representation on the full group. This shall prove to be an indispensable tool in all that comes. The most apparent reason is that it is often much easier to produce representations on smaller groups and we can use this to create representations on the larger groups.

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Induced Representations

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Let $G$ be a finite group and $H\leqslant G$. Let then $\rho:H\to\mathcal{U}\left(\mathscr{V}\right)$ be a representation of $H$. Consider then the set $\mathscr{V}^G$ with the usual inner product structure with normalization constant $|G|^{-1}$. Define then $\mathscr{X}$ to be the set $\left\{f\in\mathscr{V}^G:f(gh)=\rho_h^{-1}f(g)\text{ for all }g\in G\text{ and }h\in H\right\}$. We claim then that $\mathscr{X}\leqslant\mathscr{V}^G$. Indeed, let $f,f'\in\mathscr{X}$ and $z,z'\in\mathbb{C}$ then for any $g\in G$ and $h\in H$ one has that

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\begin{aligned}(zf+z'f')(gh)& =zf(gh)+z'f'(gh)\\ &=z\rho_{h}^{-1}f(g)+z'\rho_h^{-1}f'(g)\\ &=\rho_{h}^{-1}\left(zf(g)+z'f(g)\right)\\ &=\rho_h^{-1}\left(zf+z'f'\right)(g)\end{aligned}

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from where the conclusion follows. We next note that for any $g\in G$ and any $f\in\mathscr{X}$ the function $\psi_g(f)$ given by $(\psi_g(f))(h)=f(g^{-1}h)$ is an element of $\mathscr{X}$. Indeed, for any $k\in G$ and $h\in H$ one has that

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$(\psi_g(f))(kh)=f\left(g^{-1}(kh)\right)=f\left(\left(g^{-1}k\right)h\right)=\rho_{h}^{-1}f(g^{-1}k)=\rho_h^{-1}(\psi_g(f))(k)$

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Thus, the map $\psi:G\to\text{End}\left(\mathscr{X}\right)$ defined by $\psi(g)=\psi_g$ is well-defined. What we next claim is that $\text{im}(\psi)\subseteq\mathcal{U}\left(\mathscr{X}\right)$. Indeed, we note that for any $g\in G$ and any $f_1,f_2\in\mathscr{X}$ one has

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\displaystyle \begin{aligned}\left\langle \psi_g(f_1),\psi_g(f_2)\right\rangle &= \frac{1}{|G|}\sum_{h\in G}\left\langle (\psi_g(f_1))(h),(\psi_g(f_2))(h)\right\rangle_\mathscr{V}\\ &= \frac{1}{|G|}\sum_{h\in G}\left\langle f_1\left(g^{-1}h\right),f_2\left(g^{-1}h\right)\right\rangle_\mathscr{V}\\ &= \frac{1}{|G|}\sum_{h\in G}\left\langle f_1(h),f_2(h)\right\rangle_\mathscr{V}\\ &= \left\langle f_1,f_2\right\rangle\end{aligned}

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The last thing we claim is that in fact the mapping $\psi:G\to\mathcal{U}\left(\mathscr{X}\right)$ is a group morphism. Indeed, for any $a,b,c\in G$ and any $f\in\mathscr{X}$ one has that $\displaystyle \left(\psi_{ab}(f)\right)(c) =f\left(b^{-1}a^{-1}c\right)=(\rho_b(f))(a^{-1}c)= \left(\rho_a\left(\rho_b(f)\right)\right)(c)$. But, since $c\in G$ was arbitrary it follows that $\psi_{ab}(f)=(\psi_a\circ \psi_b)(f)$. But, since $f\in\mathscr{X}$ was arbitrary it follows that $\psi_{ab}=\psi_a\circ\psi_b$, and finally since $g,h\in G$ were arbitrary the conclusion follows.  Summing this all up:

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Theorem: Let $G$ be a finite group and $H\leqslant G$ and $\rho:H\to\mathcal{U}\left(\mathscr{V}\right)$ a representation. Then,the set $\mathscr{X}$ given by

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$\mathscr{X}=\left\{f\in\mathscr{V}^G:f(gh)=\rho_h^{-1}f(g)\text{ for all }g\in G\text{ and }h\in H\right\}$

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is a subspace of $\mathscr{V}^G$. Moreover, the map $\psi:G\to\mathcal{U}\left(\mathscr{X}\right)$ given by the formula $(\psi_g(f))(k)=f(g^{-1}k)$ is a representation of $G$.

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The representation described in this theorem as $\psi$ is called the representation induced on $G$ by $\rho$ and is generally denoted $\text{Ind}^G_H(\rho)$.

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As a final simple exercise to end this post we find a connection between the degree of $\rho$ and the degree of $\text{Ind}^G_H(\rho)$. Indeed:

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Theorem: Let $G$ be a finite group and $H\leqslant G$. Then, if $\rho:H\to\mathcal{U}\left(\mathscr{V}\right)$ is a representation of $H$ and $\text{Ind}^G_H$ is the corresponding induced representation on $G$ then $\deg\text{Ind}^G_H=\left(G:H\right)\deg\rho$.

Proof: Let $\mathscr{X}$ denote the representation space of $\text{Ind}^G_H(\rho)$ as described above. Let $\{t_j\}_{j=1}^{(G:H)}$ be a transversal of the set of cosets $G/H$. Define then the map $T:\mathscr{X}\to\mathscr{V}^{(G:H)}$ by $f\mapsto (f(t_1),\cdots,f(t_{(G:H)}))$, we claim that $T$ is an isomorphism. Indeed, we note that for any $f,f'\in\mathscr{X}$ and $z,z'\in\mathbb{C}$ one has that

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\displaystyle \begin{aligned}T(zf+z'f') &= \left((zf+z'f')(t_1),\cdots,(zf_1+z'f')(t_{(G:H)})\right)\\ &= \left(zf(t_1)+z'f'(t_1),\cdots,zf(t_{(G:H)})+z'f'(t_{(G:H)})\right)\\ &= z(f(t_1),\cdots,f(t_{(G:H)}))+z'\left(f'(t_1),\cdots,f'(t_{(G:H)})\right)\\ &= zT(f)+z'T(f')\end{aligned}

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To see that $T$ is injective note that if $f\in\ker T$ then $f(t_j)=\bold{0}$ for $j=1,\cdots,(G:H)$. But, for any $g\in G$ we know that $g=t_jh$ for some $t_j$ and some $h\in H$ and so $f(g)=f(t_jh)=\rho_h^{-1}f(t_j)=\bold{0}$. Thus, since $g$ was arbitrary it follows that $f=\bold{0}$ and so $\ker T$ is trivial. Thus, $T$ is injective. To see that $T$ is surjective we let $(v_1,\cdots,v_{(G:H)})\in\mathscr{V}^{(G:H)}$ be arbitrary. We define then $f$ by $f(g)=f(t_j h)=\rho_h^{-1}(v_j)$ where $g=t_j h$ is the unique representation of $g$ as the product of some element of $\{t_j\}_{j=1}^{(G:H)}$ and an element of $H$. Evidently $f\in\mathscr{X}$ and since $f(t_j)=v_j$ we see that $T(f)=(v_1,\cdots,f_{(G:H)})$ and thus $(v_1,\cdots,v_{(G:H)})\in\text{im}(T)$. But, since $(v_1,\cdots,v_{(G:H)})$ was arbitrary it follows that $\text{im}(T)=\mathscr{V}^{(G:H)}$. Therefore, $T$ is an isomorphism as claimed and so we may conclude that

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$\deg\text{Ind}^G_H(\rho)=\dim\mathscr{X}=\dim\mathscr{V}^{(G:H)}=(G:H)\dim\mathscr{V}=(G:H)\deg\rho$

$\blacksquare$

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References:

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

April 23, 2011 -

## 6 Comments »

1. […] 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 […]

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

2. […] Let be a finite group, , and . Then, if is our first definition of induced representation and the one developed in this post, then […]

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

3. […] we have already seen given a finite group a subgroup and a representation we can create a representation . Of course […]

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

4. […] of induced representation we derived second. A natural question is that if we can use our first equivalent form of induced representations to derive another formula for induced characters. In this post we do […]

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

5. […] Since is linear we have that   thus it suffices to prove that the containment holds. But, we know that is a character of for each and so from a previous theorem we have that . The conclusion […]

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

6. […] problem I can mention that it has many uses. For example, I have in the past discussed the notion of induced representations which can be seen as extension of scalars problem. Namely, we suppose that we have some group and […]

Pingback by Extension of Scalars and Change of Ring (Pt. I) « Abstract Nonsense | January 24, 2012 | Reply