Abstract Nonsense

Crushing one theorem at a time

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


April 23, 2011 - Posted by | Algebra, Representation Theory | , , , ,


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