Abstract Nonsense

Crushing one theorem at a time

Induced Class Functions and the Space of Integral Class Functions (Pt. II)


Point of Post: This is a continuation of this post.

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What we now noteis that the map \text{Ind}^G_H is linear. Indeed:

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Theorem: Let G be a finite group and H\leqslant G. Then the map \text{Ind}^G_H:\text{Cl}(H)\to\text{Cl}(G) is linear.

Proof: This follows immediately from the obvious fact that for any two f_1,f_2\in\text{Cl}(G) and any z,_1,z_2\in\mathbb{C} one has

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(z_1f_1+z_2f_2)^\circ=z_1f_1^\circ+z_2f_2^\circ

 \blacksquare

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 There’s no hope that \text{Ind}^G_H will be a ring morphism since a quick check shows that \text{Ind}^G_H(f)\ast\text{Ind}^G_H(\delta_e)=\left(G:H\right)\text{Ind}^G_H(f\ast \delta_e)  for any f\in\text{Cl}(G).

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That said, we clearly get the result that an isomorphic copy of the additive group \text{Cl}(H) sits inside \text{Cl}(G).

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Space of Integral Class Functions

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We define the space \text{span}_{\mathbb{Z}}\left(\text{irr}(H)\right)\subseteq as the space of integral class functions and denote it \mathbb{Z}\left(\text{irr}(H)\right). It’s evident that \mathbb{Z}\left(\text{irr}(H)\right) is a subring of \text{Cl}(H). The first thing we note about \mathbb{Z}\left(\text{irr}(H)\right) is its relation to \text{Ind}^G_H. Namely:

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Theorem: Let G be a finite group and H\leqslant G. Then, if \text{Ind}^G_H is defined as usual it’s true that \text{im}\left(\mathbb{Z}\left(\text{irr}(H)\right)\right)\subseteq\mathbb{Z}\left(\text{irr}(G)\right).

Proof: Since \text{Ind}^G_H is linear we have that  \displaystyle \text{im}\left(\text{span}_\mathbb{Z}\left(\text{irr}(H)\right)\right)=\text{span}_\mathbb{Z}\left(\text{im}\left(\text{irr}(H)\right)\right) thus it suffices to prove that the containment \text{im}\left(\text{irr}(H)\right)\subseteq\text{span}_\mathbb{Z}\left(\text{irr}(G)\right) holds. But, we know that \text{Ind}^G_H(\chi) is a character of G for each \chi\in\text{irr}(H) and so from a previous theorem we have that \text{Ind}^G_H(\chi)\in\text{span}_\mathbb{N}\left(\text{irr}(G)\right). The conclusion follows. \blacksquare

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Corollary: Let \mathscr{B}_1,\mathscr{B}_2 be some ordering of the basis \text{irr}(H) and \text{irr}(G) respectively. Then, \left[\text{Ind}^G_H\right]_{\mathscr{B}_1,\mathscr{B}_2}\in\text{Mat}_{k_1\times k_2}\left(\mathbb{Z}\right) (where k_1,k_2 are the number of conjugacy classes in H and G respectively).

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Thus, we may restrict \text{Ind}^G_H to get a mapping \mathbb{Z}\left(\text{irr}(H)\right)\to\mathbb{Z}\left(\text{irr}(G)\right). We denote this map by \bigtriangleup.

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Remark: The notation is chosen for two reasons: 1) it saves me a lot of typing (and writing at the white board) and 2) it suggests the ‘go upedness’ of the map.

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Dual Notions of Restriction

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There is a mapping \text{Cl}(G)\to\text{Cl}(H) which is dual, in a sense soon to be made precise, to \text{Ind}^G_H. Namely, we are speaking the restriction mapping \text{Res}^H_G:\text{Cl}(G)\to\text{Cl}(H) which is literally just restricting an element of f\in\text{Cl}(G) to H thus producing an element of \text{Cl}(H). Accordingly we have the dualized map of \bigtriangleup, denoted \bigtriangledown, which maps \mathbb{Z}\left(\text{irr}(G)\right)\to\mathbb{Z}\left(\text{irr}(H)\right) given by the restriction of \text{Res}^H_G to \mathbb{Z}\left(\text{irr}(G)\right). The interplay of \bigtriangleup and \bigtriangledown shall be the focus of our next few posts.

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References:

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

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April 27, 2011 - Posted by | Algebra, Representation Theory | , , , , ,

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