# Abstract Nonsense

## 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 -

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