# Abstract Nonsense

## Permutation Representation

Point of Post: In this post we discuss the notion of the permutation representation and its relation to induced characters.

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Motivation

In this post we discuss a very fruitful notion known as the permutation representation of a group $G$. In particular, we will show that given a finite group $G$ which acts on a finite set $\Omega$ there is a natural way to define a representation of $G$ on the free vector space $\mathbb{C}\left[\Omega\right]$. Namely, we will merely have $\displaystyle \pi_g\left(\sum_{x\in\Omega}z_x x\right)=\pi_g z_x(g\cdot x)$. This is a very fruitful form of representation which occurs very often, particularly in the representations of $S_n$.

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Permutation Representation

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Let $G$ be a finite group which acts on the set $X=\{x_1,\cdots,x_n\}$ with action denoted, as usual, $g\cdot x$. Consider then the free vector space $\mathbb{C}\left[X\right]$. Consider then the map $\pi:G\to\text{GL}\left(\mathbb{C}[X]\right)$ defined by

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$\displaystyle \pi_g\left(\sum_{j=1}^{n}z_j x_j\right)=\sum_{j=1}^{n}z_j (g\cdot x_j)$

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To see that in fact $\pi_g\in\text{GL}(\mathbb{C}[X])$ we merely note that $\pi_g$ permutes the basis $X$. We then define an inner product $\langle\cdot,\cdot\rangle$ on $\mathbb{C}[X]$ by declaring that $X$ is orthonormal and extending by sequilinearity. Clearly then $\pi_g$ is unitary within this inner product since it carries the orthonormal basis $X$ onto itself. Then this representation is known as the permutation representation of the action $G$ on $X$. The first obvious question is what does the character of $\pi$ look like? What we note then is that for any $x_j\in X$ the diagonal entry of $\pi_g$ associated to $x_j$ (in the ordered basis $(x_1,\cdots,x_n)$ is either $0$ if $g\cdot x_j\ne x_j$ and $1$ if $g\cdot x_j$. In other words

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$\chi(g)=\#\left\{x\in X:g\cdot x=x\right\}=\#\text{Fix}(g)$

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What we note though is this:

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Theorem: Let $G$ be a finite group which acts on the finite set $X$, suppose that $X$ breaks up into oribts as $\displaystyle \bigsqcup_{x\in\Gamma}\mathcal{O}_x$ where $\Gamma$ is transveral for the set of orbits $X/G$. Then, if $\pi$ is permutation representation of $G$ on $\mathbb{C}[X]$ for every $x\in\Gamma$, $\mathbb{C}[\mathcal{O}_x]$ is $\pi$-invariant and

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$\displaystyle \pi\cong\bigoplus_{x\in\Gamma}\pi_x$

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where $\pi_x$ is the map $G\to\mathcal{U}\left(\mathbb{C}\left[\mathcal{O}_x\right]\right):g\mapsto \left(\pi_g\right)_{\mid\mathcal{O}_x}$

Proof: The fact that each $\mathbb{C}[\mathcal{O}_x]$ is $\pi$-invariant is obvious since by definition if $g\cdot x$ is a basis vector for $\mathbb{C}[\mathcal{O}_x]$ and $k\in G$ is arbitrary then  $\pi_k(g\cdot x)=k\cdot(g\cdot x)=(kg)\cdot x\in\mathcal{O}_x$. To note the second part it suffices to note that

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$\displaystyle \mathbb{C}[X]=\bigoplus_{x\in\Gamma}\mathbb{C}\left[\mathcal{O}_x\right]$

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Corollary: Let $G$ act on $X$ and let $\pi$ be the associated permutation character. Then, if $\Gamma$ is a transversal for $X/G$ and $\chi_x=\chi_{\pi_x}$ as described above then $\displaystyle \chi_\pi=\sum_{x\in\Gamma}\chi_x$.

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References:

Fulton, William, and Joe Harris. Representation Theory: a First Course. New York: Springer-Verlag, 1991. Print.