Abstract Nonsense

Crushing one theorem at a time

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.

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May 2, 2011 - Posted by | Algebra, Representation Theory | , , ,

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