Abstract Nonsense

Crushing one theorem at a time

A ‘Lemma’ (Pt. I)


Point of post: In this post we will prove a lemma which shall prove to be absolutely necessary in proving Burnside’s lemma.

 

Motivation

The basic idea of this post is to use a variant of the left \ast-representation of G generated by an irrep \rho to create a mapping \omega:\text{irr}(G)\to\mathbb{C}. We shall then use this to show that the number \displaystyle \frac{\chi(g)}{\chi(e)}\#\left(\mathcal{C}_g\right) is algebraic. This will prove to be surprising important in our proof of Burnside’s theorem.

The Preamble

Let G be a finite group and \chi be an irreducible character of G. Let then \rho:G\to\mathcal{U}\left(\mathscr{V}\right) be any irrep which admits \chi as its character. Recall that we may then define the induced \ast-representation \rho^\ast:\mathcal{A}(G)\to\mathcal\mathcal{U}\left(\mathscr{V}\right) by \displaystyle \rho^\ast(f)=\sum_{g\in G}f(g)\rho(g). Our first claim is that \rho^\ast is irreducible. Indeed:

 

Theorem: Let \rho^\ast be defined as above. Then, \rho^\ast is irreducible.

Proof: Suppose that\mathscr{W}\leqslant\mathscr{V} is \rho^\ast-invariant then in particular we must have that \left(\rho^\ast(g_0)\right)\left(\mathscr{W}\right)\subseteq\mathscr{W} but \rho^\ast(g_0)=\rho(g_0) so that \left(\rho(g_0)\right)\left(\mathscr{W}\right)\subseteq\mathscr{W}. Since this was true for every g_o\in G it follows that \mathscr{W} is \rho-invariant and thus \mathscr{W} is either \{\bold{0}\} or \mathscr{V}. Since \mathscr{W} was abritrary the conclusion follows.\blacksquare

 

Note though from basic properties of \ast-representations we have that for any f\in\text{Cl}(G) (i.e. a class function) one has that \rho^\ast(f)\rho^\ast(a)=\rho^\ast(f\ast a)=\rho^\ast(a\ast f)=\rho^\ast(a)\rho^\ast(f) for any a\in\mathcal{A}(G).  Thus, by Schur’s lemma we may conclude that \rho^\ast(f)=z_f\mathbf{1} for some z_f\in\mathbb{C}. Since this mapping is conjugation invariant (i.e. \psi(f)=z_f\mathbf{1} for every \psi=t\rho t^{-1} for some isomorphism t) and every irrep which admits \chi as its character is conjugate it follows that \psi^\ast(\chi^{(\alpha)})=z_\alpha\mathbf{1} for every \psi which admits \chi as its character. Thus, we can define the function \omega_\chi:\text{Cl}(G)\to\mathbb{C} by \omega_\chi(f)\mathbf{1}=\rho^\ast(f) for any \rho^\ast where \rho is an irrep which admits \chi as its character. Our first theorem says that \omega_\chi is a homomorphism Indeed:

 

Remark: We now begin using \text{irr}(G) to denote \left\{\chi^{(\alpha)}:\alpha\in\widehat{G}\right\}.

 

Theorem: Let \chi\in\text{irr}(G), and let \omega_\chi be defined as above. Then \chi is a \mathbb{C}-homomorphism (i.e. it’s an associative algebra homomorphism)

Proof: Let f,f'\in\text{Cl}(G) and z,z'\in\mathbb{C}. Then, for any representation \rho admitting \chi as it’s character we have that

 

\begin{aligned}\rho^\ast(zf+z'f') &=z\rho^\ast(f)+z'\rho^\ast(f')\\ &=z\omega_\chi(f)\mathbf{1}+z'\omega_\chi(f)\mathbf{1}\\ &=(z\omega_\chi(f)+z'\omega_\chi(f'))\mathbf{1}\end{aligned}

 

and so by definition \omega_\chi(zf+z'f')=z\omega_\chi(f)+z'\omega_\chi(f'). Moreover, let f,f'\in\text{Cl}(G) then one has that

 

\begin{aligned}\rho^\ast(f\ast f') &=\rho^\ast(f)\rho^\ast(f')\\ &=\omega_\chi(f)\mathbf{1}\omega_\chi(f')\mathbf{1}\\ &=(\omega_\chi(f)\omega_\chi(f'))\mathbf{1}\end{aligned}

 

Furthermore, let f,f'\in\text{Cl}(G) then one has that

\displaystyle \begin{aligned}\rho^\ast(f\ast f') &= \rho^\ast(f)\rho^\ast(f')\\ &= \omega_\chi(f)\mathbf{1}\omega_\chi(f')\mathbf{1}\\&=(\omega_\chi(f)\omega_\chi(f'))\mathbf{1}\end{aligned}

 

and so \omega_\chi(f\ast f')=\omega_\chi(f)\omega_\chi(f') from where the conclusion follows. \blacksquare

 


In particular this tells us that \omega_\chi is entirely determined on the canonical basis \left\{\mathbf{1}_{\mathcal{C}_j}\right\}_{j=1}^k for \text{Cl}(G) where \mathcal{C}_1,\cdots,\mathcal{C}_k are the conjugacy classes of G.

 

 

Note though that there is a natural way to relate \omega_\chi\left(\bold{1}_{\mathcal{C}_j}\right) and any g\in\mathcal{C}_j. Indeed:

 

Theorem: Let G be a finite group with conjugacy classes \mathcal{C}_1,\cdots,\mathcal{C}_k then for any j\in[k] and any g_0\in\mathcal{C}_j one has that


\displaystyle \omega_\chi\left(\mathbf{1}_{\mathcal{C}_j}\right)=\frac{\chi(g)\#\left(\mathcal{C}_j\right)}{\chi(e)}


Proof: We note that by definition one has for any character \rho which admits \chi as its character

 

\displaystyle \begin{aligned}\chi(e)\omega_\chi\left(\bold{1}_{\mathcal{C}_j}\right) &= \text{tr}\left(\omega_\chi\left(\mathbf{1}_{\mathcal{C}_j}\right)\mathbf{1}\right)\\ &= \text{tr}\left(\rho^\ast\left(\mathbf{1}_{\mathcal{C}_j}\right)\right)\\ &= \text{tr}\left(\sum_{g\in G}\mathbf{1}_{\mathcal{C}_j}(g)\rho(g)\right)\\ &= \sum_{g\in G}\mathbf{1}_{\mathcal{C}_j}(g)\chi(g)\\ &= \sum_{g\in\mathcal{C}_j}\chi(g)\\ &= \#\left(\mathcal{C}_j\right)\chi(g_0)\end{aligned}

 

from where the conclusion follows. \blacksquare

 

References:

1. Isaacs, I. Martin. Character Theory of Finite Groups. New York: Academic, 1976. Print.

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March 10, 2011 - Posted by | Algebra, Representation Theory | , , , , ,

1 Comment »

  1. […] Point of post: This post is a continuation of this one. […]

    Pingback by Representation Theory: A ‘Lemma’ (pt. II) « Abstract Nonsense | March 10, 2011 | Reply


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