Abstract Nonsense

Crushing one theorem at a time

A ‘Lemma’ (pt. II)


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

The ‘Lemma’

We next study as a tool to prove our still looming lemma the multiplicative relation between the elements of the canonical basis of \text{Cl}(G). Namely, we see that:

 

Theorem: Let G be a finite group with conjugacy classes \mathcal{C}_1,\cdots,\mathcal{C}_k. Then, for any i,j\in[k] there exists c_{i,j,r}\in\mathbb{N}\cup\{0\} for every r\in[k] such that \displaystyle \mathbf{1}_{\mathcal{C}_j}*\mathbf{1}_{\mathcal{C}_k}=\sum_{r=1}^{k}c_{i,j,r}\mathbf{1}_{\mathcal{C}_r}.

Proof: Recall that \text{Cl}(G) is real the center of the group algebra \mathcal{A}(G). But, recall that the center of an algebra is a subalgebra. In particular since \text{Cl}(G) is a subalgebra of \mathcal{A}(G) we have that  \text{Cl}(G) is in particular closed under multiplication–convolution in this case. Thus, since \mathbf{1}_{\mathcal{C}_i} and \mathbf{1}_{C_j} are class functions then so is \mathbf{1}_{\mathcal{C}_i}\ast\mathbf{1}_{\mathcal{C}_j} . Thus, since \left\{\mathbf{1}_{\mathcal{C}_r}\right\}_{r=1}^k is a basis for \text{Cl}(G) we know that there does exist constants c_{i,j,r}\in\mathbb{C} for r=1,\cdots,k such that \displaystyle \mathbf{1}_{\mathcal{C}_i}\ast\mathbf{1}_{\mathcal{C}_j}=\sum_{r=1}^{k}c_{i,j,r}\mathbf{1}_{\mathcal{C}_r} and in fact (as has already been proven in general) c_{i,j,r}=\left(\mathbf{1}_{\mathcal{C}_i}\ast\mathbf{1}_{\mathcal{C}_j}\right)(g_r) for any g_r\in\mathcal{C}_r. But, by definition one has that

 

\displaystyle \begin{aligned}\left(\mathbf{1}_{\mathcal{C}_i}\ast\mathbf{1}_{\mathcal{C}_j}\right)(g_r) &= \sum_{g\in G}\mathbf{1}_{\mathcal{C}_i}\left(xg^{-1}\right)\mathbf{1}_{\mathcal{C}_j}(g)\\ &= \#\left\{(a,b)\in \mathcal{C}_i\times\mathcal{C}_j:ab=x\right\}\end{aligned}

 

which is surely a nonnegative integer. The conclusion follows. \blacksquare

 

From this obtain the following:

 

Theorem : Let G be a finite group with conjugacy classes \mathcal{C}_1,\cdots,\mathcal{C}_k and \chi\in\text{irr}\left(G\right). Then, if \omega_\chi is defined as before then \omega_\chi\left(\mathbf{1}_{\mathcal{C}_r}\right)\in\mathbb{A} where \mathbb{A} are, as usual, the algebraic integers.

Proof: We first claim that \omega_\chi\left(\mathbf{1}_{\mathcal{C}_i}\right)\omega_\chi\left(\mathbf{1}_{\mathcal{C}_j}\right) is a \mathbb{Z}-linear combination of \left\{\omega_\chi\left(\mathbf{1}_{\mathcal{C}_1}\right),\cdots,\omega_{\mathcal{C}_k}\right\}. Indeed, one has that there exists c_{i,j,r}\in\mathbb{Z} for r=1,\cdots,k such that \displaystyle \mathbf{1}_{\mathcal{C}_i}\ast\mathbf{1}_{\mathcal{C}_j}=\sum_{r=1}^{k}c_{i,j,r}\mathbf{1}_{\mathcal{C}_r}. Thus, applying \omega_\chi to both sides and recalling that it’s an algebra homomorphism we may conclude that \displaystyle \omega_\chi\left(\mathbf{1}_{\mathcal{C}_i}\right)\omega_\chi\left(\mathbf{1}_{\mathcal{C}_j}\right)=\sum_{r=1}^{k}c_{i,j,r}\omega_\chi\left(\mathbf{1}_{\mathcal{C}_r}\right) Fix then \ell\in[k]. Rephrasing the above,we know that there exists k^2 elements of \mathbb{Z}, call them c_{r,s} for r,s\in[k] such that \displaystyle \omega_{\chi}\left(\mathbf{1}_{\mathcal{C}_i}\right)\omega_\chi\left(\mathbf{1}_{\mathcal{C}_r}\right)=\sum_{s=1}^{k}c_{r,s}\omega_{\chi}\left(\mathbf{1}_{\mathcal{C}_s}\right). Let M=[c_{r,s}] and v=\left(\omega_\chi\left(\mathbf{1}_{\mathcal{C}_1}\right),\cdots,\omega_\chi\left(\mathbf{1}_{\mathcal{C}_k}\right)\right)^{\top}.  A quick check then shows that Mv=\omega_\chi\left(\mathbf{1}_{\mathcal{C}_i}\right)v  and so \omega_\chi\left(\mathbf{1}_{\mathcal{C}_i}\right) is in the spectrum of some matrix in \text{Mat}\left(\mathbb{Z}\right) and so by a prior characterization we know that \omega_\chi\left(\mathbf{1}_{\mathcal{C}_i}\right)\in\mathbb{A}. Since i\in[k] was arbitrary the conclusion follows. \blacksquare

 

From this we get the follwing:

 

Theorem (The ‘Lemma’): Let G be a finite group and \chi\in\text{irr}\left(G\right). Then, for every g\in G one has that \displaystyle \frac{\chi(g)}{\chi(e)}\#\left(\mathcal{C}_g\right) where \mathcal{C}_g is the conjugacy class of G containing g.

Proof: This follows immediately from the previous theorem and the already proven fact that \displaystyle \frac{\chi(g)}{\chi(e)}\#\left(\mathcal{C}_g\right)=\omega_\chi\left(\mathbf{1}_{\mathcal{C}_g}\right). \blacksquare

 

Corollary: Let G be a finite group and \chi\in\text{irr}(G).Then, either \chi(g) is irrational or \chi(e)\mid \chi(g)\#\left(\mathcal{C}_g\right).

 

So our ‘lemma’ turned out to be a theatrical production in and of itself.

 

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. […] know them from a previous theorem that is an algebraic integer and since is evidently an algebraic integer, and […]

    Pingback by Representation Theory: Burnside’s Theorem « Abstract Nonsense | March 11, 2011 | Reply


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