Abstract Nonsense

The Kernel of a Character

Point of post: In this post we discuss the notion of the kernel of a character in general preparation to prove some pretty serious theorems in pure group theory.

Motivation

Despite all the cool things we’ve done with representations, we must not forget that at the end of the day every representation is just a homomorphism between groups. It makes sense then that one of the most natural things to do with a representation is to look at its kernel with respect to its homomorphism qualities. It turns out though that there is a natural way to translate this notion of the kernel of a representation into the notion of the kernel of a character. This notion shall be paramount in the theory to come.

Kernel of a Character

Let $G$ be a finite group and $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ a representation. We define as usual, pursuant to the usual definitions of the kernel of a homomorphism the kernel of $\rho$, denoted $\ker\rho$ as

$\text{ }$

$\ker\rho=\left\{g\in G:\rho(g)=\mathbf{1}\right\}$

What is interesting about this whole concept is that there is way to describe $\ker\rho$ in terms of its induced character $\chi_\rho$. Indeed:

Theorem: Let $G$ be a finite group and $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ is a representation with induced character $\chi_\rho$. Then,

$\text{ }$

$\displaystyle \ker\rho=\left\{g\in G:\chi_\rho(g)=\chi_\rho(e)\right\}$

$\text{ }$

Proof: Evidently if $g\in\ker\rho$ then $\rho(g)=\mathbf{1}_{\mathscr{V}}$ and so $\chi_\rho(g)=\text{tr}(\rho(g))=\dim\mathscr{V}=\chi_\rho(e)$ from where the conclusion follows.

Conversely, suppose that $\chi_\rho(g)=\chi_\rho(e)$. Note that since $\rho(g)$ is unitary we know by the spectral theorem that $\rho(g)$ is similar to $\text{diag}(\lambda_1,\cdots,\lambda_n)$ (where $n=\dim \mathscr{V}$ and the $\lambda_j$‘s are the eigenvalues) with respect to any basis. But, we know by the assumption that

$laetx \text{ }$

$\displaystyle \sum_{j=1}^{n}\lambda_j=n$

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and since $\rho(g)$ is unitary we know that $\lambda_j\in\mathbb{S}^1$ for $j=1,\cdots,n$ and so it follows then that $\lambda_1=\cdots=\lambda_n=1$. Thus, $\rho(g)$ is similar to $\mathbf{1}$ and thus $\rho(g)=\mathbf{1}$. Thus, $g\in\ker\rho$ as desired. The conclusion follows. $\blacksquare$

It makes sense then to define the kernel of the character $\chi$, denoted $\ker\chi$, by

$\text{ }$

$\ker\chi=\left\{g\in G:\chi(g)=\chi(e)\right\}$

$\text{ }$

In particular, we know that for every character $\chi$ one has that $\ker\chi\unlhd G$. For the irreducible characters $\chi^{(\alpha)}\;\; \alpha\in\widehat{G}$ we give the special symbols $N^{(\alpha)}$ for $\ker\chi^{(\alpha)}$. What we now show is that knowing $N^{(\alpha)}$ for every $\alpha\in\widehat{G}$ enables one to know $\ker\chi$ for every character $\chi$. Indeed:

Theorem: Let $\chi$ be a character with representation as a linear combination of the irreducible characters $\displaystyle \chi=\sum_{\alpha\in\widehat{G}}m^{(\alpha)}\chi^{(\alpha)}$. Then,

$\text{ }$

$\displaystyle \ker\chi=\bigcap\left\{N^{(\alpha)}:m^{(\alpha)}>0\right\}$

$\text{ }$

Proof: Clearly if $\chi^{(\alpha)}(g)=d_\alpha$ for every $\alpha$ such that $m^{(\alpha)}>0$ one sees that

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\displaystyle \begin{aligned}\chi(g) &=\sum_{\alpha\in\widehat{G}\text{ s.t. }m^{(\alpha)}>0}m^{(\alpha)}\chi^{(\alpha)}(g)\\ &=\sum_{\alpha\in\widehat{G}\text{ s.t. }m^{(\alpha)}>0}m^{(\alpha)}\chi^{(\alpha)}(e)\\ &=\chi(e)\end{aligned}

$\text{ }$

and so $g\in\ker\chi$.

Conversely, since one evidently has that $\left|\chi^{(\alpha)}(g)\right|\leqslant d_\alpha$ for every $\alpha\in\widehat{G}$ we see that for any $g\in \ker\chi$ one has that

$\text{ }$

\displaystyle \begin{aligned}\left|\chi(g)\right| &=\left|\sum_{\alpha\in\widehat{G}\text{ s.t. }m^{(\alpha)}>0}m^{(\alpha)}\chi^{(\alpha)})(g)\right|\\ &\leqslant\sum_{\alpha\in\widehat{G}\text{ st. }m^{(\alpha)}>0}m^{(\alpha)}\left|\chi^{(\alpha)}(g)\right|\\ &\leqslant\sum_{\alpha\in\widehat{G}\text{ s.t. }m^{(\alpha)}>0}m^{(\alpha)}d_\alpha\\ &= \chi(1)\\ &= \chi(g)\end{aligned}

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from where it follows from this that $\chi^{(\alpha)}(g)$ must be real and so if $\chi^{(\alpha)}(g) for any $\alpha\in\widehat{G}$ then this would induce a strict inequality for $\chi(g)$ and $\chi(1)$. It follows that $\chi^{(\alpha)}(g)=d_\alpha$ for every $\alpha\in\widehat{G}$ such that $m^{(\alpha)}>0$. The conclusion follows. $\blacksquare$

$\text{ }$

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References:

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

March 7, 2011 -

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4. Hi, Din’t you mean |Re(\lambda)| \le 1? The norm of the eigenvalues of a unitary matrix is always equal to 1.

Comment by Soumyashant Nayak | April 24, 2011 | Reply

• Soumyashant Nayak,

Indeed. Thank you for pointing that out!

Best,
Alex Youcis

Comment by Alex Youcis | April 24, 2011 | Reply