# Abstract Nonsense

## General Characters and the Uniqueness of Decomposition Into Irreps

Point of post: In this post we use our results about irreducible characters to show that any decomposition of a representation $\rho$ into a direct sum of irreps is unique.  We do this by introducing the notion of a character for a general (not necessarily irreducible) representation.

Motivation

In our last few posts we saw that every irrep of a finite group $G$ produces a class function namely the irreducible character given by that representation. That said, one can plainly see that the methodology of producing this class function had nothing to do with the fact that the representation was irreducible(this was important to prove that any set of irreducible characters, one from each equivalency class $\alpha\in\widehat{G}$, forms a basis for the space of class functions $\text{Cl}(G)$). Thus, using the same definition we will define the character of a general representation and use this notion to show that for any arbitrary but fixed set of representative irreps $\rho^{(\alpha)}$ for each $\alpha\in\widehat{G}$ the decomposition of $\displaystyle \rho\cong\bigoplus_{\alpha\in\widehat{G}}m^{(\alpha)}\rho^{(\alpha)}$ (where $m^{(\alpha)}$ denotes how many times $\rho^{(\alpha)}$ appears in the direct sum) is unique.

General Characters

Let $G$ be a finite group. Then for any representation $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ we define the character of $\rho$, denoted $\chi_{\rho}$ to be $\displaystyle \chi_{\rho}\left(x\right)=\text{tr}\left(\rho(x)\right)$ for every $x\in G$. It’s clear that since the trace function is invariant under conjugation and basis choice one has that $\chi_\rho=\chi_{\rho'}$ whenever $\rho\cong\rho'$. In particular it’s easy to see that although we fixed some $\rho^{(\alpha)}$ representative in $\alpha\in\widehat{G}$ that any representative will give the same character (we fixed it for clerical convenience–although on occasion we shall most likely invoke this independence). Some things are, just as before, immediately clear. Namely:

Theorem: Let $G$ be a finite group and $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ any representation with character $\chi_\rho$. Then for every $g\in G$

\displaystyle \begin{aligned}&(1)\quad \chi_{\rho}(e)=\dim\left(\mathscr{V}\right)\\ &(2)\quad \chi_{\rho}\left(g^{-1}\right)=\overline{\chi_\rho\left(g\right)}\\ &(3)\quad \chi_\rho\in\text{Cl}(G)\end{aligned}

The proof of this theorem is precisely the same as in the case for irreducible characters. Another very obvious theorem (one we have implicitly used before) is:

Theorem: Let $G$ be a finite group and $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ and $\psi:G\to\mathcal{U}\left(\mathscr{W}\right)$ be two representations of $G$ with characters $\chi_\rho$ and $\chi_\psi$ respectively. Then, if $\rho\otimes\psi:G\to\mathcal{U}\left(\mathscr{U}\otimes\mathscr{V}\right)$ is the tensor product of $\rho$ and $\psi$ then $\chi_{\rho\otimes\psi}=\chi_\rho\chi_\psi$. If $\rho\oplus\chi:G\to\mathcal{U}\left(\mathscr{V}\oplus\mathscr{W}\right)$ is the direct sum of $\rho$ and $\psi$ then $\chi_{\rho\oplus\psi}=\chi_\rho+\chi_\psi$.

Proof: This follows almost immediately by looking at the matrix representations for $\rho\otimes\psi$ and $\rho\oplus\psi$ with respect to the tensor basis (lexicographically ordered) and the direct sum basis (any ordering) respectively. $\blacksquare$

Uniqueness of Decomposition

We now show that given any representation $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ and distinguished representatives $\rho^{(\alpha)}$ from each $\alpha\in\widehat{G}$ then the (guaranteed) decomposition $\displaystyle \rho\cong\bigoplus_{\alpha\in\widehat{G}}m^{(\alpha)}\rho^{(\alpha)}$ is unique. Indeed:

Theorem: Let $G$ be a finite group and $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ a representation of $G$. Then, given distinguished representatives $\rho^{(\alpha)}$ for each $\alpha\in\widehat{G}$ the decomposition of $\rho\cong\bigoplus_{\alpha\in\widehat{G}}m^{(\alpha)}\rho^{(\alpha)}$ is unique (up to reordering).

Proof: Suppose that

$\displaystyle \bigoplus_{\alpha\in\widehat{G}}n^{(\alpha)}\rho^{(\alpha)}\cong\rho\cong\bigoplus_{\alpha\in\widehat{G}}m^{(\alpha)}\rho^{(\alpha)}$

then by taking the trace we may conclude that

$\displaystyle \sum_{\alpha\in\widehat{G}}n^{(\alpha)}\chi^{(\alpha)}=\chi=\sum_{\alpha\in\widehat{G}}m^{(\alpha)}\chi^{(\alpha)}$

and since $\left\{\chi^{(\alpha)}:\alpha\in\widehat{G}\right\}$ forms an orthonormal basis for $\text{Cl}(G)$ we may conclude that $n^{(\alpha)}=m^{(\alpha)}$ for every $\alpha\in\widehat{G}$ and since everything was arbitrary the conclusion follows. $\blacksquare$

Remark: It’s also clear from this that two representations are equivalent if and only if they admit the same character.

References:

1.Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.

2. Serre, Jean Pierre. Linear Representations of Finite Groups. New York: Springer-Verlag, 1977. Print

February 27, 2011 -

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