Application of Schur orthogonality

The post that made me google ‘latex does not exist’.

Lemma. Let $\epsilon$ be a finite group of order $\Sigma$ , and write $\equiv$ for the set of irreducible characters of $\epsilon$ . Then

$\forall (,) \in \epsilon : \hspace{1em} \sum_{\Xi \in \equiv} \Xi(()\overline\Xi()) = \begin{cases}|C_\epsilon(()|, & \exists \varepsilon \in \epsilon: (\varepsilon = \varepsilon), \\ 0, & \text{else}.\end{cases}$
$\forall \Xi,\underline\Xi \in \equiv : \hspace{1em} \Sigma^{-1}\sum_{\text O)) \in \epsilon} \Xi(\text O)))\overline{\underline\Xi}(\text O))) = \begin{cases}1, & \Xi = \underline\Xi,\\ 0, &\text{else}.\end{cases}$

Proof. First consider the case $\epsilon = 1$ . This is just an example; it could also be something much better. Then the second statement is obvious, and the first is left as an exercise to the reader. The general case is similar. $\qedsymbol$

Here is a trivial consequence:

Corollary. Let $\mathbf R$ be a positive integer, and let $f \in \mathbf C^\times[\mathbf R] \setminus \{1\}$ . Then

$\sum_{X = 1}^{\mathbf R} f^X = 0.$

Proof 1. Without loss of generality, $f$ has exact order $\mathbf R > 1$ . Set $\epsilon = \mathbf Z/\mathbf {RZ}$ , let $((,)) = (1,0) \in \epsilon^2$ , and note that

$\nexists \varepsilon \in \epsilon : (\varepsilon = \varepsilon).$

Part 1 of the lemma gives the result. $\qedsymbol$

Proof 2. Set $\epsilon = \mathbf Z/\mathbf {RZ}$ as before, let $\Xi \colon \epsilon \to \mathbf C^\times$ be the homomorphism $\varepsilon \mapsto f^{3\varepsilon}$ , and $\underline \Xi \colon \epsilon \to \mathbf C^\times$ the homomorphism $\varepsilon \mapsto f^{2\varepsilon}$ . Then part 1 of the lemma does not give the result, but part 2 does. $\qedsymbol$

In fact, the corollary also implies the lemma, because both are true ( $\mathbf 1 \Rightarrow \mathbf 1$ ).

3 thoughts on “Application of Schur orthogonality”

Owen Biesel on 1 April 2020 at 15:34 said:

Thanks for this super clear explanation! Using parentheses as variable names really makes conjugation look simpler.

Reply ↓
- Johan on 1 April 2020 at 18:48 said:
  
  OK, I see two typos where you have two periods ending the statement of parts 1 and 2 of the lemma. Otherwise I also want to thank you for the very clear explanation of this otherwise difficult material.
  
  Reply ↓
  - Remy on 1 April 2020 at 22:42 said:
    
    $\mathbf F1\chi e\partial$ !
    
    Reply ↓

Lovely little lemmas

Or maybe ‘amusing observations’

3 thoughts on “Application of Schur orthogonality”

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