2
$\begingroup$

Let $G$ be a non-trivial finite group. Let $\chi$ be an irreducible character of the group $G$. Find $$\frac{1}{|G|}\sum_{g \in G} {\chi( g)}$$

I try. But I think that I am wrong. $$G=C_{i_1}\oplus C_{i_2}\oplus\ldots C_{i_k},$$ as $C_{i}$ is a cyclic group If $\chi$ is 1-dimensional character then $\sum_{g \in C_{m}} {\chi( g)}=0.$ Thus $$\frac{1}{|G|}\sum_{g \in G} {\chi( g)}= \frac{k}{|G|}.$$ Am I on the right path?

There are not answer or hints in our book

$\endgroup$
  • 2
    $\begingroup$ You're not on the right path. I suggest 1) try some examples using whatever character tables you can find 2) think about the definition of the inner product of characters $\langle \chi, \phi\rangle$ $\endgroup$ – Matthew Towers Jan 28 '14 at 23:11
  • $\begingroup$ "Thus [formula]." Where are you getting the formula? Also, whether or not we know $G$ is abelian (the gray block doesn't say even though you write a factor decomposition of $G$) is important, since it determines how tough the proof may be. $\endgroup$ – anon Jan 29 '14 at 8:52
3
$\begingroup$

Here is an extension of the comment by mt_:

Consider the definition of the inner product on the characters $\langle \chi,\psi\rangle = \frac{1}{|G|}\sum_{g\in G}\chi(g)\overline{\psi(g)}$.

Can you get the sum you are looking as an inner product between suitable characters?

$\endgroup$
  • $\begingroup$ Using character tables I assume that $\frac{1}{|G|}\sum_{g \in G} {\chi( g)}=0$ as affording irreducible representation is non-trivial and $\frac{1}{|G|}\sum_{g \in G} {\chi( g)}=1$ as affording irreducible representation is trivial. I want get the sum looking as an inner product. Сonsider representation representation $ \alpha=\rho \otimes \bar{\sigma}$. we have $$\frac{1}{|G|}\sum_{g \in G} {\chi( g)}= \frac{1}{|G|}\sum_{g\in G}\chi_{\rho}(g)\overline{\chi_{\bar{\sigma}}(g)}=0, \text{as $\rho \otimes \bar{\sigma}$ is n non-trivial}.$$ Does this make sense? $\endgroup$ – nadia-liza Jan 30 '14 at 8:18
  • $\begingroup$ I don't quite follow what you mean. The sum you are looking at is simply the inner product of $\chi$ with the trivial character. $\endgroup$ – Tobias Kildetoft Jan 30 '14 at 8:55
  • $\begingroup$ Oh, Yes! I at last understood! $\endgroup$ – nadia-liza Jan 30 '14 at 8:58
0
$\begingroup$

Apply the orthogonality relations.

Alternatively, consider an irreducible representation $\mathscr{X}$ affording $\chi$. Let $A = \sum_{g \in G} \mathscr{X}(g)$. Prove that either $A = 0$ or $\mathscr{X}$ is the trivial representation.

$\endgroup$
  • $\begingroup$ Using character tables I assume that $\frac{1}{|G|}\sum_{g \in G} {\chi( g)}=0$ as affording irreducible representation is non-trivial and $\frac{1}{|G|}\sum_{g \in G} {\chi( g)}=1$ as affording irreducible representation is trivial. I want get the sum looking as an inner product. Сonsider representation representation $ A=\rho \otimes \bar{\sigma}$. we have $$\frac{1}{|G|}\sum_{g \in G} {\chi( g)}= \frac{1}{|G|}\sum_{g\in G}\chi_{\rho}(g)\overline{\chi_{\bar{\sigma}}(g)}=0, \text{as $\rho \otimes \bar{\sigma}$ is n non-trivial}.$$ Does this make sense? $\endgroup$ – nadia-liza Jan 30 '14 at 8:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.