If $\chi$ is a nontrivial irreducible character of $G$ (a finite group), define $S_{\chi}:= \sum_{x \in G} \chi(x)$. In terms of conjugacy classes $\mathcal{C}$, this is $\sum_{\mathcal{C}} |\mathcal{C}| \chi(\mathcal{C})$. Is there a nice condition that guarantees $S_{\chi}=0$?

I've noticed that this occurs, for instance, with $S_5$. I'd love a description of this phenomenon and a proof, if possible.

  • 2
    $\begingroup$ Notice that $\sum_{g\in G} \chi(g) = |G|\langle \chi, \mathbb 1\rangle$, so $S_{\chi} = 0$ iff $\mathbb 1$ doesn't appear as a direct summand of $\chi$. $\endgroup$ – ah11950 Apr 25 '14 at 0:02
  • $\begingroup$ Of course. Wow, that was an obvious one. Sorry for the time wasted. $\endgroup$ – José Siqueira Apr 25 '14 at 0:06
  • $\begingroup$ No worries! It can be dreadfully easy to miss the most simple things at times; I certainly know it happens to me! (I've just noticed that you've assumed $\chi$ is irreducible, so obviously it's possible to make the stronger statement that $S_{\chi}=0 \iff \chi \neq \mathbb 1$...) $\endgroup$ – ah11950 Apr 25 '14 at 0:09

As @ah11950 states in the comments, $$\sum_{g\in G} \chi(g) = |G|\langle \chi, \mathbb 1\rangle,$$ so $S_{\chi} = 0$ if and only if $1$ doesn't appear as a direct summand of $\chi$.


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.