1
$\begingroup$

I'm trying to show that if $G$ is a finite odd order group then, all of its nontrivial complex representations are of complex type (i.e., it is not realisable over the reals).

(I have answered it here: If $G$ is a finite non-trivial group of odd order, it has an irreducible representation not realisable over the reals.)

Let $V$ be such a representation, and let $V^G:=\{v\in V: gv=v, \forall g \in G\}$ be its fix point set. Using some theorems in Bröcker's book about Representations of Compact Lie Groups, I can solve this problem if I show that $V^G=0$. Is that true?

$\endgroup$
  • 1
    $\begingroup$ Well, what about the trivial representation? $\endgroup$ – Matt Samuel Jan 14 '19 at 0:11
  • $\begingroup$ It should be a non-trivial representation*, I'll change @MattSamuel $\endgroup$ – Andre Gomes Jan 14 '19 at 0:12
  • $\begingroup$ I have answered the first question here: math.stackexchange.com/questions/1033844/… $\endgroup$ – Andre Gomes Jan 14 '19 at 17:24
2
$\begingroup$

Any nonzero vector $v\in V^G$ spans a $1$-dimensional trivial sub-representation. If $V$ is irreducible of dimension $>1$, then it follows that $V^G=0$. If $\dim V=1$, then $V^G\neq 0$ implies $V$ is the trivial representation.

| cite | improve this answer | |
$\endgroup$

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.