6
$\begingroup$

$\textbf{Lemma }$If $V$ is a representation of a finite group $G$, then $V$ is of real type if and only if $V$ is the complexification of a representation $V_{\mathbb{R}}$ over the field of real numbers.

Now I have to show that if $G$ is a finite non-trivial group of odd order, it has an irreducible representation not realisable over the reals. This is my attempt so far:

Since the order of $G$ is odd, it has no non-trivial elements whose order is 2. So by Frobenius-Schur we know that $1=\sum_{V}\text{dim}(V)FS(V)$ where $FS(V)$ is the Frobenius-Schur indicator, this sum can be of course be written as $$1=\sum_{V\text{ is of real type}}\text{dim}(V)-\sum_{V\text{ is of quaternionic type}}\text{dim}(V).$$ Since $G$ is non-trivial, it has by the above equality an irreducible representation which is quaternionic and by the above lemma, this one is not realisable over the reals.

My question is whether my reasoning is not flawed, especially when I am using the lemma. Any help would be appreciated. Thanks.

$\endgroup$
  • $\begingroup$ Yes, your proof is correct. You can also prove there are no real characters for an odd-order group. $\endgroup$ – Hempelicious Jan 17 at 3:23
0
$\begingroup$

I can't say if your proof is correct. But I have another one, in case it helps.

I'll assume the following theorems in the Bröck's book about Representations of Compact Lie Groups, in which $V$ is a complex representation of $G$ and $\chi_V:G\to\mathbb{C}$ its character:

Theorem II.4.11: (i) $\displaystyle\int\chi_V(g)\operatorname{dg}=\dim V^G;$

$V^G:=\{v\in V:gv=v,\,\forall g\in G\}$, is the fixed point set of $V$.

Theorem II.6.8: Let $V$ be an irreducible complex representation of $G$. Then

$$\int\chi_V(g^2)dg=\left\{\begin{array}{ll} 1&\Leftrightarrow V\quad \textrm{is of real type (realisable over reals)}\\ 0&\Leftrightarrow V\quad \textrm{is of complex type}\\ -1&\Leftrightarrow V\quad \textrm{is of quaternionic type} \end{array}\right.$$


I'll prove that every nontrivial complex irreducible representation of $G$ is of complex type (not realisable over reals and does not admit a quaternionic structure).

Indeed, since $G$ is of odd order, all of its elements has unique square root. Thus $g\mapsto g^2$ is a diffeomorphism. Then, by the change of variables, $$\int\chi_V(g^2)dg=\int\chi_V(g)dg=\dim V^G=0;$$ since every $v\in V^G$ spans a $1$-dimensional trivial sub-representation of $V$, $V^G=0$.

Therefore $V$ is of complex type.

$\endgroup$
  • $\begingroup$ You're not answering the question : the point is not to find a solution, but to check whether the given one is correct $\endgroup$ – Max Jan 14 at 18:24
  • $\begingroup$ But this answer I gave might help other people who are interested in this problem, no? $\endgroup$ – Andre Gomes Jan 14 at 19:06
  • 1
    $\begingroup$ Sure it can help; but perhaps (if you have the answer) you could start by answering whether the proof is correct or not (I don't know Frobenius-Schur indicators so I can't, unfortunately), then say "hey, here's another solution" $\endgroup$ – Max Jan 14 at 19:08
  • $\begingroup$ I also don't know if his proof is correct. But I know another proof. I'll make it explicit in the answer $\endgroup$ – Andre Gomes Jan 14 at 19:14
0
$\begingroup$

The reasoning is correct, though you could perhaps be a little more clear on how you deduce the existence of a quaternionic representation from your equality: the group is non-trivial so must have at least two irreducible representations, so the equality couldn't hold if there were no quaternionic irreducibles.

This is the sort of argument that would be reasonably omitted in many papers, but when first learning these things it helps to spell everything out and omit no arguments.

$\endgroup$

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.