Complex character of a finite group $G$

Question

We let $G$ be a finite group.

If $\chi$ is a complex character of $G$, we define $\overline{\chi}:G \to \mathbb{C}$ by $\overline{\chi}(g)=\overline{\chi(g)}$ for all $g \in G$, and define $\chi^{(2)}:G \to \mathbb{C}$ by $\chi^{(2)}(g) = \chi(g^2)$. We write $\chi_{S}$ and $\chi_{A}$ for the symmetric and alternating part of $\chi$. We note that $\chi_{S}$ and $\chi_{A}$ are characters of $G$ with $\chi^{2}=\chi_{S} + \chi_{A}$ and $\chi^{(2)}=\chi_{S} - \chi_{A}$.

First, I want to show that $\overline{\chi}$ is a character of $G$. Now, we can show that $\overline{\chi}(g)=\overline{\chi(g)}=\chi(g^{-1})$ for all $g \in G$ thus $\overline{\chi}(g)$ is a character. Is that OK?

Next, I want to show that $\chi$ is irreducible iff $\overline{\chi}$ is irreducible.

For $(\implies)$ we assume that $\overline{\chi}$ is not irreducible. Thus we must have a reducible representation $\rho:G \to GL(V)$. But then $\chi$ must also be reducible, w.r.t. to that reducible representation, which is a contradiction. $(\impliedby)$ we can show by the same argument. Tbh, it doesn't seem correct to me, I don't think that I really understand what could go wrong here.

Lastly, we let $\chi_{1}$ be the trivial character of $G$. If I understand it correctly, $\chi_{1}(g)=1$ for all $g \in G$. We want to show that $\langle \chi , \overline{\chi} \rangle= \langle \chi_{S},\chi_{1}\rangle + \langle \chi_{A}, \chi_{1} \rangle$.

We have:

$\langle \chi , \overline{\chi} \rangle = \frac{1}{|G|} \displaystyle\sum_{g \in G} \chi(g)\overline{\chi(g)} = \frac{1}{|G|} \displaystyle\sum_{g \in G} \chi(g)\chi(g^{-1})$

and now I am not sure where to go from here, I'd appreciate any hints.

Answer 1

Recall that a character $\chi$ is irreducible if and only if $\langle \chi,\chi\rangle=1$. Note then that

$$\langle \overline{\chi},\overline{\chi}\rangle =\frac{1}{|G|}\sum_{g\in G}\overline{\chi}(g)\overline{\overline{\chi}(g)}=\frac{1}{|G|}\sum_{g\in G}\overline{\chi(g)}\chi(g)$$

but

$$\langle \chi,\chi\rangle=\frac{1}{|G|}\sum_{g\in G}\chi(g)\overline{\chi(g)}$$

Thus, we see that $\langle \chi,\chi\rangle=\langle\overline{\chi},\overline{\chi}\rangle$.

Answer 2

Let me answer your question, in addition to Alex Youcis answering this. First of all I rectified some of your formulas: "We note that $\chi_{S}$ and $\chi_{A}$ are characters of $G$ with $\chi^{2}=\chi_{S} + \chi_{A}$ and $\chi^{(2)}=\chi_{S} - \chi_{A}$." Then, concerning your last formula: $$\langle \chi , \overline{\chi} \rangle = \frac{1}{|G|} \displaystyle\sum_{g \in G} \chi(g)\overline{\overline{\chi}(g)} = \frac{1}{|G|} \displaystyle\sum_{g \in G} \chi(g)\chi(g)=\frac{1}{|G|} \displaystyle\sum_{g \in G} \chi^2(g)=\langle \chi^2,\chi_1\rangle.$$ It now follows that $\langle \chi , \overline{\chi} \rangle= \langle \chi_{S},\chi_{1}\rangle + \langle \chi_{A}, \chi_{1} \rangle$, and this is what you asked for.

Finally, you asked in the comment if you showed correctly that $\overline{\chi}$ is a character. For this you start with a representation $\frak{X}$ that affords the character $\chi$ and define a new representation $\overline{\frak{X}}: G \rightarrow GL(n,\mathbb{C})$ where $n$ is the degree $\chi(1)$, by $\overline{\frak{X}}(g)=\overline{{\frak{X}}(g)}$, the complex conjugate of the matrix ${\frak{X}}(g)$. Since complex conjugation is compatible with $\frak{X}$ being a homomorphism, this is again a respresentaion and it is easy to see that for the corresponding character is $\overline{\chi}(g)=\chi(g^{-1})$ holds for all $g \in G$. And as Alex pointed out, this character $\overline{\chi}$ is irreducible.