3
$\begingroup$

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 $\chi^{(2)}=\chi_{S} - \chi_{A}$. We write

$\nu(\chi):= \frac{1}{|G|}\displaystyle\sum_{g \in G}\chi(g^2)$

for the Frobenius Schur Indicator.

First, let $\chi_{1}$ be the trivial character of $G$, i.e. $\chi_{1}(g)=1$ 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: \begin{split} \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)1\\ &= \frac{1}{|G|}\displaystyle\sum_{g \in G} (\chi_{S}+\chi_{A})(g)1 \\ &=\langle \chi_{S}+\chi_{A}, 1 \rangle \\ &= \langle \chi_{S},1 \rangle +\langle \chi_{A} , 1 \rangle \\ &= \langle \chi_{S},\chi_{1} \rangle +\langle \chi_{A} , \chi_{1} \rangle \end{split}

Is this correct?

Next, we let $\chi$ be irreducible. We want to show that $\nu(\chi) \in \{-1,1\}$ if $\chi$ is real-valued, and that $\nu(\chi)=0$ otherwise. Let us start from the 'otherwise' case first. We have:

\begin{split} \nu(\chi) &:=\frac{1}{|G|} \displaystyle\sum_{g \in G} \chi(g^2)\\ &= \frac{1}{|G|} \displaystyle\sum_{g \in G} (\chi_{S}-\chi_{A})(g) \\ &= \langle \chi_{S},\chi_{1} \rangle - \langle \chi_{A},\chi_{1} \rangle \\ &= \langle \chi , \overline{\chi} \rangle - 2\langle \chi_{A} , \chi_{1} \rangle \end{split}

and I get stuck here. I think, for the 'otherwise' case, $\langle \chi, \overline{\chi} \rangle = 0$, because we assumed that $\chi$ is irreducible and so it follows that $\overline{\chi}$ is also irreducible and we also know that the irreducible characters form an orthonormal basis (but is it for an arbitrary field?) and so it follows(?). For the real case we'd have that $\langle \chi , \overline{\chi} \rangle = \langle \chi , \chi \rangle =1 $ from irreducibility of $\chi$, but then again, I am still not sure how to deal with $1- 2\langle \chi_{A} , \chi_{1} \rangle$...I'd very much appreciate some help.

$\endgroup$
1
$\begingroup$

\begin{split} \nu(\chi) &:=\frac{1}{|G|} \displaystyle\sum_{g \in G} \chi(g^2)\\ &= \frac{1}{|G|} \displaystyle\sum_{g \in G} (\chi_{S}-\chi_{A})(g) \\ &= \langle \chi_{S},\chi_{1} \rangle - \langle \chi_{A},\chi_{1} \rangle \\ &= \langle \chi , \overline{\chi} \rangle - 2\langle \chi_{A} , \chi_{1} \rangle \end{split}

'Real' case:

$\nu(\chi)=1 -2 \langle \chi_{A}, \chi_{1} \rangle$.

Now, $\langle \chi_{S}, \chi_{1} \rangle =$ dim $Sym^2(V)^G$, $\langle \chi_{A}, \chi_{1} \rangle =$ dim $\wedge^2(V)^G$.

Moreover, we have isomorphism of $G$-representations

$Sym^2(V)^G \oplus \wedge^2(V)^G \cong Hom(V,V^*)$

So dim $Sym^2(V)^G +$ dim $\wedge^2(V)^G =$ dim $Hom(V^*,V)^G$.

By Schur's Lemma the RHS of the above is either $1$ or $0$. It is $1$ if $\rho$ on $V$ and its dual rep. on $V^*$ are equivalent. Thus for the real case we'd have

dim $Sym^2(V)^G +$ dim $\wedge^2(V)^G = 1 $ iff ($Sym^2(V)^G=1$ and $\wedge^2(V)^G=0$) or ($Sym^2(V)^G=0$ and $\wedge^2(V)^G=1$).

Thus we get

\begin{split} \nu(\chi) &=1 -2 \langle \chi_{A}, \chi_{1} \rangle \\ &=1 - 2 \text{dim} \wedge^2(V)^G \\ &= \{-1,1\} \end{split}, as required.

'Other' case:

$\rho$ and its dual are inequivalent

$\implies $ $Sym^2(V)^G +$ dim $\wedge^2(V)^G = 0 $

$\implies$ \begin{split}\nu(\chi)&=\langle \chi , \overline{\chi} \rangle -2 \langle \chi_{A}, \chi_{1} \rangle \\ &= - 2\langle \chi_{A} ,\chi_{1} \rangle \\ &= -2 \text{dim} \wedge ^2 (V)^G\\ &= 0 \end{split},

as required.

$\endgroup$
  • $\begingroup$ Yes, that's what I had in mind! $\endgroup$ – Kenny Wong Apr 10 at 22:08
  • $\begingroup$ That's brilliant! Again, thank you very much for your help. $\endgroup$ – amator2357 Apr 10 at 22:13
0
$\begingroup$

I think everything you've done is correct. There are actually two formulas that you can derive for $\nu(\chi)$ (not that it matters): $$ \nu(\chi) = \langle \chi, \bar\chi \rangle - 2\langle \chi_A, \chi_1 \rangle = 2\langle \chi_S, \chi_1 \rangle - \langle \chi, \bar\chi \rangle .$$ And you're right that, since the character is irreducible, $\langle \chi, \bar\chi \rangle$ is $1$ if the character is real, or $0$ if the character is complex.

Now, let's throw in a few more facts.

  • If $\chi$ is the character for a representation $\rho$ of $G$ on a vector space $V$, then $\chi_S$ is the character for the representation that $\rho$ induces on ${\rm Sym}^2(V)$, and $\chi_A$ is the character for the representation that $\rho$ induces on $\wedge^2 V$.

  • $\langle\chi_S, \chi_1 \rangle$ is the dimension of the $G$-invariant subspace in ${\rm Sym}^2(V)$, and $\langle \chi_A, \chi_1 \rangle$ is the dimension of the $G$-invariant subspace in $\wedge^2V$.

  • We have the decomposition ${\rm Hom}(V, V^\star) \cong {\rm Sym}^2(V) \oplus \wedge^2 V$, where ${\rm Hom}(V, V^\star)$ is the vector space of linear maps from $V$ to $V^\star$. Our representation $\rho$ induces a representation on ${\rm Hom}(V, V^\star)$, which reduces to the representations mentioned earlier when restricted to ${\rm Sym}^2(V)$ and $\wedge^2 V$.

  • Since $\rho$ is irreducible, Schur's lemma tells us the number of $G$-invariant linear maps from $V$ to $V^\star$ is either equal to $1$ (if the representation $\rho$ on $V$ and its dual representation on $V^\star$ are equivalent), or equal to $0$ (if $\rho$ and its dual are inequivalent).

  • The character of the dual representation is precisely $\bar\chi$.

This should be enough information for you to piece together a proof. Feel free to leave a comment if this is insufficient.

$\endgroup$
  • $\begingroup$ Thank you for such a great answer Kenny! I'll be spending some more on this tonight, hopefully I can piece it all together! $\endgroup$ – amator2357 Apr 10 at 16:38
  • $\begingroup$ I have posted my answer below. I'd be very grateful if you could have a look at it, whenever you can and let me know if my reasoning is correct :) $\endgroup$ – amator2357 Apr 10 at 21:36

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.