Working through the Fulton-Harris representations of $GL_2(q)$, I've encountered the following, where $-e = -I_2, \in GL_2(q), g \in GL_2(q)$:
"Since $-e$ acts as the identity or minus the identity for any irreducible representation (Schur's lemma),
$\chi(-g) = \chi(g)\cdot\chi(1)/\chi(-e)$
for any irreducible character $\chi$."
I'm a little lost here. I feel like it's staring me in the face. Any help appreciated!
I'm afraid I'm going to write something silly, but...
Suppose first that $-e$ acts as the identity. Then $-g\bigl(=(-e).g\bigr)$ acts like $g$ and so that equalityy just says that $\chi(-g)=\chi(g)$, which is true.
Otherwise, $-e$ acts like minus identity. In that case $\frac{\chi(1)}{\chi(-e)}=\frac{\dim V}{-\dim V}=-1$,where $V$ is the space $GL_2(q)$ is acting on. On the other hand, $\chi(-g)=\chi\bigl((-e).g\bigr)$ and the action go $(-e).g$ is minus the action of $g$. Therefore, $\chi(-g)=-\chi(g)$.
What bothers me is: why to use both $e$ and $1$ for the identity element?