2
$\begingroup$

Say I have a complex character $\alpha$ of a finite group $G$ with the inner product $(\alpha, \alpha) = 2$. Since the only decomposition of $2$ as a sum of squares is $2 = 1^2 + 1^2$, the representation belonging two $\alpha$ must be a direct sum of two irreducible representations.

In the particular example I am working with, it is easy to calculate that for $\chi_1$ the trivial representation, the class function $\gamma(g) = \alpha(g) - \chi_1(g)$ has inner product $(\gamma, \gamma) = 1$. Is this enough to show that $\alpha$ is the direct sum of $\chi_1$ and $\gamma$? Since it's not even guaranteed that the representation for $\alpha$ has the subrepresentation for $\chi_1$, this does not seem like something I could deduce.

What about the case that $(\gamma, \chi_1) = 0$? Since this means that $\gamma$ is orthogonal to $\chi_1$ in the vector space of class functions on $G$, and all the irreducible characters are orthogonal to each other, does this imply $\gamma$ is a character (and thus irreducible)?

I would be interested in knowing this in the most generality possible, but I can supply concrete numbers if necessary, I just omitted them as the calculations are somewhat tedious.

$\endgroup$

1 Answer 1

3
$\begingroup$

Any difference of characters is still an integral linear combination of irreducible characters. So if your $\gamma$ has $\langle \gamma, \gamma \rangle = 1$, then $\gamma$ is $\pm 1$ times an irreducible character. In your case, it cannot be $-1$, so it is in fact an irreducible character.

$\endgroup$
2
  • $\begingroup$ Could you elaborate on this? It sounds like you are using a general fact about orthonormal sets, but I am having trouble seeing how you get the implication $(\gamma, \gamma) = 1 \Rightarrow \gamma$ is $\pm 1$ an irreducible character. $\endgroup$
    – tolUene
    Jun 7, 2021 at 12:04
  • 1
    $\begingroup$ @tolUene: the irreducible representations form an orthonormal set. So if $\gamma = \sum n_i \gamma_i$ then $\langle \gamma, \gamma \rangle = \sum n_i^2$. Note also that $n_i$ are integers. $\endgroup$
    – orangeskid
    Jun 7, 2021 at 14:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .