If $\chi$ is a character of $G$, I want to show that $\alpha = \sum_{g \in G} \chi(g)$ is a non-negative integer, and that $\alpha > 0$ if and only if the function $f \colon G \to \mathbb{C}$ such that $f(g) = \chi(g) -1$ is a character of $G$. I've studied representation theory of finite groups but I am still not comfortable with proving stuff (especially when I need to show that a function is a character).

So for the first part, if $\chi$ is trivial, then $\chi(g) = 1$ for any $g \in G,$ so $\alpha = |G|$ which is certainly a non-negative integer. On the other hand , if $\chi$ is not trivial, then the inner product $\langle \chi,1 \rangle$ must be $0,$ so then $$ 0 = \langle \chi,1 \rangle = \frac{1}{|G|}\sum_{g \in G} \chi(g)\overline{1(g)} = \frac{1}{|G|}\sum_{g \in G} \chi(g),$$ where $1$ just represents the trivial character. Is this correct? The answer seems a lot simpler than what the question asks: $\alpha = 0$ for any non-trivial character and $\alpha = |G|$ for the trivial character?

If the above is right, then $\alpha > 0$ if and only if $\chi$ is the trivial character. But is $f(g) = \chi(g) - 1 = 1 -1 = 0$ even be a character? Would it be a character for the zero module? And if everything is right until now, how do I prove the converse? I have no idea what it actually means if $f$ is a character. Any help would be much appreciated!

  • $\begingroup$ You proof is not correct : you did assume that $\chi$ was irreducible. For example, if $V$ is any non-trivial irrep and $U$ is the trivial one, $\chi = \chi_{V \oplus U}$ is not trivial but $\langle \chi, 1 \rangle = 1$. $\endgroup$
    – user171326
    May 6 '17 at 17:15
  • $\begingroup$ You meant if $\chi$ is a non-trivial irreducible character $\endgroup$
    – reuns
    May 6 '17 at 17:15
  • $\begingroup$ Ah, right. So using the fact that any character can be written uniquely as $\chi = n_1\chi_1 + \cdots + n_k\chi_k$ where $n_i \geq 0$ and $\chi_i$ are irreducible characters, we can deduce that $\langle \chi, 1 \rangle = \sum_i n_i \langle \chi_i, 1 \rangle$? $\endgroup$ May 6 '17 at 17:24

Your proof is not correct, due to the reasons that have been pointed out in the comments. However, your idea can be salvaged: if $ \chi_0 $ is the trivial character, we have

$$ \frac{1}{|G|} \sum_{g \in G} \chi(g) = \frac{1}{|G|} \sum_{g \in G} \chi(g) \chi_0(g^{-1}) = \langle \chi, \chi_0 \rangle $$

Now, if we enumerate the irreducible characters of $ G $ as $ \chi_0, \chi_1, \ldots, \chi_n $, we may use complete reducibility to write

$$ \chi = a_0 \chi_0 + a_1 \chi_1 + \ldots + a_n \chi_n $$

for natural numbers $ a_i $. Then, we have that $ \alpha = |G| \langle \chi, \chi_0 \rangle = |G| a_0 > 0 $ by orthogonality of characters, so that $ a_0 > 0 $. Then, by taking the appropriate direct sum of the irreducible representations $ \rho_i $, we may find a representation with character

$$ \chi' = (a_0 - 1) \chi_0 + \ldots + a_n \chi_n = \chi - \chi_0 = \chi - 1 $$

The other direction is proved similarly.

  • 1
    $\begingroup$ Thank you for the answer! I'm just not exactly sure how to show that $\chi'$ is indeed a character? Is it because it is equivalent to the representation $\rho_0^{a_0-1} + \cdots + \rho_n^{a_n},$ which is again a representation? $\endgroup$ May 6 '17 at 17:31
  • 1
    $\begingroup$ @dhk628 Yes, exactly. $\endgroup$
    – Ege Erdil
    May 6 '17 at 17:32

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