# For a character $\chi$ for a finite group $G$, the sum of $\chi(g)$ as $g$ ranges over $G$ is a non-negative integer.

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!

• 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 $\langle \chi, 1 \rangle = 1$.
– user171326
May 6 '17 at 17:15
• You meant if $\chi$ is a non-trivial irreducible character May 6 '17 at 17:15
• 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 $\langle \chi, 1 \rangle = \sum_i n_i \langle \chi_i, 1 \rangle$? 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.

• 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? May 6 '17 at 17:31
• @dhk628 Yes, exactly. May 6 '17 at 17:32