Kernel of a (complex) character.

Let $$\psi$$ be a character in $$\textbf{Class}(G,\mathbb{C})$$. Then $$\psi = \sum\limits_{\chi \in \text{Irr}(G)} n_{\chi}(\psi) \chi$$ where $$\text{Irr}(G)$$ is a representative set of irreducible characters of $$G$$.

The following should hold: $$\ker \psi = \bigcap_{\{\chi \in \text{Irr}(G)\;|\; n_{\chi}(\psi) \geq 1\}} \ker \chi.$$

Let's denote the RHS by $$\mathfrak{S}$$. The inclusion $$\mathfrak{S} \subseteq \ker \psi$$ is clear. We aim to show the inclusion $$\ker \psi \subseteq \mathfrak{S}$$.

We want to show that if $$g$$ in $$G$$ is such that $$$$\psi(g) = \psi(1) \qquad \qquad(1)$$$$ then $$\chi(g) = \chi(1)$$ for $$\chi \in \text{Irr}(G)$$ such that $$n_{\chi}(\psi) \geq 1$$.

By the equality $$(1)$$ we have $$\sum\limits_{\chi \in \text{Irr}(G)} n_{\chi}(\psi)\chi(g) = \sum\limits_{\chi \in \text{Irr}(G)} n_{\chi}(\psi)\chi(1).$$

We know that:

1. $$|\chi(g)| \leq \chi(1)$$.
2. $$|\psi(g)| \leq \sum n_{\chi}(\psi) |\chi(g)|$$.

From this, one should be able to deduce that $$\chi(g) = \chi(1)$$. We get the following inequality

$$\psi(g) \leq |\psi(g)| \leq \sum n_{\chi}(\psi)|\chi(g)| \leq \sum n_{\chi}(\psi)\chi(1) = \psi(1).$$

Since $$n_{\chi}(\psi) \in \mathbb{Z}_{\geq 0}$$ and $$|\chi(g)| \geq 0$$, if there is some $$\chi$$ such that $$|\chi(g)| < \chi(1)$$ then we will not have equality, hence $$|\chi(g)| = \chi(1)$$ for all $$\chi$$ such that $$n_{\chi}(\psi) \geq 1$$.

I believe my reasoning is sound up to this point.

Lastly, we want to show that $$|\chi(g)| = \chi(g)$$ which would imply that $$\chi(g) = \chi(1)$$. We know that $$\chi(g) = re^{i \theta}$$ such that $$|re^{i\theta}| = |r| = \chi(1) = r$$. We want to show that $$e^{i \theta_{\chi}} = 1$$ for all $$\chi$$ such that $$n_{\chi}(\psi) \geq 1$$. Here, I am currently stuck. Any hints on how to proceed?

Edit $$1$$: Wait, I just realized something; since $$\psi(g) = \sum n_{\chi}(\psi)\chi(g) \leq |\psi(g)| \leq \sum n_{\chi}(\psi)|\chi(g)| \leq \sum n_{\chi}\chi(1) = \psi(1)$$ if there was some $$\chi(g)$$ that was negative (where $$n_{\chi}(\psi) > 0$$), then we would have a strict inequality between $$\psi(g)$$ and $$\sum n_{\chi}(\psi)|\chi(g)|$$; but then $$\psi(g) < \psi(1)$$, hence $$\chi(g)$$ must be positive for all $$g$$ such that $$n_{\chi}(\psi) \geq 1$$.

Edit $$2$$: Or well, hm, the issue is that $$\chi(g) \in \mathbb{C}$$, so my statement in edit $$1$$ does not really make sense.

• In edit 1 you've proved that $\chi(g)=|\chi(g)|$. Then in particular $\chi(g)$ is real.
– Deif
Commented May 11 at 1:08
• I don't see that I proved it? What if $\chi(g)$ is complex? "Negative" does not make sense if the complex part of $\chi(g)$ is not zero (what does make sense then is "negative real part", i.e. $\text{Re}(\chi(g)) < 0$). Commented May 11 at 1:14
• Actually, I see what the issue is, $\psi(g) \leq |\psi(g)|$ does not make sense without assuming that $\chi(g)$ is real for all $n_{\chi}(\psi) \neq 0$; so I have implicitly assumed that $\chi(g)$ is real. Commented May 11 at 1:25
• Note that you have the equality $|\sum n_{\chi}(\psi)\chi(g)|=\sum |n_{\chi}(\psi)\chi(g)|$. Then each term $n_{\chi}(\psi)\chi(g)$ is in the same direction. In particular, they are in the same direction as $n_{1_G}(\psi)1_G(g)=n_{1_G}(\psi)$ and hence each term is real.
– Deif
Commented May 11 at 1:35
• oh, yes, right. I forgot about $\mathbf{1}_G$! And what I said above was slightly incorrect, what I meant to say is: We know that $\psi(g)$ is real. What you just said shows that $\theta_X = 2\pi k$, right. It follows then that $|\chi(g)| = \chi(1) = \chi(g)$ (otherwise the inequalities don't work) so that $g \in \ker \chi$ for $\chi$ such that $n_{\chi}(\psi) \geq 1$. In particular, since each $\chi(g)$ is real, we have $\psi(g) \leq |\psi(g)| \leq \psi(1)$, so unless $\chi(g) > 0$ for each $n_{\chi}(\psi) \geq 1$, we can't have $\psi(g) = \psi(1)$. Thanks @Deif. Commented May 11 at 1:48

Given a character $$\chi$$ of the finite group $$G$$, the kernel of $$\chi$$ is the kernel of any representation $$f:G \to \mathrm{GL}(V)$$ affording $$\chi$$. The sum of characters is afforded by the direct sum of representations, and the kernel of the direct sum is the intersection of the kernels of the summands. This proves the result you want (in a very different way to yours).