# Character on conjugacy classes

Let $V_j$, $j = 1,2$ be finite dimensional representations of a group $G$. Show: $\chi_{V_j}$ is a constant on each conjugacy class of $G$, where $\chi_{V_j}$ is the character of the representation.

I've just started with group theory and have a really hard time so I'd like someone to confirm what I did so far was correct:

Per definition: $\chi_{V_j} = Tr(\rho(g))$ where $\rho$ is the grouphomomorphism $G \rightarrow GL(V)$ which represents $G$. The conjugacy class of $G$ is defined as $\{ ghg^{-1} | g \in G \}$ so I'm just plugging in:

$\chi_{V_j}(ghg^{-1}) = Tr(\rho(ghg^{-1}))$ which is the same as (because it's a group hom.) $Tr(\rho(g) \rho(h) \rho(g^{-1}))$ and since $Tr(AB) = Tr(BA)$ we get: $Tr(\rho(g) \rho(h) \rho(g^{-1})) = Tr(\rho(h) \rho(g^{-1}) \rho(g)) = Tr(\rho(h))$

Now since $h$ is a constant element of $G$ this is a constant function and I'm done.

Is this correct?

Thanks a lot in advance! Cheers

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It s correct. But it is better you say since $h$ is an arbitrary element of $G$ ... – Vahid Shirbisheh Mar 16 '13 at 14:53
OP is defining the conjugacy class in terms of a fixed $h$ while letting $g$ vary to cover the various elements in the class, so I think it's a fine way to write it. Yes, the reasoning is correct OP. – anon Mar 16 '13 at 14:56
Eh, I was trying to emphasize the constancy, hence why I wrote it this way. Anway thaks for the quick check! – Howdy Ho Mar 16 '13 at 15:03
You just started group theory and you're already attempting representation theory? That doesn't seem like a good idea, @HowdyHo. – Alexander Gruber Mar 16 '13 at 17:55
Not my idea, it's just on the curriculum. – Howdy Ho Mar 16 '13 at 18:39

You say that the conjugacy class of $G$ is defined as $\{ghg^{-1} \mid g \in G\}$. But the group as a whole does not have a conjugacy class. That is the conjugacy class of the element $h$.
Hey I don't wanna start another thread so I thought I might ask here: Show that $\chi_{V_1} = \chi_{V_2}$ if $V1$ is isomorphic to $V2$. So basically I'm supposed to show that $Tr(\rho_1(g)) = Tr(\rho_2(g))$ Since $\rho_1(g)$ is an element of $GL(V_1)$ and $\rho_2(g)$ is a element of $GL(V_2)$ I can say that $\rho_1(g) = f(\rho_2(g)$ where $f$ is the isomorphism between $V1$ and $V2$, correct? But I don't see how $Tr(\rho_2(g) = Tr(f(\rho_2))$ Is there some theorem that says an isomorphism leaves the trace untouched or something? Or am I on the wrong path here? – Howdy Ho Mar 16 '13 at 16:30
A homomorphism isn't a function $GL(V_1) \to GL(V_2)$, it's a function $V_1 \to V_2$. If it's an isomorphism you can think of it as a change of basis, which is just conjugation, which doesn't change the trace. – Jim Mar 16 '13 at 16:49
Hm okay that was my second guess, the invariance of the trace under change of basis. But how can I be sure $\rho_1$ is just $\rho_2$ in a different basis and not an entirely different function? – Howdy Ho Mar 16 '13 at 17:05