Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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

1 Answer 1

That is correct save one small point about terminology, which is probably a typo and not a misunderstanding:

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 $h$.

share|improve this answer
    
Oops yes of course. Thanks for pointing that out –  Howdy Ho Mar 16 '13 at 15:42
    
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
    
If you write out the definition of a homomorphism of linear representations, then the fact that it is an isomorphism means you will be able to move one of the matrices to the other side of the equation and get conjugation. –  Jim Mar 17 '13 at 1:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.