Re: Group character...silly question

[EMAIL PROTECTED] (Dave Rusin) wrote in message news:<[EMAIL PROTECTED]>...
> Your subject line is a little different, though: when you are looking
> at a character of a (finite) group, you're looking at a _function_
> defined on the group, namely  \chi(g) = trace( R(g) )  where  R(g)  is
> a representation matrix. That is, you haven't got just one matrix
> but lots of them. In fact, one is not usually so interested in _a_
> character of a group but rather the whole character table, which sheds
> light on the structure of the group. These pieces of information make
> a difference in your question: it's not just linear algebra now but
> rather group theory. For example, we have some theorems:

I must confess this is a bit confusing. R(g) is a matrix, each R
(representation) gives me a set of matrices R(g) for each g, and there
are several R's possible. That's OK by me. But the whole character
table...of course we are not interested in ALL the R's (if R is a
rep., so is R+R, and so on as far as I get it), so we only talk about
irreducible ones. So far so good. But the next step...there are as
many irreps as the conjugacy classes...is mindblowing. I am still
trying to understand the proof. Could you offer something alternative
to what the text books say? I mean, not the *proof*, it's there all
right, but hints to make it look plausible to myself?

> 
> 1. If  R_1  and  R_2  are two complex representations of a finite group
> whose characters are equal, then the representations are equivalent
> (that is, there is a single invertible  P  with  P R_1(g) = R_2(g) P  for
> every  P; you might say the R1's and R2's are "uniformly similar".)
> In other words: the trace is all you need to distinguish two 
> representations anyway.

This one I understand. Linear Algebra cannot get simpler than this
(without reducing to something more trivial).

> 
> 2. Any complex-valued function on  G  which is constant on conjugacy 
> classes is a linear combination of characters. In other words, the
> traces of all the characters already give you all the kinds of functions
> you could make out of the similarity classes anyway.

So what we are really after is the "partition" of the conjugacy
classes?

> 
> Taken together, (1) and (2) sort of tell us that traces are the only
> similarity invariant we need in classical representation theory.

Got it. Thanks for your (easy to swallow) reply.

> 
> dave