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a) What does it mean to say that the Character of a representation is irreducible on its own?

b) If Char($K$) is $0$ then kernel of character is a normal subgroup of G , why ??

c) Over a field of char $0$ , representation are isomorphic iff they have the same Character. why ?

What is so special about the field being Char $O$ ?

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By the way, you should accept some answers to your questions if you want your "accept rate" to be higher. And higher it is, better are the chances of getting an answer to your future questions. –  M Turgeon Apr 19 '12 at 14:35
    
link can you look up here and what you think about it on the kernel of a Character. :) –  Theorem Apr 19 '12 at 14:45

2 Answers 2

up vote 3 down vote accepted

(a) We say a character $\chi$ associated with representation $\varphi$ is irreducible if and only if the representation $\varphi$ is irreducible.

(b) In characteristic 0, the kernel of the character turns out to be exactly the kernel of the corresponding representation.

(c) This one is much more complicated. This essentially boils down to using orthogonality relations among characters. The idea is that if two characters match, then the inner product of these characters with some irreducible character will match and it turns out that this inner product is the number of times the corresponding irreducible representation appears in the representations in question. Thus because both representations decompose into the same number of copies of various irreducible representations, they must be isomorphic. To get a really good sense of what is going on here, just pick up a book on representation theory and read the chapter(s) on character theory. [For example: Martin Burrow's Representation Theory of Finite Groups (a cheap dover book) is quite accessible.]

Why is char 0 theory so different? Because much of group representation theory revolves around having an invariant inner product. The standard inner product is formed by "averaging over the group". So to compute this "average" you need to divide by the order of the group. If the group's order is not invertible, this cannot be done. So no inner product. So you can't use the standard techniques.

Do note though that characteristic p theory does go through about the same if the order of the group n is relatively prime to p. Why? Because if n and p are relatively prime, then n is invertible in a field of char p so we can form the invariant inner product etc. [All the standard techniques work.]

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@ Bill what do you mean be "averaging over the group" and the last paragraph of your post, if you could explain a bit more in detail , i would be glad ! –  Theorem Apr 19 '12 at 15:04
    
I am not sure what happened here... you gave pretty much the same answer as I did (except part (b), where I was confused by the terminology), and your answer is now the accepted one? With more upvotes? –  M Turgeon Apr 19 '12 at 15:05
    
@Vedananda Let me just refer you to these notes: maths.gla.ac.uk/~ajb/dvi-ps/groupreps.pdf Maybe I'll write up something more careful later. Look at page 20 and you'll see a summation appearing over and over again (that's averaging over the group). And again on page 37 (definition 3.11). It's a common trick but to make it work you must be able to divide by the order of $G$. –  Bill Cook Apr 19 '12 at 16:42
    
If you look into representation theory of compact Lie groups, you'll find a similar averaging trick used. But instead of summations, you have integrals. The "compact" assumption guarantees that you have finite volume and so you can divide by the volume of $G$. Since this averaging gives you an invariant Hermitian form, you can treat these (infinite) groups kind of like finite ones. –  Bill Cook Apr 19 '12 at 16:46
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(b) is not true (nor very sensical) if the characteristic divides the order of the group. The Brauer character of a representation does not determine its kernel, and the Frobenius character is even worse. Consider the regular representation of a cyclic group of order p over a field of characteristic p. The kernel of the character is the entire group, but the kernel of the rep is just the identity. –  Jack Schmidt Apr 19 '12 at 19:31

I will try to answer your questions:

  1. The only time I have seen the words "irreducible character", it was a short way of saying "character of an irreducible representation". But I could be wrong.
  2. Edit: For a character $\chi$ associated to a representation $(V,\rho)$, we define the kernel of $\chi$ to be $$\ker_\chi:=\{g\in G:\chi(g)=\chi(1_G)\}$$ (I wasn't aware of this terminology). Now, if $g\in\ker\rho$, then clearly $\chi(g)=\chi(1_G)$, even in positive characteristic. Now, if you're working over characteristic 0 and $g$ is such that $\chi(g)=\chi(1_G)$, then $g\in\ker(\rho)$, because $\rho(g)$ is semisimple of finite order, so its eigenvalues are roots of unity, and so having the same trace as the identity implies it is equal to the identity. Therefore, $\ker_\chi$ is a normal subgroup, because it is equal to the kernel of $\rho$, which is clearly a normal subgroup. Now, in characteristic $p$, I imagine you could get a $p$-dimensional representation (so in particular, $\chi(1_G)=0$), where some element $g$ has all $p$-th roots of unity as eigenvalues (there would be a problem of existence of $p$-th roots of unity here) has trace 0 without being in the kernel of $\rho$.
  3. Basically, it follows from the fact that characters coming from distinct irreducible representations will be orthogonal with respect to a suitably defined inner product on the space of class functions. Moreover, their "norm" will be equal to 1, and since the number of distinct irreducible representations of a finite group $G$ is equal to the number of conjugacy classes, it follows that the set of characters associated to the irreducible representations is an orthonormal basis (for the space of class functions). Now, using Frobenius reciprocity and the fact that any representation is semisimple will give you the result. Namely, if $V$ is a representation of $G$ with character $\chi$, and $\chi_1,\ldots,\chi_r$ are the characters of the irreducible representations, then $\langle\chi,\chi_j\rangle$ will give you the multiplicity of $\chi_j$ in $V$ (by Frobenius). These multiplicities uniquely determine the representation $V$ (by semisimplicity).

Now, in all I have said, we don't really need the field to be of characteristic zero. The most important thing is that the characteristic of your field of definition should not divide the order of the group (otherwise, you won't have semisimplicity). You might need you field to be algebraically closed to prove that characters uniquely define representations, but I am not sure about that.

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Otherwise ok. But I think that the kernel of a character $\chi$ (or of a representation $\rho$) consists of the elements $g\in G$ such that $\chi(g)=\chi(1)$. Such elements $g$ do form a normal subgroup: Because $\chi(g)$ is the sum of $\chi(1)$ roots of unity (=the eigenvalue of $\rho(g)$), this can happen only, if they are all equal to one. As $g$ acts semisimply, we must then have that $\rho(g)$ is the identity matrix. So $$ \chi(g)=\chi(1)\Leftrightarrow g\in \ker\rho. $$ –  Jyrki Lahtonen Apr 19 '12 at 14:47
    
As Jyrki mentions, you've got the definition of the kernel of a representation wrong. For example, see en.wikipedia.org/wiki/Character_theory –  Bill Cook Apr 19 '12 at 14:49
    
@ Jyrki what do you mean by saying that 'g acts semisimply'! –  Theorem Apr 19 '12 at 14:56
    
@Bill , i think what jyrki has defined is right , isn't it ? –  Theorem Apr 19 '12 at 14:58
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@Vedananda Yes. If you look at the Wikipedia page on character theory, you'll find the same definition there. It seems like an odd definition at first, but the whole point is to make the character's kernel match the representation's kernel. –  Bill Cook Apr 19 '12 at 16:34

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