What Changes in Representation Theory Over Non-Complex Fields?

[Hello Kitty]

I'm done a basic course on representation theory and character theory of finite groups, mainly over a complex field. When the order of the group divides the characteristic of the field clearly things are very different.

What I'd like to learn about is what happens when the field is not complex but still quite well-behaved. In particular if we have an algebraically closed field whose characteristic doesn't divide the order of the group what changes?

The reason I ask is that there doesn't seem to be a very good treatment of this in any of the books I've seen. Can anyone offer any suggestions? I guess I could start from scratch and go though all the proofs in the complex case from the bottom up checking whether they still hold, but it would be nice to have a reference.

Are there any major pitfalls when trying to transfer the theory from the complex case?

 

[matt grime]

In particular if we have an algebraically closed field whose characteristic doesn't divide the order of the group what changes?

Nothing. That's why there are no treatises on the subject - the only added complication is it not being over C.

 

[Hello Kitty]

So the character degrees over \overline{\mathbb F_p} are the same as for \mathbb C provided p \not\vert \ |G|?

 

[matt grime]

Look at your notes - where does it use that the characteristic of C is zero? It will only use that it is co-prime to |G|.

The method of going from C to char p was given by Brauer in the 50s. Reps over C are actually realizable over the algebraic integers, A. Pick a maximal ideal containing the prime ideal (p) in A, and reduce modulo this ideal. This yields the projective modules over the field of char p, which are all the modules if p is coprime to |G|.

 

[nehamakhijani]

I have a question in representation theory. There is a result that says that if I have a linear character of a subgroup H of a group G with kernel K, then the induced character is irreducible iff (H,K) is a Shoda pair.

The proof uses the fact that
If, chi(ghg-1)=chi(h) for all h in H ∩ g(-1)Hg, then
[H,g]∩H ⊂ K.

I am not able to prove this one...can sumbody help??