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I am currently studying representation theory and am struggling with the concepts which relate modules and representations. The specific question I am looking at right now is this:

FS_4 Representation

but while specific help would obviously be greatly appreciated, I could really do with Idiot's Help to start me off getting my head around the general theory. I'm currently feeling very stupid.

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Are you familiar with irreducible representations (aka simple $FS_4$-module in this case), and that you can use characters to check that something is irreducible? –  user27126 Nov 11 '12 at 2:07

1 Answer 1

Here's help for getting off the ground with general theory that addresses your title question:

As you probably know, for an $R$ module $M$, all of the module axioms together are equivalent to the existence of a ring homomorphism from $R$ into $End(M)$. When we're talking about representations over fields, $M$ is also (usually) a finite dimensional vector space, so $End(M)=End(F^n)=M_n(F)$ where $n$ is the dimension of the representation as a vector space.

Now group representations are usually first defined as a group homomorphism $\rho:G\rightarrow GL(n,F)$. This kind of resembles the ring homomorphism above, but right now it's only homomorphism of groups. The trick is that since $\rho:G\rightarrow GL(n,F)\subset M_n(F)$, it extends to be a ring homomorphism from group ring $F[G]\rightarrow M_n(F)$.

At this point you have the extension $\hat{\rho}:F[G]\rightarrow End(M)$, and guess what: that means the extension $\hat{\rho}$ provides an $F[G]$ module structure to $M$!

The connection also works conversely: if you are given an $F[G]$ module, then by just restricting it to $G$ inside of $F[G]$, you get the group homomorphism from $G$ into $GL(n,F)$. This is the correspondence of representations with $F[G]$ modules.


To solve the specific question, you can check if for every $m, n\in M$, there exists $r\in F[G]$ such that $mr=n$ (that would mean $M$ is simple.) But I think Sanchez's suggestion above about characters is better (if you know it).

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