Schur functors are pairwise non isomorphic

77 Views Asked by At

In Fulton-Harris Part I Weyl's construction there a characterization of some of the irreducible representation of $GL(V)$, with $V$ a finite complex vector space. In particular, Theorem $6.3$ point $(4)$ states that $S_{\lambda}V$ with $\lambda$ partition of $d$ are irreducible. For this it sufficient to know that Lemma $6.22$ can be applied.

I'd like to know whether I could infer that $S_{\lambda}V$ is not isomorphic to $S_{\mu}V$ when $\lambda \ne \mu$. I have the sentence written on my notes taken during class after the statement $(3)$ ( which states that $\chi_{S_{\lambda}V}(g) = s_{\lambda}(x_1,\cdots,x_k)$, where $s_{\lambda}$ is the schur symmetric functions and $x_1,\cdots,x_k$ are the eigenvalues of $g \in GL(V)$).

What I don't get is how the characters could be enough since here we are talking about obejcts having a $B = Hom_{S_d}(V^{\otimes d},V^{\otimes d})$ module structure while, as far as I see completly invariance of characters could be used only for representations of (finite groups in particular) symmetric group $S_d$ (since a reason could be that $s_{\lambda}$ are the characters of the irreducible representions of $S_d$). There is any relation between being an $S_d$-representation and $B$ representation which could give what I'm looking for, i.e $S_{\lambda}V$ is not isomorphic to $S_{\mu}V$ when $\lambda \ne \mu$.

Any help of solution will be appreciated.

1

There are 1 best solutions below

2
On BEST ANSWER

The character can be defined for finite-dimensional representations of an arbitrary group $G$ (or more generally for finite-dimensional modules over an arbitrary algebra), and it's an invariant of that representation in this generality, so for an arbitrary group $G$, if two representations have different characters they cannot be isomorphic. This does not require the stronger result for finite groups that if two representations have the same character then they are isomorphic (which is generally false for infinite groups and for algebras); it is much easier and true in much more generality.

So yes, it suffices to know that the Schur functions are different.