Past exam of the representation theory of semisimple Lie algebras course of the University of Oxford / 2010 / 1 / c / Solution

Let $V\ne 0$ be finite-dimensional. Choose an $H$-eigenvector and apply $e$ repeatedly until the last nonzero vector $v$ in that chain. Such a last vector exists because $e$ raises the $H$-eigenvalue by two and $H$ has only finitely many eigenvalues. Then $ev=0$ and $Hv=lv$ for some $l$.

The nonzero $f$-chain cannot continue indefinitely, since its weights are distinct. Part (a), with its zero endpoint included, gives $l\in {\mathbb{Z}}_{\ge 0}$ and the basis $v,fv,\dots ,f^{l}v$. The raising formula in part (a) shows its span is a submodule isomorphic to $A_l$.

To prove $A_l$ irreducible, a nonzero submodule stable under diagonal $H$ contains a weight vector: polynomial interpolation in the distinct eigenvalues projects onto individual representation weight lines. Repeated raising reaches $v_0$, and repeated lowering then reaches every $v_k$. Thus the submodule is all of $A_l$. Distinct $l$ give different dimensions $l+1$, so the modules are nonisomorphic. Every finite-dimensional irreducible module contains one of these nonzero simple submodules and hence equals it. ThereforeThe zero module has no nonzero simple submodule and is dealt with as the empty direct sum in part (d).