Past exam of the Lie algebras course of the University of Oxford / 2011 / 2 / iv

On a monomial $X^a{Y}^b$, direct differentiation givesFor example $(EF-FE)(X^a{Y}^b)=[a(b+1)-b(a+1)]X^a{Y}^b=(a-b)X^a{Y}^b$. The same calculation with $H$ givesThese are exactly the brackets of $h,e,f$, so the linear map $\rho$ respects every bracket and is a Lie algebra representation.

For $m\ge 0$, let $W_m$ be the space of homogeneous polynomials of total degree $m$. Each operator preserves total degree. Every polynomial is a finite sum of homogeneous polynomials, soIn $W_m$, the monomials $v_j=X^{m-j}{Y}^j$ have distinct weights $m-2j$ and exactly the actions described in part (i). To see simplicity directly, a nonzero invariant subspace is invariant under $H$ and hence contains one of its weight components: polynomial interpolation in $H$ projects onto each of its distinct eigenspaces. Starting with a nonzero multiple of $v_j$, repeated application of $E$ reaches $v_0$, and repeated application of $F$ then reaches all $v_j$. All coefficients used are nonzero in characteristic zero. Therefore $W_m$ is simple and isomorphic to $L_m$.

ThusThis is an algebraic direct sum, not an infinite product or completion. Its simple submodules are precisely the $W_m$: they are pairwise nonisomorphic and each occurs once, so projection onto homogeneous degrees and Schur's lemma exclude diagonal copies between distinct degrees. Complete reducibility here follows from the explicit decomposition, rather than applying a finite-dimensional theorem directly to the infinite-dimensional $W$.