= Solution
On a monomial $X^aY^b$, direct <differentiation> gives
$$
E(X^aY^b)=bX^{a+1}Y^{b-1},\quad
F(X^aY^b)=aX^{a-1}Y^{b+1},\quad
H(X^aY^b)=(a-b)X^aY^b.
$$
For example $(EF-FE)(X^aY^b)=[a(b+1)-b(a+1)]X^aY^b=(a-b)X^aY^b$. The same calculation with $H$ gives
$$
[H,E]=2E,\qquad[H,F]=-2F,\qquad[E,F]=H.
$$
These are exactly the brackets of $h,e,f$, so the <linear map> $\rho$ respects every bracket and is a <Lie algebra representation>.
For $m\ge0$, 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, so
$$
W=\bigoplus_{m=0}^{\infty}W_m.
$$
In $W_m$, 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 $L_m$.
Thus
$$
\boxed{\mathbb C[X,Y]=\bigoplus_{m\ge0}\operatorname{span}\{X^m,X^{m-1}Y,\ldots,Y^m\}
\cong\bigoplus_{m\ge0}L_m.}
$$
This 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 $W$.
Back to article page
