Past exam of the lie algebras course of Part C of the mathematics course of the University of Oxford / 2023 / 2 / c

There is a sign error in the printed assertion under the standard sl2 triple convention. Work first with the operators exactly as printed. Since the $v_i$ form a Hamel basis, their prescribed images extend to linear maps on $V$; each vector is a finite linear combination of basis vectors.

For $i\ge 1$, the operator commutators with $h$ areAt $i=0$, the first identity holds because both sides vanish; the second holds because $hf(v_0)=-2v_1$ and $fh(v_0)=0$. On the other hand, for $i\ge 1$,At $i=0$, both $ef(v_0)=e(v_1)=0$ and $fe(v_0)=0$, so this identity also holds. Thus the actual relations areIn particular, $[e,f]v_1=2v_1\ne -2v_1=h{v}_1$ in characteristic zero; the requested equality cannot hold as printed. Settingcorrects the sign and gives $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$. These are the standard sl2 triple relations, so the standard basis of the sl2 Lie algebra acts on $V$ by $E,H,F$, giving a Lie algebra representation.

For completeness, these operators are linearly independent: applying $aE+bH+cF=0$ to $v_0$ gives $c=0$; applying it to $v_1$ then gives $b=0$; applying it to $v_2$ gives $a=0$. Hence the printed operators span a Lie subalgebra isomorphic to the sl2 Lie algebra, even though their printed normalization fails. Replacing the coefficient $i-1$ in the printed definition of $e(v_i)$ by $1-i$ is one equivalent way to repair the statement.

Use the corrected Lie algebra representation $E=-e$, $H=h$, $F=f$ from part (i). A vector subspace is invariant under $E,H,F$ exactly when it is invariant under the printed $e,h,f$, since changing the sign of a linear map does not change its invariant subspaces. DefineThis is a nonzero proper submodule: $H$ preserves each $k{v}_i$, $F$ increases the index, $E(v_1)=0$, and $E(v_i)=-(i-1)v_{i-1}\in W$ for $i\ge 2$.

We must rule out submodules consisting only of mixtures of weight vectors. Let $U\ne 0$ be any submodule, and choose $u={\sum }_{j\in S}{a}_j{v}_j\in U$ with $S$ finite and every $a_j\ne 0$. The eigenvalues $-2j$ of $H$ are distinct. For $i\in S$, the Lagrange interpolation polynomialsatisfies $p_i(H)u=a_i{v}_i$. Since $U$ is invariant under $H$, it is invariant under every polynomial in $H$, so $v_i\in U$.

If $i=0$, repeated applications of $F$ give $F^m{v}_0=m!v_m$, and $U=V$. If $i\ge 1$, repeated applications of $E$ give a nonzero multiple of $v_1$, because the successive coefficients $-(i-1),\dots ,-1$ are nonzero in characteristic zero. Repeated applications of $F$ to $v_1$ then give every $v_j$ with $j\ge 1$, so $W\subseteq U$.

Finally, if such a $U$ contains any vector outside $W$, that vector has a nonzero coefficient of $v_0$; the same polynomial interpolation extracts $v_0$, forcing $U=V$. Hence the complete list of submodules is $0,W,V$. In particular, $W$ is a simple module. The quotient module $V/W$ has basis ${\overline{v}}_0$; $E{\overline{v}}_0=H{\overline{v}}_0=0$ and $F{\overline{v}}_0={\overline{v}}_1=0$. It is the one-dimensional trivial Lie algebra representation.

The module extension does not split. If $V=W\oplus U$ for a submodule $U$, then $U$ would be a one-dimensional trivial Lie algebra representation, since it maps isomorphically to $V/W$. But $\ker H=k{v}_0$ and $F{v}_0=v_1\ne 0$, so no such $U$ exists. ThusHere semisimple module means a direct sum of modules that are simple modules, equivalently a module in which every submodule has a complementary submodule. This is the nonsplit zero-weight Verma extension for sl2: rescaling to $w_i=i!v_i$ gives $F{w}_i=w_{i+1}$ and $E{w}_i=i(1-i)w_{i-1}$. Its infinite-dimensional vector space is outside the hypotheses of the Weyl complete reducibility theorem.