Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 102 / 1

Write $e,h,f$ for the standard generators of the sl2 Lie algebra. For the representation $\varphi$, use the normalizationThis is the quadratic Casimir element, and by assumption it commutes with every $\varphi (x)$.

Schur lemma says that an endomorphism of a finite-dimensional irreducible complex representation which commutes with the representation is a scalar. Hence $\Omega =c{I}_V$ when $V$ is irreducible.

Let $v$ be a highest-weight vector of highest weight $m$, so $ev=0$ and $hv=mv$. Since $[e,f]=h$,ThereforeIt follows from scalarity thatThis is the Casimir eigenvalue for sl2 in the chosen normalization.

The one-dimensional quotient $V/W$ is trivial because every one-dimensional representation vanishes on the derived algebra $[{\mathfrak{sl}}_2,{\mathfrak{sl}}_2]={\mathfrak{sl}}_2$. Choose $v\in V$ mapping to $1\in V/W$. Thenis a $1$-cocycle:We show that it is a coboundary.

Decompose $W$ into generalized eigenspaces of its Casimir element $\Omega$. These are subrepresentations because $\Omega$ is central. On a generalized eigenspace with nonzero eigenvalue, $\Omega$ is invertible. If $(x_i)$ and $(x^i)$ are dual bases of ${\mathfrak{sl}}_2$ for the Killing form, putInvariance of the Killing form and the cocycle identity give the standard Casimir calculationThus on every nonzero generalized eigenspace, $c(x)=x({\Omega }^{-1}u)$.

On the zero generalized eigenspace, every irreducible composition factor has zero Casimir eigenvalue. By part i and the Classification of finite-dimensional sl2 representations, each such factor is trivial. In a basis adapted to a composition series, the image of ${\mathfrak{sl}}_2$ is therefore strictly upper triangular and hence solvable. Since ${\mathfrak{sl}}_2$ is simple and non-solvable, that image is zero. The cocycle then vanishes because it kills $[{\mathfrak{sl}}_2,{\mathfrak{sl}}_2]$.

Combining the generalized eigenspaces gives $w\in W$ such that $c(x)=xw$ for every $x$. Hence $v-w$ is invariant, andis a decomposition into subrepresentations. This proves the codimension-one case of the Weyl complete reducibility theorem.

The relation $[h,e]=2e$ forces $e{w}_i$ to have weight $a+2(i+1)$, sofor scalars $c_i$. The relation $[e,f]=h$ becomesWith $c_0=b$, this recurrence has the unique solutionfor every $i\in \mathbb{Z}$. Direct substitution also verifies $[h,f]=-2f$ and $[h,e]=2e$, so these formulas define the unique required sl2 Lie algebra action. They form an Intermediate-series sl2 module.

Let $0\ne U\subseteq W$ be a subrepresentation and choosewith finite support. The $h$-eigenvalues $a+2r$ are pairwise distinct. By Lagrange interpolation, there is a polynomial $P$ which is one at one chosen eigenvalue appearing in $u$ and zero at all the others. Then $P(h)u$ is a nonzero scalar multiple of one basis vector $w_i$. Since $U$ is invariant under $h$, it contains $w_i$.

Ifthen the coefficient $c_j=b-ja-j(j+1)$ vanishes. Consequentlyis stable under $h$, $f$, and $e$: the only raising operation that could leave it is $e{w}_j$, and that is zero. Thus $W$ is reducible.

Conversely, let $U$ be a nonzero subrepresentation. By part ii it contains some $w_i$, and repeated application of $f$ gives every $w_r$ with $r\le i$. If every $c_r$ is nonzero, repeated application of $e$ also gives every $w_r$ with $r>i$, so $U=W$. Hence a proper nonzero subrepresentation exists exactly when some $c_j=0$, or