Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 102 / 2

Choose the Borel subalgebra $\mathfrak{b}=\mathfrak{t}\oplus {\mathfrak{n}}^{+}$ determined by the positive roots. Regard the one-dimensional space ${\mathbb{C}}_{\lambda }$ as a $\mathfrak{b}$-module on which ${\mathfrak{n}}^{+}$ acts by zero and $h\in \mathfrak{t}$ acts by $\lambda (h)$. The Verma module isThe Poincare-Birkhoff-Witt theorem identifies it as a vector space with $U({\mathfrak{n}}^{-})$ acting on a highest-weight vector $v_{\lambda }$.

For each positive root $\alpha$, arbitrary powers of a negative-root vector contribute the geometric series $1+e^{-\alpha }+e^{-2\alpha }+\cdots$. Consequently the formal character of a weight module isThis product is interpreted in the completion of the group algebra in the negative-root direction; the PBW basis proves that every coefficient is the correct finite weight space dimension.

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The weights of $M_{\lambda }$ lie below $\lambda$ in the positive-root order, and every weight space is finite-dimensional. A nonzero submodule $V$ is stable under the Cartan subalgebra, so it is a direct sum of its weight spaces. Choose a maximal weight $\mu$ occurring in $V$. Every positive-root operator would raise its weight; maximality therefore makes it kill any nonzero $v\in V_{\mu }$. Thus $v$ is a singular vector.

The Casimir element is central and acts throughout $M_{\lambda }$ byThe same element acts on the highest-weight vector $v$ of weight $\mu$ byBoth are the action of one operator on the same module, so the scalars agree and

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For the sl2 Lie algebra, let $v_{\lambda }$ be the highest-weight vector. The Poincare-Birkhoff-Witt theorem gives the basis $(f^r{v}_{\lambda }{)}_{r\ge 0}$, and the defining Lie brackets implyA positive-degree basis vector is singular exactly when $r=\lambda +1$ is a positive integer. Therefore $M_{\lambda }$ is irreducible when $\lambda \notin {\mathbb{Z}}_{\ge 0}$.

If $\lambda =m\in {\mathbb{Z}}_{\ge 0}$, the vector $f^{m+1}{v}_{\lambda }$ has weight $-m-2$ and generates a submodule isomorphic to $M_{-m-2}$. The latter is irreducible because $-m-2\notin {\mathbb{Z}}_{\ge 0}$. Every nonzero submodule contains a singular vector by the preceding part, and the displayed coefficient shows that this is the only possible proper singular vector. Henceis the unique proper nonzero submodule, as summarized by the Reducibility of an sl2 Verma module.

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