The relation $[h,e]=2e$ forces $ew_i$ to have weight $a+2(i+1)$, so
$$
ew_i=c_iw_{i+1}
$$
for scalars $c_i$. The relation $[e,f]=h$ becomes
$$
(c_{i-1}-c_i)w_i=(a+2i)w_i.
$$
With $c_0=b$, this recurrence has the unique solution
$$
\boxed{c_i=b-ia-i(i+1)}
$$
for 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>.

Solution

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

{c}

An intermediate-series $\mathfrak{sl}_2$-module has one-dimensional weight spaces indexed by all integers. In the basis $(w_i)_{i\in\mathbb Z}$, one useful normalization is
$$
hw_i=(a+2i)w_i,\qquad fw_i=w_{i-1},\qquad
ew_i=(b-ia-i(i+1))w_{i+1}.
$$
It is irreducible exactly when every raising coefficient is nonzero.


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