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Over a field of characteristic zero, a locally nilpotent operator has an algebraically defined exponential, because the sum is finite for each vector. Its inverse is $\exp (-T)$. The nilpotence bound may depend on the vector. In an sl2 Lie algebra Verma module, the raising operator is locally nilpotent but the lowering operator is not, so the latter's exponential does not define an endomorphism of the ordinary algebraic module.

// Target: lie-algebra.bigb

Let $v_0$ be the highest vector of the Verma module $M(n)$ and set $v_k=f^k{v}_0$. The Poincaré-Birkhoff-Witt theorem implies that $(v_k{)}_{k\ge 0}$ is a basis, so none of these vectors vanishes. The actions areThe raising operator is a locally nilpotent operator: on every vector, repeated application of $e$ eventually removes all its finitely many basis terms. Thus its algebraic exponential of a locally nilpotent operator is defined, and $\exp (e)v_0=v_0$. But the middle factor in the proposed product would have to send this vector towhich has infinitely many nonzero, linearly independent basis coefficients. An element of the algebraic Verma module is a finite linear combination of the $v_k$, so this expression is not in $M(n)$. In particular the lowering operator is not locally nilpotent.

Even for integral $n\ge 0$, $f^{n+1}{v}_0$ is nonzero in the Verma module; it becomes zero only in its finite-dimensional irreducible quotient $L(n)$. Hence $s$ is not defined as the displayed composition of endomorphisms of the algebraic Verma module. Working in a completion would be a different problem and is not part of the module specified here.