Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 3 / 5

Use the standard sl2 Lie algebra relations $[h,e]=2e$, $[h,f]=-2f$ and $[e,f]=h$. The classification of finite-dimensional sl2 representations says that every finite-dimensional complex module is a direct sum of $L(m)$, $m\in {\mathbb{Z}}_{\ge 0}$, where $L(m)$ has weights $m,m-2,\dots ,-m$. The raising operator $e$ and lowering operator $f$ move weights by two and minus two. On $L(m)$ each therefore has $(m+1)$st power zero. There is consequently one integer $N$ such that $e^N=f^N=0$ on all of $V$.

The matrix exponentials are thus finite polynomials in nilpotent operators:Each is a well-defined endomorphism, and its inverse is obtained by negating the argument. Their product is therefore a linear automorphism,This Weyl reflection lift in an sl2 representation is algebraically defined by finite sums; no choice of topology or convergence of an infinite series is needed.

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.

The classification of finite-dimensional sl2 representations identifies $L(n)$ with ${\mathrm{Sym}}^n{\mathbb{C}}^2$. In the defining two-dimensional representation,so $e^2=f^2=0$ andOn the $n$th tensor power, the infinitesimal operator associated with $e$ is the sum of its actions on the individual factors. These actions commute, so exponentiating their sum is the tensor product of the individual exponentials; the same holds for $f$. Restricting to the symmetric power consequently identifies the operator $s$ on $L(n)$ with ${\mathrm{Sym}}^{n}S$. It follows thatIndeed $-I$ contributes one minus sign from each of the $n$ tensor factors. The Weyl reflection lift in an sl2 representation can therefore square to minus the identity, even though its image in the Weyl group is an involution.