= Solution
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 $L(m)$, $m\in\mathbb Z_{\geq0}$, where $L(m)$ has <weights> $m,m-2,\ldots,-m$. 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>:
$$
\exp(e)=\sum_{j=0}^{N-1}\frac{e^j}{j!},\qquad\exp(-f)=\sum_{j=0}^{N-1}\frac{(-f)^j}{j!}.
$$
Each is a well-defined endomorphism, and its inverse is obtained by negating the argument. Their product is therefore a linear automorphism,
$$
\boxed{s=\exp(e)\exp(-f)\exp(e)\in\operatorname{GL}(V).}
$$
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.
Back to article page
