The Weyl group of $\widehat{\mathfrak{sl}}_2$.

Question

On page 5 of this paper, example 3.1, it is said that the Weyl group of $\widehat{\mathfrak{sl}}_2$ is $$ W= \langle s_1, s_2 \mid s_1^2 = s_2^2 = 1 \rangle. $$ Why the Weyl group of $\widehat{\mathfrak{sl}}_2$ is $W$? Thank you very much.

Answer 1

Do you ask for a proof adapted to this special case?

In general, if you have a symmetrizable generalized Cartan matrix $A\in\text{Mat}_{n\times n}({\mathbb Z})$, then the Weyl group of the Kac-Moody algebra ${\mathfrak g}(A)$ associated with $A$ admits a description as a Coxeter group as follows: It has generators $s_1,...,s_n$, and relations $s_i^2=e$ and $(s_i s_j)^{m(i,j)}=e$ for all $i\neq j$, where $m(i,j)=2,3,4,6,\infty$ according to whether $A_{ij} A_{ji}$ is $0,1,2,3$ or bigger. Here, the generators $s_i$ correspond to the simple reflections in the Cartan subalgebra of ${\mathfrak g}(A)$, and the difficult part is to establish that the above relations already generate all relations between the simple reflections.

In the case of $\widehat{{\mathfrak s}{\mathfrak l}}_2$, $A_{12}A_{21}=4$, so $m(1,2)=\infty$, and your Weyl group is just $\langle s_1,s_2\ |\ s_1^2=s_2^2=e\rangle$.