most recent 30 from http://mathoverflow.net 2013-06-20T03:01:35Z http://mathoverflow.net/feeds/question/92794 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92794/exact-sequence-of-weyl-groups Rajkarov 2012-04-01T05:40:15Z 2012-04-01T16:40:54Z

<p>If we note $A_{k}$ the category of affine algebraic groups defined over $k$ and $\mathcal{G}$ the category of finite groups, we have a functor $W:A_{k}\longrightarrow \mathcal{G}$, where $W(G)$ is the Weyl group of an algebraic group $G$. Is taht functor is exact ? </p>

http://mathoverflow.net/questions/92794/exact-sequence-of-weyl-groups/92822#92822 Mikhail Borovoi 2012-04-01T16:40:54Z 2012-04-01T16:40:54Z

<p>Yes, it is exact.</p> <p>For an algebraically closed field $k$ and a connected reductive $k$-group $G$, one can construct a canonical based root datum of $G$, so we obtain a canonically defined Weyl group $W(G)$. I am not sure that $G\mapsto W(G)$ is a functor on the category of connected reductive $k$-groups and homomorphisms of $k$-groups. However, it is certainly a functor on the category of connected reductive $k$-groups and <em>normal</em> homomorphisms. A homomorphism of connected reductive $k$-groups is called normal if its image is a normal subgroup.</p> <p>If we have a short exact sequence of connected reductive $k$-groups, then we obtain an induced short exact sequence of semisimple groups of adjoint type, which clearly splits, so we obtain a split short exact sequence of Weyl groups. In other words, if we have a short exact sequence $1\to G_1\to G_2\to G_3\to 1$ of connected reductive $k$-groups, then $W(G_2)=W(G_1)\times W(G_3)$.</p>