3
$\begingroup$

Let $G$ a connected reductive split group over $k=\bar{k}$, $(B,T)$ a split Borel pair. Let $F:=k((t)))$. Let $\tilde{W}$ the extended Weyl group, $\tilde{W}=N_{G}(T(F))/T(O)$.

By Iwasawa decomposition,

$G(F)=\coprod\limits_{w\in\tilde{W}}IwI$,

where $I$ is the Iwahori subgroup associated to $B$.

We denote by $\overline{IwI}/I$ the closure of $IwI/I$ in $G(F)/I$. In general, this closure is singular and if we assume that $G$ is semsimple and simply connected, then we know that $\tilde{W}$ is generated by simple roots and we can solve the singularities with a Demazure resolution by writing $w=s_{1}\dots s_{l}$ with $s_{i}$ simple reflexions.

Nevertheless, when $G$ is only split conneceted reductive, then $\tilde{W}$ is no more generated by simple reflexions. My question is then how do we solve the singularities of $\overline{IwI}/I$?

$\endgroup$

1 Answer 1

2
$\begingroup$

If $G$ isn't simply connected, then $G(F)$ isn't connected, so $G(F)/I$ isn't connected. Your question is about resolving those $\overline{IwI}/I$ that live in components other than the one containing the basepoint $I/I$.

On such a component $C$, there will still be a closed $I$-orbit $X_C$, automatically smooth. (In the familiar case, $X_C = \{ I/I \}$.) Generalize the Bott-Samelson-Demazure-Hansen resolution to $P_{\alpha_1} \times^I \cdots \times^I P_{\alpha_k} \times^I X_C \to C$. The LHS is smooth, and can be made to resolve any choice of $\overline{IwI}/I$, through suitable choice of $\vec \alpha$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.