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Let X a curve over an algebraically closed k. Fix $x$ and $y$ two distinct closed points of X. Let G be a connected reductive group over k.

We denote Spec $\hat{\mathcal{O}}_{X,x}$ the formal neighborhood around $x$.

Let $I_{x}\subset G(\hat{\mathcal{O}}_{X,x})$ be the Iwahori subgroup on x.

Let $g\in G(X-y)$. We have in particular that $G(X-y)\subset G(\hat{\mathcal{O}}_{X,x})$

Can we find an element $k\in G(X-x)\cap I_{y}$ such that $kg\in I_{x}$?

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  • $\begingroup$ It depends on how you're choosing the Iwahori subgroups $I_x$ and $I_y$. (You say "the" Iwahori, but that's as bad as "the" Borel.) Since $g$ is regular at $x \in X - y$, all elements of $G(k(X))$ lying in $I_xg^{-1}$ (viewed inside $G({\rm{Frac}}(O^{\wedge}_x))$) are regular at $x$. Hence, the hypothetical $k$ is regular at $x$, so it is everywhere regular and hence constant (over the ground field denoted as $k$...). So $k$ lies in whatever Borel $B_y$ of $G$ "corresponds" to $I_y$ and you're asking if $g(x) \in B_y B_x$. If $B_y$ and $B_x$ are opposite Borels, you win; if equal, you lose... $\endgroup$
    – user29283
    Commented Nov 24, 2012 at 3:05
  • $\begingroup$ Thanks. And for n points is it still true? Say, is there a choice of Iwahoris $I_{x_{1}}$,..., $I_{x_{n}}$ such that : $\forall (h_{2},....,h_{n})\in G(X-x_{2})\times...\times G(X-x_{n})$ there exists $(l_{2},...,l_{n})$ such that: (i) $\forall i$, $l_{i}\in G(X-x)\cap I_{x_{i}}$ (ii) l_{2}h_{2}=....=l_{n}h_{n} (iii)l_{i}h_{i}\in I_{x_{1}} ? $\endgroup$
    – prochet
    Commented Nov 24, 2012 at 3:32
  • $\begingroup$ Correction to the precedent comment in (i) And for n points is it still true? Say, is there a choice of Iwahoris $I_{x_{1}}$,..., $I_{x_{n}}$ such that : $\forall$ $(h_{2},....,h_{n})\in G(X−x_{2})\times...\times G(X−x_{n})$ there exists $(l_{2},...,l_{n})$ such that: $\\$ (i) $\forall i, l_{i}\in G(X-\{x_{1},...,x_{n}\}\cap I_{x_{i}}$ $\\$ (ii)$ l_{2}h_{2}=....=l_{n}h_{n}$ $\\$ (iii) $l_{i}h_{i}\in I_{x_{1}}$ ? $\endgroup$
    – prochet
    Commented Nov 24, 2012 at 3:40
  • $\begingroup$ Perhaps you should rename your ground field, since you use $k$ later. $\endgroup$
    – S. Carnahan ♦
    Commented Nov 24, 2012 at 6:37
  • $\begingroup$ In fact, I'm not sure to understand why, in the case where $B_{x}$ and $B_{y}$ are oppposite, $g(x)\in B_{x}B_{y}$? $\endgroup$
    – prochet
    Commented Nov 24, 2012 at 12:38

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