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Let $x_{1}$, $x_{2}$, $x_{3}$ three disctinct closed points of a curve $X$ over an algebraically closed field k.

Let G a connected reductive group and $\mathfrak{g}$ his Lie algebra. I fix a Borel $B_{x_{1}}$ and let $I_{x_{1}}$ the corresponding Iwahori.

For $l\in\mathfrak{g}(X-x_{3})\cap Lie(I_{x_{1}})$, is there a choice of a Borel $B_{x_{2}}$ such that there exists $u\in\mathfrak{g}(X-x_{3})\cap Lie(I_{x_{1}})$ and $v\in\mathfrak{g}(X-x_{3})\cap Lie(I_{x_{2}})$ such that:

$l+u=v$

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I may be missing something, but let $u=0$ and $B_{x_2}$ be a Borel whose Lie algebra contains $l(x_2)$. –  Pavel Safronov Nov 29 '12 at 0:24
    
sorry, there is a problem of quantificators. The exact sentence should be? Is there a choice of a Borel $B_{x_{2}}$ such that for all $l\in\mathfrak{g}(X-x_{3})$ there exists u.... –  prochet Nov 29 '12 at 1:25
    
It seems you can still take $u = -l$. –  Pavel Safronov Nov 29 '12 at 1:57
    
Forget about what I have said. My question is rather. Do we have for appropriate Iwahori $I_{x_{1}}$ $I_{x_{2}}$ $I_{x_{3}}$ a decomposition $\mathfrak{g}(k((t_{3}))=Lie (I_{x_{3}})+\mathfrak{g}(X-x_{3})\cap Lie(I_{x_{1}}\cap Lie(I_{x_{2}})$ Here I chose the local uniformiser $t_{3}$ s.t $\mathfrak{g}(k((t_{3})))\subset\mathfrak{g}(X-x_{3})$ –  prochet Nov 29 '12 at 2:53
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