MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:


share|cite|improve this question
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.