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Let $G$ a semisimple group and $B$ a Borel subgroup. We denote by $Bun_{G}$ the stack of G-bundles.

Is it true that a certain open subset $Bun_{B,r}$ maps smoothly to $Bun_{G}$?

My question comes from Lemma 14 .2.1 from http://arxiv.org/pdf/math/0611323.pdf

but I'm not sure to understand well.

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You need to formulate the question more precisely - in this way it is obvious (any map between reduced algebraic stacks locally of finite type is generically smooth). –  Alexander Braverman Jun 13 '13 at 0:36
    
I fixed the typo in your title, which you can also do if you notice them in future. –  David Roberts Jun 13 '13 at 1:02
1  
If $H$ is a subgroup of $G$, the map from the stack of $H$-bundles to the stack of $G$-bundles is a fiber bundle with fiber $G/H$. –  Angelo Jun 13 '13 at 4:17
    
I don't understand the purpose of the lemma 14.2.1 then, why restricting to an open subset if everything is already smooth with connected fibers? –  prochet Jun 13 '13 at 6:11
    
Angelo: what you wrote is wrong (it is only true for bundles over a point). –  Alexander Braverman Jun 13 '13 at 15:57

1 Answer 1

up vote 4 down vote accepted

The short answer is the following: for a $G$-torsor $E_G$ over $C$ and for the associated projective scheme $E_{G,B} := E_G/B$, a lift $E_B$ of $E_G$ to a $B$-torsor over $C$ is the same thing as a section $\sigma:C\to E_{G,B}$ of the projection $\pi:E_{G,B}\to C$. Via infinitesimal deformation theory of the Hilbert scheme, this section is unobstructed if $H^1(C,\sigma^*(\Omega_\pi)^\vee)$ is zero. Finally, $\sigma^*(\Omega_\pi)^\vee$ turns out to be $E_B \times^B \text{Lie}(U^{-})$. In fact, since $E_B$ has a further reduction of structure group to a maximal torus $T$, i.e., $E_B$ equals $B \times^T E_T$ for a $T$-torsor $E_T$, the bundle $E_B\times^B \text{Lie}(U^{-})$ turns out to equal $E_T\times^T \text{Lie}(U^{-})$, which splits as a direct sum of invertible sheaves on $C$ (because every representation of $T$ is a direct sum of characters). Thus $E^T\times^T \text{Lie}(U^{-})$ has vanishing $h^1$ if and only if each of these summands has vanishing $h^1$. That is precisely the condition imposed by Gaitsgory and Nadler to define the open subset $\text{Bun}_{B,r}$ inside $\text{Bun}_B$.

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very enlightening thank you! –  prochet Jun 19 '13 at 17:22

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