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.



Smooth map to the stack of G-bundles 
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 $\text{Bun}\_B$. 


