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