I have a couple questions regarding the proof of Proposition $3$ (see page $10-11$ of arxiv.org/abs/math/0102039) in Bezrukavnikov's paper "Quasi-exceptional sets and equivariant coherent sheaves on the nilpotent cone" . For simplicity, assume $G$ is simply connected, simple algebraic group (so that $G$ is its own universal cover). Also, $\alpha$ is a simple root, let $P\_{\alpha}$ be the associated minimal parabolic and $u\_a$ its unipotent radical; $G/B, G/P\_{\alpha}$ are the flag variety and a partial flag variety with projection map $p\_{\alpha}: G/B \rightarrow G/P\_{\alpha}$; $\lambda, \lambda' \in \Lambda$ lie in the weight lattice and satisfy $s\_{\alpha}(\lambda) = \lambda - n \alpha, s\_{\alpha}(\lambda') = \lambda' - (n-1) \alpha$; and $O\_{G/B}(\lambda')$ for instance is the line bundle associated to $\lambda'$. Set $\tilde{\mathcal{N}} = G \times\_B \mathfrak{u}$ to be the Springer resolution ($u$ is unipotent radical, as usual). 


Question $1$: Set $V\_{\lambda'} = p\_{\alpha}^\* p\_{\alpha \*} O\_{G/B}(\lambda')$. 


On the first paragraph of pg $11$ it is claimed that there is a $G$-invariant filtration on $V\_{\lambda'}$ with the quotients being $\mathcal{O}\_{G/B}(\lambda'-i \alpha)$ for $0 \leq i \leq n-1$; why is this? 


Question $2$: Set $\tilde{\mathcal{N}\_{\alpha}} = G \times\_B u\_a$ (I am under the impression this is the same as what Roman defines on pg 11; if not, please correct me). How is the following exact sequence obtained (equation $23$ in the paper)? 
$0 \rightarrow O\_{N'}(\alpha) \rightarrow O\_{N'} \rightarrow O\_{N'\_{\alpha}} \rightarrow 0$ 


[In the above equation, I have used a ' to replace the tilde, since I cannot get Latex to work.]


Question $3$: Consider the natural projection $\pi': \tilde{\mathcal{N}\_{\alpha}} \rightarrow T^{\*}(G/P\_\alpha)$.


The fibers are clearly projective lines. If $p': \tilde{\mathcal{N}\_{\alpha}} \rightarrow G/B$ is the projection, in the last line of the proof (just before Prop $4$), it seems to be claimed that $p'^\*(V\_{\lambda'}(\lambda - \lambda'-\alpha))$ is isomorphic to a direct sum of copies of $\mathcal{O}\_{\mathbb{P}^1}(-1)$ when restricted to any fibre of $\pi'$. Why is this? 


EDIT: In all of the above, I had to simplify many of the Latex formulas to make them work (e.g. replace tilde by '). I think everything should be readable now, at the cost of bad notation. 



Pushforwards/pullbacks of some line bundles on (partial) flag varieties
For question 1, let $L\_\alpha$ denote the Levi of the parabolic $P\_\alpha$; then $L\_\alpha$ has derived group $SL\_2$. Let $B\_\alpha$ denote the Borel of $L\_\alpha$ such that $B \cap L\_\alpha = B\_\alpha$. Note that $V\_{\lambda'} = p\_\alpha^\* O\_{G/B}(V\_\alpha(\lambda'))$, where $V\_\alpha(\lambda')$ is the $L\_\alpha$ representation induced from $\lambda'$. That is, $V\_\alpha(\lambda') = Ind\_{B\_\alpha}^{L\_\alpha} (\chi\_{\lambda'}) = Ind\_B^{P\_\alpha} (\chi\_{\lambda'})$, where $\chi\_{\lambda'}$ is the 1-dimensional representation corresponding to the weight $\lambda'$. Since $L\_\alpha$ has derived group $SL\_2$, the weights of $V\_\alpha(\lambda')$ are precisely $\lambda'- i \alpha$, $0 \leq i \leq n-1$. Hence $V\_\alpha(\lambda')$ has (considered as a $B$-module now by extending trivially over the unipotent radical of $P\_\alpha$) a $B$-equivariant filtration with quotients $\chi\_{\lambda' - i \alpha}$, $0 \leq i \leq n-1$, so that $V\_{\lambda'}$ has the desired $G$-equivariant filtration.


For question 3, the fibers of $\pi'$ are all isomorphic to $L\_\alpha / B\_\alpha$, the flag variety of $L\_\alpha$; and this flag variety is isomorphic to $\mathbb P^1$. Since $(\lambda - \lambda' - \alpha)(\alpha^\vee) = -1$, under the identification of $L\_\alpha / B\_\alpha$ with $\mathbb P^1$, we see that the restriction of $p'^\* O\_{G/B}( \lambda - \lambda' - \alpha )$ to the fibers of $\pi'$ has degree -1, as desired. (I assume you want $p'^\* O\_{G/B}( \lambda - \lambda' - \alpha )$ here and not $p'^\* V\_{\lambda'}(\lambda - \lambda' - \alpha)$, since $V\_{\lambda'}(\lambda - \lambda' - \alpha)$ is not a bundle on $G/B$).


