Questions on the proof of Beilinson-Bernstein localization theorem I am trying to understand the Beilinson-Bernstein localization theorem (following the book by Hotta, Takeuchi and Tanisaki).  I got stuck at the following two steps.  Any help will be greatly appreciated.
For an integral weight $\lambda$, I will write $\mathcal L(\lambda)$ for the sheaf associated to the line bundle $G \times_B \mathbb C_{\lambda}$ on the flag variety $X=G/B$, where $\mathbb C_{\lambda}$ means the one-dimensional $B$-representation on which the torus $H$ in $B$ acts by $\lambda$.  I will write $D_{\lambda}$ for the sheaf of differential operators acting on sections of the sheaf $\mathcal L(\lambda+\rho)$, where $\rho$ is the Weyl vector.


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*On page 282 (of the book by HTT), they asserted that if $\mathcal M$ is a $D_{\lambda}$-module, then $\mathcal M \otimes_{\mathcal O_X} \mathcal L(\mu)$ is a $D_{\lambda+\mu}$-module.  I am confused by how $D_{\lambda+\mu}$ acts on $\mathcal M \otimes_{\mathcal O_X} \mathcal L(\mu)$.

*On page 281 they introduced a filtration on the trivial bundle $X \times L^-(\nu)$, where $L^-(\nu)$ stands for the lowest weight $G$-representation with lowest weight $\nu$, as follows.  First filter $L^-(\nu)$ by $L^-(\nu)=L^1 \supset L^2 \supset \cdots \supset L^r =0$, where each quotient $L^i/L^{i+1}$ is a $B$-representation associated to some weight $\mu_i$.  Then filter $X\times L^-(\nu)$ by $X \times L^-(\nu) = U^1 \supset U^2 \supset \cdots \supset U^r$, where $U^i = \{(gB, l): l \in g.L^i\}$.  This induces a filtration on $\mathcal O_X \otimes_{\mathbb C} L^-(\nu)$ and, then, on $\mathcal M \otimes_{\mathbb C} L^-(\nu)$.  Is this filtration on $\mathcal M \otimes_{\mathbb C} L^-(\nu)$ preserved by the action of the center of the universal enveloping algebra of the Lie algebra $\mathfrak g$ of $G$?


Thanks a lot!
 A: For (2), first note that we have an isomorphism $G/B \times L^-(\nu) \to G \times^B L^-(\nu)$ given by $(gB,v) \to (g, g^{-1}v)$. Under this isomorphism, we can identify $\mathcal{U}^i$ with $G \times^B L^i$ (following the notation in the proof). 
To show that each $\mathcal{M} \otimes_{\mathcal{O}_X}\mathcal{V}^i$ is closed under the action of $\mathfrak{g}$, consider the map of sheaves $\mathcal{M} \otimes_{\mathcal{O}_X} \mathcal{V}^i \to \mathcal{M} \otimes_{\mathcal{O}_X}\mathcal{L}(\mu_i)$ induced by the identity on $\mathcal{M}$ and the map $\mathcal{V}^i \to \mathcal{L}(\mu_i)$ of sections induced by maps on total spaces $G \times^B L^i \to G \times^B L^i/L^{i + 1}$. This map and the identity map are both maps of $\mathfrak{g}$-modules, the latter because the maps $G/B \times L^-(\nu) \to G \times^B L^-(\nu)$ and $G \times^B L^i \to G \times^B L^i/L^{i + 1}$ are $G$-equivariant. 
Furthermore, the kernel of the map $\mathcal{M} \otimes_{\mathcal{O}_X} \mathcal{V}^i \to \mathcal{M} \otimes_{\mathcal{O}_X}\mathcal{L}(\mu_i)$ is $\mathcal{M} \otimes_{\mathcal{O}_X} \mathcal{V}^{i+1}$. Thus $\mathcal{M} \otimes_{\mathcal{O}_X} \mathcal{V}^{i+1}$ is the kernel of a map of $\mathfrak{g}$-modules, and therefore is closed under the action of $\mathfrak{g}$ (meaning for every open $W \subset G/B, (\mathcal{M} \otimes_{\mathcal{O}_X} \mathcal{V}^{i+1})(W)$ is closed under the $\mathfrak{g}$ action.)
