Question:Question: From following some example computations online (e.g. Romanov - Four examples of Beilinson-Bernstein localization), it seems like this same sheaf, $(\iota_w)_\star\mathcal{O}_{X_w}$, considered as a twisted D-module over the twisted sheaf $\mathcal{D}_X^{-\lambda}$, should correspond to the dual Verma $M_{(ww_0)\circ\lambda}$ (here $\lambda$ is shifted-dominant). In the case of $\mathfrak{sl}_2$ it seems you can sometimes just sort of do the twisted D-action directly on this sheaf, but in general it is completely mysterious to me where the twisted D-action on $(\iota_w)_\star\mathcal{O}_{X_w}$ comes from, especially since the sheaf itself is not changing and there does not appear to be a twisted structure on $\mathcal{O}_{X_w}$ (indeed the symbol $\mathcal{D}_{X_w}^\lambda$ doesn't seem defined since there is no action of $G$ on $X_w$). So where does this twisted action come from? So where does this twisted action come from? (In the case of integral $\lambda$ one has the equivalence of modules over $\mathcal{D}_X$ and $\mathcal{D}_X^\lambda$, but I do not think this is true in general as nonintegral blocks of category $\cal O$ can be ill-behaved.) The possibilities to me are that either the structure sheaves $\mathcal{O}_{X_w}$ somehow have some hidden `twisted' structure with respect to $\lambda$, or that the pushforward $f_\star$ has some twisted variant (in which case I would have to redefine the side-switching modules, the transfer modules, etc. etc., all over again for the twisted case), or that somehow any $\mathcal{D}_X$-module can be granted a $\mathcal{D}_X^\lambda$-module for general $\lambda$. Or maybe some combination of the above.
Unnecessary example (just for added detail):
For $\mathfrak{sl}_2$, $X=\mathbb{P}^1$. My choice of coordinates is such that $D(x)$ contains the point $z=y/x=0$ corresponding to the standard Borel. Let $\iota$ denote the closed embedding of the point $z=0$ (the Bruhat cell for $w=1$) and $\jmath$ denote the open embedding of $D(y)$ (the Bruhat cell for $w=s$). Then $\iota_\star\mathcal{O}_{z=0}$ should correspond to the antidominant Verma, and indeed its global sections is $\mathbb{C}[\partial_z]\delta_0$. The trivial central character case is clear to me, so let's take the twist. By checking locally the action of $h=2z\partial_z+n$ (here I am taking differential operators on $\mathcal{L}(n)=\mathcal{O}_{\mathbb{P}^1}(-n)$, where $-n$ being shifted-dominant and shifted-regular implies $n\le 0$), we see $h\cdot\partial_z^k\delta_0=(2z\partial_z+n)\partial_z^k\delta_0=2(\partial_z^{k+1}z-(k+1)\partial_z^k)\delta_0+n\partial_z^k\delta_0=(n-2-2k)\partial_z^k\delta_0$, so that e.g. the `highest-weight vector' $\partial_z^0$ has weight $n-2$. As the nonpositive integer $n$ varies this gives the integral antidominant Vermas (the $e,f$ actions can be checked similarly). So in this case the twisted action still works on the nose, no modifications necessary.
However, things no longer work so nicely for the dominant Verma module, corresponding to $\mathbb{D}_{\mathbb P^1}\jmath_\star \mathcal{O}_{D(y)}$. This sheaf (I believe) has global sections $\mathbb{C}[\partial_z]$, considered first as a (right-)submodule of $\mathbb{C}[z,\partial_z]/\partial_z z\mathbb C[z,\partial_z]$ and then side-switched to a left module (note well that the thing we quotient out by is a right subobject). Let again $\lambda=n$ label the differential operator side, so that $-n$ is shifted-dominant and shifted-regular, so that $n\le 0$. Checking again the action of $h=2z\partial_z+n$, we see $h\cdot \partial_z^k=\partial_z^k\cdot(-\partial_z\cdot 2z+n)=-2\partial_z(z\partial_z^k+k\partial_z^{k-1})+n\partial_z^k=(n-2k)\partial_z^k$. By looking at $k=0$ we see this is not the dominant Verma, as $n$ is nonpositive. So whatever the action is, it is no longer the naive one. One runs into other problems when one tries to compute $e\cdot \partial_z^0$ also, since $e=z^2\partial_z+nz$ and $z$ doesn't make sense in $\mathbb{C}[\partial_z]$ (this is not a problem for $n=0$). So what is going on here?
