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Apologies for the basic question, but my experience on math.stackexchange tells me that this will go unanswered there.

Background: In the principal block, the dual Verma modules (with highest weight $w\circ(-2\varrho)$, i.e. the shifted Weyl action of $w$ applied to the (shifted-)antidominant weight) under Beilinson–Bernstein localization correspond to $(\iota_w)_\star\mathcal{O}_{X_w}$, where $\iota_w\colon X_w\to X$ is the locally closed embedding of the Bruhat cell into the flag variety and $f_\star$ denotes the D-module pushforward (given by the regular pushforward of $\mathcal{O}$-modules composed with tensoring by a transfer module). Evidently this should work for arbitrary (shifted-dominant and shifted-regular on the $\mathfrak{g}$-side of the story) weights $\lambda$ in general. I am defining the twisted D-sheaf (for arbitrary weights) $\mathcal{D}_X^\lambda$ as the universal enveloping algebroid modded out by the ideal generated by $\widetilde{\xi}-\widetilde{\lambda}(\widetilde{\xi})$ for $\widetilde\xi\in\widetilde{\mathfrak{b}}$, which for integral $\lambda$ corresponds to differential operators on the line bundle $\mathcal{L}(\lambda)$, which for $\mathfrak{sl}_2$ is $\mathcal{L}(n)=\mathcal{O}(-n)$; the exact signs/shifts that should appear here vary depending on your conventions.

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? (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?

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  • $\begingroup$ One can think of twisted D-modules as D-modules on the $T$-bundle $G/N \to G/B$ with fixed monodromy on each fiber. Since the Bruhat cells are affine, every $T$-bundle over them trivializes, so one can think of pushforward as going from ordinary D-modules to twisted ones. Obviously, this depends on a trivialization for each cell, so I'm not sure that's the right way to think about it $\endgroup$
    – Exit path
    Commented Mar 13, 2023 at 3:13
  • $\begingroup$ Sorry, I’m not sure I understand, could you please elaborate? I know that $\lambda$-twisted D-modules on $G/B$ are the same as weakly $B$-equivariant D-modules on $G$ with the difference between the equivariant and the D-module structures being $-\lambda$, but I'm not sure how to do this for the Bruhat cells (indeed, what are twisted D-modules over Bruhat cells?). Also, can one actually compute this action from your point of view, say for this $\mathfrak{sl}_2$ example? I feel like I need to actually be able to compute it here to understand it. $\endgroup$
    – Fan Zhou
    Commented Mar 17, 2023 at 21:58
  • $\begingroup$ Twisted D-modules make sense on any space equipped with a torus bundle (see the accepted answer from this question). To twist D-modules on Bruhat cells just pull $G/N \to G/B$ back along the inclusion. To compute for $SL_2$, in principle you just need a concrete model of the line bundle as well as trivializations over both cells $\endgroup$
    – Exit path
    Commented Mar 17, 2023 at 23:48
  • $\begingroup$ For every morphism of smooth varieties $f: Y \to X$ and every twisted sheaf of differential operator (tdo) $\mathcal D$ on $X$, one can define a tdo $\mathcal D^f$ on $Y$, the "pullback" of $\mathcal D$. Then one defines pushforward of $\mathcal D^f$-modules as $\mathcal D$-modules. In the current setting $Y = X_w$, $f = \iota_w$, $\mathcal D = \mathcal D_X^\lambda$, and you can check relatively easily that $\mathcal D_X^{\lambda, \iota_w}$ is naturally isomorphic to the ordinary differential operators $\mathcal D_{X_w}$. $\endgroup$ Commented Mar 21, 2023 at 18:17
  • $\begingroup$ When we are defining $\iota_{w\star} \mathcal O_{X_w}$ as a $\mathcal D_X^\lambda$-module, we are secretly viewing $\mathcal O_{X_w}$ as a $\mathcal D_X^{\lambda,\iota_w}$-module and use this structure to define push foward. The resulting direct image has the same underlying $\mathcal O_X$-module structure (at least locally). It would probably be helpful to look at the definition of direct image functor for twisted modules. $\endgroup$ Commented Mar 21, 2023 at 18:25

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