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What does translation through the wall correspond to under Beilinson Bernstein localization?

More precisely I am interested in the following:

There is a well known equivalence between the principal block of category $\mathcal O$ and perverse sheaves on the flag manifold, constructible along $B$ orbits:

$$\mathcal O_0 \cong \mathcal P_{(B)}(G/B)$$

Now for a singular integral weight $\lambda$ one can consider the translation through the wall functor $$ \theta_\lambda:\mathcal O_0 \rightarrow \mathcal O_\lambda \rightarrow \mathcal O_0$$

What does it correspond to under the above equivalence? My naive guess/hope would be, that it is given by convolution with the sheaf corresponding to $\theta_\lambda (L_e)$ where $L_e$ is the antidominant simple. Is this correct? If so, is there a geometric way to construct this sheaf?

PS: I am aware that there are descritions of the translation functors using slightly more elaborate version of localization, for example in this paper by Beilinson Ginzburg. However I would prefer to keep the above setup.

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I wonder whether such an interpretation can make more natural the fact that the wall-crossing functor is independent of the choice of $\lambda$ (thought of as lying in just the $s$-wall of the fundamental Weyl chamber). In Jantzen's original treatment of translation functors it was hard to understand this independence directly. I think the standard proof relies on the study of projective functors by S. Gelfand and J. Bernstein. – Jim Humphreys Dec 21 '12 at 1:17
Jim: Perhaps I should make this a separate question. But don't Soergel's arguments (which I am implicitly using to justify my answer below) show this? Under Soergel's functor to combinatorics $\mathbb{V}$, translation across the wall corresponds to (roughly) restriction/induction for the coinvariant algebra. The latter depending only on the stabilizer (namely $s$) of $\lambda$. Or am I confused? Of course, $\mathbb{V}$ is not an equivalence but it is full and faithful on maps between projectives/tiltings, but this should be enough to show the desired independence? No? – Reladenine Vakalwe Dec 21 '12 at 1:39
RV: The question asked concerns only integral weights, where it seems to be enough to use the classification of projective functors to see that $\lambda$ doesn't matter. (But the original setting for translation functors is very elementary, so going this far already increases the sophistication.) Soergel's deeper methods seem essential, however, when you also consider non-integral weights (and introduce an "integral" Weyl subgroup): for instance, it's nontrivial to show that everything just depends on that small Weyl group. – Jim Humphreys Dec 21 '12 at 14:25

I am guessing the following is well known to you/not what your are looking for, but nonetheless:

Let $s$ be a simple reflection, $P_s$ the corresponding minimal parabolic, $\pi_s\colon G/B\to G/P_s$ the projection. Translation across the $s$-wall `corresponds' to $\pi_s^*\pi_{s*}$. I use quotation marks because as stated this is clearly not true (translation across the wall is t-exact, $\pi_s^*\pi_{s*}$ is certainly not). However, $\pi_s^*\pi_{s*}$ does correspond to translation across the wall under Koszul duality. This is also the same as convolving with the $IC$-complex corresponding to $s$.

Morally (as you point out), reflection across the wall should correspond to convolving with the corresponding tilting. But there is an annoying issue here: tiltings are not $B$-equivariant. Similar problem occurs if instead of convolution using equivariant derived categories you try to use the standard Fourier-Mukai formalism and try to use an object on $G/B\times G/B$ as a kernel. However, there is a fix that comes at some technical expense. Namely, Bezrukavnikov and Yun's free monodromic sheaves The idea actually goes back to the paper of Beilinson and Ginzburg that you cite (look at Section 5).

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