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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
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1 Answer 1

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 http://arxiv.org/abs/1101.1253. The idea actually goes back to the paper of Beilinson and Ginzburg that you cite (look at Section 5).

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