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Suppose that $\mathfrak{g}$ is a finite dimensional, complex, semisimple Lie algebra. Let $\mathcal{O}$ be the BGG category over $\mathfrak{g}$.

Tilting module theory play an important role in the study $\mathcal{O}$. They are sometimes projective covers and hence injective since the duality functor preserve tilting modules. Is it true that every tilting module has simple socle?

Thanks!

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    $\begingroup$ So clearly you need to restrict to indecomposable tilting modules. Also, I assume that by the duality functor preserving tilting modules you mean that tilting modules are self-dual (rather than just the class of tilting modules being preserved). $\endgroup$
    – Tobias Kildetoft
    May 6, 2016 at 6:33

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EDIT: Following a conversation with Ivan Losev, the situation is clearer now. Consider the principal block of $\mathcal{O}$. Recall two facts:

1) the socle of any Verma module is $L_{w_0}$,

2) taking the socle is a left exact functor.

Thus the socle of any object with Verma flag (in particular a tilting module) is isomorphic to a direct sum of copies of $L_{w_0}$.

Now if $T_x$ is an indecomposable tilting module we have

$\dim Hom(L_{w_0}, T_x) = P_{id, xw_0}(1)$

where $P_{y,z}$ is a Kazhdan-Lusztig polynomial. Thus the socle is simple if and only if $P_{id, xw_0} = 1$ which is the case if and only if the Schubert variety $xw_0$ is rationally smooth.

The formula is a consequence of the more general formula (which Peter reminded me of):

$dim Hom(\Delta_x, T_y) = P_{w_0x,w_0y}(1)$

See e.g. "Tilting exercises" or Soergel's papers on tilting modules.

(I deleted the longer version of the answer, because this seems much cleaner than earlier attempts.)

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    $\begingroup$ I support this answer. We know $T_{w_0}$ is tilting, simple, Verma and dual Verma. More generally, there is a formula for the space of homomorphisms between a Verma and a tilting in terms of the value of a Kazhdan-Lusztig polynomial at 1. See Soergel's arxiv.org/abs/math/0604590 (Theorem 4.4), although I guess a proof is sketched here. The statement of this theorem in Humphreys' BGG Category O book appears to be missing some $w_0$s. (Geordie I'm sure you know all this, but I think this might be useful for other readers) $\endgroup$
    – Peter McNamara
    May 12, 2016 at 0:16
  • $\begingroup$ @Peter: If you can specify where there is a problem with $w_\circ$ in my treatment, I can check it more carefully and add a correction to my online list of revisions. The notation here is definitely tricky, with different conventions used in the literature. $\endgroup$
    – Jim Humphreys
    May 15, 2016 at 14:04

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