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?


  • 1
    $\begingroup$ 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 '16 at 6:33

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.)

  • 1
    $\begingroup$ 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 '16 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 '16 at 14:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.