Let $G$ be a semisimple linear algebraic group, $P$ be a parabolic subgroup and $w$ be an element of the Weyl group of $G$. I want to calculate the Picard group of the Schubert variety $X_P(w):=\overline{BwP/P} \subset G/P$. I'm particularly interested in the case of a maximal parabolic subgroup, but I suppose the general case is not much harder. In that case, I read without proof, that $\text{Pic}(X_P(w)) \cong \mathbb{Z}$ (of course if $w$ is non trivial). I would also hope, that there is a very ample generator in that case?

I have no problems to calculate the divisor class group, which is freely generated by the divisorial Schubert subvarieties. So the problem lies mainly in the non-smooth case.

A good reference on this topic would also be nice.

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    $\begingroup$ There is a simple description of the Picard group of a Schubert variety in the full flag variety in type A in Proposition 2.2.8 of Brion's lecture notes "Lectures on the geometry of flag varieties," which is available online. He refers to Mathieu's paper "Formules de caractères pour les algèbres de Kac-Moody générales" for the general case, so that paper has the answer you're looking for. Unfortunately that paper doesn't appear to be online, but I may have time later to look at it and post a full answer. $\endgroup$ Apr 14, 2011 at 16:33
  • $\begingroup$ Mathieu's "paper" is actually a monograph. $\endgroup$ Apr 14, 2011 at 16:50

1 Answer 1


Alex Yong and I work this out in our paper for the case of the Borel subgroup, but I'm pretty sure it's the same for every parabolic.

When is a Schubert variety Gorenstein?, Advances in Math. 207 (2006), 205-220.

Please note our conventions in that paper are backwards from yours in that our Schubert varieties are $\overline{B_-wB/B}$.

We don't say explicitly what happens for groups other than GL_n, but you can do the same thing using the appropriate Monk-Chevalley formula for the group in question.

EDIT: More details upon glancing at my own paper... Mathieu (reference in comments) shows that every line bundle on a Schubert variety is the restriction of a line bundle on the homogeneous space. (Actually, iirc the proof in the finite dimensional case predates Mathieu.) This means the Picard group of the Schubert variety will be the same as that of the homogeneous space unless some nontrivial line bundle restricts to a trivial one. For $G/P$ where $P$ is a maximal parabolic, this only happens if your Schubert variety is a point.

What we do is actually work out the Picard group as a subgroup of the class group.

  • $\begingroup$ The crucial point seems to be the fact, that all lines bundles come from retriction of the homogeneous space. I'm still trying to understand, in which cases the restriction of a line bundle becomes trivial. But now I have something to start from. The ampleness for $P$ maximal should come from the corresponding fact on the homogeneous space. $\endgroup$ Apr 14, 2011 at 17:23
  • $\begingroup$ IIRC the answer is as follows: the line bundles are indexed by simple negative roots not in the parabolic. A line bundle indexed by $\alpha$ restricts to the trivial one on $X_w$ if reduced word decompositions of $w$ do not include $s_\alpha$. (The set of $\alpha$ appearing in a reduced word decomposition for $w$ is the same for all reduced word decompositions.) $\endgroup$ Apr 14, 2011 at 20:55
  • $\begingroup$ Indeed; one way to think about it is that we know the map $c_1:Pic\to H^2$ is an isomorphism for $G/B$, and this restriction fact says that the same is true for $X_w$. Since $X_w$ is a union of $G/B$'s cells it's easy to compute the map $H^2(G/B) \to H^2(X_w)$ and it has the kernel you describe. $\endgroup$ Mar 29, 2015 at 12:35

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