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.



Picard group of Schubert varieties
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.


