Schubert variety $V$ is a special type of (possibly singular) subvarieties of a Grassmannian. Since the Grassmannians are Kähler manifolds (in fact projective varieties) are we able to conclude that any smooth Schubert subvariety of a Grassmannian is in fact also a Kähler manifold?



Are smooth Schubert varieties Kähler?
Yes, smooth closed subvarieties of projective varieties are projective, and hence Kahler. Smooth Schubert varieties are very rare though, see the sources below for a description of them:


*Ryan, Kevin M.*, [**On Schubert varieties in the flag manifold of Sl(n,({\mathbb C}))**](http://dx.doi.org/10.1007/BF01450738), Math. Ann. 276, 205-224 (1987). [ZBL0579.14045](https://zbmath.org/?q=an:0579.14045).


*Wolper, James S.*, [**A combinatorial approach to the singularities of Schubert varieties**](http://dx.doi.org/10.1016/0001-8708(89)90048-0), Adv. Math. 76, No. 2, 184-193 (1989). [ZBL0705.14048](https://zbmath.org/?q=an:0705.14048).


*Gasharov, V.; Reiner, V.*, [**Cohomology of smooth Schubert varieties in partial flag manifolds**](http://dx.doi.org/10.1112/S0024610702003605), J. Lond. Math. Soc., II. Ser. 66, No. 3, 550-562 (2002). [ZBL1064.14056](https://zbmath.org/?q=an:1064.14056).


