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4 votes

Is the upper boundary of a Schubert variety Cartier?

The answer to the first question is No. Up to a twist by the restriction of $\mathcal{L}_{-\rho}$, this divisor is the canonical divisor, so the question is equivalent to whether the Schubert variety …
Alexander Woo's user avatar
8 votes
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Schubert varieties which admit small resolutions of singularities

For the Hermitian symmetric $G/P$, Nicolas Perrin explicitly classified all the Schubert varieties admitting a small resolution. (More explicitly, he classifies all the minimal models and quotes a th …
Alexander Woo's user avatar