Is the upper boundary of a Schubert variety Cartier?

Question

On $G/B$, the divisor $\bigcup_\alpha X_{r_\alpha}$ is Cartier (where $X_w := \overline{B_- w B}/B$, and $\alpha$ varies over simple roots), not least because $G/B$ is smooth.

Is the same true for the divisor $\bigcup_{w' \gtrdot w} X_{w'} \subset X_w$, where $\gtrdot$ is the covering relation in strong Bruhat order?

Definitely some combination $\sum_{w'\gtrdot w} c_{w'} [X_{w'}]$, all $c_{w'} > 0$, is Cartier; restrict a $G$-equivariant ample line bundle $\mathcal L_\lambda$ from $G/B$ and consider the unique $T$-weight section $\sigma$ that doesn't vanish at $wB/B \in X_w$. This vanishes along the set $\bigcup_{w'\gtrdot w} X_{w'}$ and for $w' = w r_\beta$, its order of vanishing $c_{w'}$ will be $\langle \lambda, \check\beta \rangle$.

If the first question fails, how could one find the minimal Cartier combinations $\sum_{w'\gtrdot w} c_{w'} [X_{w'}]$?

Answer 1

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 is Gorenstein.

In my paper with Alex Yong on the Gorenstein Schubert varieties, there probably is enough combinatorics to tell you how to work out the answer to the second question in type A. (AFAIK, no explicit answer is known.) For other types, you need to use the appropriate Chevalley formula instead of the Monk formula to tell you what the Cartier divisors are, and the combinatorics are not as nice.