# Is the upper boundary of a Schubert variety Cartier?

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 $\sum_{w'\gtrdot w} c_{w'} [X_{w'}]$?

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.