In the paper: Pattern Avoidance and Rational Smoothness of
Schubert Varieties, Sara C. Billey, Advances in Mathematics 139, 141-156(1998), https://www.sciencedirect.com/science/article/pii/S0001870898917443/pdf?md5=43f7ddfffa6e4e285eec1f7183f736c8&pid=1-s2.0-S0001870898917443-main.pdf,

in Theorem 1.1 about rational-smoothness, the Author takes $W$ to be Weyl subgroup of any Semisimple Lie group, not necessarily simply-connected, and in fact in the next page, they do consider Schubert varieties of Type B as subvarieties of flag varieties $SO_{2n+1}(\mathbb C)/ B(\mathbb C)$, where $SO_{2n+1}$ are known to be **not simply connected**.

However, chasing the reference the author provides for Theorem 1.1, we get to the paper: The Bruhat graph of a Coxeter group, a conjecture of Deodhar, and rational smoothness of Schubert varieties, J.B. Carrell, Proc. Symposia Pure Math. 56 (Part 1) (1994), 53-61, https://books.google.com/books?id=YZ0ECAAAQBAJ&lpg=PA53&ots=m3XRziUL2r&lr&pg=PA58#v=onepage&q&f=false, the author of that article does start with a **simply connected, simple algebraic group** (see Section 3, page 57 and 58).

Can someone please explain to me then how does this apparent ambiguity be taken care of then ? How does the situation of the two papers reconcile ?

Further questions, as originated from the discussion in comments with Sam Hopkins below: If $W$ is a Weyl group of classical Type B or D, and we consider Schubert varieties corresponding to elements $w\in W$, does the question of rational-smoothness depend on whether our ambient group is $SO(n)$ or Spin group ? Does the question of smoothness depend on the ambient group ?

(I am not really bothered about Type A or C because they are pretty much always taken to be $Sl(n)$ and $Sp_{2n}$ respectively, both of which are simply connected)

Thanks

rationally smoothis first discussed. Of course, the later contributions of Billey, Deodhar and Carrell are essential, too. The answer by Spice seems to be useful. The emphasis on simply connected groups is mainly conventional, I think. $\endgroup$