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)


  • $\begingroup$ @SamHopkins: Thanks for the reference ... so, does it matter in what ambient groups do we consider our Weyl groups and Schubert varieties ? Do we get the same results for rational-smoothness and/or smoothness irrespective of whether we consider $SO_{2n+1}$ or Spin$_{2n+1}$ ? $\endgroup$
    – user102248
    May 8 '19 at 0:25
  • $\begingroup$ @SamHopkins: I see, well thank you very much for your help ... I will edit my question to ask the further point whether the ambient group makes any difference or not, in case someone comes up with a precise answer or point towards some precise reference ... $\endgroup$
    – user102248
    May 8 '19 at 0:42
  • 1
    $\begingroup$ (In light of the answer of LSpice I'm deleting my comments, but you still might be interested in the follow-up work of Billey-Postnikov arxiv.org/abs/math/0205179.) $\endgroup$ May 8 '19 at 1:00
  • 1
    $\begingroup$ @SamHopkins, unfortunately the Markdown parser isn't very smart about where URLs end, and it ate the following period. Billey and Postnikov - Smoothness of Schubert varieties via patterns in root systems. $\endgroup$
    – LSpice
    May 8 '19 at 19:09
  • 1
    $\begingroup$ @user102248: It's worth looking at the Appendix to the fundamental paper of Kazhdan and Lusztig (Invent. Math. 53 (1979), 165-184). where the notion of rationally smooth is 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. $\endgroup$ May 8 '19 at 22:06

Expanded with lots of references at @JimHumphreys's request. Borel is Borel - Linear algebraic groups (2nd edition). Springer is Springer - Linear algebraic groups (2nd edition).

Let $G$ be reductive. A maximal torus $T$ of $G$ is connected soluble, hence contained in a maximal such subgroup $B$, i.e., a Borel subgroup; but, since all Borel subgroups are conjugate (Springer Theorem 6.2.7(iii)), every Borel subgroup contains a maximal torus of $G$. Since every maximal torus of $G$ contains the centre $Z(G)$ of $G$ (Springer Proposition 8.1.8(i)), so too does every Borel subgroup of $G$. Thus the flag variety $G/B$ of $G$ equals $G_{\text{ad}}/B_{\text{ad}}$, where $G_{\text{ad}} = G/Z(G)$ is the adjoint quotient of $G$ and $B_{\text{ad}}$ is the image of $B$ in the adjoint quotient, which is also a Borel subgroup of $G_{\text{ad}}$ (Springer Corollary 6.2.8); that is, the flag variety of $G$ equals that of $G_{\text{ad}}$. Similarly, the Weyl group $\operatorname W(G, T) = \operatorname N_G(T)/T$ equals $\operatorname N_{G_{\text{ad}}}(T_{\text{ad}})/T_{\text{ad}}$, where $T_{\text{ad}}$ is the image of $T$ in $G_{\text{ad}}$, which is a maximal torus in $G_{\text{ad}}$ (Borel Proposition 11.14(1)); that is, the Weyl group of $G$ equals that of $G_{\text{ad}}$. The same argument shows that, for any $w \in \operatorname W(G, T) = \operatorname W(G_{\text{ad}}, T_{\text{ad}})$, the image of $B w B$ in $G/B = G_{\text{ad}}/B_{\text{ad}}$ (a Schubert cell) is the same as that of $B_{\text{ad}}w B_{\text{ad}}$, so that their closures (Schubert varieties) are also equal.

Since $G$ and its derived group have the same adjoint quotient (Borel Proposition 14.2), this shows that all of the above objects are the same for $G$ as for its derived group. Since the simply connected cover of the derived group of $G$ is a central extension of the derived group (Springer Exercise 10.1.4(1)), hence also has the same adjoint quotient, all of the above objects are the same for the simply connected cover as for $G$.

  • 1
    $\begingroup$ These are useful comments, but I wonder if you can document them a bit? $\endgroup$ May 8 '19 at 21:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.