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Let $G$ be a split reductive algebraic group (over a local field if you like), $B$ be a fixed Borel subgroup, and $P$ be a fixed standard parabolic subgroup. Let $W$ be the Weyl group of $G$. For $w\in W$, denote $C(w)=BwB$.

Given Weyl elements $w>w'>w_1$ (Bruhat order), if $C(w)\subset Pw_1P$, do we know that $C(w')\subset Pw_1P$? It's like to ask whether there are holes in $Pw_1P$. On the other hand, this amounts to asking if $Pw_1P\cap \Omega_{w_1}$ is closed in $\Omega_{w_1}$, where $$\Omega_{w_1}:=\coprod_{w\ge w_1}C(w).$$

Any comments and references are welcome. Thanks in advance.

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    $\begingroup$ I think the condition is that $BwB \subset P w_0 P$ if and only if $w = w_1 w_0 w_2$ for some $w_1, w_2 \in W_I$. Here $W_I$ is the Weyl group of a standard Levi of $P$. See mathoverflow.net/questions/118974/… for a related question. This should give the answer to your explicit question, with a bit of work with minimal-length representatives, I think. $\endgroup$
    – Marty
    Apr 11, 2017 at 0:18
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    $\begingroup$ @Marty's thought is correct; this is Corollaire 5.20 of Borel and Tits (MR). $\endgroup$
    – LSpice
    Apr 11, 2017 at 1:13
  • $\begingroup$ @LSpice thanks for the comments and the reference. I am more curious if that explicit question has already an answer in the literature. $\endgroup$
    – Q. Zhang
    Apr 11, 2017 at 1:45
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    $\begingroup$ The formulation of the question is too loose, since you apparently want the group to be split (over some field which isn't important here). Also, it's helpful to give the Weyl group a label such as $W$; then "a Weyl element $w$" just means $w \in W$. The standard use of the symbol $w_0$ is for the "longest element" of $W$ (relative to some fixed set of simple reflections). Finally, it's enough to specify that $P$ is a standard parabolic $P_I$ for a subset $I$ of the simple reflections. $\endgroup$
    – Jim Humphreys
    Apr 11, 2017 at 16:08
  • $\begingroup$ @JimHumphreys Professor Humphreys, thanks for your comment. I changed the formulation a little bit accordingly. $\endgroup$
    – Q. Zhang
    Apr 12, 2017 at 15:35

1 Answer 1

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Your condition $C(w)\subseteq Pw_1P$ implies $PwP=Pw_1P\Rightarrow\overline{PwP}=\overline{Pw_1P}$.

Moreover $w\ge w'\Rightarrow\overline{BwB}\supseteq\overline{Bw'B}\Rightarrow\overline{PwP}\supseteq \overline{Pw'P}$.

Similarly $w'\ge w_1\Rightarrow\overline{Bw'B}\supseteq\overline{Bw_1B}\Rightarrow\overline{Pw'P}\supseteq \overline{Pw_1P}$.

Together, one gets $\overline{Pw'P}=\overline{Pw_1p}$.

But then $Pw'P=Pw_1P$ and in particular $C(w')\subseteq Pw_1P$.

Remark: One can show stronger that the set of $w$ with $C(w)\subseteq Pw_1P$ is the double coset $W_Iw_1W_I$ and that this double coset forms a Bruhat interval, i.e., there are $w_{min},w_{max}\in D$ such that $w\in D\Leftrightarrow w_{min}\le w\le w_{max}$. After a very cursory search I found it without proof in the paper "Parabolic double cosets in Coxeter groups" by Billey et al. Proposition 2.

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  • $\begingroup$ Thanks for your answer. Could your kindly explain further how $\overline{BwB}\supset \overline{Bw'B}$ implies $\overline{PwP}\supset \overline{Pw'P}$ in the second line? Thanks. $\endgroup$
    – Q. Zhang
    Apr 14, 2017 at 17:06
  • $\begingroup$ $PwP=P(BwB)P$ is dense in $P\overline{BwB}P$. Hence $\overline{PwP}=\overline{P\overline{BwB}P}\supseteq\overline{P\overline{Bw'B}P}=\overline{Pw'P}$. $\endgroup$
    – Friedrich Knop
    Apr 14, 2017 at 17:31

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