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Let $G$ be a semisimple algebraic group over $\mathbb C$, $T$ be a maximal torus and $B$ be a Borel subgroup of $G$ containing $T$. Let $R^+$ be the set of positive roots with respect to $B$. Let $Q$ be a parabolic subgroup containing $B$ corresponding to a subset $\{\alpha_1, \alpha_2, \cdots ,\alpha_k \}$ of the set of simple roots $\{\alpha_1, \alpha_2, \cdots ,\alpha_n \}$.

Then Bruhat decomposition of $G/Q$ is given by $G/Q=\cup_{w \in W^Q}BwQ/Q$, where $W^Q$ is the Weyl group of $Q$.

$C_Q(w):=BwQ/Q$ is called a Schubert cell and its closure $X_w$ in $G/Q$ is called the Schubert variety associated to $w$. Let $B^-$ be the Borel subgroup of $G$ opposite to $B$. Then $B^-vQ/Q$ is called the opposite cell and its closure $X^v$ in $G/Q$ is called the opposite Schubert variety associated to $v$.

When $Q=B$ then $C_B(w)= \prod_{\{\alpha \in R^+: w^{-1}(\alpha) <0 \}}U_{\alpha}$, where $U_{\alpha}$ is the root subgroup corresponding to $\alpha$.

My questions are the following:

  1. What is the expression for $C_Q(w)$ in terms of root subgroups for $Q \neq B$.

  2. Let $v < w$ in Bruhat order. Then how does an element in $BwQ/Q \cap B^-vQ/Q$ look like in the form of a matrix.

  3. Lets take $G=SL_6$, $B=$ the subgroup of upper triangular matrices and $Q$ be the maximal parabolic corresponding to the simple root $\alpha_2$, $w=s_2s_1s_5s_4s_3s_2$ and $v=s_3s_2$. If $x \in BwQ/Q \cap B^-vQ/Q$, then what is the matrix form of $x$ ?

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  • $\begingroup$ Your parabolic seems to be denoted by both Q and P. $\endgroup$
    – Tobias Kildetoft
    Commented Feb 2, 2015 at 7:56
  • $\begingroup$ You have denoted both $B$-orbit and $B_-$-orbit closures by $X_u$. I prefer to denote the $B$-orbit closures by $X^u$, so when you want $X^a_b = X^a \cap X_b$, one has $a\geq b$. $\endgroup$
    – Allen Knutson
    Commented Feb 2, 2015 at 12:17
  • $\begingroup$ I did that as you suggested. I already assumed that $v < w$. $\endgroup$
    – Ram
    Commented Feb 2, 2015 at 12:59
  • $\begingroup$ @Ram: Probably too many questions here to get a single correct answer. Anyway, in #2, what is meant by "in the form of a matrix"? And why not use $P$ rather than $Q$? (And does the characteristic of the field matter?) $\endgroup$
    – Jim Humphreys
    Commented Feb 2, 2015 at 14:02
  • $\begingroup$ I mean as an element of $SL_{n+1}$, which co-ordinates are zero and which are non-zero. I am assuming that the ground field is $\mathbb C$. $\endgroup$
    – Ram
    Commented Feb 3, 2015 at 7:43

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