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Let $G$ be a reductive group over an algebraically closed field. Fix a maximal torus $T$ and a Borel subgroup $B$ containing $T$. Let $(W,S)$ be the Coxeter system associated to $(B,T)$, where $S$ denotes the set of simple reflections. For $I\subset S$, let $W_I$ be the subgroup generated by $I$ and let $P_I$ denote the corresponding standard parabolic subgroup of $G$.

For $I,J\subset S$, the generalized Bruhat decomposition gives a natural bijection $$P_I \backslash G / P_J \simeq W_I \backslash W / W_J.$$ As a consequence, any pair $(P,Q)$ respectively of type $I$ and $J$ is conjugate to a pair $(P_I,{}^wP_J)$ for a unique $w \in {}^IW^J$ (the set ${}^IW^J$ corresponds to the elements $w\in W$ which are $I$-reduced-$J$).

For $w \in {}^IW^J$, let $\mathcal O_{I,J}(w)$ denote the $G$-orbit of $(P_I,{}^wP_J)$. It can alternatively be described using cosets as $$\mathcal O_{I,J}(w) = \{(gP_I,hP_J)\in G/P_I \times G/P_J \,|\, g^{-1}h \in P_IwP_J\}.$$ As a variety, it is irreducible and smooth.

What is $\dim(\mathcal O_{I,J}(w))$ ?

I believe that the answer should be $\ell(w) + \dim(G/P_{I\cap wJw^{-1}})$ because it would be coherent with the dimension of Deligne-Lusztig varieties (in case $k = \overline{\mathbb F_p}$). However, even though the literature on Bruhat decomposition is quite extensive, I have found myself unable to find a reference for this computation.

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  • $\begingroup$ Whatever the answer is, it had better be invariant under replacing $w$ by an element in its $(W_I, W_J)$-double coset, so we might as well assume that $w$ is the longest element in its double coset. Then I think that $I \cap w J w^{-1}$ is empty, so you are conjecturing that the dimension is $\ell(w) + \operatorname{dim}(G/B)$. That is, you're proposing that the dimension is the same as that of $\mathcal O_{\emptyset, \emptyset}(w)$. (Just some observations, not a proof.) $\endgroup$
    – LSpice
    Commented Jul 6, 2022 at 1:53
  • $\begingroup$ @LSpice Since I assume my element $w$ to be $I$-reduced-$J$, I think it is already uniquely determined inside its double coset $W_IwW_J$. In fact, it has minimal length. $\endgroup$
    – Suzet
    Commented Jul 6, 2022 at 2:14
  • $\begingroup$ Ah, I had missed that condition. But doesn't that mean that $I \cap w J w^{-1}$ is empty? $\endgroup$
    – LSpice
    Commented Jul 6, 2022 at 3:16
  • $\begingroup$ @LSpice (I deleted my previous comment because it was wrong). I don't think that $I\cap wJw^{-1}$ is necessarily empty. Consider the case $W = \mathfrak S_n$ with simple reflections $s_i = (i, i+1)$ for $1\leq i \leq n-1$. Let $I = J = \{s_2,\ldots ,s_{n-1}\}$ and let $w = s_1$. Then $s_1Is_1 = \{s_1s_2s_1, s_3, \ldots ,s_{n-1}\}$. In particular $I \cap s_1Is_1$ contains $s_3,\ldots ,s_{n-1}$. $\endgroup$
    – Suzet
    Commented Jul 6, 2022 at 4:20
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    $\begingroup$ Ah, so we are saying different things. I was saying that one should take $w$ to be maximal length, and I misread you to be saying the same. I agree that the minimal-length $w$ will usually give an intersection $I \cap w J w^{-1}$. In fact, I think (without having thought very hard) that $\ell(w) + \dim(G/P_{I \cap w J w^{-1}})$ might be independent of the choice of representative $w$, so that it's right for all choices or wrong for all choices. I am suggesting that thinking of the longest element might make it easiest to prove. $\endgroup$
    – LSpice
    Commented Jul 6, 2022 at 13:18

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