Let $G$ be a quasisplit connected reductive group over a $p$adic field $k$. Identify $G$ with its rational points. Let $B$ be a Borel subgroup of $G$ containing a maximal torus $T$, both defined over $k$. For $w \in N_G(T)$, consider the double cosets $BwB$, which are locally closed submanifolds of $G$. Is $B$ the only closed double coset? Is there some general description of the closure of a Bruhat cell?

2$\begingroup$ For the final question over any field equipped with a nontrivial absolute value (or even a wider class of Hausdorff topological fields), see 21.28 in Borel's Linear Algebraic Groups (2nd ed.). $\endgroup$– nfdc23Mar 12, 2018 at 8:03

1$\begingroup$ @D_S: Your first line is confusing to me, since it's misleading to identify an algebraic group with its points over a field which isn't algebraically closed. The group scheme formulation is generally preferable in any case. $\endgroup$– Jim HumphreysMar 12, 2018 at 20:25
1 Answer
In general, and this holds for algebraically closed field as well as local field of charracteristic 0, at least in the split case, the description of orbit closures is as follows:
Let $\omega \in W_G$ be a weyl element, and write $\omega = \sigma_{\alpha_1} ... \sigma_{\alpha_k}$ as a reduced expression in simple reflections, i.e.
$\ell(\omega) = k$.
Let $P_{\alpha_i}$ denote the parabolic subgroup containing $T$ of the form
$SL_2^\alpha B$ where $SL_2^\alpha$ is the $SL_2$triple with roots
$\alpha, \alpha$. Then, the orbit closure is given by
$\overline{B \omega B} = P_{\alpha_1} ... P_{\alpha_k}B$. In particular,
$B$ is the unique closed orbit, and the orbit closure contains exactly those orbits $B \omega' B$ such that $\omega'$ is obtained from $\omega$ by excising several reflections from the reduced expression.

2$\begingroup$ As mentioned to me by Uri Bader, the order I described in my answer on the $\omega$s is called Bruhat order en.wikipedia.org/wiki/Bruhat_order $\endgroup$ Mar 12, 2018 at 7:55

$\begingroup$ Note that the question is formulated in the quasisplit (not the split) case. (Also, I think your "arising" should be something like "excising".) $\endgroup$ Mar 12, 2018 at 20:27

$\begingroup$ thanks, I edited my answer. For some reason I assumed that the torus in the problem is split. $\endgroup$ Mar 12, 2018 at 20:37