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.). $\endgroup$ – nfdc23 Mar 12 '18 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 Humphreys Mar 12 '18 at 20:25
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 $\endgroup$ – S. carmeli Mar 12 '18 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".) $\endgroup$ – Jim Humphreys Mar 12 '18 at 20:27

$\begingroup$ thanks, I edited my answer. For some reason I assumed that the torus in the problem is split. $\endgroup$ – S. carmeli Mar 12 '18 at 20:37