TITLE: $G(k)/H(k)$ as a submanifold of $G/H(k)$
QUESTION [8 upvotes]: Let $k$ be a local field (if necessary, assume characteristic zero).  In general, if $X$ is a smooth variety of finite type over $k$ of dimension $n$, then the set of $k$-rational points $X(k)$ is an analytic manifold over $k$ of dimension $n$.  I was thinking about the passage $X \mapsto X(k)$ from smooth varieties to manifolds in the context of coset spaces, and had a couple of questions.
Let $H$ be a closed subgroup of a linear algebraic group $G$ over $k$.  Assume $H$ is defined over $k$.  Then the coset space $G/H$ has the structure of a smooth quasi-projective variety over $k$, which tells us that $G/H(k)$ is an analytic manifold over $k$ of dimension $n = \operatorname{Dim}G - \operatorname{Dim}H$.
On the other hand, $H(k)$ is a closed subgroup of the analytic Lie group $G(k)$, and the quotient $G(k)/H(k)$ has the structure of an $n$ dimensional analytic manifold (the quotient map $G(k) \rightarrow G(k)/H(k)$ is a principal fiber bundle with $H(k)$ as a fiber).  This just follows from the general theory of Lie groups.
Of course, $G(k)/H(k)$ injects into $G/H(k)$, but they don't have to be equal.  They are the same if  $H^1(\operatorname{Gal}(k_s/k),H) $ is trivial.
What can we say about the inclusion $G(k)/H(k) \rightarrow G/H(k)$ from the perspective of manifolds?  Is the inclusion an analytic map?  Does it make $G(k)/H(k)$ into an open submanifold?  How different can these two manifolds be?

REPLY [4 votes]: From the number of votes, it seems that the useful comment of Laurent Moret-Bailly has been under appreciated, so I thought it would be useful to explicitly record what the main theorem in the linked paper says (in community wiki mode, since this is really his answer).
Theorem 1.2 of GGMB14 (in the special case of $k$ a non-archimedean local field) The map $G(k)/H(k)\to (G/H)(k)$ 
$\bullet$ is always a homeomorphism onto its image,
$\bullet$ has closed image if $H$ satisfies the condition ($\ast$),
$\bullet$ has an open image if $H$ is smooth.
The condition ($\ast$) appearing in this theorem is technical (see Definition 2.4.3 of the paper), but $H$ satisfies the condition ($\ast$) if it has either one of the following properties: smooth, unipotent, commutative, or being a normal subgroup of a smooth group. Interestingly, there are examples of $H$ not satisfying $(\ast)$ for which the map $G(k)/H(k)\to (G/H)(k)$ has non-closed image (see example 7.1 of the paper, taking for example $k=\mathbf{F}_p(\!(T)\!)$).
Note that in characteristic $0$, (affine) group schemes (of finite type) are always smooth by a result of Cartier. Also, this puts in perspective the comment of YCor on the non-oppeness of $\text{SL}_p(K)\to \text{PGL}_p(K)$ for $K = \mathbf{F}_p(\!(T)\!)$. Finally, let me remark that the implicit function theorem should prove in all characteristic the openness of $G(k)/H(k)\to (G/H)(k)$ when $H$ is smooth (as suggested in Venkataramana's answer in characteristic $0$).