Closed subgroups of a $p$-adic algebraic group

Question

Let $F$ be a finite extension of $\mathbb{Q}_p$ and $G$ the group of rational points of an algebraic group defined over $F$, endowed with the natural topology. Any Zariski closed subgroup $H \subset G$ is $\text{exp } \mathfrak{h}$ for some subalgebra $\mathfrak{h}$ of the Lie algebra $\mathfrak{g}$ of $G$, but what about subgroups which are closed in the $p$-adic topology? For instance, the compact open subgroups of $G$ are very important, and I wonder if such a subgroup arises by exponentiating a compact open subgroup of $\mathfrak{g}$ (i.e. an $\mathcal{O}_F$-lattice) closed under the Lie bracket. Is the situation better when $G$ is unipotent?

Answer 1

Let $G$ be a $p$--adic Lie group. There is a 1:1 correspondence between $p$-adically closed subgroups up to finite index of $G$ and Lie subalgebras of its Lie algebra (which is a Lie algebra over $\mathbb Q_p$). The correspondence works just as in the classical setting. This is shown in a paper by Mattuck from the fifties, and in much greater generality in Lazard's thesis (beware, beware).

Attention: The image of a closed subgroup $H \subseteq G$ under the logarithm map is not always a $\mathbb Z_p$ submodule of the Lie algebra. Example: take $p=2$ and $H \subseteq \mathrm{GL}_3$ the unipotent radical of the Borel. Then $\mathrm{log}(H(\mathbb Z_p))$ is not stable under $+$.

You cannot hope for such a correspondence for strictly all closed subgroups. Although the exponential map is a homeomorphism locally around zero, it can in general not be extended to a surjective map. There is already a problem with $\mathbb Z_p^\ast$.

Also, this does not make much sense over extensions of $\mathbb Q_p$. If $F$ is some nontrivial finite extension of $\mathbb Q_p$, then $F$ viewed as a Lie group under addition has many closed subgroups (all the $\mathbb Q_p$-linear subspaces), but the Lie algebra (which is also $F$) has no proper $F$--linear subalgebra.