p-adic Lie groups vs. algebraic groups over $\mathbb{Q}_p$ I am somewhat confused about the following two concepts and the relations between them-
One concept is a Lie group $G$ over the $p$-adic field. This is defined in a similar fashion to a (real) Lie group - using atlases of charts, while making sure to work in the analytic (=defined via a power series) category.
The other concept is the group $H=\mathbb{H}(k)$ of $k$ points for a $k$-defined algebraic group $\mathbb{H}$, for $k=\mathbb{Q}_p$. As any algebraic group, $\mathbb{H}$ is a matrix group, and hence so is $H$ as well.
Every group of the second type can be considered as a group of the first kind under an appropriate atlas (as is the case over any local field).
On the other hand, for $k=\mathbb{R}$ for example it is known (if im not mistaken?) that every semisimple Lie group with trivial center, and every compact group is in fact the connected component of the $\mathbb{R}$-points of some algebraic group defined over $\mathbb{R}$.
My question is - do similar results hold over $\mathbb{Q}_p$? Obviously we cannot depend on taking the connected component... In other words, when is a $p$-adic Lie group in fact a algebraic? What about a semi-simple $p$-adic Lie group (i.e. having a semi-simple Lie algebra)?
Thank you in advance
 A: I was asked by the person who asked the question to provide an answer. Here it goes.
You should keep in mind the example of $\text{PGL}_n(\mathbb{Z}_p)$ which is both compact and semisimple (in the sense that its Lie-algebra is semisimple) but it is not isomorphic (as a p-adic Lie group) to an algebraic group (hereafter, the $\mathbb{Q}_p$ points of a $\mathbb{Q}_p$-algebraic group).
However, $\text{PGL}_n(\mathbb{Z}_p)$ is isomorphic to an open subgroup of $\text{PGL}_n(\mathbb{Q}_p)$ which is an algebraic group, and there is a general phenomenon here.
Claim:
Let $G$ be a semisimple p-adic group with trivial center. Then there exists an algebraic group $H$ such that $G$ is isomorphic to an open subgroup of $H$. Moreover, this open subgroup is either compact or of a finite index in $H$.
Proof:
Define $H$ to be the group of automorphisms of the Lie-algebra of $G$.
Then $H$ is algebraic and the adjoint map $G\to H$ is injective by the triviality of the center. By semisimplicity, every derivation is inner, hence $G\to H$ induces an isomorphism on the level of Lie algebra. We conclude that the image of $G$ is a submanifold of the same dimension of $H$, hence it is open. An open subgroup is closed, of course, and it follows that $G$ is isomorphic to its image (this is a general fact about Polish groups). The fact that an open subgroup of $H$ must be either open or of finite index could be deduced from Howe-Moore Theorem (considering $\ell^2(H/G)$).
