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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

the_lar
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2 Answers2

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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 $\ell^2(H/G)$).

Uri Bader
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Given the question was posed in 2013 and answered in 2017, perhaps it is okay to object to the answer in 2024.

The kernel of the adjoint map is not the center, but the quasi-center. Uri's argument should be correct if the center is replaced by the quasi-center.

Recently, Caprace-Minasyan-Osin constructed simple $p$-adic Lie groups for which the adjoint map is trivial, in particular they have abelian Lie algebra. Those are very far from algebraic.

Now there is quite some research about the original question, perhaps I should best refer to the article by Benoist-Quint titled "How far are p-adic Lie groups from algebraic groups?"

WaLe
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