Timeline for What is the difference between p-adic Lie groups and linear algebraic groups over p-adic fields?
Current License: CC BY-SA 3.0
14 events
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| Jan 30, 2015 at 18:17 | comment | added | YCor | (...) and the subgroup $\mathbf{Q}_p^*\times\{1\}$ is closed in the Zariski topology while its image is not. Thus for a $p$-adic Lie group that can be "made" algebraic (i.e. isomorphic to some $\mathbb{G}(\mathbf{Q}_p)$), the notion of algebraic subgroup is not intrinsic to ambient topological group. (For this reason Venkataramana's answer makes little sense to me, since $\mathbf{Z}_p$ is not algebraic in a canonical way.) | |
| Jan 30, 2015 at 18:13 | comment | added | YCor | On the other hand when we have a group $G$ as above, we have to be careful saying that $G$ is algebraic because there are several non-equivalent ways... for instance the Zariski topology is not uniquely determined by the structure of topological group. For instance if we consider $G=(\mathbf{Q}_p^*)^2$, it is isomorphic to $(\mathbf{Z}_p^*\times\mathbf{Z})^2$, and thus if we think of its "natural" Zariski topology, there are automorphisms of topological groups (fixing both $\mathbf{Z}_p^*$ and switching both $\mathbf{Z}$), that are not continuous for the Zariski topology (...) | |
| Jan 30, 2015 at 18:09 | comment | added | YCor | It's true that for every $p$-adic algebraic group $\mathbb{G}$, the group of points $G=\mathbb{G}(\mathbf{Q}_p)$ is a $p$-adic Lie group. See Bourbaki. But plenty of $p$-adic Lie groups don't arise this way, e.g. all discrete groups are $p$-adic Lie groups, and there are many other examples. | |
| Jan 30, 2015 at 15:37 | vote | accept | m07kl | ||
| Jan 30, 2015 at 15:26 | vote | accept | m07kl | ||
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| Jan 30, 2015 at 15:23 | vote | accept | m07kl | ||
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| Jan 30, 2015 at 15:06 | answer | added | Jim Humphreys | timeline score: 14 | |
| Jan 30, 2015 at 13:17 | answer | added | ACL | timeline score: 7 | |
| Jan 30, 2015 at 11:36 | comment | added | m07kl | @74230: Thanks for comments. I refer to T.A. Springer's book for linear algebraic groups over any fields and Peter Schneider's book for p-adic Lie groups. | |
| Jan 30, 2015 at 11:32 | history | edited | m07kl | CC BY-SA 3.0 |
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| Jan 30, 2015 at 7:13 | comment | added | user74230 | It would immensely clarify your question if you say what you think the definition of "linear algebraic group over a field $k$" means, and mention your background in algebraic geometry (over non-algebraically closed fields, and with schemes). Also, the notion of a $p$-adic analytic group manifold much predates Peter Schneider; what reference do you have in mind? | |
| Jan 30, 2015 at 3:34 | answer | added | Venkataramana | timeline score: 10 | |
| Jan 30, 2015 at 3:26 | comment | added | user74230 | Even over $\mathbf{R}$ this fails: the natural map ${\rm{SL}}_n \rightarrow {\rm{PGL}}_n$ of linear algebraic groups over $\mathbf{R}$ (or any field) is a degree-$n$ isogeny, so not an isomorphism, but on $\mathbf{R}$-points it is an isomorphism of Lie groups. The formation of Lie algebra naturally commutes with analytification of smooth group schemes over any field $F$ complete for a nontrivial absolute value (likewise for the tangent space at an $F$-point on a smooth $F$-scheme). Smooth $F$-groups and analytification in the sense of $F$-points are related but no in sense "the same". | |
| Jan 30, 2015 at 0:21 | history | asked | m07kl | CC BY-SA 3.0 |