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In flipping through the Springer lecture notes on Serre's 1964 'Lie Algebras and Lie Groups' lectures at Harvard, I found this pair of suprising results (page 157):

Let $G$ be a locally compact group. Then

  1. (Gleason-Montgomery-Zippin-Yamabe) G is a real Lie group iff it does not contain arbitrarily small subgroups (i.e., there exists a neighbourhood of the identity containing no nontrivial subgroup).
  2. (Lazard) G is a $p$-adic Lie group iff it contains an open subgroup $U$ such that $U$ is a finitely generated pro-$p$-group with $[U,U] \subset U^{p^2}$.

Are there further results that tell us when $G$ is a Lie group over $K$, $K = \mathbb{C}$ or $[K: \mathbb{Q}_p] < \infty$?

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  • $\begingroup$ Cross-listed on math.se: math.stackexchange.com/questions/442793/… $\endgroup$
    – Joshua Seaton
    Commented Jul 20, 2013 at 12:44
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    $\begingroup$ You also might wanna have a look at this amazing breakthrough arxiv.org/abs/1112.2324 $\endgroup$
    – Dan Sălăjan
    Commented Jul 20, 2013 at 13:08
  • $\begingroup$ For quadratic $K/\mathbf{Q}_p$ with nontrivial automorphism $z\mapsto z'$, $a\in K^{\times}$ not Galois-invariant mod squares, and odd $m\ge 3$, let $q=ax_1^2+x_2^2+\dots+x_{2m}^2$ and $q'=a'x_1^2+x_2^2+\dots+x_{2m}^2$. The ratio of discriminants is nontrivial mod squares, so $q$ and $q'$ are not homothetic, so SO$(q)$ and SO$(q')$ are not isomorphic as algebraic $K$-groups, so also for their adjoint quotients since $m$ is odd. So by Zariski closure with analytic and algebraic Ad$_G$, SO$(q)(K)$ and SO$(q')(K)$ are not isomorphic $K$-analytically, but they are isomorphic topologically! $\endgroup$
    – user36938
    Commented Jul 20, 2013 at 14:33
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    $\begingroup$ The point of the previous comment is that in general the $K$-analytic structure on a $K$-analytic Lie group can fail to be determined by the underlying $p$-adic Lie group (or equivalently the underlying topological group) when $K \ne \mathbf{Q}_p$. So asking for a characterization of the existence of a $K$-analytic structure in terms of the underlying topological group may be rather delicate matter. $\endgroup$
    – user36938
    Commented Jul 20, 2013 at 14:37

2 Answers 2

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Every locally compact and locally contractible topological group is a Lie group (Hofmann-Neeb arXiv:math/0609684).

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  • $\begingroup$ Thank you for the answer. However - although it is indeed cool and in the same vein as the result that initially excited me - this is clearly not an answer to the question I posed; I asked for similar conditions conditions to those above for Lie groups over $\mathbb{C}$ and extensions of $\mathbb{Q}_p$. $\endgroup$
    – Joshua Seaton
    Commented Jul 31, 2013 at 1:48
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    $\begingroup$ I thought the original post asked for "further results that tell us when G is a Lie group". $\endgroup$
    – Linus
    Commented Jul 31, 2013 at 6:21
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I like the following theorem of Gleason and Palais (http://vmm.math.uci.edu/PalaisPapers/OnAClassOfTransformGrps%28Gleason%29.pdf): Let $G$ be a locally arcwise connected topological group in which some neighborhood of the identity admits a continuous one-to-one map into a finite dimensional metric space. Then $G$ is a (real) Lie group.

This result follows from a criterion they prove in the same paper which says that a locally arcwise connected topological group is a Lie group provided that its compact metrizable subspaces are of bounded dimension. As they put it, "the criterion is remarkable in that local compactness is not one of the assumptions."

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  • $\begingroup$ Thank you for the answer. However - although it is indeed cool and in the same vein as the result that initially excited me - this is clearly not an answer to the question I posed; I asked for similar conditions conditions to those above for Lie groups over $\mathbb{C}$ and extensions of $\mathbb{Q}_p$. $\endgroup$
    – Joshua Seaton
    Commented Jul 31, 2013 at 1:48

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