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Consider a $p$-adic Lie group $G$. My question is if the pro-finite completion $\hat{G}$ is a $p$-adic Lie group. First we note that since $$\hat{G}=\text{lim}_{N\subset G} G/N$$ where the limit ranges over all normal subgroups of finite index, and each $G/N$ is $p$-adic Lie group. So therefore the question reduces to the question if such a limit is a $p$-adic Lie group.

curious math guy
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    That being said, a lot is known about $p$-adic Lie groups, and the general philosophy is that locally, they are "very" pro-$p$. The standard sources for this are Bourbaki, Serre, Lazard (http://www.numdam.org/item/?id=PMIHES_1965__26__5_0), Dixon-Mann-Sautoy-Segal's "Analytic Pro-p-Groups", and Schneider's "p-Adic Lie Groups". – Torsten Schoeneberg Jul 31 '20 at 05:49
  • @TorstenSchoeneberg "finite rank" has no meaning here, I ought to have written "finite index"! – curious math guy Jul 31 '20 at 12:10
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    @TorstenSchoeneberg The fact that not all $p$-adic Lie groups are compact doesn't seem to be an issue to me. I'm just asking if a specific compact group is a $p$-adic Lie group, not if all $p$-adic Lie groups are a profinite completion. Or did you have another thing in mind? – curious math guy Jul 31 '20 at 12:33
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    Right, I misread the question there, sorry. But then (sorry for being pedantic), the question in boldface is not the one you actually ask either, because you're not interested in general inverse limit of $p$-adic Lie groups, but only in the very specific ones which come from the finite quotients of a given $p$-adic Lie group. (Otherwise the inverse limit of $... \rightarrow \mathbb Q_p^3 \rightarrow \mathbb Q_p^2 \rightarrow \mathbb Q_p$, projecting to the first $n-1$ components, seems to be a counterexample.) – Torsten Schoeneberg Jul 31 '20 at 16:27
  • @TorstenSchoeneberg no problem. I was hoping the more general, bold question was true, but thanks for providing a counter-example. I'll change the question accordingly! – curious math guy Jul 31 '20 at 16:29

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If we take the multiplicative group $G= \mathbb Q_p^\times$, then $\hat G \simeq \widehat{\mathbb Z} \times \mathbb Z_p^\times$ where the first factor is the (additive group of the) profinite completion of $\mathbb Z$ (by local CFT, this group is actually isomorphic to the Galois group of the maximal abelian extension $\mathbb Q_p^{ab} \mid \mathbb Q_p$ and hence of central interest). That first factor is well known to be isomorphic to the direct product of all additive groups of the $\ell$-adic integers for all (!) primes $\ell$, i.e. we get

$$\hat G \simeq \mathbb Z_p^\times \times\prod_{\ell \text{ prime}} \mathbb Z_\ell $$

Now I cannot shake a rigorous proof out of my sleeve right now, but I would be very surprised if this thing (well, the part $\prod_{\ell \text{ prime} \neq p} \mathbb Z_\ell$) is a $p$-adic Lie group.

On the other hand, I have a strong feeling that for compact $G$, we might be more lucky via the sources given in the comments.

Torsten Schoeneberg
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