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I have seen and heard the statement that the $p$-adic and $l$-adic topologies are incompatible. I would appreciate a proof or references supporting this statement. More precisely, I am interested in the proof of the following statement:

Let $G$ and $H$ be $p$-adic and $l$-adic Lie groups respectively, $l \neq p$. Then any homomorphism between them is locally constant.

I am quite certain that this can be found in text book, if so references are welcome.

Thanks!

Watson
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random123
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  • The answer here is useful: https://math.stackexchange.com/questions/182078/ – Watson Dec 05 '18 at 17:30
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    If $f: G \to H$ is a continuous morphism, then $f(G)$ is compact in the Hausdorff space $H$, hence closed. But a closed subgroup of a pro-$\ell$ is pro-$\ell$. Also, $f(G)$ is pro-p. Finally, notice that if $f(G)$ is both pro-$p$ and pro-$\ell$, then every open normal subgroup has quotient being a finite p-group and an $\ell$-group... it should be helpful. – Watson Dec 05 '18 at 17:37
  • @Watson Thanks for wonderful answer. Since I am new to this, let me just ask, we can assume $G$ and $H$ are pro $p$ and $l$ groups since the question is local in nature and I can always go to open subgroups? Thanks again. Also would you consider making this into an answer or perhaps post the comment itself as an answer? – random123 Dec 06 '18 at 03:33

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Let $p \neq \ell$ be distinct prime numbers, $G$ be a $p$-adic Lie group and $H$ be an $\ell$-adic Lie group. Let $f: G \to H$ is a continuous group morphism. We show that $f$ is locally constant. Since $f$ is a group morphism, it is sufficient to show that there is an open neighbourhood $U$ of $1_G$ such that $f$ is constant on $U$.

In general, a $p$-adic Lie group is Hausdorff (by definition) and locally compact (since $\Bbb Q_p$ is), and has an open neighbourhood of the identity which is a pro-$p$-group (see corollary 8.33 in Analytic Pro-P Groups $^{[1]}$), i.e. for every open normal subgroup $N\triangleleft G$, the quotient $G/N$ is a (finite) $p$-group.

Let $W \subset H$ be an open neighbourhood of $1_H$ which is a pro-$\ell$-group, and $V \subset G$ be an open neighbourhood of $1_G$ which is a pro-$p$-group Let $U = f^{-1}(W) \cap V$, which is an open (hence closed) subgroup of $V$. A closed subgroup of a pro-$p$-group is a pro-$p$-group (prop. 2.2.1 a) in Ribes, Zalesskii, Profinite groups). Hence $U$ is a pro-$p$-group, which maps via $f$ to the pro-$\ell$-group $W$, that is $f(U) \subset W$.


Thereby, we may assume that $f: G \to H$ is a continuous morphism from a pro-$p$-group to a pro-$\ell$-group. We wish to show that $f(G) = \{1_H\}$, i.e. $f$ is constant.

In that setting, $f(G)$ is a compact subspace of the Hausdorff space $H$, hence closed. But a closed subgroup of a pro-$\ell$-group is a pro-$\ell$-group. Also, $f(G)$ is a pro-$p$-group. Finally, notice that if $f(G)$ is both pro-$p$ and pro-$\ell$, then every open normal subgroup $N \leq f(G)$ has quotient being both a finite $p$-group and an $\ell$-group, which implies $f(G) / N = \{1\}$.

But the identity element of $f(G)$ admits a fundamental system of open normal subgroups, which are all equal to $f(G)$ as we just saw above. Then the topology on $f(G)$ is trivial, and also Hausdorff, hence $f(G)$ is a singleton. This finishes the proof.


$^{[1]}$ I'm not sure that this is an easy result. The main steps seems to be theorem 8.29 on the one hand: every $p$-adic analytic group has an open subgroup which is a "standard group", which is locally homeomorphic to $p\Bbb Z_p^r$ for some $r \geq 0$ (see definition 8.22), and theorem 8.31 on the other hand.

Watson
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  • Unfortunately, I don't know any reference which states explicitely that every continuous morphism from a $p$-adic Lie group to an $\ell$-adic Lie group is locally constant when $p \neq \ell$. – Watson Dec 08 '18 at 12:45
  • Thanks a lot for posting an answer. It is of great help especially the details and the precise references provided in the answer. – random123 Dec 08 '18 at 12:56
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    This is a wonderful proof. –  Apr 26 '20 at 07:02