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I am looking for a version of Ehresmann's theorem for analytic manifolds over the $p$-adic numbers $\mathbb{Q}_p$ or, more generally, local fields. I follow the conventions from Serre's book "Lie algebras and Lie groups" concerning analytic manifolds over local fields.

Recall that Ehresmann's theorem states that a proper submersion between smooth manifolds is a locally trivial fibration.

Does a version of this hold for analytic manifolds over $\mathbb{Q}_p$? Namely, is a proper submersion between analytic manifolds over $\mathbb{Q}_p$ a locally trivial fibration?

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  • $\begingroup$ More generally, one can ask whether a submersion in the p-adic world is locally like a projection on charts. I do ask: what's the definition of a submersion in that setting? Is it still that the tangent mapping is surjective at each point, or that it has a continuous section? $\endgroup$
    – David Roberts
    Commented Sep 5, 2016 at 10:46
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    $\begingroup$ @Giulio: although the (usual) proof of Ehresmann's theorem doesn't work in the real-analytic case, and it is easy to make counterexamples to a complex-analytic analogue of the statement, what is a counterexample to the real-analytic analogue of the statement? (In various references it is asserted that the statement is false for the real-analytic setting, but counterexamples are not given.) Note that by work of Grauert and Morrey, real-analytic structures are more "floppy" than one might have expected from analogy with the complex-analytic case (due to a non-obvious link with Stein spaces). $\endgroup$
    – nfdc23
    Commented Sep 5, 2016 at 15:13
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    $\begingroup$ @Giulio: Also, since the question concerns the naive notion of $p$-adic analytic manifold that is totally disconnected (in contrast with the rigid-analytic setting that has a good notion of connectedness and behaves more like the complex-analytic case), it may be that for some silly reasons the answer is affirmative by an elementary bootstrap from the submersion theorem (combined with local compactness considerations) since any open cover a closed ball has a refinement consisting of finitely many pairwise disjoint closed balls... $\endgroup$
    – nfdc23
    Commented Sep 5, 2016 at 16:11
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    $\begingroup$ @user40276: Every compact (naive) $p$-adic manifold is just a finite disjoint union of copies of $\mathbf{Z}_p$. The methods over $\mathbf{R}$ don't provide good guidance for what to expect in the (totally disconnected) $p$-adic case (not using rigid-analytic spaces). $\endgroup$
    – nfdc23
    Commented Sep 7, 2016 at 6:08
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    $\begingroup$ @nfdc23: Since there is no difficulty with partitions of unity in the setting of "naive" $p$-adic manifolds, is there some other reason why the usual proof of Ehresmann's theorem doesn't work? $\endgroup$
    – naf
    Commented Sep 8, 2016 at 7:18

1 Answer 1

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It seems to me that the following lines essentially show that the answer is yes. Let $f\colon X\to Y$ be a proper submersion of $p$-adic manifolds, we want to show that $f$ is locally trivial on the target: every point $y$ has an open neighborhood $V$ such that $f_V\colon f^{-1}(V)\to V$ is isomorphic to the projection $p_2\colon U\times V\to V$ of a product.

By the inverse function theorem, $f$ is locally trivial on the source: for every point $x\in X$, there exists an open neighborhood $W$ of $x$ such that $f|_W\colon W\to f(W)$ is isomorphic to a product.

Let $y\in Y$ and let us cover the compact manifold $f^{-1}(y)$ by a finite family $(W_i)$ of open subsets on which $f$ is isomorphic to a product $U_i\times V_i \to V_i$. Let us refine it to a disjoint finite covering, still denoted by the same letter, of the same form. The union $W=\bigcup W_i$ is an open neighborhood of $f^{-1}(y)$. Since $f$ is proper, there exists an open neighborhood $V$ of $y$ such that $W$ contains $f^{-1}(V)$. Replacing $V_i$ by $V$, we are reduced to the case $V_i=V$ for all $i$. It is now visible that $f^{-1}(V)$ is isomorphic to the projection of the product $U\times V\to V$, where $U$ is defined as the disjoint union of the $U_i$.

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  • $\begingroup$ Merci Antoine . $\endgroup$ Commented Sep 11, 2016 at 19:35

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