Consider $K/ \mathbb{Q}_p$ a finite extension of the field of $p$-adic numbers. If for every such an extension $ f_K: K \to K$ is continuous can we extend these functions to $\mathbb{C}_p$? My idea was that since $x \in \mathbb{C}_p$ then $x= lim_{n \to \infty}x_n$ where $x_n$ is an element of a finite extension on $\mathbb{Q_p}$ then $$\mathbb{C}_p \subset \prod_{K/Q_p finite}K$$ the map given by the sequence $F=(f_K)_K$ is continuous since all the components are continuous. And so $F_{|C_p}$ is also continuous.

This method could work?

  • $\begingroup$ Do you mean that for each finite $K\vert \Bbb Q_p$ contained in $\Bbb C_p$ you have a continuous $f_K$, such that if $K_1 \subseteq K_2$ then the restriction of $f_{K_2}$ to $K_1$ is $f_{K_1}$? Then I would think yes, there is a continuous $F$ on $\Bbb C_p$ whose restriction to $K$ is $f_K$, for all $K$. However, I do not think your "inclusion in product" formula is valid, and I don't understand what "map given by the sequence" should mean. Rather, you can indeed define $F$ by $F(x) = \lim F(x_n)$ for $x_n \to x$, just have to show this is independent from the choice of the sequence $x_n$. $\endgroup$ – Torsten Schoeneberg Feb 27 at 4:48
  • $\begingroup$ @TorstenSchoeneberg How do you extend $f_L(x) = f (\frac{1}{[\mathbb{Q}_p(x):\mathbb{Q}_p]} Tr_{\mathbb{Q}_p(x)/\mathbb{Q}_p}(x)), x \in L$ to $\overline{\mathbb{Q}_p}$ and $\mathbb{C}_p$ ? $\endgroup$ – reuns Feb 27 at 17:39
  • $\begingroup$ @reuns: Good point. I'll think about it and take back my "I would think yes" from the comment for the time being. The main point I wanted to make is that the question seems not well-posed, and that the attempt in there does not make sense to me. $\endgroup$ – Torsten Schoeneberg Feb 27 at 20:20
  • $\begingroup$ @reuns: Well if those maps are compatible in the sense of my comment, they already well-define a map on $\overline{\Bbb Q_p} = \bigcup_{\Bbb Q_p \subseteq K \subset \overline {\Bbb Q_p}, K\vert \Bbb Q_p \text{ finite}} K$. What I realise now as indeed unclear, in the general situation and your example, is whether that map is continuous. $\endgroup$ – Torsten Schoeneberg Feb 27 at 22:41

Your question seems not well-posed to me. User reuns seems to interpret it as:

Let $K$ be one fixed finite extension of $\Bbb Q_p$. If $f: K \rightarrow K$ is continuous, is there a continuous map $\hat{f}: \Bbb C_p \rightarrow K$ such that $\hat{f}_{\vert K} = f$?

and in his answer simultaneously restricts to the special case $\Bbb Q_p$ for the domain of $f$, but generalises to an arbitrary topological space $X$ for the codomain of $f$ and $\hat{f}$. Both the restriction and the generalisation are harmless though, and indeed this boils down to the question whether there is a continuous map $\phi: \Bbb C_p\rightarrow K$ (or in the special case $\rightarrow \Bbb Q_p$) with $\phi_{\vert K} = id_K$, which I think he indeed constructs. So the answer to that interpretation of the question is yes.

I however interpret the question differently, namely as:

For each finite extensions $K\vert \Bbb Q_p$ (contained in $\Bbb C_p$), let a continuous map $f_K : K \rightarrow K$ be given. Then is there a continuous map $F: \Bbb C_p \rightarrow \Bbb C_p$ such that $F_{\vert K} = f_K$ for all $K$ as above?

The answer to this is no, for two reasons.

  • Quite obviously the given maps $f_K$ need to be compatible in the sense that for every $K_1, K_2$, we need $f_{K_1 \vert (K_1 \cap K_2)} = f_{K_2 \vert (K_1 \cap K_2)}$ (or something similar).
  • Now if the condition in 1. is satisfied, then indeed the maps do define a map $f$ on $\overline{\Bbb Q_p} = \bigcup_{K\vert \Bbb Q_p \text{ finite}} K$, and maybe that is what you have in mind when you write that product in the OP. I am not sure if that map necessarily is continuous on $\overline{\Bbb Q_p}$; however, even if it is continuous, I think that such a map does not necessarily extend to $\Bbb C_p$. A counterexample is:
    • Let $c \in \Bbb C_p \setminus \overline{\Bbb Q_p}$ such that there exists a sequence $(x_n)_n$ with $x_n \to c$ and $v_p(x_n-c) \in \Bbb Z$ for all but finitely many $n$. Then for all $K$ as above, set $f_K(x) = \phi (\dfrac{1}{x-c})$, where $\phi: \Bbb C_p \rightarrow \Bbb Q_p$ is the map constructed in reuns' answer. While the $f_K$ give a map $f$ on $\overline{\Bbb Q_p}$ as above, we have $\lim_{n\to \infty} v_p(f(x_n)) = -\infty$, which means $f$ cannot be extended to $c$.
  • $\begingroup$ Thank you @Torsten Schoeneberg for the answer, now I understand the problem, for me it is enough to extend the map from $\mathbb{Q}_p \to \mathbb{Q}_p$ to $\mathbb{C}_p$ I'll try to prove it. $\endgroup$ – andres Mar 9 at 0:45

If $f$ is continuous $\mathbb{Q}_p \to X$ then $f \circ \phi$ is continuous $\mathbb{C}_p\to X$ and extends $f$.

Where $\phi$ sends $x \in \mathbb{C}_p$ to the closest point in $\mathbb{Q}_p$ :

let $u(x) = \sup_{t \in \mathbb{Q}_p} v_p(x-t)$ and $$\phi(x) = \cases{ x \text{ if } x\in \mathbb{Q}_p \\ 0 \text{ if } v_p(x)=u(x) \not \in \mathbb{Z} \\ p^{v_p(x)} \min \{n \in \mathbb{Z}_{\ge 0}, v_p(x-p^{v_p(x)} n) = u(x)\} \text{ otherwise}}$$

Since $v_p(x-\phi(x)) = u(x)$ then $v_p(\phi(x)-\phi(y)) \ge v_p(x-y)$

  • $\begingroup$ Why is the composition of those projections continuous? Also, to extend to $\mathbb C_p$ you need $\phi$ to be uniformly continuous, why does that hold? $\endgroup$ – Wojowu Feb 27 at 21:38
  • 1
    $\begingroup$ (ignoring just about everything you said, existence of such a projection follows from nonarchimedean Hahn-Banach theorem, since $\mathbb Q_p$ is spherically complete; I meant to post this as an answer to your question, but you've deleted it before I could post.) $\endgroup$ – Wojowu Feb 27 at 21:40
  • $\begingroup$ @Wojowu Tks, your idea of using Hahn-Banach is good. What about this one ? Is it possible to make $\phi$ a $\mathbb{Q}_p$-linear map ? $\endgroup$ – reuns Feb 28 at 0:11

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