1
$\begingroup$

Given a non-archimedean field $\mathbb K$, there is a natural map $$ \mathrm{val}: (\mathbb K^*)^n\to\mathbb R^n$$ (See Section 4 of Gubler's paper). Gubler mentions there $\mathrm{val}$ is a continuous map. So, for a rational polytope $\Delta\subset\mathbb R^n$, the preimage $\mathrm{val}^{-1}(\Delta)$ should be closed since so is $\Delta$. But, it is known (like Proposition 4.1 there) that the preimage $\mathrm{val}^{-1}(\Delta)$ is actually a Weierstrass domain and seems to be open. Anyway, I get confused about the topology.

The very question I would like to ask is as follows: Gubler's Proposition 4.4 (e) states that if $\Delta'\subset \Delta$ is another rational polytope, then

$\mathrm{val}^{-1}(\Delta')\subset\mathrm{val}^{-1}(\Delta)$ is an open immersion if and only if $\Delta'$ is a closed face of $\Delta$.

The "if" part is ok for me, but I have trouble in understanding the "only if" part. Imagine $\Delta$ is very large so that $\mathrm{val}^{-1}(\Delta)$ is approximately the whole torus $(\mathbb K^*)^n$, and we already know $\mathrm{val}^{-1}(\Delta')$ is a Weiestrass domain in $(\mathbb K^*)^n$. So we should expect it to be also open in $\mathrm{val}^{-1}(\Delta)$. But this contradicts to the above statement. Where is my mistake?

Maybe this is a stupid question and I am not an expert, but I get really confused. Thank you for any help!

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ You got the statement wrong: the open immersion is not the map between the Weierstrass domains, but the induced map between their reductions. $\endgroup$
    – Jérôme Poineau
    Commented Aug 30, 2019 at 6:31
  • $\begingroup$ @JérômePoineau Thanks! So, let put it in a simpler way: if $\Delta'\subset \Delta$ is not a closed face, then is $\mathrm{val}^{-1}(\Delta')$ an open subspace of $\mathrm{val}^{-1}(\Delta)$? $\endgroup$
    – Hang
    Commented Aug 30, 2019 at 14:11
  • 1
    $\begingroup$ You should make clear which space and which topology you use: the Berkovich analytification of $\mathbb{G}_m^n$? Then, I do not see how $\textrm{val}^{-1}(\Delta')$ could be ever be open except when it is the whole thing. Do you have any example? $\endgroup$
    – Jérôme Poineau
    Commented Aug 30, 2019 at 15:41
  • $\begingroup$ @JérômePoineau Let's say we are in the category of the classical rigid analytic space and let's even consider the special case $\Delta'=\{0\}$. It seems strange to me why $\mathrm{val}^{-1}(\Delta')$ is not open in $\mathrm{val}^{-1}(\Delta)$, because both are open in the whole $\mathbb G_m^n$, if we consider subspace topology (or maybe not). $\endgroup$
    – Hang
    Commented Aug 30, 2019 at 16:21
  • 1
    $\begingroup$ OK. I had the Berkovich picture in mind where affinoid domains are closed but not open in general. Anyway, $\mathrm{val}^{-1}(\Delta') \subset \mathrm{val}^{-1}(\Delta')$ is an embedding of an affinoid domain, so it would indeed be an open immersion in rigid geometry. $\endgroup$
    – Jérôme Poineau
    Commented Aug 30, 2019 at 21:03

0

Reset to default

You must log in to answer this question.