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It is known that in non-archimedean world there is also a GAGA-functor from the category of $K$-schemes of locally finite type to the category of rigid $K$-spaces. Here $K$ is a field with a non-trivial non-archimedean absolute value. Given a scheme $X$, we call the resulting rigid space $X^{an}$ a rigid analytification.

For example, the rigid analytification of the affine scheme $\mathbb A_K^n$ is given by the admissible covering $\mathbb A_K^{n,an}=\cup_{k\ge 0} \mathrm{Sp} T_n(r^k)$ for some fixed $r>1$. Here for any $s>1$ we denote $T_n(s)=K\langle s^{-1}x_1, \dots, s^{-1} x_k\rangle$.

On the other hand another example we often encounter is the $n$-torus $\mathbb G_m^{n,an}$ where $\mathbb G_m^n=\mathrm{Spec} K[x_1^\pm, \dots, x_n^\pm]$. My question is that:

Is there is a similar explicit description of the analytification $\mathbb G_m^{n,an}$?

Naively I guess the building blocks should look like $\mathrm{Sp} K \langle s^{-1}x_1, sx_1^{-1}, \dots, s^{-1}x_n, sx_n^{-1} \rangle$ or $\mathrm{Sp} K \langle s^{-1}x_1, s^{-1} x_1^{-1}, \dots, s^{-1}x_n, s^{-1} x_n^{-1} \rangle$. Which one is right? Or is there a better way to explain?

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    $\begingroup$ It should be easiest to see that the first is right by viewing $\mathbb G_m$ as the closed set in $\mathbb A^2$ defines by the equation $xy=1$. Then restricting your original open cover gives a cover of the torus. $\endgroup$
    – Will Sawin
    Jan 20, 2019 at 18:11
  • $\begingroup$ Thank you. If I identify $\mathbb G_m^{n,an}$ with $(K^\times)^n$ then what subset of $(K^\times)^n$ does $\mathrm{Sp} ( K \langle s^{-1} x_i, sx_i^{-1} \rangle )$ correspond? I guess it should be the subset with $|s^{-1} x_i| \le 1$ and $|sx_i^{-1}|\le 1$ i.e. $\{x_i=s \mid \forall i\}$, is this correct? $\endgroup$
    – Hang
    Jan 20, 2019 at 18:23
  • $\begingroup$ @WillSawin By restricting $K\langle s^{-1}x, s^{-1}y \rangle$ to $xy=1$ do you mean simply replace $y$ by $x^{-1}$? But this gives $K\langle s^{-1}x, s^{-1} x^{-1} \rangle$ which is the second one not the first one. Probably I mess up again. $\endgroup$
    – Hang
    Jan 20, 2019 at 20:43
  • $\begingroup$ @Hang You are right, I was getting confused by the notation. My argument certainly suggests that the second one is the right one. $\endgroup$
    – Will Sawin
    Jan 20, 2019 at 21:08
  • $\begingroup$ You can write $\mathbb{G}_m^{an}$ as an increasing union of annuli, i.e. $\mathbb{G}_m^{an} = \bigcup_{r>1} \text{Sp }(K<r^{-1}T,r^{-1}T^{-1}>)$. See e.g. Chapter V, Section 1 of W. Cherry's thesis [wcherry.math.unt.edu/pubs/dissertation.pdf]. This should readily generalize to $\mathbb{G}_m^{n,an}$. $\endgroup$
    – Jackson Morrow
    Aug 28, 2019 at 18:23

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