# Explicit description of rigid analytification of torus

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?

• 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. – Will Sawin Jan 20 at 18:11
• 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? – Hang Jan 20 at 18:23
• @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. – Hang Jan 20 at 20:43
• @Hang You are right, I was getting confused by the notation. My argument certainly suggests that the second one is the right one. – Will Sawin Jan 20 at 21:08
• 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 $\mathbb{G}_m^{n,an}$. – Jackson Morrow Aug 28 at 18:23