Let $K$ be an algebraically closed complete non-archimedean field and consider the unit ball $\mathbb{B}^{1}_{K}=Sp(K\langle t\rangle)$. We have an action of $\mathbb{Z}/2\mathbb{Z}$ on $\mathbb{B}^{1}_{K}$ given by sending the non-trivial element in $\mathbb{Z}/2\mathbb{Z}$ to the unique automorphism of $K\langle t\rangle$ which sends $t$ to $-t$. Denote this automorphism by $\xi:\mathbb{B}^{1}_{K}\rightarrow \mathbb{B}^{1}_{K}$. We know that the points in $\mathbb{B}^{1}_{K}$ are given by elements in $K^{\circ}$, and the action of $\xi$ sends a point $\lambda\in\mathbb{B}^{1}_{K}$ to $-\lambda\in\mathbb{B}^{1}_{K}$. Thus, the only fixed point of $\xi$ is the origin. In other words, consider the cartesian diagram: enter image description here it follows that $Y$ is just a point, which corresponds to the origin. Now let $r$ be the functor which sends rigid analytic $K$-varieties to tfp adic spaces over $Spa(K,K^{\circ})$. Namely, $r(Sp(A))=Spa(A,A^{\circ})$ for any affinoid algebra $A$. According to 1.1.13 page 43 in "Étale cohomology of rigid analytic varieties and adic spaces" the functor $r$ commutes with fiber products. Hence, the fixed locus of the automorphism $\xi$ when regarded as an automorphism of $r(\mathbb{B}^{1}_{K})$ should just be $r(Y)=Spa(K,K^{\circ})$ which is just a classical point in $r(\mathbb{B}^{1}_{K})$. However, in example 2.20 in "Perfectoid spaces", all adic points in $r(\mathbb{B}^{1}_{K})$ are classified. It is clear that the action of $\xi$ fixes all points of type $(2)$ and $(3)$ centered arround the origin. For example, the Gauss point sends a power series $f=\sum a_{n}t^{n}$ to $max\{\vert a_{n}\vert \}$. As $\xi(f)=\sum (-1)^{n}a_{n}t^{n}$, it is clear that the Gausspoint is a fixed point of the morphism $\xi:r(\mathbb{B}^{1}_{K})\rightarrow r(\mathbb{B}^{1}_{K})$. I would like to understand what is wrong about my calculations.

Context: I would like to know if it is true that if an action of a finite group $G$ on an affinoid $K$-variety is free (in the sense that any $g\in G$ which fixes a point in $X$ must be the identity) then the induced action on $r(X)$ is also free (in the same sense as above). My idea for this is that the action is free if and only if for every $g\in G$ the space $X^{g}$ (constructed as in the diagram above ) is empty. As the functor $r$ respects fiber products then if $X^{g}$ is empty it should follow that $r(X^{g})$ is also empty, hence the action of $G$ on $r(X)$ should be free. I don't see any problems with this argument. However, this seems to contradict the example above. Namely, the action of $\mathbb{Z}/2\mathbb{Z}$ on $\mathbb{B}^{1}_{K}$ is not free, but it is when restricted to the boundary $Sp(K\langle t,t^{-1}\rangle)$. However, the induced action on $r(Sp(K\langle t,t^{-1}\rangle))$ fixes the Gauss-point as shown above.

Scholze, Peter, Perfectoid spaces, Bonn: Univ. Bonn, Mathematisch-Naturwissenschaftliche Fakultät (Diss.). 51 p. (2011). ZBL1296.14020.

Huber, Roland, Étale cohomology of rigid analytic varieties and adic spaces, Aspects of Mathematics. E30. Wiesbaden: Vieweg. x, 450 p. (1996). ZBL0868.14010.


1 Answer 1


The issue with your calculation is, to put it bluntly, that fiber products of adic spaces are weird, and more specifically, they cannot be computed naively at the level of points - categorically speaking, the forgetful functor from adic spaces to topological spaces doesn't preserve fiber products. This is something that is not unique to the setting of adic spaces, and occurs in the exact same way at the level of classical varieties and schemes - let me elaborate on this case as it is conceptually simpler.

You can consider a similar action of $\newcommand{\Spec}{\operatorname{Spec}}\mathbb Z/2$ on $\mathbb A^1_K=\Spec K[t]$, induced by mapping $t$ to $-t$ (where $K$ is some field of characteristic not $2$). You will notice that at the level of topological spaces, there are two fixed points: the classical point corresponding to $t=0$, as well as the generic point of the scheme. However, the pullback of the analogous diagram formed by $\mathbb A_1^K$ and $\mathbb A_1^K\times_K \mathbb A_1^K$ is going to consist of just the single classical point - so what goes wrong?

The rough idea is that while the generic point is topologically a single point, it can still admit nontrivial actions while thought of over $K$, as the field $K(t)$ admits nontrivial automorphisms over $K$. To make it all concrete, let us consider the embedding of the generic point $\Spec K(t)\hookrightarrow\Spec K[t]$ and how it composes with the two maps into $\Spec K[t]\times_K\Spec K[t]\cong\Spec K[t_1,t_2]$: the two maps are induced by homomorphisms $\Spec K[t_1,t_2]\to\Spec K[t]$ given by, respectively, mapping $t_1,t_2\mapsto t$ and mapping $t_1\mapsto t,t_2\mapsto -t$. Now we blatantly see that these maps do not agree after composing with the inclusion $K[t]\hookrightarrow K(t)$, and so the fiber product you consider will not include the point $\Spec K(t)$ (while, on the other hand, after composing with the map $K[t]\to K[t]/(t)\cong K$ they agree, so it will include the closed point $t=0$).

Going back to adic spaces - they usually do not have generic points in quite the same sense as schemes, but I still find it to be convenient intuition to think of points of type (2) and (3) as "generic points" of disks that define them. In particular, more formally, their residue fields will still be nontrivial (transcendental) extensions of $K$, and so can also be acted on nontrivially, and similarly those higher type points will be excluded from the fiber product you consider.

  • 2
    $\begingroup$ Maybe worth remarking that for this example of the $t \mapsto -t$ action on $\mathbb{A}^1$ there is no potential for confusion if we think in terms of functors of points: as long as the ground field $K$ has characteristic $\neq 2$ we see that over any $K$-algebra $R$ the $t \mapsto -t$ action on $\mathbb{A}^1(R) \cong R$ has $0$ as the only fixed point. The functor of points, unlike the Zariski spectrum, respects fiber products! $\endgroup$ Nov 4, 2022 at 23:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.