## Superfectoid spaces!?

Note (6/21): There is an issue with the proof of Lemma 1.7 below, but I’m pretty confident Kiran and I will find a fix.  In any case, you should never take blog math too seriously!

In this post I want to announce some new foundational developments in p-adic geometry. More precisely, I want to introduce and motivate the category of sousperfectoid spaces defined below. This is a certain intrinsically defined full subcategory of analytic adic spaces over ${\mathrm{Spa}\,\mathbf{Z}_{p}}$ containing all perfectoid spaces and all smooth rigid analytic spaces. These spaces turn out to be extremely well behaved with respect to étale morphisms and pro-finite étale morphisms, and certain other favorable properties of sousperfectoid spaces allow us to non-trivially transfer consequences of the almost purity theorem from perfectoid spaces to smooth rigid spaces, with some striking consequences.

Some version of this blog post will appear in the next revision of my period maps paper; a more systematic development of sousperfectoid spaces and their applications (of which many are expected) will appear in a joint article of myself and Kedlaya. Since these spaces may be of general interest to the community, however, I wanted to describe them in public sooner rather than later; hence this post.

Nb. in what follows, KL1 and KL2 refers to Kedlaya-Liu’s Relative padic Hodge theory I & II papers, SW refers to Scholze-Weinstein’s paper on p-divisible groups, and HK refers to my forthcoming paper with Kedlaya.

OK, let’s do some math! In what follows, all Tate rings will be complete Tate rings over ${\mathbf{Z}_{p}}$ in which ${p}$ is topologically nilpotent. If ${A}$ is a uniform Tate ring, we always regard it as being normed by its spectral norm.

The key ring-theoretic definition is as follows.

Definition 1.1. A Tate ring ${A}$ is sousperfectoid if there exists a perfectoid Tate ring ${\tilde{A}}$ and a morphism ${f:A\rightarrow\tilde{A}}$ of Tate rings which admits a continuous (or equivalently, bounded) ${A}$-linear splitting ${\sigma:\tilde{A}\rightarrow A}$.

The motivation for the name should be clear (if you speak a little French, at least): ${A}$ lives under some perfectoid Tate ring in a meaningful way. Admittedly, I also like the auditory suggestion of SUPERfectoid. Note that any perfectoid Tate ring is sousperfectoid.

Proposition 1.2. Any sousperfectoid Tate ring ${A}$ is stably uniform. In particular, any Huber pair ${(A,A^{+})}$ with ${A}$ sousperfectoid is sheafy.

Proof: Let ${f:A\rightarrow\tilde{A}}$ be as in the definition, so ${\tilde{A}}$ is perfectoid, and in particular stably uniform. The result now follows from the following observation of Kedlaya-Liu (KL1, Remark 2.8.12): if ${B}$ is a stably uniform Tate ring and ${A\rightarrow B}$ is a morphism of Tate rings which admits a continuous ${A}$-linear splitting, then ${A}$ is stably uniform as well. $\Box$

Proposition 1.3. If ${A}$ is sousperfectoid, then each of the following rings is sousperfectoid as well:

i. The coordinate ring ${\mathcal{O}_{X}(U)}$ for ${U\subset X=\mathrm{Spa}(A,A^{\circ})}$ any rational subset.

ii. Any finite étale ${A}$-algebra ${B}$.

iii. ${A\left\langle T\right\rangle }$ and ${A\left\langle T^{1/p^{\infty}}\right\rangle }$.

iv. ${\mathrm{Cont}(S,A)}$ for any profinite set ${S}$.

With regards to ii., we remind the reader that if ${A}$ is any uniform Tate ring, then any finite étale ${A}$-algebra ${B}$ inherits a canonical compatible structure of uniform Tate ring such that ${A\rightarrow B}$ is continuous: this follows immediately upon combining (KL1, Lemma 2.8.14) and (KL1, Prop. 2.8.16(b)). We also point out that for a general stably uniform Tate ring ${A}$, the question of whether the stable uniformity (or even just the sheafyness) of ${A}$ enjoys permanence properties analogous to ii. and iii. here is wide open.

Proof: ${ }$Choose ${\tilde{A},f}$, and ${\sigma}$ witnessing the sousperfectoid nature of ${A}$ as in Definition 1.1.

For i., set ${Y=\mathrm{Spa}(\tilde{A},\tilde{A}^{\circ})}$, so ${f}$ induces a map ${\pi:Y\rightarrow X}$. One easily checks that ${\sigma}$ extends to a bounded splitting of the map ${\mathcal{O}_{X}(U)\rightarrow\mathcal{O}_{Y}(\pi^{-1}U)}$ induced by ${f}$. Since ${Y}$ is affinoid perfectoid, ${\mathcal{O}_{Y}(\pi^{-1}U)}$ is a perfectoid Tate ring, and so the result follows.

For ii., we simply observe that ${\sigma}$ extends uniquely to a bounded ${B}$-linear splitting of the obvious map

$\displaystyle B=B\otimes_{A}A\overset{1\otimes f}{\rightarrow}B\otimes_{A}\tilde{A}$

induced by ${f}$, while on the other hand ${B\otimes_{A}\tilde{A}}$ is finite étale over ${\tilde{A}}$ and thus perfectoid by the almost purity theorem.

For iii., note that ${\sigma}$ extends uniquely to a bounded ${A\left\langle T^{1/p^{\infty}}\right\rangle }$-linear splitting of the obvious map

$\displaystyle A\left\langle T^{1/p^{\infty}}\right\rangle \rightarrow\tilde{A}\left\langle T^{1/p^{\infty}}\right\rangle$

induced by ${f}$, and that ${\tilde{A}\left\langle T^{1/p^{\infty}}\right\rangle }$ is perfectoid, so ${A\left\langle T^{1/p^{\infty}}\right\rangle }$ is sousperfectoid. To descend to ${A\left\langle T\right\rangle }$, observe that ${A\left\langle T\right\rangle \rightarrow A\left\langle T^{1/p^{\infty}}\right\rangle }$ admits a canonical bounded ${A\left\langle T\right\rangle }$-linear splitting, so we get a bounded splitting of the composite map

$\displaystyle A\left\langle T\right\rangle \rightarrow A\left\langle T^{1/p^{\infty}}\right\rangle \rightarrow\tilde{A}\left\langle T^{1/p^{\infty}}\right\rangle$

as well.

Finally, iv. is easy and left to the reader (hint: for fixed ${S}$, ${\mathrm{Cont}(S,A)}$ is functorial in${A}$). $\Box$

Now we globalize the notion of being sousperfectoid in the obvious way.

Definition 1.4. A sousperfectoid space is an adic space ${X}$ which admits an open cover by affinoid adic spaces ${U_{i}=\mathrm{Spa}(A_{i},A_{i}^{+})\subset X}$ with each ${A_{i}}$ a sousperfectoid Tate ring.

By Proposition 1.3.i, any open subspace of a sousperfectoid space is sousperfectoid. We also caution the reader that if ${(D,D^{+})}$ is a sheafy Tate-Huber pair and ${X=\mathrm{Spa}(D,D^{+})}$ is an affinoid adic space which is sousperfectoid in this sense, it’s not clear whether the (completion of the) underlying Tate ring ${D}$ is necessarily sousperfectoid (though we don’t know any counterexample).

So far, this is mathematics which has been in my head since December 2015.  The dam broke last Friday, though, when I realized how to prove Lemma 1.5 and Lemma 1.7 below.

Recall that for any adic space ${X}$, Kedlaya-Liu have defined an étale site ${X_{\mathrm{et}}}$ (KL1, Def. 8.2.19) and a pro-étale site ${X_{\mathrm{proet}}}$ (KL1, Def. 9.1.4). In general, the objects of ${X_{\mathrm{et}}}$ are only preadic spaces étale over ${X}$, and the objects of ${X_{\mathrm{proet}}}$ are certain (equivalence classes of) projective systems of preadic spaces étale over ${X}$. (Since preadic spaces will only play an intermediate role in the following discussion, I won’t review them here; cf. (KL1, §8.2).) For sousperfectoid spaces, however, the étale and pro-étale sites turn out to be extremely well-behaved, as we now demonstrate.

Lemma 1.5. Let ${X}$ be a sousperfectoid space, and let ${Y}$ be a preadic space equipped with an étale morphism ${f:Y\rightarrow X}$. Then ${Y}$ is an honest adic space, and moreover ${Y}$ is sousperfectoid.

Proof: The claim is local on ${Y}$, so we may assume that ${f}$ factors as ${f=f_{1}\circ f_{2}\circ f_{3}}$ where the ${f_{i}}$‘s are morphisms of preadic spaces (a priori) such that ${f_{1}:X_{1}\rightarrow X}$ is an open immersion, ${f_{2}:X_{2}\rightarrow X_{1}}$ is finite étale, and ${f_{3}:Y\rightarrow X_{2}}$ is an open immersion. Now ${X_{1}}$ is open in the sousperfectoid space ${X}$, and hence is honest and sousperfectoid. Arguing locally on ${X_{1}}$, Proposition 1.3.ii and (KL1, Lemma 8.2.17(a)) together imply that ${X_{2}}$ is honest and sousperfectoid. But then ${Y}$ is open in ${X_{2}}$, so we’re done. $\Box$

Proposition 1.6. Let ${K}$ be a nonarchimedean field which is sousperfectoid. Then any smooth rigid analytic space over ${\mathrm{Spa}\, K}$ is sousperfectoid.

There is also an obvious relative version of this result, whose statement and proof we leave to the interested reader.

Proof: Since ${K}$ is sousperfectoid by assumption, repeated use of Proposition 1.3.iii shows that any unit polydisk ${\mathbf{B}_{K}^{n}=\mathrm{Spa}\, K\left\langle T_{1},\dots,T_{n}\right\rangle }$ is sousperfectoid.

Now let ${X}$ be any smooth rigid space over ${\mathrm{Spa}\, E}$, so we can choose a covering of ${X}$ by smooth affinoids ${U_{i}}$ together with étale maps ${f_{i}:U_{i}\rightarrow\mathbf{B}_{K}^{n_{i}}}$. Since ${f_{i}}$ is étale with sousperfectoid target, the previous proposition implies that each ${U_{i}}$ is sousperfectoid as well, so ${X}$ is sousperfectoid as desired. $\Box$

In (HK) we’ll show that every nonarchimedean field ${K}$ (with ${p\in K}$ topologically nilpotent) is sousperfectoid; the idea is that ${K\rightarrow\widehat{\overline{K}}}$ admits a continuous splitting, which is not entirely obvious.

Next we bootstrap from étale to pro-étale morphisms. The key lemma is as follows.

Lemma 1.7. Fix a sousperfectoid Tate ring ${A}$, and let ${(A_{i})_{i\in I}}$ be a directed system of Tate rings finite étale over ${A}$. Let ${\widehat{A_{\infty}}}$ denote the completion of ${A_{\infty}=\lim_{\substack{\rightarrow\\ i\in I} }A_{i}}$ for the topology making ${A_{\infty}^{\circ}=\lim_{\substack{\rightarrow\\ i\in I} }A_{i}^{\circ}}$ open and bounded. Then ${\widehat{A_{\infty}}}$ is sousperfectoid.

Proof: Choose a perfectoid Tate ring ${\tilde{A}}$, a continous ring map ${f:A\rightarrow\tilde{A}}$, and a bounded ${A}$-linear splitting ${\sigma:\tilde{A}\rightarrow A}$. Set ${\tilde{A}_{i}=A_{i}\otimes_{A}\tilde{A}}$, so ${\tilde{A}_{i}}$ is perfectoid by almost purity. Furthermore, ${\sigma}$ extends canonically as before to a bounded ${A_{i}}$-linear splitting ${\sigma_{i}:\tilde{A}_{i}\rightarrow A_{i}}$ of the evident map ${f_{i}:A_{i}\rightarrow\tilde{A}_{i}}$, compatibly with varying ${i}$. Passing to the direct limit, we get the same properties for the evident maps

$\displaystyle f_{\infty}:A_{\infty}\rightarrow\tilde{A}_{\infty}=\lim_{\substack{\rightarrow\\ i} }\tilde{A}_{i}$

and

$\displaystyle \sigma_{\infty}:\tilde{A}_{\infty}\rightarrow A_{\infty},$

where we give ${\tilde{A}_{\infty}}$ the obvious topology; in particular, ${\sigma_{\infty}}$ is bounded since it carries the open bounded subring ${\mathrm{im}(A_{\infty}^{\circ}\otimes_{A^{\circ}}\tilde{A}^{\circ})\subset\tilde{A}_{\infty}^{\circ}}$ into the bounded subset ${A_{\infty}^{\circ}\sigma(\tilde{A}^{\circ})\subset A_{\infty}}$. Since everything is continous we may extend ${f_{\infty}}$ and ${\sigma_{\infty}}$ uniquely to the completions of these rings, getting analogous maps

$\displaystyle \hat{f}_{\infty}:\widehat{A_{\infty}}\rightarrow\widehat{\tilde{A}_{\infty}}$

and

$\displaystyle \hat{\sigma}_{\infty}:\widehat{\tilde{A}_{\infty}}\rightarrow\widehat{A_{\infty}}.$

But ${\widehat{\tilde{A}_{\infty}}}$ is perfectoid, so these maps verify the sousperfectoid nature of ${\widehat{A_{\infty}}.}$ $\Box$

The next lemma is an easy weakening of (SW, Lemma 2.4.5), and accounts for all uniqueness statements we’ll make concerning inverse limits of adic spaces.

Lemma 1.8. Let ${(X_{i})_{i\in I}}$ be a filtered inverse system of adic spaces with qcqs transition maps, and let ${X}$ be an adic space with a compatible family of morphisms ${f_{i}:X\rightarrow X_{i}}$ such that

$\displaystyle X\sim\lim_{\substack{\leftarrow\\ i\in I} }X_{i}$

in the sense of (SW, Def. 2.4.1). If ${X}$ is stably uniform, then

$\displaystyle \mathrm{Hom}(Y,X)=\lim_{\leftarrow}\mathrm{Hom}(Y,X_{i})$

for any stably uniform adic space ${Y}$. In particular, a stably uniform adic space ${X}$ with ${X\sim\lim_{\substack{\leftarrow\\ i\in I} }X_{i}}$ is unique up to unique isomorphism.

Theorem 1.9. Let ${X}$ be a sousperfectoid space. Let ${Y}$ be any object in ${X_{\mathrm{proet}}}$, so we may choose a pro-étale presentation ${Y=\lim_{\substack{\leftarrow\\ i\in I} }Y_{i}}$ of ${Y}$ as a cofiltered inverse limit of (pre)adic spaces étale over ${X}$; note that by Lemma 1.5 each ${Y_{i}}$ is a sousperfectoid adic space étale over ${X}$. Then there is a sousperfectoid space ${\hat{Y}}$ over ${X}$ with compatible maps ${f_{i}:\hat{Y}\rightarrow Y_{i}}$ such that

$\displaystyle \hat{Y}\sim\lim_{\substack{\leftarrow\\ i\in I} }Y_{i}.$

The space ${\hat{Y}}$ is unique up to unique isomorphism, independently of the choice of pro-étale presentation of ${Y}$. The functor ${Y\mapsto\hat{Y}}$ defines a fully faithful embedding of ${X_{\mathrm{proet}}}$ into the slice category ${\mathrm{Adic}_{/X}}$ of adic spaces over ${X}$.

Proof: Taking into account Lemma 1.5 and Lemma 1.7, this immediately follows from Lemma 1.8 by an easy gluing argument; we leave the details to the interested reader. $\Box$

On account of this theorem, when ${X}$ is sousperfectoid there is no harm in conflating a formal pro-system ${Y\in X_{\mathrm{proet}}}$ with the associated space ${\hat{Y}/X}$; we will often write ${Y}$ instead of ${\hat{Y}}$.

Combining this theorem with Proposition 1.6 shows that the pro-étale site of any smooth rigid space ${X}$ over ${\mathrm{Spa}\, K}$ (for any nonarchimedean field ${K}$) is extremely well-behaved. In particular, we get the following theorem as a special case of Theorem 1.9.

Theorem 1.10. Fix a nonarchimedean field ${K}$ in which ${p}$ is topologically nilpotent, and let ${(X_{i})_{i\in I}}$ be a filtered inverse system of smooth rigid analytic spaces over ${\mathrm{Spa}\, K}$ with finite étale transition maps. Then there exists an adic space ${X_{\infty}}$ together with compatible morphisms ${f_{i}:X_{\infty}\rightarrow X_{i}}$ to the tower of ${X_{i}}$‘s such that

$\displaystyle X_{\infty}\sim\lim_{\substack{\leftarrow\\ i\in I} }X_{i}.$

The space ${X_{\infty}}$ is sousperfectoid. In particular, ${X_{\infty}}$ has an open covering by affinoid adic spaces associated with stably uniform Tate rings, and is unique up to unique isomorphism; furthermore, ${X_{\infty}}$ has a well-behaved étale site and pro-étale site, and it represents the functor ${\lim_{\leftarrow i}\mathrm{Hom}(-,X_{i})}$ on stably uniform adic spaces.

This result has many applications. For example, it instantly gives an infinite-level adic Shimura variety ${\mathcal{S}_{K^{p}}}$ with

$\displaystyle \mathcal{S}_{K^{p}}\sim\lim_{\leftarrow K_{p}}\mathcal{S}_{K^{p}K_{p}}$

for any Shimura datum ${(\mathbf{G},X)}$ and any (reasonable) tame level ${K^{p}}$. It also shows that the infinite-level Rapoport-Zink space ${\mathcal{M}_{\infty}}$ constructed in (SW, Theorem D) is an honest adic space.

Next we record a necessary and sufficient condition for a Tate ring to be sousperfectoid which doesn’t mention perfectoid rings.

Definition 1.11. A morphism ${A\rightarrow B}$ of Tate rings is faithfully profinite étale if ${B}$ is isomorphic to the completion of some filtered direct limit ${B'=\lim_{\substack{\rightarrow\\ i\in I} }B_{i}}$ of faithfully finite étale ${A}$-algebras for the spectral seminorm (or equivalently, for the topology making ${\lim_{\substack{\rightarrow\\ i} }B_{i}^{\circ}}$ an open and bounded subring of ${B'}$).

Note that if ${A\rightarrow B}$ is faithfully profinite étale, then ${A}$ is uniform if and only if ${A\rightarrow B}$ is injective.

Proposition 1.12. Let ${A}$ be a uniform Tate ring. Then ${A}$ is sousperfectoid if and only if every faithfully profinite étale morphism ${A\rightarrow B}$ admits a continuous ${A}$-linear splitting.

Proof: “If” is easy, since one can find such an ${A\rightarrow B}$ with ${B}$ perfectoid. “Only if” will be proved in (HK). $\Box$

The following result makes it easy to find examples of uniform Tate rings which are not sousperfectoid.

Proposition 1.13. If ${A}$ is any sousperfectoid Tate ring, then ${H^{0}(\mathrm{Spa}(A,A^{\circ})_{\mathrm{proet}},\widehat{\mathcal{O}})\cong A}$. In particular, ${A}$ is seminormal.

Proof: The former will appear in (HK). For the latter, use (KL2, Corollary 2.7.5). $\Box$

I want to give a few more constructions and examples involving sousperfectoid spaces. Note that if ${X}$ is sousperfectoid, then by Proposition 1.3.iii the relative polydisk

$\displaystyle \mathbf{B}_{X}^{n}=X\times_{\mathrm{Spa}\,\mathbf{Z}_{p}}\mathrm{Spa}\,\mathbf{Z}_{p}\left\langle T_{1},\dots,T_{n}\right\rangle$

is well-defined and sousperfectoid for any ${n\geq0}$. This suggests the following definition.

Definition 1.14. A morphism ${f:Y\rightarrow X}$ of sousperfectoid spaces is smooth (resp. pro-smooth) if we can find open coverings ${X=\cup X_{i\in I}}$ and ${Y=\cup_{i\in I}f^{-1}(X_{i})=\cup_{j\in J_{i},i\in I}Y_{ij}}$ such that for all ${i\in I}$ and ${j\in J_{i}}$, the restriction ${f|_{Y_{ij}}:Y_{ij}\rightarrow X_{i}}$ factors as a composition

$\displaystyle Y_{ij}\overset{g_{ij}}{\rightarrow}\mathbf{B}_{X_{i}}^{n}\overset{\mathrm{pr}}{\rightarrow}X$

for some ${n=n_{ij}}$ where ${\mathrm{pr}}$ is the obvious projection and ${g_{ij}}$ is étale (resp. ${g_{ij}}$ realizes ${Y_{ij}}$ as ${\hat{U}_{ij}}$ for some ${U_{ij}\in\mathbf{B}_{X_{i},\mathrm{proet}}^{n}}$).

A morphism ${f:Y\rightarrow X}$ of sousperfectoid spaces is (pro-)smooth proper if it is (pro-)smooth, quasicompact and universally specializing.

When ${X}$ is a sousperfectoid rigid space, this specializes to the usual notions of smooth and smooth proper morphisms in rigid geometry. It’s easy to check that smooth and pro-smooth morphisms are well-behaved with respect to fiber products and are stable under base change along an arbitrary morphism ${f:W\rightarrow X}$ of sousperfectoid spaces.

Example 1.15.  Let ${X}$ be a smooth rigid space over a nonarchimedean field of characteristic zero, and let ${A\rightarrow X}$ be a family of abeloid spaces, i.e. a family of smooth proper (connected) rigid analytic groups over ${X}$. Then the multiplication-by-p maps on ${A}$ are finite étale, so we get the universal cover

$\displaystyle \tilde{A}=\lim_{\substack{\leftarrow\\ \times p} }A$

as a sousperfectoid space over ${X}$, and the natural morphism ${\tilde{A}\rightarrow X}$ is pro-smooth proper. When ${X=\mathrm{Spa}(C,\mathcal{O}_{C})}$ is a geometric point and ${A}$ has good reduction, it’s not hard to check that ${\tilde{A}}$ is a perfectoid space. In particular, ${\tilde{A}\rightarrow X}$ typically has perfectoid geometric fibers. However, the total space of ${\tilde{A}}$ need not be perfectoid.
Note that we can also form the “physical” Tate module

$\displaystyle \mathcal{T}_{p}A=\lim_{\substack{\leftarrow\\ n} }A[p^{n}]$

as a sousperfectoid space pro-finite étale over ${X}$.

${\phantom{}}$

Example 1.16 (The dual of Example 1.15.) We can define the notion of a family of abeloid varieties over any sousperfectoid space ${X}$: it is a smooth and proper morphism ${f:A\rightarrow X}$ of sousperfectoid spaces with connected geometric fibers, together with morphisms

$\displaystyle m:A\times_{X}A\rightarrow A,\; i:A\rightarrow A,\; e:X\rightarrow A$

making ${A}$ into a commutative group object over ${X}$. Using this notion, one can show that infinite-level adic Shimura varieties associated with PEL Shimura data actually represent an obvious functor on the category of sousperfectoid spaces.
More generally, we can make sense of families of smooth (or smooth proper) rigid spaces over an arbitrary sousperfectoid base.

Feel free to start improving your life with sousperfectoid spaces today!

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### 4 Responses to Superfectoid spaces!?

1. lucqin says:

I’d like to know how to write down the math formulas on this blog. Thanks!

• arithmetica says:

??

2. in example 1.15, why do you need A to have good reduction in order to prove \tilde{A} is a perfectoid space?

• arithmetica says:

The idea is that the tilt of \tilde{A} can be identified, under the tilting equivalence between perfectoid spaces over C and C^flat, with the perfection of the abelian scheme A’ = A_0 x_{O_C} (O_C/p), where A_0 denotes the Neron model of A over O_C. More precisely, the generic fiber of the formal scheme over O_C^flat defined by \lim_{<– Frob} A' is clearly a perfectoid space over C^flat, and (less clearly) it untilts to \tilde{A}. This argument is sketched in Scholze's ICM article, for example. The result may well be true without the good reduction assumption, though!