Multiple times in talks about condensed mathematics (e.g. the Masterclass talks, Clausen's RAMpAGe talk), it is stated that the derived structure sheaf given by the condensed formalism "fixes" the non-sheafiness of non-sheafy adic spaces, and if the space is sheafy then "rational localizations are what you would expect" and the structure sheaf should be "concentrated in degree zero" in some infinity-categorical sense.

I am wondering about the opposite direction: could the condensed formalism be used as a criterion (or even characterization) of when an adic space is sheafy?

If the derived nature of the condensed formalism truly "fixes" the structure sheaf, then it seems to me like there should be some precises statement of the following form:

Conjecture. If the "condensed version" of the Huber pair $(A,A^+)$ is "concentrated in degree zero", then $(A,A^+)$ is sheafy.

Does such a statement exist? Is it trivial once one understands the construction? Or does it take work to prove, or is it even open?


1 Answer 1


Thanks for the question! One interpretation of the conjecture is true. Let me elaborate. The following results are kind of implicit in some discussion towards the end of www.math.uni-bonn.de/people/scholze/Analytic.pdf (see especially Proposition 13.16, Propositon 14.7, and some surrounding discussion), although some relevant computations are not explicitly done there, but in the Master Thesis of Grigory Andreychev (I hope it will be public soonish):

  1. To any Huber pair $(A,A^+)$, that is a pair of a certain kind of (always assumed complete here) topological ring $A$ together with an open and integrally closed subring $A^+\subset A$ consisting of powerbounded elements, one can associate an analytic ring $(A,A^+)_{\blacksquare}$. This is Proposition 13.16. Analytic rings are pairs of a condensed ring $B$ together with a notion of "completeness" for condensed $B$-modules: See Lecture 7 of www.math.uni-bonn.de/people/scholze/Condensed.pdf for the "non-animated" version (and Lecture 12 of Analytic.pdf for a more general version). As remarked in Remark 13.17, Andreychev has proved that this has the property that $(A,A^+)_{\blacksquare}[S]$ is concentrated in degree $0$ for any profinite set $S$, so it's an analytic ring in the "non-animated" sense (where everything is a usual condensed ring, and condensed module).

  2. This functor from Huber pairs to analytic rings is fully faithful. Again, this is part of Proposition 13.16. It is in this sense that the analytic geometry defined in these lectures extends the category of adic spaces.

  3. The functor $(A,A^+)\mapsto (A,A^+)_{\blacksquare}$ always has underlying condensed ring the condensed ring $\underline{A}$ corresponding to the topological ring $A$, but the notion of completeness for modules depends on the subring $A^+\subset A$. From this perspective, the seemingly obscure conditions on $A^+$ have a very natural interpretation, see Remark 13.18: Talking about such subrings of $A$ is just one way to talk about the associated analytic ring structures (other subrings of $A$ can also lead to such analytic ring structures, but for all such analytic ring structures there would be a different choice of subring that satisfies the conditions imposed on $A^+$).

  4. The functor $(A,A^+)\mapsto (A,A^+)_{\blacksquare}$ extends to more general pairs of a condensed animated ring $A$ (where "animated" = "simplicial, up to homotopy"), satisfying some conditions, and equipped with a similar subalgebra $A^+\subset \pi_0 A$. (Technically, it is enough if $A$ is nuclear over $\mathbb Z[[X_1,\ldots,X_n]]$ for some $n$.)

  5. For any rational open subset $U\subset \mathrm{Spa}(A,A^+)$, one can naturally define a condensed animated $A$-algebra $\mathcal O_X(U)$ together with a subalgebra $\mathcal O_X^+(U)\subset \pi_0 \mathcal O_X(U)$, fitting into the class implicit in 4). In particular, there is an associated analytic ring $(\mathcal O_X(U),\mathcal O_X^+(U))_{\blacksquare}$. This is the localization of the analytic ring $(A,A^+)_{\blacksquare}$ to the preimage of $U$ in $\mathrm{AnSpec}((A,A^+)_{\blacksquare})$ via the map of Proposition 14.7.

  6. The association $U\mapsto \mathcal O_X(U)$ defines a sheaf of condensed animated $A$-algebras on $\mathrm{Spa}(A,A^+)$, for any Huber pair $(A,A^+)$. This is a special case of Proposition 12.18.

  7. If $(A,A^+)$ satisfies any of the classical criteria of being sheafy, or generally if $A$ is Tate and sheafy, then $\mathcal O_X(U)$ is concentrated in degree $0$ and comes from the usual structure sheaf of Huber rings. This is Proposition 14.7. (The general Tate case uses some results of Kedlaya. It may be that by recent work of Ramero http://math.univ-lille1.fr/~ramero/CoursAG.pdf, Chapter 12, who extends some of Kedlaya's work to the case of general Huber rings, simply asking $(A,A^+)$ to be sheafy is enough.)

  8. Now conversely, if the sheaf $\mathcal O_X(U)$ of condensed animated $A$-algebras happens to be concentrated in degree $0$, and also to be quasiseparated, then it agrees with the (presheaf of condensed rings associated to the) presheaf of Huber rings defined by Huber, which is thus sheafy.

That final point 8) is, I think, a partial answer to your question. At least if $A$ is Tate, it shows that sheafyness is equivalent to the assertion that for all rational $U\subset \mathrm{Spa}(A,A^+)$, the condensed animated $A$-algebra $\mathcal O_X(U)$ is concentrated in degree $0$ and quasiseparated. One could wonder whether the "... and quasiseparated" ending is necessary; my gut feeling is that it is necessary.

Upshot: Huber was working in the context of complete topological rings (satisfying some conditions), and had to insist that his structure presheaf stays in the same realm. If you allow yourself more flexibility, in particular a notion of completeness that does not entail separatedness, and a formalism of topological rings that allows for higher homotopy groups (i.e., animated rings), then one can define a better version of his presheaf, that is always a sheaf. (See the beginning of Lecture 11 in Analytic.pdf, especially page 73, for some intuitive explanations.) Then sheafyness of Huber pairs is simply the question whether this more general construction stays in the classical realm, which fortunately happens so far in all cases of interest. I should mention that Bambozzi-Kremnizer have recently reached similar results, using different foundations, see arXiv:2009.13926. (In their approach, the role of $A^+$ is less clear.)

Edit: Let me actually be much more explicit about all of this. Consider a rational subset $U\subset \mathrm{Spa}(A,A^+)$, explicitly given as the locus $U=\{|f_1|,\ldots,|f_n|\leq |g|\neq 0\}$, for $f_1,\ldots,f_n,g\in A$ generating an open ideal. In that case, Huber's ring $\mathcal O_X^H(U)$ is given by $$ A\langle T_1,\ldots,T_n\rangle[\tfrac 1g]/\overline{(f_1-gT_1,\ldots,f_n-gT_n)}. $$ Intuitively, this is just expressing that on this subset $g$ is invertible, and $T_i=\tfrac{f_i}g$ has absolute value $\leq 1$, so we can allow all convergent power series in the $T_i$. To stay in the setting of complete topological rings, it is necessary to take the quotient by the closure of this ideal.

On the other hand, a theorem of Kedlaya (https://kskedlaya.org/papers/aws-notes.pdf, Theorem 1.2.7) shows that if $\mathcal O_X^H$ is a sheaf and $A$ is Tate, then it is not necessary to take the closure. Moreover, his results show that the sequence $(f_1-gT_1,\ldots,f_n-gT_n)$ is (Koszul-)regular. In other words, in this case $\mathcal O_X^H(U)$ is computed by the Koszul complex $$ A\langle T_1,\ldots,T_n\rangle[\tfrac 1g]/^{\mathbb L}(f_1-gT_1,\ldots,f_n-gT_n), $$ where I write, for an $A$-module $M$ and elements $a_1,\ldots,a_n\in A$, $$ M/^{\mathbb L}(a_1,\ldots,a_n) $$ for the (homological) Koszul complex $$ 0\to M\xrightarrow{(a_1,\ldots,a_n)} M^n\to \ldots\to M^n\xrightarrow{(a_1,\ldots,a_n)} M\to 0. $$

What condensed mathematics allows you to do is to consider the derived quotient $$ A\langle T_1,\ldots,T_n\rangle[\tfrac 1g]/^{\mathbb L}(f_1-gT_1,\ldots,f_n-gT_n) $$ as the correct answer in general. For this, you have to consider it simultaneously as endowed with (something like) a topology, as a complex, and as a (commutative) algebra. Thus, you have to mix higher (i.e. homotopical) algebra with topology, and this is what condensed mathematics can easily accomodate. In fact, these are just the condensed animated rings I've been talking about above. (Animated rings are, at least in characteristic 0, just the "commutative algebras in the derived category", correctly understood; making them condensed amounts to putting something like a topology on them, in particular on their homotopy groups.)

In particular, what my answer above means is the following, at least if $A$ is Tate: $\mathcal O_X^H$ is a sheaf if and only if for all $U$ as above, the sequence $(f_1-gT_1,\ldots,f_n-gT_n)$ is Koszul-regular and generates a closed ideal. This is in fact already a theorem of Kedlaya, I believe.

  • $\begingroup$ Do you tacitly assume in the fourth point that your "animated Huber pair" is complete? And this does not depend on the choice of generators of the defining ideal? $\endgroup$
    – Z. M
    May 7, 2021 at 12:30
  • $\begingroup$ I guess $A$ should be $A^+$-solid, yes. This condition does not depend on any choices. (There's not really any defining ideal anymore; you could say a defining ideal is one generated by $X_1,\ldots,X_n$ for any map $\mathbb Z[[X_1,\ldots,X_n]]\to A$ making $A$ nuclear. But this doesn't make so much sense as I'm not sure how to define rings of definition, and the ideal of definition should be an ideal in a ring of definition...) $\endgroup$ May 7, 2021 at 16:36
  • $\begingroup$ No, I meant the choice of maps $\mathbb Z[[\{X_i\}]]\to A$, i.e. a priori this condition does not seem to intrinsic to $(A,A^+)$. $\endgroup$
    – Z. M
    May 7, 2021 at 17:13
  • $\begingroup$ Of course being nuclear over $\mathbb Z[[X_1,\ldots,X_n]]$ depends on the map $\mathbb Z[[X_1,\ldots,X_n]]\to A$. But classically if say $A$ is already a ring of definition, you want to ask that it has a finitely generated ideal of definition, for which again you implicitly choose a map $\mathbb Z[[X_1,\ldots,X_n]]\to A$ (sending the $X_i$ to generators of an ideal of definition). $\endgroup$ May 7, 2021 at 19:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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