spectral spaces; snark

(Update 12/24: Unfortunately the argument below doesn’t work as written. I overlooked the following detail in my “proof” that X^{wl} \to X is open: if S \to S' is a surjective continuous map of finite T_0 spaces, it’s not necessarily true that S^Z \to S'^Z is surjective. For instance, if S'= \{ \eta \rightsquigarrow s \} is the spectrum of a DVR and S=\{ \eta, s \} is S' equipped with the discrete topology, then S'^Z has three points but S=S^Z has two.

For the argument below to work, it would be enough to know that for any open subset V \subset X^{wl}, its image in X contains an open subset. Is this true?

In any case, the Corollary is still true, although by a totally different argument.)

*****

I spent about six hours yesterday and today proving the following thing.

Lemma. Let X be a spectral space, and let Z be a closed generalizing nowhere-dense subset of X. Then Z is nowhere-dense for the constructible topology on X, i.e. Z doesn’t contain any nonempty constructible subset of X.

This has the following concrete consequence, which is what I really needed.

Corollary. Let (A,A^+) be some Tate-Huber pair with adic spectrum X=\mathrm{Spa}(A,A^+), and let Z \subset X be a Zariski-closed nowhere-dense subset. Suppose U_1 and U_2 are quasicompact open subsets of X such that U_1 \cap (X - Z) = U_2 \cap (X - Z). Then U_1 = U_2.

Proof of Corollary. We need to check that V= (U_1 \cup U_2 ) - (U_1 \cap U_2) is empty. But V is a constructible subset of X contained in Z, so this is immediate from the lemma.

Amusingly, even though this corollary is pretty down-to-earth, I only managed to prove it by proving the lemma, and I only managed to prove the lemma by exploiting the structure of the w-localization X^{wl} of X. Is there a more direct approach? Am I missing something obvious?

(Sketch of actual argument: the profinite set of closed points X^{wl}_{c} maps homeomorphically onto X equipped with the constructible topology, so if V \subset X is constructible it is clopen when viewed as a subset of X^{wl}_{c}. The key point is then to check that Z is nowhere-dense when viewed as a subset of X^{wl}_c. This can be done, using that the natural surjection t: X^{wl} \to X is open and that t^{-1}(Z) (which is then closed, generalizing and nowhere-dense in X^{wl}, the last point by openness of t) is the preimage of its image in \pi_0(X^{wl}) \cong X^{wl}_c.

The openness of t doesn’t seem to be stated in the literature, but it can be deduced from the proof of Lemma 2.1.10 in Bhatt-Scholze, using the fact that it’s obviously true for finite T_0 spaces.)


You may have noticed that RIMS is hosting a series of four workshops next year under the umbrella of a “RIMS Research Project” entitled Expanding Horizons of Inter-universal Teichmuller Theory.  The first of the workshops looks pretty reasonable, the other three not so much.  In case you’re wondering (as I did) how much money RIMS is ponying up for this, it seems to be capped at 5 million yen, or about $41k (according to e.g. this document). This doesn’t seem like very much money to support four workshops; I guess some funding is also coming from that infamous EPSRC grant.

Anyway, when you’re inside a black hole, your horizons might seem quite expansive indeed, but I doubt you’ll have much luck convincing others to join you.

w-local spaces are amazing

Let X be a spectral space.  Following Bhatt-Scholze, say X is w-local if the subset X^c of closed points is closed and if every connected component of X has a unique closed point.  This implies that the natural composite map X^c \to X \to \pi_0(X) is a homeomorphism (cf. Lemma 2.1.4 of BS).

For the purposes of this post, a w-local adic space is a qcqs analytic adic space whose underlying spectral topological space is w-local.  These are very clean sorts of spaces: in particular, each connected component of such a space is of the form \mathrm{Spa}(K,K^+), where K is a nonarchimedean field and K^+ is an open and bounded valuation subring of K, and therefore has a unique closed point and a unique generic point.

I’ve been slowly internalizing the philosophy that w-local affinoid perfectoid spaces have a lot of amazing properties.  Here I want to record an example of this sort of thing.

Given a perfectoid space \mathcal{X} together with a subset T \subseteq |\mathcal{X}|, let’s say T is perfectoid (resp. affinoid perfectoid) if there is a pair (\mathcal{T},f) where \mathcal{T} is a perfectoid space (resp. affinoid perfectoid space) and f: \mathcal{T} \to \mathcal{X} is a map of adic spaces identifying |\mathcal{T}| homeomorphically with T and which is universal for maps of perfectoid spaces \mathcal{Y} \to \mathcal{X} which factor through T on topological spaces. Note that if the pair (\mathcal{T},f) exists, it’s unique up to unique isomorphism.

Theorem. Let \mathcal{X} be a w-local affinoid perfectoid space. Then any subset T of X = |\mathcal{X}| which is closed and generalizing, or which is quasicompact open, is affinoid perfectoid.

Proof when T is closed and generalizing. The key point here is that the map \gamma: X \to \pi_0(X) defines a bijection between closed generalizing subsets of X and closed subsets of the (profinite) space pi_0(X), by taking preimages of the latter or images of the former. To check that this is true, note that if T is closed and generalizing, then its intersection with a connected component X' of X being nonempty implies (since T is generalizing) that T \cap X' contains the unique rank one point of X'. But then T \cap X' contains all specializations of that point (since T \cap X' is closed in X'), so T \cap X' = X', so any given connected component of X is either disjoint from T or contained entirely in T.  This implies that T can be read off from which closed points of X it contains.  Finally, one easily checks that \gamma(T) is closed in \pi_0(X), since \pi_0(X) is profinite and \gamma(T) is quasicompact.  Therefore T = \gamma^{-1}(\gamma(T)).

Returning to the matter at hand, write \gamma(T) as a cofiltered intersection of qc opens V_i \subset \pi_0(X), i \in I. But qc opens in \pi_0(X) are the same as open-closed subsets, so each V_i pulls back to an open-closed subset U_i \subset X, and its easy to check that any such U_i comes from a unique rational subset \mathcal{U}_i \subset \mathcal{X}.  Then \mathcal{T} := \lim_{\leftarrow i \in I} \mathcal{U}_i is the perfectoid space we seek.

Proof when T is quasicompact open. 

First we prove the result when X is connected, i.e. when \mathcal{X} = \mathrm{Spa}(K,K^+) as above.  We claim that in fact T is a rational subset of X. When T is empty, this is true in many stupid ways, so we can assume T is nonempty. Since T is a qc open, we can find finitely many nonempty rational subsets \mathcal{W}_i = \mathrm{Spa}(K,K^{+}_{(i)}) \subset \mathcal{X} such that T = \cup_i |\mathcal{W}_i|.  But the \mathcal{W}_i‘s are totally ordered, since any finite set of open bounded valuation subrings of K is totally ordered by inclusion (in the opposite direction), so T = |\mathcal{W}| where \mathcal{W} is the largest \mathcal{W}_i.

Now we turn to the general case. For each point x \in \pi_0(X), we’ve proved that T \cap \gamma^{-1}(x) is a rational subset (possibly empty) of the fiber \gamma^{-1}(x).  Since \gamma^{-1}(x) = \lim_{\substack{\leftarrow}{V_x \subset \pi_0(X) \mathrm{qc\,open}, x\in V_x}} \gamma^{-1}(V_x) and each \gamma^{-1}(V_x) is the topological space of a rational subset \mathcal{U}_x of \mathcal{X}, it’s now easy to check* that for every x and for some small V_x as above, there is a rational subset \mathcal{T}_x \subset \mathcal{U}_x such that |\mathcal{T}_x| = U_x \cap T. Choose such a V_x for each point in \pi_0(X).  Since \pi_0(X) = \cup_x V_x, we can choose finitely many x‘s \{x_i\}_{i\in I} such that the V_{x_i}‘s give a covering of \pi_0(X).  Since each of these subsets is open-closed in \pi_0(X), we can refine this covering to a covering of \pi_0(X) by finitely many pairwise-disjoint open-closed subsets W_j, j \in J where W_j \subseteq V_{x_{i(j)}} for all j and for some (choice of) i(j) \in I. Then \gamma^{-1}(W_j) again comes from a rational subset \mathcal{S}_j of \mathcal{X}, so the intersection |\mathcal{T}_{x_{i(j)}}| \cap \gamma^{-1}(W_j) comes from the rational subset \mathcal{T}_j := \mathcal{T}_{x_{i(j)}} \times_{\mathcal{U}_{x_{i(j)}}} \mathcal{S}_j of X, and since |\mathcal{T}_j| = T \cap \gamma^{-1}(W_j) by design, we (finally) have that \mathcal{T} = \coprod_{j} \mathcal{T}_j \subset \coprod_{j} S_j = \mathcal{X} is affinoid perfectoid. Whew! \square

*Here we’re using the “standard” facts that if X_i is a cofiltered inverse system of affinoid perfectoid spaces with limit X, then |X| = \lim_{\leftarrow i} |X_i|, and any rational subset W \subset X is the preimage of some rational subset W_i \subset X_i, and moreover if we have two such pairs (i,W_i) and (j,W_j) with the W_{\bullet}‘s both pulling back to W then they pull back to the same rational subset of X_k for some large k \geq i,j.

Let T be a subset of a spectral space X; according to the incredible Lemma recorded in Tag 0A31 in the Stacks Project, the following are equivalent:

  • T is generalizing and pro-constructible;
  • T is generalizing and quasicompact;
  • T is an intersection of quasicompact open subsets of X.

Moreover, if T has one of these equivalent properties, T is spectral. (Johan tells me this lemma is “basically due to Gabber”.) Combining this result with the Theorem above, and using the fact that the category of affinoid perfectoid spaces has all small limits, we get the following disgustingly general statement.

Theorem. Let \mathcal{X} be a w-local affinoid perfectoid space. Then any generalizing quasicompact subset T \subset |\mathcal{X}| is affinoid perfectoid.

By an easy gluing argument, this implies even more generally (!) that if T \subset |\mathcal{X}| is a subset such that every point t\in T has a qc open neighborhood U_t in |\mathcal{X}| such that T \cap U_t is quasicompact and generalizing, then T is perfectoid (not necessarily affinoid perfectoid).  This condition* holds, for example, if T is locally closed and generalizing; in that situation, I’d managed to prove that T is perfectoid back in May (by a somewhat clumsy argument, cf. Section 2.7 of this thing if you’re curious) after Peter told me it was so.  But the argument here gives a lot more.

*Johan’s opinion of this condition: “I have no words for how nasty this is.”