Artin-Grothendieck vanishing, again

A few years ago I started thinking about whether there was a natural rigid analytic version of the Artin-Grothendieck vanishing theorem. Last summer this grew into an obsession, and I managed to prove some general results. In particular, I showed that if $X$ is an affinoid rigid space over a complete algebraically closed field, AND $X$ comes via base change from an affinoid defined over a discretely valued subfield, then $H^i(X,\mathbf{Z}/n)=0$ for all $i > \mathrm{dim}(X)$ and all $n$ prime to the residue characteristic. I also proved a similar result with a non-constant coefficient sheaf, assuming moreover that the base field is of characteristic zero. This all got written up here.

Now, the hypothesis of definability over a discretely valued field is clearly stupid and shouldn’t be there, but I wasn’t able to remove it. So I was extremely happy this morning when Akhil Mathew and Bhargav Bhatt sent me an expanded version of their paper on arc-descent, in which they give a beautiful proof of rigid analytic Artin-Grothendieck vanishing without any superfluous assumptions. Their arguments are phrased in terms of algebraic geometry, rather than rigid analysis; in this post I want to recast (mostly for my own benefit I guess) the essential point of their argument in rigid analytic language.

The key is to prove the following.

Theorem (Bhatt-Mathew). Let $\mathrm{Spa}A$ be an affinoid rigid space over a complete algebraically closed nonarchimedean field $K$. Set $\Lambda = \mathbf{Z}/n$ where $n$ is any integer prime to the residue characteristic. Then $R\Gamma(\mathrm{Spa}A,\Lambda) \in D^{\leq \mathrm{dim}A}(\Lambda)$.

This implies the characteristic zero case of Conjecture 1.2 in my paper.

The proof proceeds in three steps.

Step One: Treat the case where $\mathrm{Spa}A$ is smooth. This was already done by Berkovich in the 90’s and I’ll take it for granted, although BM give their own nice argument for it. (Both arguments eventually appeal to the classical Artin-Grothendieck vanishing theorem.)

Step Two: Prove the weaker statement that $R\Gamma(\mathrm{Spa}A,\Lambda) \in D^{\leq 1+\mathrm{dim}A}(\Lambda)$ in general.

For this we use induction on $\mathrm{dim}A$. I’ll assume for simplicity that $K$ has characteristic zero. Without loss of generality we can assume that $A$ is reduced. Then by excellence of affinoid algebras, we can pick some non-zero-divisor $f \in A$ such that $A[1/f]$ is regular. Fix a nonzero nonunit $\pi \in \mathcal{O}_K$, and for any $n \geq 1$ consider the rational subsets $U_n = \{ x\,with\,|f(x)| \geq |\pi|^n \}$ and $V_n = \{x\,with\,|f(x)| \leq |\pi|^n \}$ inside $\mathrm{Spa}A$. Set $W_n = U_n \cap V_n$, so we get a Mayer-Vietoras distinguished triangle

$R\Gamma(\mathrm{Spa}A,\Lambda) \to R\Gamma(U_n,\Lambda)\oplus R\Gamma(V_n,\Lambda) \to R\Gamma(W_n,\Lambda)\to$

for any $n$. Note that $U_n$ and $W_n$ are smooth affinoids, so their etale cohomology is concentrated in degrees $\leq \mathrm{dim}A$ by Step One. Therefore, truncating the above Mayer-Vietoras sequence we get a quasi-isomorphism

$\tau^{\geq \mathrm{dim}A+2}R\Gamma(\mathrm{Spa}A,\Lambda) \simeq \tau^{\geq \mathrm{dim}A+2}R\Gamma(V_n,\Lambda)$

for any $n$. Moreover, $\mathrm{Spa}(A/f) \sim \lim_{n} V_n$ in the sense of adic spaces, which implies that the etale cohomology of the left-hand side is the colimit of the etale cohomologies of the right-hand sides. Therefore, passing to the colimit over $n$, the previous quasi-isomorphism gives a quasi-isomorphism

$\tau^{\geq \mathrm{dim}A+2} R\Gamma(\mathrm{Spa}A,\Lambda) \simeq \tau^{\geq \mathrm{dim}A+2}R\Gamma(\mathrm{Spa}(A/f),\Lambda)$.

But now we win, because $A/f$ is an affinoid of dimension $\dim(A)-1$, so by the induction hypothesis its etale cohomology is concentrated in degrees $\leq \mathrm{dim}A$.

Step Three. Bootstrap from the result of Step Two by a trick. More precisely, let $X=\mathrm{Spa}A$ and $\Lambda=\mathbf{Z}/n$ be as in the statement of the main theorem. By Step Two, we just have to show that $H^{\mathrm{dim}+1}(X,\Lambda)=0$. By another application of Step Two, the complex $R\Gamma(X,\Lambda) \otimes_{\Lambda}^{\mathbf{L}} R\Gamma(X,\Lambda)$ has cohomology in degree $2\mathrm{dim}A+2$ given by $H^{\mathrm{dim}A+1}(X,\Lambda)^{\otimes 2}$, and its enough to show that the latter module is zero. But

$R\Gamma(X,\Lambda) \otimes_{\Lambda}^{\mathbf{L}} R\Gamma(X,\Lambda) \simeq R\Gamma(X \times X,\Lambda)$

by the Kunneth formula*, and $X \times X$ is a $2\mathrm{dim}A$-dimensional affinoid, so its cohomology is concentrated in degrees $\leq 2\mathrm{dim}A+1$ by yet another application of Step Two. This gives the result.

*The necessary result is that if $X$ and $Y$ are $K$-affinoid spaces, then $R\Gamma(X \times Y, \Lambda) \simeq R\Gamma(X,\Lambda) \otimes_{\Lambda}^{\mathbf{L}} R\Gamma(Y,\Lambda)$. I’m not sure if this is in the literature; Bhargav and Akhil prove (an algebraic form of) it in their paper. However, it is easy to deduce this from the results in Huber’s book. The point is that $X, Y$ have canonical adic compactifications $\overline{X},\overline{Y}$, and etale cohomology (with constant coefficients) doesn’t change if you replace $X$ or $Y$ by its compactification. But then $\overline{X}$ and $\overline{Y}$ are proper over $\mathrm{Spa}K$ (in the sense of Huber’s book), so it’s easy to show that

$R\Gamma(\overline{X} \times \overline{Y},\Lambda) \simeq R\Gamma(\overline{X},\Lambda) \otimes_{\Lambda}^{\mathbf{L}} R\Gamma(\overline{Y},\Lambda)$

by the usual combination of proper base change and the projection formula.

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.”