## p-adic Kahler manifolds

In complex geometry, the most interesting class of complex manifolds is probably the Kahler class. In the non-archimedean world, say over a fixed p-adic base field $K$, the analogue of a compact complex manifold is a smooth proper rigid analytic space. In some ways, these are already surprisingly “close” to being Kahler – in particular, the Hodge-de Rham spectral sequence of such a space always degenerates at $E_1$. However, Hodge symmetry can definitely fail. A standard example is the non-archimedean Hopf surface $X = \mathbf{A}^2_{K} \smallsetminus \{ (0,0) \} / p^{\mathbf{Z}}$ where $p^n$ acts through diagonal multiplication. By a fun direct calculation, one checks that $H^0(X,\Omega^1_X)=0$ and $H^1(X,\mathcal{O}_X) = K$, so Hodge symmetry fails in degree one.

We now see a natural question: is there is some non-archimedean analogue of the Kahler condition which restores Hodge symmetry? Two years ago, Shizhang Li hit upon the following candiate condition:

A smooth proper rigid space $X$ satisfies (*) if it admits a formal model $\mathfrak{X}$ over $\mathcal{O}_K$ whose special fiber is projective (as opposed to merely proper).

Using fantastic ideas due to Shizhang, we managed to prove the following suggestive result.

Theorem. Let $X$ be a smooth proper rigid space satisfying (*). Then $h^{1,0}(X) = h^{0,1}(X)$.

Of course, one can then guess that (*) implies Hodge symmetry in all degrees. This speculation seems to have caught the imagination of others in the field, but until recently I personally regarded it as not much more than wishful thinking. However, my perspective completely changed a month ago, when I learned from Shizhang that, according to Robert Friedman, the archimedean analogue of “(*) implies Hodge symmetry” is a theorem! More precisely, we have the following result:

Theorem. Let $D$ be the complex disk, with $D^\times =D \smallsetminus \{0 \}$ the punctured disk. Let $f:Y \to D$ be a proper map of complex analytic spaces. Suppose that $f^{-1}(D^\times) \to D^\times$ is a submersion, and that the central fiber $Y_0=f^{-1}(0)$ is the analytification of a projective (and not necessarily smooth) algebraic variety. Then for all $t \in D^\times$ with $|t| \ll 1$, the fiber $Y_t$ satisfies Hodge symmetry and Hodge-de Rham degeneration.

Of course, the analogy is that $\mathfrak{X} \to \mathrm{Spf} \mathcal{O}_K$ is analogous to $Y \to D$, and $X$ is analogous to the “nearby” fibers $Y_t$ with $0<|t| \ll 1$.

The proof of this theorem uses the full power of mixed Hodge theory. In fact the claim about Hodge-de Rham degeneration is exactly Corollary 11.24 in the book of Peters-Steenbrink. Hodge symmetry is even more subtle, and the argument for this doesn’t seem to be written down anywhere; Friedman explained it to Shizhang, who explained it to me, but the details entailed such a horrible explosion of gradings, filtrations, and multi-indices that I can’t hope to reproduce it here.

Anyway, I’m now completely convinced that Shizhang’s condition (*) implies Hodge symmetry in all degrees, and that this is truly the “right” p-adic analogue of the Kahler condition.

## sheafiness; perversity

$\bullet$ One of the main annoyances in the theory of adic spaces is that, for a given Huber pair $(A,A^+)$, the structure presheaf on $\mathrm{Spa}(A,A^{+})$ is not always a sheaf. One usually remedies this by restricting attention to various classes of Huber rings, e.g. strongly Noetherian Tate rings, perfectoid Tate rings, Noetherian adic rings, etc. However, the following class of rings doesn’t seem to be addressed in the literature:

Definition. Let $A$ be an adic ring with finitely generated ideal of definition $I \subset A$. We say $A$ is strongly Noetherian outside $I$ if, for all $n \geq 0$, the scheme $\mathrm{Spec}\,A\left\langle X_1,\dots,X_n \right\rangle \smallsetminus V(IA\left\langle X_1,\dots,X_n \right\rangle )$ is Noetherian.

Here $A\left\langle X_1,\dots,X_n \right\rangle$ denotes the usual ring of restricted power series. Note that if $A$ is a Tate ring and $(A_0,(\varpi))$ is any couple of definition, then $A$ is strongly Noetherian if and only if $A_0$ is strongly Noetherian outside $(\varpi)$. I should also point out that the condition of being strongly Noetherian outside $I$ is already considered in a very interesting paper of Fujiwara-Gabber-Kato; they use the terminology “topologically universally rigid-Noetherian”, but I prefer my terminology on account of the previous sentence. Anyway, the following conjecture seems reasonable:

Conjecture. If $A$ is strongly Noetherian outside $I$, the structure presheaf on $\mathrm{Spa}(A,A)$ is a sheaf.

This implies that any strongly Noetherian Tate ring is sheafy (which of course is already known), but it also implies e.g. that if $A$ is topologically finitely presented over $\mathcal{O}_K$ for some nonarchimedean field $K$, then $A$ is sheafy. Sheafiness in the latter situation is known when $K$ is discretely valued, but to the best of my knowledge it’s open for general $K$.

I’m sure this conjecture is within reach, and maybe it’s an easy exercise for experts on sheafiness (*cough* Kiran *cough*). Note that FGK already proved some interesting consequences of this definition, which are probably relevant to proving this conjecture. Precisely, they show that if $A$ is strongly Noetherian outside $I$, then:
1. The $I$-power-torsion submodule of any finitely generated $A$-module is killed by a power of $I$.
2. If $N \subset M$ is any inclusion of $A$-modules, with $M$ finitely generated, then the subspace topology on $N$ induced by the $I$-adic topology on $M$ coincides with the $I$-adic topology on $N$.

$\bullet$ Let $j:U \to X$ be some immersion of varieties over a separably closed field. Everyone knows that the intermediate extension functor $j_{!\ast}$ on perverse sheaves (say with coefficients in $\Lambda = \mathbf{Q}_{\ell}$) is pretty great: it’s totally canonical, it commutes with Verdier duality, it preserves irreducibility, it preserves monic and epic maps of perverse sheaves, etc.

Recently I noticed that if $f: Z \to X$ is any map of varieties, with $X$ smooth and $Z$ irreducible, there’s still a natural functor $f^{\ast !}:\mathrm{Perv}(X) \to \mathrm{Perv}(Z)$ which commutes with Verdier duality. To define this functor, note that for any $f$ and any $\mathcal{F} \in D^b_c(X)$, there is a natural map $f^\ast \mathcal{F} \otimes Rf^! \Lambda \to Rf^! \mathcal{F}$, obtained by adjunction from the composite map $Rf_!( f^\ast \mathcal{F} \otimes Rf^! \Lambda) \cong \mathcal{F} \otimes Rf_! Rf^! \Lambda \to \mathcal{F}$ (the first isomorphism here is the projection formula). Since $X$ is smooth, the dualizing complex $\omega_X$ is just $\Lambda[2 \dim X]$, so then $Rf^! \Lambda = Rf^!\omega_X[-2 \dim X] = \omega_Z[-2 \dim X]$. Thus we get a natural map $\alpha: f^\ast \mathcal{F} \otimes \omega_Z[-2 \dim X] \to Rf^! \mathcal{F}$.

Next, note that the complex $\omega_Z$ is concentrated in degrees $[-2 \dim Z,0]$, and in the lowest of these degrees it’s just the constant sheaf, i.e. $\tau^{ \leq -2 \dim Z} \omega_Z \cong \Lambda[2 \dim Z]$. In particular, there is a canonical map $\Lambda[2 \dim Z] \to \omega_Z$. Shifting by $-2 \dim X$ and tensoring with $f^\ast \mathcal{F}$ gives a map $\beta: f^\ast \mathcal{F}[2 \dim Z - 2 \dim X] \to f^\ast \mathcal{F} \otimes \omega_Z[-2 \dim X]$. Putting things together, we get a natural map $\alpha \circ \beta: f^\ast \mathcal{F}[2 \dim Z - 2 \dim X] \to Rf^! \mathcal{F}$. Set $c= \dim X - \dim Z,$ so after shifting this becomes a natural map

$\gamma=\gamma_{\mathcal{F}}: f^{\ast}\mathcal{F}[-c] \to Rf^!\mathcal{F}[c]$.

This shifting has the advantage that Verdier duality exchanges the functors $f^{\ast}[-c]$ and $Rf^![c]$ on $D^b_c$, and one can check that the Verdier dual of $\gamma_{\mathcal{F}}$ identifies with $\gamma_{\mathbf{D}\mathcal{F}}$.

Definition. The functor $f^{\ast !}: \mathrm{Perv}(X) \to \mathrm{Perv}(Z)$ sends any $\mathcal{F}$ to the image of the map $^p\mathcal{H}^0(\gamma):$ $^p\mathcal{H}^0(f^{\ast} \mathcal{F}[-c]) \to$ $^p\mathcal{H}^0(Rf^!\mathcal{F}[c])$.

Here of course $^p\mathcal{H}^0(-)$ denotes the zeroth perverse cohomology sheaf.

Exercise. Show that $f^{\ast !}(\Lambda[\dim X]) \cong \mathcal{IC}_Z$.

It might be interesting to compute this functor in some other examples. Note that it can be quite stupid: if $f$ is a closed immersion (with $c > 0$) and $\mathcal{F} \in \mathrm{Perv}(X)$ is already supported on $Z$, then $f^{ \ast !} \mathcal{F} = 0$. On the other hand, if $f$ is smooth and surjective, then $f^{\ast !} \cong f^{\ast}[-c] \cong Rf^![c]$ is faithful.

## Report from Oberwolfach

Just returned from a workshop on “Arithmetic of Shimura varieties” at Oberwolfach. Some scattered recollections:

• Gabber wasn’t there, but there were some Gabberesque moments anyway. In particular, during Xuhua He’s talk, Goertz observed that a point is an example of a Deligne-Lusztig variety, so any variety is a union of Deligne-Lusztig varieties. Gotta be careful…
• The food was about the same as usual. Worst Prize was tied between two dishes: a depressing vegetable soup which somehow managed to be flavorless and bitter simultaneously, and a dessert which looked like a lovely innocent custard but tasted like balsamic vinegar. The best dishes were all traditional German fare.
• Best Talks (in no particular order): Jean-Stefan Koskivirta, Miaofen Chen, Ben Howard, Timo Richarz.
• Apparently this paper can be boiled down to a page or two.
• There was (not surprisingly) some late-night discussion of the Stanford Mystery. [Redacted] proposed a theory so mind-bogglingly outrageous that it certainly won’t fit in this margin.
• “Fun was never really my goal.” – A representative UChicago alum.
• On Thursday it snowed, and a snowball fight broke out after dinner. This was a lot of fun, but I’m still glad we didn’t follow Pilloni’s suggestion of a match between Team Europe (Pilloni, Stroh, Morel, Anschutz, Richarz, Mihatsch, etc.) and Team USA (me).
• Here’s an innocent problem which turns out to be pretty tricky. Let $X$ be a (separated, smooth) rigid analytic space over $\mathbf{Q}_p$, and let $Y \to X$ be a map from a perfectoid space which is a $\underline{G}$-torsor for some profinite group $G$. In shorthand, you should think that $X =" Y/G$ with G acting freely (this is all literally true in the category of diamonds). It’s easy to cook up examples of this scenario: for instance, you can take $X=\mathrm{Spa}\mathbf{C}_p \left\langle T^{\pm 1} \right\rangle$ and $Y=\mathrm{Spa}\mathbf{C}_p \left\langle T^{\pm 1/p^\infty} \right\rangle$, so then $Y \to X$ is a torsor for the group $\mathbf{Z}_p$. However, there are also much more complicated examples which arise in nature. In particular, if $X$ is a Rapoport-Zink space or abelian-type Shimura variety at some finite level, and $Y$ is the associated infinite level perfectoid guy over it, then we’re in the situation above, with $G$ open in the $\mathbf{Q}_p$-points of some auxiliary reductive group.

Anyway, supposing we’re in the situation above, we can ask the following complementary questions:
Q1. Suppose that $Y$ is affinoid perfectoid. Does this imply that $X$ is an affinoid rigid space?
Q2. Suppose that $X$ is an affinoid rigid space. Does this imply that $Y$ is affinoid perfectoid?

It seems like both of these questions are actually really hard! For Q1, we can (by assumption) write $Y=\mathrm{Spa}(A,A^+)$ for some perfectoid Tate-Huber pair $(A,A^+)$, and then one might guess that $X$ coincides with $X'=\mathrm{Spa}(A^G,A^{+G})$. There is certainly a map $X \to X'$, but now one is faced with the problem of showing that $A^G$ is “big enough” for this map to be an isomorphism. This can be reduced to any one of a handful of auxiliary problems, but they all seem hard (at least to me). For instance, as a warmup one could try to prove either of the following implications:

W1. Under the hypothesis of Q1, $H^1(X,\mathcal{O}_X)$ vanishes.
W2. Under the hypothesis of Q1, $H^1(X,\mathcal{O}_{X}^+)$ is killed by a fixed power of $p.$

Both of these conclusions would certainly hold if we already knew that $X$ was affinoid: the first is just (a consequence of) Tate acyclicity, while the fact that $H^1(X,\mathcal{O}_{X}^+)$ is killed by some power of $p$ for smooth affinoids is a non-trivial theorem of Bartenwerfer. But I have totally failed to prove either W1 or W2.

In any case, the essential point with Q1 seems to be the following. If $H$ is some open subgroup, then $(A^+/p^n)^H$ will always have plenty of elements, and indeed taking the direct limit as $H$ shrinks recovers $A^+/p^n$. But the obstruction to lifting an element of $(A^+/p^n)^H$ to an element of $(A^+)^H$ is a torsion class in $H^1(H,A^+)$, and the latter group seems hard to control.

For Q2, there is maybe a slightly clearer path through the forest: it would follow from the following conjecture, which I explained during my talk in the workshop.

To set things up, let $(A,A^+)$ be any uniform Tate-Huber pair over $(\mathbf{Q}_p,\mathbf{Z}_p)$, and let $X=\mathrm{Spa}(A,A^+)$ be the associated pre-adic space. Let $X_v$ denote the site given by perfectoid spaces over $X$ with covers given by v-covers, and let $\mathcal{O}$ and $\mathcal{O}^+$ be the obvious structure sheaves on $X_v$. Set $\breve{A}^+ = H^0(X_v,\mathcal{O}^+)$ and $\breve{A} = \breve{A}^+ [1/p] = H^0(X_v,\mathcal{O})$, so the association $(A,A^+) \mapsto (\breve{A}, \breve{A}^+)$ is an endofunctor on the category of uniform Tate-Huber pairs over $\mathbf{Q}_p$. One can check that breve’ing twice is the same as breve’ing once, and that the natural map $(A,A^+) \to (\breve{A},\breve{A}^+)$ induces an isomorphism of diamonds. If $A$ is a smooth (or just seminormal) affinoid $K$-algebra for some $K/\mathbf{Q}_p$, or if $A$ is perfectoid, then breve’ing doesn’t change $A$.

Conjecture. Let $(A,A^+)$ be a uniform Tate-Huber pair over $\mathbf{Q}_p$ such that every completed residue field of $\mathrm{Spa}(A,A^+)$ is a perfectoid field. Then $\breve{A}$ is a perfectoid Tate ring.

Aside from disposing of Q2, this conjecture would settle another notorious problem: it would imply that if $A$ is a uniform sheafy Huber ring and $\mathrm{Spa}(A,A^+)$ is a perfectoid space, then $A$ is actually perfectoid.

It may be instructive to see an example of a non-perfectoid (uniform) Tate ring which satisfies the hypothesis of this conjecture. To make an example (with $p>2$), set $A=\mathbf{C}_p \left\langle T^{1/p^\infty} \right\rangle$, and let $B=A[\sqrt{T}]$ with the obvious topology. Set $C=\mathbf{C}_p \left\langle T^{1/2p^\infty} \right\rangle$, so there are natural maps $A \to B \to C$. Then $A$ and $C$ are perfectoid, but $B$ isn’t: the requisite $p$-power roots of $\sqrt{T}$ mod $p$ don’t exist. Nevertheless, every completed residue field of $B$ is perfectoid (exercise!), and the map $B \to C$ induces an isomorphism $\breve B \cong C$.

OK, this bullet point turned out pretty long, but these things have been in my head for the last couple months and it feels good to let them out. Besides, Yoichi Mieda asked me about Q1 during the workshop, so despite the technical nature of these questions, I might not be the only one who cares.

• Oberwolfach continues to be one of the best places in the world to do mathematics.

Thanks to the organizers for putting together such an excellent week!

## Two questions and a story

$\bullet$ Let $f$ be some cuspidal Hecke eigenform, with associated Galois representation $\rho_{f}:G_{\mathbf{Q}}\to \mathrm{GL}_2(\overline{\mathbf{Q}_p})$. A notorious conjecture of Greenberg asserts that if $\rho_{f}|G_{\mathbf{Q}_p}$ is abelian (i.e. is a direct sum of characters), then $f$ is a CM form, or equivalently $\rho_f$ is induced from a character. At some point I was talking about this with Barry Mazur, and he suggested a possible generalization:

Let $\rho:G_{\mathbf{Q}} \to \mathrm{GL}_n(\overline{\mathbf{Q}_p})$ be an irreducible geometric Galois representation. Suppose that $\rho|G_{\mathbf{Q}_p}$ is a direct sum of characters. Is $\rho$ induced from a character?

Emerton found a nice argument which proves Greenberg’s conjecture conditionally on some kind of $p$-adic variational Hodge conjecture. Is there any similar evidence for this question in some higher-dimensional cases?

$\bullet$ Do separated etale maps of schemes satisfy effective descent with respect to fpqc covers? This is known if one restricts to quasi-compact separated etale maps. An analogous result is true for perfectoid spaces.

$\bullet$ Here’s a funny story I heard from Glenn Stevens a while back:

At some point in the early ’90s, before he announced his proof of Fermat, Wiles came to Boston and gave a seminar talk at BU. He spoke about what is now known as the Greenberg-Wiles duality formula. However, he didn’t mention his main motivations for this formula. The upshot is that Stevens came away from the talk with the sad feeling that Wiles had lost his touch.

## Modern fictions

On the website for the journal Algebra and Number Theory, one finds the following remarkable statement:

ANT publishes high-quality articles of interest to a broad readership, at a level surpassing all but the top four or five mathematics journals.

I have no problem with ANT, but shoving this self-assessment down people’s throats strikes me as deeply silly. We can probably all agree that the top five journals are (in some order) JAMS, Annals, Inventiones, Publ. IHES, and Acta Mathematica. So ANT is claiming to surpass Compositio, Duke, Ann. Sci. ENS, Cambridge J. Math, etc. Hmm…

Thanks to [redacted] for pointing this out to me.

## 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.

## The Newton stratification is true

Let $G$ be a connected reductive group over $\mathbf{Q}_p$, and let $\mu$ be a $G$-valued (geometric) conjugacy class of minuscule cocharacters, with reflex field $E$. In their Annals paper, Caraiani and Scholze defined a very interesting stratification of the flag variety $\mathcal{F}\ell_{G,\mu}$ (regarded as an adic space over $E$) into strata $\mathcal{F}\ell_{G,\mu}^{b}$, where $b$ runs over the Kottwitz set $B(G,\mu^{-1})$. Let me roughly recall how this goes: any (geometric) point $x \to \mathcal{F}\ell_{G,\mu}$ determines a canonical modification $\mathcal{E}_x \to \mathcal{E}_{triv}$ of the trivial $G$-bundle on the Fargues-Fontaine curve, meromorphic at $\infty$ and with “mermorphy $\mu$” in the usual sense. On the other hand, Fargues proved that $G$-bundles on the curve are classified up to isomorphism by $B(G)$, and then Caraiani-Scholze and Rapoport proved that $\mu$-meromorphic modifications of the trivial bundle are exactly classified by the subset $B(G,\mu^{-1})$ (CS proved that only these elements can occur; R proved that all of these elements occur). The Newton stratification just records which element of this set parametrizes the bundle $\mathcal{E}_x$.

The individual strata are pretty weird. For example, if $G=GL_n$ and $\mu=(1,0,\dots,0)$, then $\mathcal{F}\ell_{G,\mu} \simeq \mathbf{P}^{n-1}$ and the open stratum is just the usual Drinfeld space $\Omega^{n-1}$, but the other strata are of the form $\Omega^{n-i-1} \times^{P_{n-i,i}(\mathbf{Q}_p)} GL_n(\mathbf{Q}_p)$, where $P_{n-i,i}$ is the evident parabolic in $GL_n$ and the action on $\Omega^{n-i-1}$ is via the natural map $P_{n-i,i}(\mathbf{Q}_p) \twoheadrightarrow GL_{n-i}(\mathbf{Q}_p)$. Qualitatively, this says that they’re unions of profinitely many copies of lower-dimensional Drinfeld spaces. In particular, the non-open strata are not rigid analytic spaces. There are also examples of strata which don’t have any classical rigid analytic points. However, the $\mathcal{F}\ell_{G,\mu}^{b}$‘s are always perfectly well-defined from the topological or diamond point of view.

Anyway, I’m getting to the following thing, which settles a question left open by Caraiani-Scholze.

Theorem. Topologically, the Newton stratification of $\mathcal{F}\ell_{G,\mu}$ is a true stratification: the closure of any stratum is a union of strata.

The idea is as follows. After base-changing from $E$ to the completed maximal unramified extension $E'$ (which is a harmless move), there is a canonical map $\zeta: \mathcal{F}\ell_{G,\mu,E'} \to \mathrm{Bun}_{G}$ sending $x$ to the isomorphism class of $\mathcal{E}_x$. Here $\mathrm{Bun}_{G}$ denotes the stack of $G$-bundles on the Fargues-Fontaine curve, regarded as a stack on the category of perfectoid spaces over $\overline{\mathbf{F}_p}$. This stack is stratified by locally closed substacks $\mathrm{Bun}_{G}^{b}$ defined in the obvious way, and by construction the Newton stratification is just the pullback of this stratification along $\zeta$. Now, by Fargues’s theorem we get an identification $|\mathrm{Bun}_{G}| = B(G)$, so it is completely trivial to see that the stratification of $\mathrm{Bun}_{G}$ is a true stratification (at the level of topological spaces). We then conclude by the following observation:

Proposition. The map $\zeta$ is universally open.

The idea is to observe that $\zeta$ factors as a composition of two maps $\mathcal{F}\ell_{G,\mu,E'} \to [\mathcal{F}\ell_{G,\mu,E'}/\underline{G(\mathbf{Q}_p)}] \to \mathrm{Bun}_{G}$. Here the first map is a $\underline{G(\mathbf{Q}_p)}$-torsor by construction, so it’s universally open by e.g. Lemma 10.13 here. More subtly, the second map is also universally open. Why? Because it is cohomologically smooth in the sense of Definition 23.8 here; universal openness then follows by Proposition 23.11 in the same document.

For the cohomological smoothness claim, take any affinoid perfectoid space with a map $T \to \mathrm{Bun}_{G}$, corresponding to some bundle $\mathcal{F} / \mathcal{X}_T$. After some thought, one works out the fiber product $X = T \times_{\mathrm{Bun}_{G}} [\mathcal{F}\ell_{G,\mu,E'}/\underline{G(\mathbf{Q}_p)}]$ “explicitly”: it parametrizes untilts of $T$ over $E'$ together with isomorphism classes of $\mu^{-1}$-meromorphic modifications $\mathcal{E}\to \mathcal{F}$ supported along the section $T^{\sharp} \to \mathcal{X}_T$ induced by our preferred untilt, with the property that $\mathcal{E}$ is trivial at every geometric point of $T$. Without the final condition, we get a larger functor $X'$ which etale-locally on $T$ is isomorphic to $T \times_{\mathrm{Spd}(\overline{\mathbf{F}_p})} \mathcal{F}\ell_{G,\mu^{-1},E'}^{\lozenge}$. (To get the latter description, note that etale-locally on $T$ we can trivialize $\mathcal{F}$ on the formal completion of the curve along $T^{\sharp}$, and then use Beauville-Laszlo to interpret the remaining data as a suitably restricted modification of the trivial $G$-torsor on $\mathrm{Spec} \mathbf{B}_{dR}^{+}(\mathcal{O}(T^{\sharp}))$. This is a Schubert cell in a Grassmannian. Then use Caraiani-Scholze’s results on the Bialynicki-Birula map.) Anyway anyway, after a little more fiddling around the point is basically that the projection $X' \to T$ is cohomologically smooth because it’s the base change of a smooth map of rigid spaces. By Kedlaya-Liu plus epsilon, the natural map $X \to X'$ is an open immersion, so $X \to T$ is cohomologically smooth. Since $T$ was arbitrary, this is enough.