## Three things I learned from colleagues this semester

1 (from Patrick Allen) Let $F$ be a number field, and let $\pi$ be a cohomological cuspidal automorphic representation of some $\mathrm{GL}_n(\mathbf{A}_F)$.  Suppose that $\rho_\pi : G_{F,S} \to \mathrm{GL}_n(\overline{\mathbf{Q}_p})$ exists and satisfies local-global compatibility at all places, and that $H^1_f(F, \mathrm{ad}\,\rho_\pi) = 0$ as predicted by Bloch-Kato.  Then the following are equivalent:

a) $H^2(G_{F,S}, \mathrm{ad}\,\rho_\pi) = 0$, as predicted by Jannsen’s conjecture;

b) $H^1(G_{F,S}, \mathrm{ad}\,\rho_\pi)$ has the right dimension;

c) The product of restriction maps $H^1_f(F,\mathrm{ad}\,\rho_\pi(1))\to \prod_{v|p} H^1_f(F_v,\mathrm{ad}\,\rho_\pi(1))$ is injective.

The equivalence of a) and b) follows from Tate’s global Euler characteristic formula, but their equivalence with c) was news to me.  The question of whether or not c) holds came up incidentally in my work with Jack on Venkatesh’s conjecture, so it was very pleasing to learn that it follows from Bloch-Kato + Jannsen.

2 (from Keerthi Madapusi Pera) If $\mathbf{G} / \mathbf{Q}_p$ is semisimple and simply connected, and isotropic (i.e. contains some $\mathbf{Q}_p$-split torus), then $\mathbf{G}(\mathbf{Q}_p)$ has no proper finite-index subgroups.

3 (from Stefan Patrikis) Let $\pi$ be as in 1) again. There are two number fields naturally associated with $\pi$ (besides $F$): the field $\mathbf{Q}(\pi)$ generated by its Hecke eigenvalues, and the “reflex field” $E\subseteq F$ of its cohomological weight.  Is there any chance that $E$ is always a subfield of $\mathbf{Q}(\pi)$?, I asked SP.  Yes, said he.

## Artin vanishing is false in rigid geometry

Sorry for the lack of blogging.  It’s been a busy semester.

Let $k$ be an algebraically closed field, and let $X$ be a $d$-dimensional affine variety over $k$.  According to a famous theorem of Artin (Corollaire XIV.3.5 in SGA 4 vol. 3), the etale cohomology groups $H^i_{\mathrm{et}}(X,G)$ vanish for any $i > d$ and any torsion abelian sheaf $G$ on $X_{\mathrm{et}}$. This is a pretty useful result.

It’s natural to ask if there’s an analogous result in rigid geometry.  More precisely, fix a complete algebraically closed extension $k / \mathbf{Q}_p$ and a $d$-dimensional affinoid rigid space $X=\mathrm{Spa}(A,A^\circ)$ over $k$.  Is it true that $H^i_{\mathrm{et}}(X,G)$ vanishes for (say) any $i>d$ and any $\mathbf{Z}/n\mathbf{Z}$-sheaf $G$ on $X_{\mathrm{et}}$ for $n$ prime to $p$?

I spent some time trying to prove this before realizing that it fails quite badly.  Indeed, there are already counterexamples in the case where $X=\mathrm{Spa}(k \langle T_1,\dots,T_d \rangle,k^\circ \langle T_1, \dots, T_d \rangle)$ is the $d$-variable affinoid disk over $k$.  To make a counterexample in this case, let $Y$ be the interior of the (closed, in the adic world) subset of $X$ defined by the inequalities $|T_i| < |p|$ for all $i$; more colloquially, $Y$ is just the adic space associated to the open subdisk of (poly)radius $1/p$. Let $j: Y \to X$ be the natural inclusion.  I claim that $G = j_! \mathbf{Z}/n\mathbf{Z}$ is then a counterexample.  This follows from the fact that $H^i_{\mathrm{et}}(X,G)$ is naturally isomorphic to $H^i_{\mathrm{et},c}(Y,\mathbf{Z}/n\mathbf{Z})$, together with the nonvanishing of the latter group in degree $i = 2d$.

Note that although I formulated this in the language of adic spaces, the sheaf $G$ is overconvergent, and so this example descends to the Berkovich world thanks to the material in Chapter 8 of Huber’s book.

It does seem possible, though, that Artin vanishing might hold in the rigid world if we restrict our attention to sheaves which are Zariski-constructible.  As some (very) weak evidence in this direction, I managed to check that $H^2_{\mathrm{et}}(X,\mathbf{Z}/n \mathbf{Z})$ vanishes for any one-dimensional affinoid rigid space $X$.  (This is presumably well-known to experts.)

## Riemann-Roch sur la courbe

Let $C/\mathbf{Q}_p$ be a complete algebraically closed extension, and let $X = X_{C^\flat}$ be the Fargues-Fontaine curve associated with $C^\flat$.  If $\mathcal{E}$ is any vector bundle on $X$, the cohomology groups $H^i(X,\mathcal{E})$ vanish for all $i>1$ and are naturally Banach-Colmez Spaces for $i=0,1$.  Recall that the latter things are roughly “finite-dimensional $C$-vector spaces up to finite-dimensional $\mathbf{Q}_p$-vector spaces”. By a hard and wonderful theorem of Colmez, these Spaces form an abelian category, and they have a well-defined Dimension valued in $\mathbf{N} \times \mathbf{Z}$ which is (componentwise-) additive in short exact sequences.  The Dimension roughly records the $C$-dimension and the $\mathbf{Q}_p$-dimension, respectively.  Typical examples are $H^0(X, \mathcal{O}(1)) = B_{\mathrm{crys}}^{+,\varphi = p}$, which has Dimension $(1,1)$, and $H^1(X,\mathcal{O}(-1)) = C/\mathbf{Q}_p$, which has Dimension $(1,-1)$.

Here I want to record the following beautiful Riemann-Roch formula.

Theorem. If $\mathcal{E}$ is any vector bundle on $X$, then $\mathrm{Dim}\,H^0(X,\mathcal{E}) - \mathrm{Dim}\,H^1(X,\mathcal{E}) = (\mathrm{deg}(\mathcal{E}), \mathrm{rk}(\mathcal{E}))$.

One can prove this by induction on the rank of $\mathcal{E}$, reducing to line bundles; the latter were classified by Fargues-Fontaine, and one concludes by an explicit calculation in that case.  In particular, the proof doesn’t require the full classification of bundles.

So cool!

## 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.”

## What does an inadmissible locus look like?

Let $H/ \overline{\mathbf{F}_p}$ be some p-divisible group of dimension d and height h, and let $\mathcal{M}$ be the rigid generic fiber (over $\mathrm{Spa}\,\breve{\mathbf{Q}}_p$) of the associated Rapoport-Zink space. This comes with its Grothendieck-Messing period map $\pi: \mathcal{M} \to \mathrm{Gr}(d,h)$, where $\mathrm{Gr}(d,h)$ is the rigid analytic Grassmannian paramatrizing rank d quotients of the (covariant) rational Dieudonne module $M(H) /\breve{\mathbf{Q}}_p$. Note that $\mathrm{Gr}(d,h)$ is a very nice space: it’s a smooth connected homogeneous rigid analytic variety, of dimension d(h-d).

The morphism $\pi$ is etale and partially proper (i.e. without boundary in Berkovich’s sense), and so the image of $\pi$ is an open and partially proper subspace* of the Grassmannian, which is usually known as the admissible locus. Let’s denote this locus by $\mathrm{Gr}(d,h)^a$. The structure of the admissible locus is understood in very few cases, and getting a handle on it more generally is a famous and difficult problem first raised by Grothendieck (cf. the Remarques on p. 435 of his 1970 ICM article). About all we know so far is the following:

• When d=1 (so $\mathrm{Gr}(d,h) = \mathbf{P}^{h-1}$) and $H$ is connected, we’re in the much-studied Lubin-Tate situation. Here, Gross and Hopkins famously proved that $\pi$ is surjective, not just on classical rigid points but on all adic points, so $\mathrm{Gr}(d,h)^a = \mathrm{Gr}(d,h)$ is the whole space. This case (along with the “dual” case where h>2,d=h-1) turns out to be the only case where $\mathrm{Gr}(d,h)^a = \mathrm{Gr}(d,h)$, cf. Rapoport’s appendix to Scholze’s paper on the Lubin-Tate tower.
• When $H \simeq \mathbf{G}_m^{d} \oplus (\mathbf{Q}_p/\mathbf{Z}_p)^{h-d}$, i.e. when $H$ has no bi-infinitesimal component, it turns out that $\mathrm{Gr}(d,h)^a = \mathbf{A}^{d(h-d)}$ is isomorphic to rigid analytic affine space of the appropriate dimension, and can be identified with the open Bruhat cell inside $\mathrm{Gr}(d,h)$. This goes back to Dwork, who proved it when d=1,h=2. (I don’t know a citation for the general result, but presumably for arbitrary d,h this is morally due to Serre-Tate/Katz?)
• In general there’s also the so-called weakly admissible locus $\mathrm{Gr}(d,h)^{wa} \subset \mathrm{Gr}(d,h)$, which contains the admissible locus and is defined in some fairly explicit way. It’s also characterized as the maximal admissible open subset of the Grassmannian with the same classical points as the admissible locus. In the classical rigid language, the map $\mathrm{Gr}(d,h)^a \to \mathrm{Gr}(d,h)^{wa}$ is etale and bijective; this is the terminology used e.g. in Rapoport-Zink’s book.
• In general, the admissible and weakly admissible loci are very different.  For example, when $H$ is isoclinic and (d,h)=1 (i.e. when $M(H)$ is irreducible as a $\varphi$-module), $\mathrm{Gr}(d,h)^a$ contains every classical point, and $\mathrm{Gr}(d,h)^{wa} = \mathrm{Gr}(d,h)$, so the weakly admissible locus tells you zilch about the admissible locus in this situation (and they really are different for any $1 < d < h-1$).

That’s about it for general results.

To go further, let’s switch our perspective a little. Since $\mathrm{Gr}(d,h)^a$ is an open and partially proper subspace of $\mathrm{Gr}(d,h)$, the subset $|\mathrm{Gr}(d,h)^a| \subseteq |\mathrm{Gr}(d,h)|$ is open and specializing, so its complement is closed and generalizing.  Now, according to a very general theorem of Scholze, namely Theorem 2.42 here (for future readers, in case the numbering there changes: it’s the main theorem in the section entitled “The miracle theorems”), if $\mathcal{D}$ is any diamond and $E \subset |\mathcal{D}|$ is any locally closed generalizing subset, there is a functorially associated subdiamond $\mathcal{E} \subset \mathcal{D}$ with $|\mathcal{E}| = E$ inside $|\mathcal{D}|$. More colloquially, one can “diamondize” any locally closed generalizing subset of $|\mathcal{D}|$, just as any locally closed subspace of $|X|$ for a scheme $X$ comes from a unique (reduced) subscheme of $X$.

Definition. The inadmissible/nonadmissible locus $\mathrm{Gr}(d,h)^{na}$ is the subdiamond of $\mathrm{Gr}(d,h)^{\lozenge}$ obtained by diamondizing the topological complement of the admissible locus, i.e. by diamondizing the closed generalizing subset $|\mathrm{Gr}(d,h)^a|^c \subset |\mathrm{Gr}(d,h)| \cong |\mathrm{Gr}(d,h)^{\lozenge}|$.

It turns out that one can actually get a handle on $\mathrm{Gr}(d,h)^{na}$ in a bunch of cases!  This grew out of some conversations with Jared Weinstein – back in April, Jared raised the question of understanding the inadmissible locus in a certain particular period domain for $\mathrm{GL}_2$ with non-minuscule Hodge numbers, and we managed to describe it completely in that case (see link below). Last night, though, I realized we hadn’t worked out any interesting examples in the minuscule (i.e. p-divisible group) setting! Here I want to record two such examples, hot off my blackboard, one simple and one delightfully bizarre.

Example 1. Take h=4, d=2 and $H$ isoclinic. Then $|\mathrm{Gr}(d,h)^a|^c$ is a single classical point, corresponding to the unique filtration on $M(H)$ with Hodge numbers $0,0,1,1$ which is not weakly admissible. So $\mathrm{Gr}(d,h)^a = \mathrm{Gr}(d,h)^{wa}$ in this case.

Example 2. Take h=5, d=2 and $H$ isoclinic\$.  Now things are much stranger.  Are you ready?
Theorem. In this case, the locus $\mathrm{Gr}^{na}$ is naturally isomorphic to the diamond $(X \smallsetminus 0)^{\lozenge} / \underline{D^\times}$, where $X$ is an open perfectoid unit disk in one variable over $\breve{\mathbf{Q}}_p$ and $D=D_{1/3}$ is the division algebra over $\mathbf{Q}_p$ with invariant 1/3, acting freely on $X \smallsetminus 0$ in a certain natural way. Precisely, the disk $X$ arises as the universal cover of the connected p-divisible group of dimension 1 and height 15, and its natural $D$-action comes from the natural $D_{1/15}$-action on $X$ via the map $D_{1/3} \to D_{1/3} \otimes D_{-2/5} \simeq D_{-1/15} \simeq D_{1/15}^{op}$.

This explicit description is actually equivariant for the $D_{2/5}$-actions on $X$ and $Gr$. As far as diamonds go, $(X \smallsetminus 0)^{\lozenge}/\underline{D^{\times}}$ is pretty high-carat: it’s spatial (roughly, its qcqs with lots of qcqs open subdiamonds), and its structure morphism to $\mathrm{Spd}\,\breve{\mathbf{Q}}_p$ is separated, smooth, quasicompact, and partially proper in the appropriate senses. Smoothness, in particular, is meant in the sense of Definition 6.1 here (cf. also the discussion in Section 4.3 here). So even though this beast doesn’t have any points over any finite extension of $\breve{\mathbf{Q}}_p$, it’s still morally a diamondly version of a smooth projective curve!

The example Jared and I had originally worked out is recorded in section 5.5 here. The reader may wish to try adapting our argument from that situation to the cases mentioned above – this is a great exercise in actually using the classification of vector bundles on the Fargues-Fontaine curve in a hands-on calculation.

Anyway, here’s a picture of $(X \smallsetminus 0)^{\lozenge} / \underline{D^{\times}}$, with some other inadmissible loci in the background:

*All rigid spaces here and throughout the post are viewed as adic spaces: in the classical language, $\mathrm{Gr}(d,h)^a$ does not generally correspond to an admissible open subset of $\mathrm{Gr}(d,h)$, so one would be forced to say that there exists a rigid space $\mathrm{Gr}(d,h)^a$ together with an etale monomorphism $\mathrm{Gr}(d,h)^a \to \mathrm{Gr}(d,h)$. But in the adic world it really is a subspace.

## Arithmetische geometrie

Just attended a week-long meeting at Oberwolfach on arithmetic geometry.

• “So did you do this computation like Gauss, or did you use a computer?” – Gabber to Katz

• “Let the indices work it out themselves!” – Janssen

• “Shouwu, either you’re going to answer my question, or I’m going to hand you over to Ofer!” – Kisin

• Katz (telling a story at the beginning of his talk): “… So anyway, after Spencer returned to Princeton, this is how he described the math department at Stanford [where he had just been a professor for a couple years]: ‘At Stanford, they’re still studying the topology of the unit disk!’ ”
Conrad (from the audience): “Those days are over.”

• “We use what I wrote.” – Janssen reassuring Gabber

• “So Peter, why did you turn down the breakthrough prize? [pause] I’m only asking because I’m drunk!”

• Anon.: “So Ofer, do you come here much?”
Gabber: [looks down at table, silently moves his finger across it in stepwise motion for 30 seconds] “Seventeen times.”

• Two common referees for technical papers on Shimura varieties: Frobenius and Verschiebung.

• Me (after writing down the “new” definition of a diamond): “Is that OK, Peter?”
Scholze (from the back row): “Looks good!”

• Zhang: “So Mochizuki is like the Buddha.  He writes his ideas.  He is satisfied.  If you want to understand them, you visit him, you ask him questions, he gives you a little idea, you go away and study.  You have to be a monk.  Have a monk’s approach.”
Anon.: “Unfortunately, there aren’t very many good monks.”

• A “symplectic lifting whatever shit”. Apparently they’re defined in Kai-Wen Lan’s thesis?

• Gabber was NOT happy when he heard about Mochizuki’s Gaussian integral analogy.

• While eating the horrible bread casserole thing, which Kedlaya, Lieblich and I had mangled pretty badly while serving ourselves:
Lieblich:”What is this supposed to BE?”
Kedlaya: “Some kind of croque madame?”
Nizioł: “Yes, a croque madame.  But I think you guys croqued it.”