Nowhere-vanishing sections of vector bundles

Let X/k be a proper variety over some field, and let \mathcal{E} be a vector bundle on X.  The functor of global sections of \mathcal{E}, i.e. the functor sending a scheme f:S \to \mathrm{Spec}\,k to the set H^0(S \times_{k} X, (f \times \mathrm{id})^{\ast} \mathcal{E}), is (representable by) a nice affine k-scheme, namely the scheme \mathcal{S}(\mathcal{E}) = \mathrm{Spec}(\mathrm{Sym}_k H^0(X,\mathcal{E})^{\vee}). Let \mathcal{S}(\mathcal{E})^{\times} \subset \mathcal{S}(\mathcal{E}) denote the subfunctor corresponding to nowhere-vanishing sections s \in H^0(S \times_k X, (f \times \mathrm{id})^{\ast} \mathcal{E}). We’d like this subfunctor to be representable by an open subscheme. How should we prove this?

Let p: \mathcal{S}(\mathcal{E}) \to \mathrm{Spec}\,k be the structure map. The identity map \mathcal{S}(\mathcal{E}) \to \mathcal{S}(\mathcal{E}) corresponds to a universal section s^{\mathrm{univ}} \in H^0(\mathcal{S}(\mathcal{E}) \times_k X, (p \times \mathrm{id})^{\ast}\mathcal{E}). Let Z\subset |\mathcal{S}(\mathcal{E}) \times_k X| denote the zero locus of s^{\mathrm{univ}}. This is a closed subset. But now we observe that the projection \pi: \mathcal{S}(\mathcal{E}) \times_k X \to \mathcal{S}(\mathcal{E}) is proper, hence universally closed, and so |\pi|(Z) is a closed subset of |\mathcal{S}(\mathcal{E})|.  One then checks directly that \mathcal{S}(\mathcal{E})^{\times} is the open subscheme corresponding to the open subset |\mathcal{S}(\mathcal{E})| \smallsetminus |\pi|(Z), so we win.

I guess this sort of thing is child’s play for an experienced algebraic geometer, and indeed it took Johan about 0.026 seconds to suggest that one should try to argue using the universal section.  I only cared about the above problem, though, as a toy model for the same question in the setting of a vector bundle \mathcal{E} over a relative Fargues-Fontaine curve \mathcal{X}_S. In this situation, \mathcal{S}(\mathcal{E}) is a diamond over S^\lozenge, cf. Theorem 22.5 here, but it turns out the above argument still works after some minor changes.

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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.”