Hard to believe

In this brief post, I want to draw attention to an amazing theorem which deserves to be well-known.  Probably many readers are familiar with Nagata’s compactification theorem: if S is any qcqs scheme and f: X \to S is a separated morphism of finite type, then f can be factored as an open embedding followed by a proper morphism. This is a by-now-classical result, and has many applications.

Less well-known, however, is the following result of Temkin (cf. Theorem 1.1.3 here):

Temkin’s Factorization Theorem. Let f:X \to Y be any separated morphism of qcqs schemes. Then f can be factored as an affine morphism followed by a proper morphism.

Telling other people about this theorem is an amusing experience. Invariably, their first reaction is that it simply cannot be true, and that the inclusion map i: \mathbf{A}^2 - \{ 0,0 \} \to \mathbf{A}^2 should give a counterexample. But then they realize (or I point out) that i can be factored as p \circ j, where p: X \to \mathbf{A}^2 is the blowup of \mathbf{A}^2 at the origin and j is the natural (affine!) open immersion of \mathbf{A}^2 - \{ 0,0 \} into X. Then they are convinced.

Unrelated: JW pointed out to me that I am now a professional writer of appendices. Maybe this should worry me?

Some historical snippets

Probably everyone is familiar with the MacTutor History of Mathematics website. While browsing through their additional material, I came across some wonderful things:

  • Hardy writing to Veblen from Princeton, ca. 1928:
    “…However, I suppose my present passion for the soda fountain will abate by degrees.” … “I do find myself regretting that Wiener’s not here: but no doubt if he were I should very quickly revise my opinion.”
  • Pedoe on Hodge: “One fine morning Hodge and I were inside the grounds of Pembroke College when we met J A Todd, an excellent geometer, the author of a fine textbook on projective geometry, a University lecturer – and a pipe smoker who spent more time striking matches than actually smoking. As we stood talking, Todd struck match after match and dropped them on the ground at Hodge’s feet. Hodge, as the Acting Bursar, was responsible for the proper maintenance of the grounds of Pembroke, so as Todd dropped each match, Hodge bent down to pick it up. Todd, who wore eyeglasses with strong lenses, was completely unaware of what was going on. The spectacle of the very thin Todd unconsciously dropping matches, and the rotund Hodge bending down every few seconds – while becoming more and more exasperated – is one I shall never forget.”

    “Hodge became Master of Pembroke and President of the Royal Society. He was very shrewd and usually tactful, but had definite ideas on certain matters. He thought, for example, that a mathematical paper should be just that, with no embellishment. When Patrick Du Val, a contemporary of Donald Coxeter, a good geometer, and a lover of the arts, submitted a paper to the Cambridge Philosophical Society for publication, with a quotation from Dante following its title, Hodge suggested that this was not “appropriate.” He was badly flustered when a furious Du Val withdrew the paper.”
  • Schwartz on Maurice Audin’s thesis
  • Hermann Weyl’s speech at Emmy Noether’s funeral
  • Hardy again, this time on Waring’s problem
  • Dedekind attending a lecture course by Gauss in the winter of 1850: “…The lecture room, separated from Gauss’ office by an anteroom, was quite small. We sat at a table which had room for three people comfortably at each side, but not for four. Gauss sat opposite the door at the top end, at a reasonable distance from the table, and when we were all present, the two who came in last had to sit quite close to him with their notebooks on their laps. Gauss wore a lightweight black cap, a rather long brown coat and grey trousers. He usually sat in a comfortable attitude, looking down, slightly stooped, with his hands folded above his lap. He spoke quite freely, very clearly, simply and plainly; but when he wanted to emphasise a new point of view, for which he used a particularly characteristic word, then he would raise his head, turn to one of those sitting beside him, and gazed at the student with his beautiful, penetrating blue eyes during his emphatic speech. That was unforgettable. …”
  • Thue on mathematics in Berlin in 1891-92: “Fuchs, whom I heard lecture on analytical mechanics, did not at first make much of an impression on me. The concepts he employed were, as far as I could see, surrounded by a mist of vagueness. When I heard him in a seminar, however, I got a strong presentiment that he can excel when he wants to do so. He lectures with his eyes shut and looks thoroughly tired and peevish. He can also be rather absent-minded. I remember how he was once talking about differentials, and quite unconsciously he picked up a handful of bits of chalk which he waved in illustration before our wondering eyes. Afterwards he carefully laid his differentials down again on his desk, with his eyes still closed. Professor Fuchs, like Kronecker, is a very prepossessing man, but not overly talkative. I was at a ball at his home this winter. It was a delightful affair. We danced so energetically that the floor cracked in a couple of places.”

    “The mathematical seminar down here functions in much the same way as yours does in Oslo. It is an established university institution. Fuchs and Kronecker preside in turn. Meetings are held between 5 and 7. No report is circulated. I have requested Kronecker to permit my highly attractive voice to be heard at the aforementioned place, but so far he hasn’t paid any attention. …”

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.