Why is it not (that) crazy.

Not-fun-fact: de Rham was born in the French-speaking area of switzerland, so I’ve probably called his name wrong for my whole life (should I change it?). Damn it!

This starts as a discussion with a friend. We were discussing a surprising result of Liu–Zhu about rigidity of local systems on a smooth rigid space being de Rham.

Let X be a smooth rigid space over K, a p-adic field. Recall that a \mathbb{Q}_p-local system (which will be denoted as \mathbb{L}) is a sheaf on the pro-étale site of X which (pro-étale-)locally looks like a finite dimensional \mathbb{Q}_p-vector space times X. When X is a point, this recaptures the definition of a p-adic Galois representation (which will be denoted as V) of a p-adic field. There’s a natural generalization of “de Rham representation” which is called a “de Rham \mathbb{Q}_p-local system” (okay, let me not recall anymore since I’m feeling lazy. From now on, let’s pretend the readers know the “modern” p-adic Hodge theory after Scholze. Previous statement is true if there’s no reader at all.). There is an amazing/surprising theorem of Liu–Zhu saying that if such a local system is “de Rham” at one point, then it is in fact de Rham everywhere. Both my friend and I knew this statement, and we had a discussion today for why it is not that crazy.

Recall that one has a morphism between sites: \nu \colon X_{pro\acute{e}t} \to X_{\acute{e}t}. In the representation situation (i.e. X is a point), one knows that there is an injection (V \otimes B_{dR})^{Gal} \otimes B_{dR} \to (V \otimes B_{dR}) and de Rham condition means this is an isomorphism. Now one has that \nu_{*} (\mathbb{L} \otimes \mathcal{O}\mathbb{B}_{dR}) is a filtered coherent sheaf and its fibres will have rank at most that of \mathbb{L}, those points where the fibres having rank equals to that of \mathbb{L} are presumably the “de Rham locus”. Then by semi-continuity, we convince ourselves that “being de Rham” is a (Zariski)-closed condition.

On the other hand, we have \nu^* \nu_* (\mathbb{L} \otimes \mathcal{O}\mathbb{B}_{dR}) \otimes_{\nu^* \mathcal{O}_X}\mathcal{O}\mathbb{B}_{dR} \to \mathbb{L} \otimes \mathcal{O}\mathbb{B}_{dR} which presumably is also injective. If one interprete the former as a sub-vector-bundle inside the latter, if one views \mathcal{O}\mathbb{B}_{dR} as some sort of structure sheaf, then the semi-continuity would tell us the locus where this is an isomorphism is an open condition. Therefore we also convince ourselves that “being de Rham” is a (pro-étale)-open condition.

Of course we didn’t read the proof (and perhaps we should), so what I said here perhaps is completely a BS.

BTW, anyone knows how to type accents in LaTeX in wordpress?


A counterexample


Let $latex C/mathbf{Q}_p$ be a complete algebraically closed nonarchimedean field extension, and let $latex X$ be any proper rigid space over $latex C$. Let $latex mathbf{L}$ be any $latex mathbf{Z}_p$-local system on $latex X_{mathrm{proet}}$. By the main results in Scholze’s p-adic Hodge theory paper, the pro-etale cohomology groups $latex H^i_{mathrm{proet}}(X,mathbf{L})$ are always finitely generated $latex mathbf{Z}_p$-modules. It also seems likely that Poincare duality holds in this setting (and maybe someone has proved this?).

Suppose instead that we’re given a $latex mathbf{Q}_p$-local system $latex mathbf{V}$. By analogy, one might guess that the cohomology groups $latex H^i_{mathrm{proet}}(X,mathbf{V})$ are always finitely generated $latex mathbf{Q}_p$-vector spaces. Indeed, this (and more) was claimed as a theorem by Kedlaya-Liu in a 2016 preprint. However, it is false. The goal of this post is to work out an explicit counterexample.

So, consider $latex X=mathbf{P}^1$ as a rigid space over $latex C$. This is the target of the…

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produce primitive p-th root of unity

Dwork is one of the mathematicians that I admire so much.

Recently I learned how he produced primitive p-th roots of unity in \mathbb{C}_p “by hand”, which is interesting enough for me to write it down here.

Let \pi \in \mathbb{C}_p satisfies \pi^{p-1} = -p. Consider the following function \theta(z) = Exp(\pi z - \pi z^p). Then after plugging \{1, \zeta, \zeta^2, \ldots, \zeta^{p-2}\} in \theta(z), we will miraculously get all the primitive p-th roots of unity in \mathbb{C}_p. Here \zeta is a primitive (p-1)-st root of unity. The function \theta(z) bears the name “Dwork’s exponential”, I believe.

To prove the above statement, it suffices to prove it for z=1. So we need to prove \theta(1)^p = 1 yet \theta(1) \not= 1.

One thing to notice at the first place is that the convergence range of exponential function is the open disc of radius |\pi|, and the polynomial \pi z - \pi z^p has norm exactly |\pi| on the closed disc of radius 1. So the value \theta(1) cannot be computed by plugging 1 into the polynomial, which is 0, and then plug the output in the exponential function.

However, \theta(z)^p = Exp(\pi z - \pi z^p)^p = Exp(p\pi z - p \pi z^p). Now the polynomial p (\pi z - \pi z^p) has smaller norm on the closed disc, therefore we may compute the value at 1 by means in the previous paragraph. Hence we see that \theta(1)^p = 1.

Now we need to see that \theta(1) \not= 1. In the following we will show that \theta(1) - 1 = \pi \mod(\pi^2). To achieve this, let us rewrite \theta(z) = E_p(z) F(z). Here E_p(z) = Exp(x + x^p/p + x^{p^2}/p^2 + \ldots), and F(z) = Exp(-x^{p^2}/p^2 - x^{p^3}/p^3 - \ldots), where x = \pi z. The function E_p(z) is called the Artin-Hasse exponential. There’s a lemma of Dwork telling us that E_p(z) \in \mathbb{Z}_p[[\pi z]], hence E_p(1) = 1 + \pi \mod (\pi^2). On the other hand it’s easy to show directly that F(1) = 1 \mod(\pi^2). Put everything together, we see that \theta(1) - 1 = (E_p(1) F(1)) - 1 = (1+\pi) - 1 = \pi \mod(\pi^2) which is what we want to show.

In fact, before doing everything we probably have to show the convergence radius of \theta(z) is strictly bigger than 1. This actually easily follows from the estimate of E_p(z) \text{ and } F(z).

From the above construction, we see that there is a bijection between sets from (p-1)-st roots of -p and primitive p-th roots of unity. This is of course well known, since if we let w = \zeta_p - 1, then we’ll have w^{p-1} = -(p + \binom{p}{2} w + \ldots + p w^{p-2}).

Why are they NOT simply connected?

OK, it’s been so long since I last posted anything here.


This is an expository notes for myself. In the beginning of this semester, I asked myself to give a talk on Johan’s paper in his seminar. It’s really fun, and I focused on the “easiest” case of it. Things are already very interesting at this stage. Hope you enjoy it as half (for otherwise you would be too excited to sleep today, hahaha) as I do.

My research somehow stucks, therefore I did this activity. Now I shall go back and drill on something else.

记一个理想主义者–心中永远的坐标–黄兆镇 The Most dedicated and influential math teacher in Zhejiang Univ.

Brings me to the old days when I was an silly undergrad (am still silly though).





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An ongoing write-up

Johan’s question seminar

Check out my ongoing write-up for an informal seminar. The theme of this seminar is that participants ask some little easy question which is somewhat fun and then we vote meet once a week to solve one or two of them. The reason to do so is to “make sure everyone get out of their bed once a week”.

I didn’t write down all the discussions of questions, due to my lack of attention, patience and time. Here are the questions of which I’ve written down some discussions:

(1) Given an étale map between varieties with constant number of fibers, is it necessarily a finite étale map?

(2) Is a quotient of locally ringed space by a finite group action always a locally ringed space?

(3) Why is there no section from PGL_n to GL_n?

(4) Is it true that any small perturbation of a diagonalizable matrix in GL_n(Z_p) with distinct eigenvalues diagonalizable? (Note that here diagonalization should happen within Z_p)

(5) How many lines of the 27 on a REAL cubic surface real (what are the possibilities)?

In the following weeks I will keep writing down further discussions of other fun little questions (hopefully), so check out the link above later~

A random thought

“I think a blessing/curse of our professions is that the lab is in our mind.” I don’t know who said that, but he’s damn right…

Today I want to talk about a wrong belief I held for the last 2 years. I thought that a variety X over k is projective if and only if each of it’s irreducible components is projective until very recently. I held the wrong belief because the following exercise in Hartshorne’s famous <Algebraic Geometry>: Let X be a scheme over k, a line bundle on X is ample if and only if the restriction on each irreducible components is ample. An problem would be that even on each irreducible component we have an ample line bundle we can not always patch them together (although we can always patch in 1-dimensional case). I came to the following (counter)example in my research.

Let’s take two pieces of P^2 blow up at one point, call them X and Y. There is a ordinary(a usual line not passing thru the point we blown up) line and an exceptional divisor on this blow up of P^2. Let’s identify the exceptional divisor on X with an ordinary line on Y  and identify the exceptional divisor on Y with an ordinary line on X. Call the result variety Z, which has two projective irreducible components. One can easily see that any two ample line bundles on X and Y can never be patched together (leave as an exercise to reader(s?)). Does anyone besides me had the same wrong belief before reading this post (Does anyone actually read this post?)? Or is it just me?

This example isn’t coming from nowhere, it is actually a reduction of some special non-archimedean Hopf surface. After a tedious computation one can show some reductions of some more general non-archimedean Hopf surfaces, whose irreducible components are blow up of Hirzebruch surfaces (toric surfaces), are also non-projective.