flat first order thickenings of projective line

First let me clarify the meaning of this title. In deformation theory, people cares about the situation where you have a flat family of schemes X \to S and a thickening S' of the base and asks whether you can extend the X to a flat family over S'. In this sense, one can show that there is not interesting first order thickening of a projective line. However, in this post, I want to discuss schemes of the following type: it is a finite type proper scheme X \to Spec(k) over a field such that (1) X_{red} \cong \mathbb{P}^1; (2) the ideal sheaf \mathcal{I} of nilpotent functions is square-zero, in particular \mathcal{I} maybe viewed as a coherent sheaf on \mathbb{P}^1, hence the following condition would make sense; (3) lastly, I want \mathcal{I} to be a line bundle on \mathbb{P}^1, justifying the title of this blog. Let’s denote the degree of \mathcal{I} by n (because why not).

These guys arise naturally in algebraic geometry. For example, take a smooth rational curve embedded in some smooth algebraic surface, then take its first order thickening. In this situation, the ideal sheaf \mathcal{I}, defined as in the first paragraph, as a line bundle on \mathbb{P}^1 is just \mathcal{O}(-n) where n is the self intersection number of the rational curve we begin with. In fact, embed \mathbb{P}^1 as zero-section of the total space of \mathcal{O}(-n) shows that any line bundle can show up in this fashion. The examples constructed via line bundles are lame in the sense that they admit projection back to their reduced closed subscheme (aka \mathbb{P}^1).

One can ask several basic questions: can we classify these guys? Do they all arise as a first order thickening of rational curve in a surface? Are there examples which do not admit projection back to \mathbb{P}^1?

In fact, one can classify them quite easily. On X, we have a short exact sequence of sheaves: 0 \to \mathcal{I} \to \mathcal{O}_X \to \mathcal{O}_{\mathbb{P}^1} \to 0. Choose our favorite affine covering of the scheme X = U \cup V, so that on each charts \mathcal{O}_{\mathbb{P}^1} is given by a polynomial algebra. On each piece, we may choose a splitting \mathcal{O}_X \simeq \mathcal{O}_{\mathbb{P}^1}[\epsilon]/(\epsilon^2). On the overlap, we have glueing information given by T = S^{-1} + \eta \cdot f(S) and \epsilon = \eta \cdot S^n, where S & T are the coordinates on each affine line and f(S) is a Laurent polynomial. One can easily see that here n is just the degree of \mathcal{I}. If one changes the choices made above, the Laurent polynomial f will be changed in degrees \geq -2 and \leq n parts (computation omitted). This discussion tells us, for each n there is exactly an \mathbb{A}^{d(n)} family of such X‘s with \mathcal{I} \cong \mathcal{O}_X, where d(n) = max\{ 0, -n-3 \}. The origin corresponds exactly to the X‘s admitting a section back to \mathbb{P}^1.

On the other hand \mathbb{A}^{d(n)} may be viewed as H^1(\mathcal{O}(n+2)) = H^1(\mathcal{O}(n) \otimes T_{\mathbb{P}^1}) = Ext^1(\Omega^1_{\mathbb{P}^1/k}, \mathcal{I}), given an element of the former space is equivalent to giving an X with \mathcal{I} of degree n. How to get the corresponding extension class (as a vector bundle on \mathbb{P}^1)? The answer is to look at the cotangent bundle. We have sequence of maps \mathbb{P}^1 \to X \to Spec(k), giving rise to a short exact sequence: 0 \to \mathcal{I}/\mathcal{I}^2 = \mathcal{I} \to \Omega_{X/k} \otimes_{\mathcal{O}_X} \mathcal{O}_{\mathbb{P}^1} \to \Omega^1_{\mathbb{P}^1/k} \to 0.

Lastly, what about realizability by surfaces? It turns out that these can always be realized as first order thickening of rational curves inside a rational surface! We may assume that n \leq -4. Consider the rational normal curve of degree m in \mathbb{P}^m, its normal bundle is \mathcal{O}(m+2)^{\oplus (m-1)}. There is an \mathbb{A}^{m-1} family of embeddings \mathcal{O}(m+2) into the normal bundle. On every point of the rational normal curve, this normal bundle determines a normal direction, we can take the “union” of these normal lines to get a rational surface on which lies the rational normal curve, then we may form the first order thickening. Let m = -n-2, then we exactly get an \mathbb{A}^{d(n)} family. This must be the family we are looking for, because what else could it be (to put it in my advisor’s wording)? Ok, one may verify this by considering the tangent-normal sequence (I made this up, but you probably should know what I meant).

The reason why I think about all of these was because in Johan’s problem, we were trying to show the first order thickening of a conic in \mathbb{P}^2 doesn’t admit a section back to the conic.

At this point, I think I probably made a mistake. I think the Laurent polynomial can actually be adjusted by a scalar, which means the discussion of moduli above probably should be quotient out by a \mathbb{G}_m-action… Should be more careful with *omitted computations* next time.

Advertisements

Random fun fact with two totally different proofs

Here’s a fun fact: let X be a smooth proper surface over a perfect field k of positive characteristic p > 0, if there is no global 2-form on X, then the Hodge-to-de-Rham spectral sequence (for X) degenerates at E_1 page. Below we will give two proofs that are totally different which builds on many (very) nontrivial results.

Here’s one proof. Look at the conjugate spectral sequence. It starts at E_2 page and is contained in the [0,2] \times [0,2] box. By Serre duality,  both E_2^{0,2} and E_2^{2,0} are zero. Therefore by plain formalism of spectral sequence, it degenerates at E_2 page. Recall that the Hodge-to-de-Rham spectral sequence (for X) degenerates at E_1 page is equivalent to the conjugae spectral sequence degenerates at E_2 page, therefore we are done.

Here’s another proof. Again by Serre duality and the configuration of the spectral sequence, it suffices to show that the differential map H^1(\mathcal{O}_X) \to H^1(\Omega^1_X) is zero. By Serre duality, we see that H^2(\mathcal{O}_X) = 0 which implies that the Picard variety of X is reduced. Therefore the pullback map H^1(\mathrm{Alb}(X), \mathcal{O}) \to H^1(X,\mathcal{O}_X) is an isomorphism, where \mathrm{Alb}(X) is the Albanese variety of X. Since the differentials on any abelian variety is zero, we get what we want.

Maybe I should propose a vote with the question “which proof is better” and let people choose between “the first one”/”the second one”/”neither because you use a steam-hammer to crack nuts, fool!”.

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

totallydisconnected

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…

View original post 1,124 more words

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.

http://math.columbia.edu/~shanbei/non%20simply%20connected

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).

流星小屋

我非老黄的学生,因为老黄的学生都是好学生,都是“黄派中人”,而我只是上过老黄的课和讨论班的蹭课的学生。但在ZJU的岁月中谁给我的影响最大?不是老杨也不是老李,而是老黄。

一直想写点关于老黄的文字,一直拖到今日。在年初看到facebook中居然有老黄的专页,几个月后居然自己成为了专页的管理者,再一问老黄近况,发现老黄已经退休了,想来该写点东西了。

在写自己眼中的老黄之前,先转载别人眼中的老黄。自己也再读一遍。

View original post 204 more words