# Yet another example

Some while ago, I learned that one can have a pair of unramified liftings of a finite group representation in positive characteristic with the property that their geometric generic fibres are different.

I started recording this example, but got constantly interrupted by all these graduation -and-travel-and-moving nonsenseseseses. Finally, I finished the writing, which is only 2 pages… And here you are.

My original motivation of finding this kind of example was to hope that in the end I can produce a geometric example of similar flavor, i.e. different unramified liftings. Anyways, I couldn’t quite make it to work. Hopefully I will come back on this.

# an example I learned yestoday

Okay, yestoday is officially a word that stands for yesterday and today.

Anyways, I learned this cool example at this year’s AWS. It is the work of a project group associated with Matthew Morrow.

First off, some background. We all know that a ring homomorphism $A \to B$ is called formally étale if the infinitesimal lifting criterion is satisfied (see https://stacks.math.columbia.edu/tag/00UQ). Note that in the criterion, you are testing liftability against *first order* thickenings. You may wonder about the following: what if we replace the condition of $I$ being square zero by asking $I$ being locally nilpotent, that is we only require every element of $I$ being nilpotent without having a bound of the nilpotency. For an obvious reason, let’s call the corresponding morphism *being ln-formally étale*.

Two people in Morrow’s project group produced the following example: for any algebra $A$, there is an $A$ algebra $B$ with a splitting surjection $f \colon B \to A$ such that $f$ is ln-formally étale, but NOT flat. In fact, later on we will also see that the cotangent complex of $f$ vanishes as well.

Now let me get to the construction. Consider the ordered group $\Lambda := \mathbb{Z}[1/2]^{\mathbb{N}}$, where we put the lexicographical order. Just to make sure that we are on the same page: $(1,0,0,\ldots) > (0,1,0,\ldots)$. Then let us consider the group algebra $A[\Lambda]$. There is a natural valuation on this algebra sending $[\lambda] \to \lambda$. Take the “ring of integers” of this valuation, which is our $B$. This is a valuation $A$-algebra, with a splitting map back to $A$ (quotienting all the $[\lambda]$‘s). Okay, the claim is that this morphism is ln-formally étale and non-flat. To see the former, just note that $(1,0,0,\ldots) = 2^n \cdot (1/2^n,-1,0,\ldots) + 2^n \cdot (0,1,0,\ldots)$ and “shift this equality to the right” and stare at the diagram long enough. To see the later, we need to show the kernel of $f$ is not a pure ideal (see https://stacks.math.columbia.edu/tag/04PQ). And it is again an easy exercise to show that the ideal $([(1,0,0,\ldots)])$ is NOT equal to the kernel times $([(1,0,0,\ldots)])$ by using the valuation we start with.

We can also see that the cotangent complex $\mathbb{L}_{A/B} \simeq 0$. To see this, use the triangle for $A \to B \to A$, along with the following facts:

(0) $\mathbb{L}_{A/A} \simeq 0$;

(1) $B$ is a filtered colimit of smooth $A$-algebras, hence $\mathbb{L}_{B/A}$ is qis to a flat $B$-module sitting in degree 0 and;

(2) $B \to A$ is a surjection with the kernel squaring to itself, hence $\mathbb{L}_{A/B}$ is supported in degree at most $-2$.

Cool example, huh?

Here is one thing that I’ve been thinking about (with other students in our team) during this winter school (which is another student’s project given by Morrow at this winter school). Say $R$ is a local Artinian $\mathbb{Q}$-algebra $R$ with residue field $\mathbb{Q}$. Take any nonzero nilpotent $f \in R$. Is it true that $df \not= 0 \in \Omega^1_{R/\mathbb{Q}}$? More generally, is it true that the de Rham complex of $R$ over $\mathbb{Q}$ is quasi-isomorphic to $\mathbb{Q}[0]$?

# a brain teaser

Consider the group $G = PSL_2(\mathbb{F}_5)$, by which I mean quotient of $SL_2(\mathbb{F}_5)$ by its center. So this group has order $60$. If an interested reader looks at the classification of groups of order $60$:”apparently”, says this reader, “this group is isomorphic to $A_5$!”

Now $A_5$ naturally acts on a set of 5 elements. On the other hand, $G$ naturally acts on a set of 6 elements, namely the $\mathbb{F}_5$ points of $\mathbb{P}^1$.

The brain teaser is: how to find a natural action of $A_5$ on a set of 6 elements, and how to find a natural action of $G$ on a set of 5 elements? I couldn’t find these in a nice way. For instance, we can let $G$ act on (the projectivization of) degree 4 homogeneous polynomials (with coefficients in $\overline{\mathbb{F}_5}$). The orbit of $x^4 - 2\sqrt{2} x^2 y^2 + y^4$ consists of $5$ elements (okay, let me not prove this here). So this must be it. But this is not satisfying. So the brain teaser is really: how can you realize these actions in a nice way?

# Good reduction or not, that is the question

Ok, admittedly this title is not the best one that I can come up with. I just read (small part of) this paper. The authors constructed a family of degenerating Calabi–Yau 3-folds (over a punctured disc), with trivial monodromy action and have a (potentially) semistable reduction which is not a smooth filling.

If I understand it correctly, the authors believe that this family has no (potentially) good reduction. Namely, however we pull this family back along maps from another punctured disc to the current one given by “$z^n$“, it won’t admit a smooth filling.

Now the (philosophical) question is, how can we tell if a smooth variety over the fraction field of a DVR does NOT have a smooth integral model over that DVR?

I know this (kind of fake) example: if we take a non-archimedean Hopf surface, then its étale cohomology groups are unramified (if you use l-adic coefficients) or Crystalline (if you use p-adic coefficients). But it doesn’t have a smooth reduction. Why? Because the weight is wrong. Ok, so this example is lame.

So I think there is a task I can give myself: to find a smooth projective variety over a non-archimedean field such that (1) its étale cohomology groups are “good reduction” meaning unramified/Crystalline of the correct weight and, (2) it does not admit any potential good reduction. I assume it would be easy to find a potential example, like the example discussed in the above link, but it perhaps is quite difficult to prove that such a thing is an actual example.

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Update on Dec. 8th: Bhargav reminded me that a curve of compact type is one such example. Meanwhile Daniel said that one can detect (this example) by looking at the monodromy action on the (algebraic) fundamental group of this curve and decided it doesn’t have (potential) good reduction. On the other hand, Dingxin pointed the existence of this paper of Robert Friedman. This paper exhibits an example of degenerating family of quintic surfaces over a punctured disk, which doesn’t have a (potential) good reduction, with the property that a finite power of the monodromy action is homotopic to identity!

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In memory of Michael Zhao.

# Explicit surfaces in characteristic p with non-reduced Picard.

In this notes of Jesse Kass, an example due to Igusa is discussed. One takes an ordinary elliptic curve $E$ over characteristic $2$ with an actual $2$-torsion point $P$, and form the quotient of $E \times E'$ by a (fixed point free) action of $\mathbb{Z}/2 = \langle \sigma \rangle$ where the automorphism $\sigma$ acts by $\sigma(x,y) = (x+P,-y)$ and here $E'$ is another elliptic curve (could be the same as $E$). The point is that this quotient is then a smooth projective surface in characteristic $2$ with non-reduced Picard scheme.

One can simply make similar examples in other characteristics too. For instance, in characteristic $3$, take ordinary $E$ with an actual $3$-torsion point. Then take $E'$ to be the supersingular elliptic curve with an automorphism by “$\omega$“. Do the similar construction above, and run the same argument, you can prove that the Picard scheme has tangent space of dimension $2$ whereas the Albanese has dimension $1$ (use $\omega^2 + \omega + 1 = 0$).

As for characteristic $p \geq 5$, take ordinary $E$ with an actual $p$-torsion as usual. Then let $C$ be the hyperelliptic curve given by $y^2 = x^p - x$, which admits the automorphism which fixes $y$ and sends $x$ to $x+1$. This is an automorphism of order $p$. So do the similar construction, by Hochschild–Serre spectral sequence, you will find that the tangent space of the Picard scheme is at least $2$. However, its Albanese has dimension $1$ again, this is because that our automorphism acts on $H^1_{crys}(C)[\frac{1}{p}]$ via those non-trivial $p-1$ characters of order $p$, hence there is no invariants.

So I think these are concrete examples of elliptic surfaces with wild fibres, as discussed in Liedtke’s notes.

# 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.