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.


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