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In section III.3.4 of Eisenbud & Harris's "The Geometry of Schemes," we/they construct an infinite family of double structures on $\mathbb{P}^1 \subset \mathbb{P}^3$ that are distinguished from each other by their genus. Here is the construction (let's just worry about $\mathbb{C}$ for now):

Let $d$ be a non-negative integer, and let $S = \mathbb{C}[u,v,x,y]/(x^2, xy, y^2, u^dx - v^dy)$.

Then $X_d = Proj$ $S$ is a double line with arithmetic genus $-d$.

Eisenbud and Harris then go on to say that "every projective double line of genus $-d$, with $d \geq 0$, is isomorphic to $X_d$."

My question is: where can I find a proof of this statement?

More generally: if you fix a curve $C$ of genus $g$, how can I describe the moduli of genus $d$ double curves lying over top of $C$?

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1 Answer 1

up vote 6 down vote accepted

(Note: I don't assume $C$ is a curve, or even that it is smooth.)

Suppose $C'$ is a 2-fold thickening of $C$. Then $C'$ has the same underlying topological space as $C$. On that space, we have a short exact sequence of $\newcommand{\O}{\mathcal O}\O_{C'}$-modules $$\newcommand{\L}{\mathcal L}\tag{$\dagger$} 0\to \L\to \O_{C'}\to \O_C\to 0 $$ with $\O_{C'}\to \O_C$ a ring homomorphism.

Since $C'$ is a square-zero thickening, the sheaf of ideals $\L$ squares to zero, so it is actually an $\O_C$-module. Since $C'$ is a 2-fold thickening, $\L$ is actually a line bundle on $C$ (thus the suggestive notation). In fact, a 2-fold thickening of $C$ is no more than an extension of sheaves of rings $\O_{C'}\to \O_C$ whose kernel is a line bundle. An isomorphism of thickenings will respect the map to $\O_C$, but need not induce the identity map on $\L$, so we care about parameterizing such extensions of algebras "up to scalar" in some sense.

Given a line bundle $\L$, the set of extensions of algebras of the form $(\dagger)$ is parameterized by $Ext^1_{\O_C}(L_C,\L)$, where $L_C$ is the cotangent complex of $C$. This is a special case of Theorem 1.2.3 in Chapter III of Illusie's Complexe Cotangent et Deformations I. The "up to scalar" in the previous paragraph should correspond to looking for elements of this $Ext^1$ up to scalar. If $C$ is smooth, then $L_C\cong \Omega^1_C$ is locally free, so $Ext^1(L_C,\L)\cong H^1(C,\mathcal{T}\otimes \L)$, where $\mathcal T = (\Omega^1_C)^\vee$ is the tangent bundle of $C$.

Upshot: A 2-fold thickening is given by choosing some $\L\in Pic(C)$ and some element (up to scalar) of $Ext^1(L_C,\L)$. Note that in such a thickening $C'$, $\L$ is the conormal bundle of $C$ in $C'$.

In the case $\newcommand{\P}{\mathbb P}C=\P^1$ and $\L=\O(d)$ ($d\ge 0$), we have that $\mathcal T=\O(2)$, and $H^1(\P^1,\O(2+d))=0$, so there is a unique extension for a given $d$. If I haven't made a mistake in this answer, there should be two non-isomorphic thickenings of $\P^1$ with conormal bundle $\L=\O(-4)$, and an infinite number of non-isomorphic thickenings with conormal bundle $\L=\O(d)$ if $d<-4$.

Special case: The element $0\in Ext^1(L_C,\L)$ corresponds to the first infintessimal neighborhood of $C$ in the total space $\newcommand{\V}{\mathbb V}\V(\L^\vee)$. Here, $C$ is thought of as the zero section of $\V(\L^\vee)$.

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A beautifully written answer, Anton ! –  Georges Elencwajg Mar 13 '11 at 9:35

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