## 53.25 Vector fields

In this section we study the space of vector fields on a curve. Vector fields correspond to infinitesimal automorphisms, see More on Morphisms, Section 37.9, hence play an important role in moduli theory.

Let $k$ be an algebraically closed field. Let $X$ be a finite type scheme over $k$. Let $x \in X$ be a closed point. We will say an element $D \in \text{Der}_ k(\mathcal{O}_ X, \mathcal{O}_ X)$ fixes $x$ if $D(\mathcal{I}) \subset \mathcal{I}$ where $\mathcal{I} \subset \mathcal{O}_ X$ is the ideal sheaf of $x$.

Lemma 53.25.1. Let $k$ be an algebraically closed field. Let $X$ be a smooth, proper, connected curve over $k$. Let $g$ be the genus of $X$.

1. If $g \geq 2$, then $\text{Der}_ k(\mathcal{O}_ X, \mathcal{O}_ X)$ is zero,

2. if $g = 1$ and $D \in \text{Der}_ k(\mathcal{O}_ X, \mathcal{O}_ X)$ is nonzero, then $D$ does not fix any closed point of $X$, and

3. if $g = 0$ and $D \in \text{Der}_ k(\mathcal{O}_ X, \mathcal{O}_ X)$ is nonzero, then $D$ fixes at most $2$ closed points of $X$.

Proof. Recall that we have a universal $k$-derivation $d : \mathcal{O}_ X \to \Omega _{X/k}$ and hence $D = \theta \circ d$ for some $\mathcal{O}_ X$-linear map $\theta : \Omega _{X/k} \to \mathcal{O}_ X$. Recall that $\Omega _{X/k} \cong \omega _ X$, see Lemma 53.4.1. By Riemann-Roch we have $\deg (\omega _ X) = 2g - 2$ (Lemma 53.5.2). Thus we see that $\theta$ is forced to be zero if $g > 1$ by Varieties, Lemma 33.44.12. This proves part (1). If $g = 1$, then a nonzero $\theta$ does not vanish anywhere and if $g = 0$, then a nonzero $\theta$ vanishes in a divisor of degree $2$. Thus parts (2) and (3) follow if we show that vanishing of $\theta$ at a closed point $x \in X$ is equivalent to the statement that $D$ fixes $x$ (as defined above). Let $z \in \mathcal{O}_{X, x}$ be a uniformizer. Then $dz$ is a basis element for $\Omega _{X, x}$, see Lemma 53.12.3. Since $D(z) = \theta (dz)$ we conclude. $\square$

Lemma 53.25.2. Let $k$ be an algebraically closed field. Let $X$ be an at-worst-nodal, proper, connected $1$-dimensional scheme over $k$. Let $\nu : X^\nu \to X$ be the normalization. Let $S \subset X^\nu$ be the set of points where $\nu$ is not an isomorphism. Then

$\text{Der}_ k(\mathcal{O}_ X, \mathcal{O}_ X) = \{ D' \in \text{Der}_ k(\mathcal{O}_{X^\nu }, \mathcal{O}_{X^\nu }) \mid D' \text{ fixes every }x^\nu \in S\}$

Proof. Let $x \in X$ be a node. Let $x', x'' \in X^\nu$ be the inverse images of $x$. (Every node is a split node since $k$ is algebriacally closed, see Definition 53.19.10 and Lemma 53.19.11.) Let $u \in \mathcal{O}_{X^\nu , x'}$ and $v \in \mathcal{O}_{X^\nu , x''}$ be uniformizers. Observe that we have an exact sequence

$0 \to \mathcal{O}_{X, x} \to \mathcal{O}_{X^\nu , x'} \times \mathcal{O}_{X^\nu , x''} \to k \to 0$

This follows from Lemma 53.16.3. Thus we can view $u$ and $v$ as elements of $\mathcal{O}_{X, x}$ with $uv = 0$.

Let $D \in \text{Der}_ k(\mathcal{O}_ X, \mathcal{O}_ X)$. Then $0 = D(uv) = vD(u) + uD(v)$. Since $(u)$ is annihilator of $v$ in $\mathcal{O}_{X, x}$ and vice versa, we see that $D(u) \in (u)$ and $D(v) \in (v)$. As $\mathcal{O}_{X^\nu , x'} = k + (u)$ we conclude that we can extend $D$ to $\mathcal{O}_{X^\nu , x'}$ and moreover the extension fixes $x'$. This produces a $D'$ in the right hand side of the equality. Conversely, given a $D'$ fixing $x'$ and $x''$ we find that $D'$ preserves the subring $\mathcal{O}_{X, x} \subset \mathcal{O}_{X^\nu , x'} \times \mathcal{O}_{X^\nu , x''}$ and this is how we go from right to left in the equality. $\square$

Lemma 53.25.3. Let $k$ be an algebraically closed field. Let $X$ be an at-worst-nodal, proper, connected $1$-dimensional scheme over $k$. Assume the genus of $X$ is at least $2$ and that $X$ has no rational tails or bridges. Then $\text{Der}_ k(\mathcal{O}_ X, \mathcal{O}_ X) = 0$.

Proof. Let $D \in \text{Der}_ k(\mathcal{O}_ X, \mathcal{O}_ X)$. Let $X^\nu$ be the normalization of $X$. Let $D' \in \text{Der}_ k(\mathcal{O}_{X^\nu }, \mathcal{O}_{X^\nu })$ be the element corresponding to $D$ via Lemma 53.25.2. Let $C \subset X^\nu$ be an irreducible component. If the genus of $C$ is $> 1$, then $D'|_{\mathcal{O}_ C} = 0$ by Lemma 53.25.1 part (1). If the genus of $C$ is $1$, then there is at least one closed point $c$ of $C$ which maps to a node on $X$ (since otherwise $X \cong C$ would have genus $1$). By the correspondence this means that $D'|_{\mathcal{O}_ C}$ fixes $c$ hence is zero by Lemma 53.25.1 part (2). Finally, if the genus of $C$ is zero, then there are at least $3$ pairwise distinct closed points $c_1, c_2, c_3 \in C$ mapping to nodes in $X$, since otherwise either $X$ is $C$ with two points glued (two points of $C$ mapping to the same node), or $C$ is a rational bridge (two points mapping to different nodes of $X$), or $C$ is a rational tail (one point mapping to a node of $X$). These three possibilities are not permitted since $C$ has genus $\geq 2$ and has no rational bridges, or rational tails. Whence $D'|_{\mathcal{O}_ C}$ fixes $c_1, c_2, c_3$ hence is zero by Lemma 53.25.1 part (3). $\square$

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