The Stacks project

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). $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0E66. Beware of the difference between the letter 'O' and the digit '0'.