I'm reading Mumford's & Oda's Algebraic Geometry II and I'm confused about explanations on geometric intuition of sections $H^0(\Theta_X, X)$ of the tangent sheaf on page 287:

Let $X$ a geometrically irreducible smooth and proper curve of genus $g$ over a field $k$ and let $\Theta_X := \mathcal{Hom}(\Omega_X ^1 , \mathcal{O}_X) \cong \mathcal{O}_X(-K)$ the tangent sheaf to $X$. applying Riemann-Roch we obtain $$\dim_k H^0(\Theta_X, X)=$$

\begin{cases} 3 & g=0, \text{ } i.e. X \cong \mathbb{P}^1_k \\ 1 & g=1 \\ 0 & g \ge 2 \end{cases}

The author makes some comments on geometrical interpretation of these sections, which I don't understand:

...In fact, the three sections of $\Theta$ when $X \cong \mathbb{P}^1_k$ come from the infinitesimal section of the $3$-dimensional group scheme $PGL_{2,k}$ acting on $\mathbb{P}^1_k$; the one section of $\Theta$ when $g = 1$ comes from the infinitesimal action of $X$ on itself (make sense, since $g=1$ says that $X$ is elliptic curve; i.e. $X$ obtains structure of a group scheme) , and the absence of sections when $g > 1$ is reflected in the fact that the group of automorphisms of such curves is finite. ...

could anybody explain how these sections concretly geometrically arise for $g=0,1$ or respectively why in case $g>1$ the finiteness of automorphism group of $X$ inply that there a no such sections.

the matter is that we need to construct somehow a certain bundle $V$ over $X$ with projection $p:V \to X$. such that the sections $s \in H^0(\Theta_X, X)$ give rise for 'geometric' sections $s:X \to V$ with $p \circ s=id_X$.

my first guess is that since $\Theta_X$ is invertible, we can construct a bundle $V:= \mathbb{P}(\Theta_X)= Proj(Sym(\Theta_X))$ and each section $s \in H^0(\Theta_X, X)$ induces a 'geometric' section $s:X \to V=\mathbb{P}(\Theta_X)$. the problem that I don't see how this construction is related to the described actions by $PGL_{2,k}$ or $X$ itself as described in the book. therefore I have a lot of doubts that the construction I tried to explain coinsides with the one sketched in the book. could anybody describe the construction the authors have sketched in the text above giving geometric interpretation of sections $H^0(\Theta_X, X)$?

  • $\begingroup$ If X has dimension 1, Theta_X has rank 1 so P(Theta_X) would have 0-dimensional fibers and be equal to X. $\endgroup$
    – Qfwfq
    Commented Oct 7, 2019 at 20:53
  • $\begingroup$ that's the argument, that the author definitely not considered $ \mathbb{P}(\Theta_X)$. do you know a reference, where the construction the author's have in mind is explained in detail? $\endgroup$
    – user267839
    Commented Oct 7, 2019 at 21:01

1 Answer 1


Here's a corrected version of my comment (which I have deleted since there was an error):

"The basic idea that underlies their description is that given a manifold $X$ with a Lie group of automorphisms $G$, every tangent vector at the origin of $G$ (ie an element of the Lie algebra of $G$) induces a vector field $V$ on $X$ in the obvious way - namely given $z\in Lie(G)$, at every point $x\in X$, the tangent vector $V_x$ is represented by the curve $\{exp(tz)(x)\}_{t\in (-1,1)}\subset X$. In the cases described in the book, they're claiming that this association yields an isomorphism between the appropriate spaces."

What follows are some more details:

I learned to think about the tangent sheaf from studying deformation theory (mostly from Hartshorne's book).

Here is some definitions and facts:

Let $X_0$ be a scheme over a field $k$. Let $D$ denote the dual numbers $k[t]/(t^2)$. A deformation of $X_0$ over $D$ is by definition a pair $(X,i)$, where $X$ is a scheme flat over $D$ and $i$ is an isomorphism: $$i : X_0\stackrel{\sim}{\longrightarrow} X\times_D k$$ A morphism between two deformations $(X,i)\rightarrow (X',i')$ is a morphism $f : X\rightarrow X'$ whose restriction to the central fiber over $k$ is an isomorphism respecting $i,i'$ (if $f_0$ is the restriction of $f$ to the central fiber then we require that $f_0\circ i = i'$).

Thus, we have a category of deformations of $X_0$ over $D$, so we may speak of automorphisms, and so on. Note that an automorphism of a deformation is just an automorphism of the $D$-scheme which induces the identity on its central fiber over $k$. Such automorphisms are called "infinitesimal automorphisms".

Here is a theorem:

Theorem: Let $X_0$ be a scheme over $k$ (maybe one needs some finiteness assumptions, but I don't think so), and let $(X,i)$ be a deformation of $X_0$ over $D$. Then the automorphism group of $(X,i)$ is naturally isomorphic to $H^0(X_0,\mathcal{T}_{X_0})$, where $\mathcal{T}_{X_0}$ is the (relative) tangent sheaf of $X_0/k$ (this is your $\Theta_X$).

When $X_0$ is affine, this is the statement of exercise 5.2 in Hartshorne's book "Deformation theory" (In his notation, $T^0(B/k,B)$ is the tangent module of the $k$-algebra $B$, written in this way due to his use of the "$T^i$" functors - see Chapter 4). For general $X$, the tangent modules associated to the open affines of $X_0$ glue to form the tangent sheaf, and once you understand how elements of the tangent module (ie, a vector field) is identified with an automorphism, then it becomes clear that this gluing respects the identification between vector fields and automorphisms.

Taking the theorem for granted, we find that the automorphism group of any two deformations of $X_0$ are the same. Thus, it suffices to consider the trivial deformation $X := X_0\times_k D$. For simplicity let's now assume that $X_0$ is projective over $k$. To the scheme $X_0/k$ we may associate the automorphism group sheaf of $X_0$, which is the contravariant functor from the category of $k$-schemes to the category of groups which associates to every $k$-scheme $S$ the group $Aut_S(X_0\times_k S)$. This functor is a sheaf relative to pretty much any Grothendieck topology you might have heard of - specifically, it will be a sheaf for any "subcanonical" topology. Let's call this sheaf/functor $A_X$. When $S = \text{Spec }D$, the inclusion map $\text{Spec }k\hookrightarrow\text{Spec }D$ yields a restriction map $A_X(D)\rightarrow A_X(k)$, which is a homomorphism of groups, and the kernel of this map is precisely the group of automorphisms of the trivial deformation $X = X_0\times_k D$. On the other hand, under our assumptions on $X_0$ being projective over $k$, $A_X$ is representable - that means that there is a group scheme $G$ over $k$ and an isomorphism of functors $A_X\cong Hom_k(*,G)$ on the category of $k$-schemes. In general this representability holds when $X_0$ is flat and projective over $k$ (of course flatness is automatic for us since we're over a field). Thus, we now find that the automorphism group of the trivial deformation $X$ is the kernel of the map $$Hom_k(\text{Spec }D,G)\rightarrow Hom_k(\text{Spec }k,G)$$ but this kernel is just another definition of the tangent space of $G$ at the origin, which is nothing but the Lie algebra of $G$.

Thus, in short, for pretty much any $k$-scheme $X_0$, sections of the tangent sheaf are the same as infinitesimal automorphisms of your scheme, and at least if $X_0$ is projective then this is in turn the same as tangent vectors of the automorphism group scheme at the identity. Here projectivity can be replaced by something which is "suitably nice". Perhaps other experts can comment on what suffices for "suitably nice".

In your situation, $\mathbb{P}^1$ has a 3-dimensional group of automorphisms, whence a three-dimensional space of global vector fields. Genus 1 curves carry a 1-dimensional group of automorphisms, whence a 1-dimensional space of global vector fields, and genus $\ge 2$ curves have a finite group of automorphisms, whence a 0-dimensional space of global vector fields.

  • $\begingroup$ Thank you a lot for your detailed answer. one aspect I not understand: I learned that if $G$ is a Lie-group with Lie-algebra $\mathfrak{g}=Lie(G)$ (that equals the tangent space at the origin $e$ of $G$). then every element $ z \in Lie(G)$ induces a curve $exp_z: [-1,1] \to G, t \to exp_z(tz)$. evaluation at $t=0$ gives map $exp: Lie(G) \to G$. the point which I not understand why your curve $exp_z$ map to the manifold $X$ back and not to the Lie group of the automorphisms $G \subset Aut(X)$? do I missing something? $\endgroup$
    – user267839
    Commented Oct 8, 2019 at 14:50
  • $\begingroup$ @TimGrosskreutz Sorry I meant to write $exp(tz)(x)$. Basically you push the point $x$ along by the infinitesimal automorphism $exp(tz)$ for $t\in(-1,1)$, and that gives you a small curve in a neighborhood of $x$ which represents the desired tangent vector. $\endgroup$
    – Will Chen
    Commented Oct 8, 2019 at 16:03
  • $\begingroup$ another detail confuses me a bit. going through the argumentation we obtain information about tangent space of automorphism groups. and the fact the the tangent sheaf of our curve $X$ of genus $g >1$ is zero says that the automorphism group scheme $G_X$ has dimension zero, i.e. it is discrete space. why the author concludes that automorphism group $G_X$ is finite for $g>1$? discrete spaces are in general obviosly not finite. $\endgroup$
    – user267839
    Commented Oct 8, 2019 at 17:24
  • $\begingroup$ @TimGrosskreutz I don't have your book in front of me, but from the text you quoted in your OP, the authors say "...is reflected in the fact that the group of automorphisms of such curves is finite". That doesn't sound like they're concluding finiteness from 0-dimensionality. They're just saying that the 0-dimensionality is reflected in the fact that it's finite. After all you are of course correct - discreteness does not imply finiteness in general, but it may be interesting to note that in algebraic geometry, noetherian + 0-dimensional implies the underlying top. space is finite discrete... $\endgroup$
    – Will Chen
    Commented Oct 8, 2019 at 17:49
  • $\begingroup$ great, then this indeed works completly without Hurwitz's automorphisms theorem (up to now I thought that Hurwitz' was the only standard way to show that the automorphism group of curves with $g >1$ is finite). again, thank you a lot for explanations! $\endgroup$
    – user267839
    Commented Oct 8, 2019 at 18:06

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