Curves have finite automorphism group (proof by D. Litt)

Question

I am trying to understand the proof of Proposition 1 from Daniel Litt's one pager CURVES HAVE FINITE AUTOMORPHISM GROUP:

Let $X$ be a smooth projective curve of genus $g \ge 2$, over a field $k$.

Proposition 1. The automorphism group of $X$ is finite.

Proof of Proposition. wlog $k = \bar{k}$ (???). Choose $n$ so that $\omega_X ^{\otimes n}$ is very ample (recall $\omega_X$ is canonical sheaf of $X$), and let

$$\iota: X \to \mathbb{P}\Gamma(X,\omega_X ^{\otimes n}) = \mathbb{P}^N _k $$

be the natural closed embedding. The action of $\text{Aut}(X)$ (which we view as an abstract group -this proof will not use the representability of the automorphism functor) extends to an action on $\mathbb{P}\Gamma(X,\omega_X ^{\otimes n})$ via pullback of differentials; (???) $\iota$ is equivariant for this action. As the action of $Aut(X)$ on $X$ is faithful, we obtain an injection $$\kappa:Aut(X) \to PGL(\Gamma(X,\omega_X ^{\otimes n})).$$

Let $G$ be the closure of the image of $Aut(X)$ in $PGL(\Gamma(X,\omega_X ^{\otimes n}))$: Suppose for the sake of contradiction that $Aut(X)$ is infinite; then $G$ has positive dimension (and also acts on $X$ ???). Consider non-zero $v \in T_e G$. We claim that the vector field on $X$ induced by $v$ is non-zero (in characteristic zero this is clear, but in characteristic $p$ an argument is required - see the remark at the bottom of the page).

Indeed, suppose $v \in G(k[\epsilon]/\epsilon^2)$ acted trivially on $X_{k[\epsilon]/\epsilon^2}$ . Then the action on $\mathbb{P}\Gamma(X,\omega_X ^{\otimes n})_{\epsilon]/\epsilon^2}$ (induced by pullback) would also be trivial -but it's easy to see that any non-zero tangent vector in $T_e PGL(V)$ acts non-trivially on $\mathbb{P}V_{\epsilon]/\epsilon^2}$. So we've obtained a non-zero element $v \in H^0(X, T_X)$. But $T_X = \omega_X^{\vee}=\mathcal{Hom}(\omega_X, \mathcal{O}_X)$ has negative degree by the assumption on the genus, so $H^0(X; T_X) = 0$; this is the desired contradiction.

recall: $deg(\omega_X)=2g-2$, therefore $deg(T_X)=2-2g <0$.

Q:

1. why we can assume $k=\bar{k}$?

2. why the action of $\text{Aut}(X)$ can be extended to action on $\mathbb{P}\Gamma(X,\omega_X ^{\otimes n})$ via 'pullback of differentials'? I not understand what this 'pullbak of differentials' has to do with extension of the group action.

3. what is the action of the closure $G$ on $X$ concretly? how acts elements of $G \backslash \kappa(Aut(X))$ on $X$?

For question (2) I think that that by differentials Litt refers to sections of sheaf $\omega_{\mathbb{P}\Gamma(X,\omega_X ^{\otimes n})}$ and $\iota$ induces maps of $s \in H^0(\omega_{\mathbb{P}\Gamma(X,\omega_X ^{\otimes n}})$ to pullbacks $\iota^*s \in H^0(\omega_X)$ in functorial sense. what has this to do with extension of the action of $\text{Aut}(X)$ to $\mathbb{P}\Gamma(X,\omega_X ^{\otimes n})$?

Answer 1

For question 1, if $a$ is an autormorphism of $X$ then it determines an autormorphism of $X_{\overline{k}}=X\otimes_k\operatorname{Spec} \overline{k}$ as $a\times id$. It is straightforwards to check that this is an injective homomorphism of groups $Aut(X)\hookrightarrow Aut(X_\overline{k})$, so if the latter is finite, the former must be as well.

For question 2, let $f$ be a map $X\to Y$. Then for a sheaf $\mathcal{F}$ on $Y$ and a section $s$ of $\mathcal{F}$ over $U\subset Y$ open, we have a map $\Gamma(U,f^*\mathcal{F})\to\Gamma(f^{-1}(U),\mathcal{F})$ given by sending $s(-)$ to $s(f(-))$. In our case, we take $X=Y=U$, $\mathcal{F}=\omega_X^{\otimes n}$ and $a$ an automorphism of $X$. It's then clear that this is an endomorphism of $\Gamma(X,\omega_X^{\otimes n})$, and we can show that it's an automorphism by showing that the pullback associated to $a^{-1}$ gives a two-sided inverse. So we have an action of $Aut(X)$ on $\Gamma(X,\omega_X^{\otimes n})$ and as it is $k$-linear, thus on the projective space of this vector space. The reason the author uses "extends" here is that they're upgrading the action from $X$ to a bigger space that $X$ embeds in.

For question 3, the action of $G$ is defined on the projective space, and saying that it acts on $G$ just means that every element of $G$ sends $X$ to $X$. One nitpick here is that $G\hookrightarrow PGL(N+1)$ is already closed, since $Aut(X)$ is an algebraic group for any $X$ which is projective and the image of any map of algebraic groups is automatically closed. So there's no reason to actually take the closure here.

Edit: this final bit is false in full generality, see for instance John Lesieutre: A projective variety with discrete, non-finitely generated automorphism group. Invent. Math. 212 (2018), no. 1, 189–211. Instead, we can show directly that $G$ preserves $X$ if some dense subset of $G$ does. Define $X\times G\to \Bbb P^N$ as the restriction of $\Bbb P^N\times G\to \Bbb P^N$, then note that the preimage of $X\subset \Bbb P^N$ must be closed and contain $X\times Aut(X)$. But as $X$ is projective, $X\to \operatorname{Spec} k$ is universally closed, so the projection $X\times G\to G$ is closed. So the subset of $G$ preserving $X$ is closed and contains a dense set and thus must be all of $G$.