Automorphisms of curves and Hurwitz-Riemann formula

Question

Let $k$ be a base field algebraically closed and of zero characteristic.

Let $C$ be a smooth projective curve and $G$ a finite group of automorphisms of it. Let $C*$ be a smooth projective curve whose function field is isomorphic to $k(C)^G$.

Question 1: how the genus $g(C*)$ is related to $g(C)$ and $G$?

I know the answer is an application of the Hurwitz-Riemann formula, but I am really having a hard time applying the formula in this situation. So I welcome very much a detailed explanation.

Question 2: I am interested in understanding the finite subgroups of automorphisms of a smooth projective curve $C$.

In case $C$ is rational, I understand the situation very well. I am looking for referentes (books, surveys, etc) for the cases

a) C is an elliptic curve. If assuming $k$ to be $\mathbb{C}$ simplifies things, I am satisfied to restrict attention to complex curves.

b) $g(C) > 1$. I am aware that there are few automorphisms in this case. Also if $g(C) \geq 3$, the general curve will have no non-trivial automorphism.

Answer 1

This answer is my attempt to address OP's questions. Please let me know if there are mistakes and any additions are appreciated.

Question 1: Let $C^G$ be the curve obtained from the covering $C\to C^G$ which is the quotient map. In particular, it is a separable morphism of curves. When $G$ is a finite group, this makes sense for any smooth projective curve. The degree of this covering map is equal to the cardinality/order of $G$ so $\phi:C\to C^G$ is a degree $|G|$ finite morphism. Applying Riemann-Hurwitz (since $\phi$ is finite and separable), we have the relationship $$ 2g(C)-2=|G|\cdot (2g(C^G)-2)+\operatorname{deg}R $$ where $R$ is the ramification divisor. The question now is to determine the degree of the ramification divisor i.e. the number of ramified points of $C^G$ and the ramification index of each such point. If one takes the interpretation that ramified points are those points $p\in C^G$ in which $|\phi^{-1}(p)|<\operatorname{deg}(\phi)$, then the fibres of $\phi^{-1}(p)$ must contain a point fixed by the a nontrivial element of $G$.

Suppose $q\in \phi^{-1}(p)$ is fixed by an element $g\in G$ which is nontrivial. Then $q,gq,g^2q,\dots,g^{ord(g)-1}q$ are all equal and map to $p$. It is clear then that the ramification index of $q$ is the order of $g$.

Example in the comments: If you are willing to work over $\mathbb{C}$, then an elliptic curve is the same as a torus $\mathbb{C}/\Lambda$. The action of $\mathbb{Z}_2$ I gave in the comments is determined in coordinates as $(x,y)\to (-x,-y)$ for $x+iy\in E$. If you actually draw out the fundamental parallelogram and count which points are fixed by this action, you can count there are four points $q_1,q_2,q_3,q_4\in E$ which ramify and so these four have ramification indices $2$. Applying the formula, $$ 0=2g(E)-2=2(2g(\mathbb{P}^1)-2)+\operatorname{deg}R $$ where $R=2q_1+2q_2+2q_3+2q_4$. Since $g(\mathbb{P}^1)=0$, everything checks out.

Observations: The ramification index of a point $q\in C$ is always going to divide the order of $G$. The number of points of $C^G$ which are ramified (i.e. have fibre of size less than $|G|$) is not able to be determined in general and depends on the example in mind.

Question 2: The answer splits into three parts which you have observed.

1. $g(C)=0$. Then $C\cong \mathbb{P}^1$ and the automorphism group is $PGL_2(k)$.
2. If $g(C)=1$, then $E:=C$ is an elliptic curve. There is no need to restrict to $\mathbb{C}$ here but if you do, then it makes the answer much cleaner. In any case, Silverman's The Arithmetic of Elliptic Curves has the answer:

Theorem 10.1. Let $E / k$ be an elliptic curve. Then its automorphism group $\operatorname{Aut}(E)$ is a finite group of order dividing 24. More precisely, the order of $\operatorname{Aut}(E)$ is given by the following table: $$ \begin{array}{|c|l|c|} \hline \# Aut (E) & {j(E)} & \operatorname{char}(k) \\ \hline \hline 2 & j(E) \neq 0,1728 & - \\ \hline 4 & j(E)=1728 & \operatorname{char}(k) \neq 2,3 \\ \hline 6 & j(E)=0 & \operatorname{char}(k) \neq 2,3 \\ \hline 12 & j(E)=0=1728 & \operatorname{char}(k)=3 \\ \hline 24 & j(E)=0=1728 & \operatorname{char}(k)=2 \\ \hline \end{array}. $$

This answer is somewhat dishonest. The automorphisms Silverman considers are invertible isogenies so must respect the group law. But it is known that any morphism $E\to E$ will be a homomorphism plus a translation. So up to translation the automorphism group of $E$ is known.

More specific information is known in the case of $\operatorname{char}(k)=0$ so if you're happy with working over $\mathbb{C}$, then the answer depends on the $j$-invariant. For $j(E)\neq 0,1728, j(E)=0, j(E)=1728$ the automorphism group is $\mu_2,\mu_4,\mu_6$ respectively where $\mu_n$ denotes the group of $n$th roots of unity. The reason for this notation is that the automorphism group respects the Galois group action, but this irrelevent to the story.

3. The story for curves of higher genus is trickier. It is known due to Hurwitz (I recall this fact following from the Riemann-Hurwitz formula and is an exercise in IV.2 of Hartshorne) that for $g(C)\geq 2$, the automorphism group $|\operatorname{Aut}(C)|\leq 84(g(C)-1)$ is finite. Some special cases are known: for a hyperelliptic curve, see Determining the Automorphism Group of a Hyperelliptic Curve by Tanush Shaska, and for genus $3$ curve, see Automorphisms groups of genus 3 curves by Francesc Bars. Actually, this second paper contains a lot of facts in the direction of your questions, so maybe this paper is enough to satisfy Question 2 of yours on its own.