I am currently reading chapter 3 of Arithmetic of Elliptic Curves by Silverman. We know that elliptic curves can be viewed as a Riemann surface of genus 1, and there is a well-known group structure on it. I wonder if we can do the same for arbitrary Riemann surfaces of genus 1.
The notes of Terence Tao for 246C (Notes 1, Exercise 43) suggests that this can be done by showing for $R$ a point on $X$, and $P$, $Q$ points on $X$, we have some points $P+Q$ such that $$(P)+(Q)-(P+Q)-(R)$$ is a principal divisor. I am not sure how to show this using Riemann-Roch, and how can we use this to get the desired group structure. Thanks in advance.
With regards to using Riemann-Roch to define a group law on a genus one Riemann surface, I think the argument that Terence Tao had in mind is something like the following.
Lemma: If $D$ is a degree zero divisor, then there exists a unique point $P \in X$ such that $D$ is linearly equivalent to $(P) - (\infty)$.
Proof: Applying Riemann-Roch to the divisor $D + (\infty)$, we get $$ L(D + (\infty)) - L(K - D - (\infty)) = \text{deg}(D + (\infty)) + 1 - g.$$ We have $\text{deg}(D) = 0$ by assumption, so $\text{deg}(D + (\infty)) = 1$. Also, $X$ is an elliptic curve, so $K = 0$ and $g = 1$. Thus the Riemann-Roch equation simplifies to $$ L(D + (\infty)) - L(- D - (\infty)) = 1.$$ The divisor $- D - (\infty)$ has negative degree. (Its degree is $-1$.) So $L(- D - (\infty)) = 0$. Therefore, $$ L(D + (\infty)) = 1.$$ This means that there exists a unique effective divisor that is linearly equivalent to $D + (\infty)$. Now any divisor that is linear equivalent to $D + (\infty)$ must have the same degree as $D + (\infty)$, and the degree of $D + (\infty)$ is $1$. A divisor on $X$ is an effective divisor of degree $1$ if and only if it is of the form $(P)$ for some point $P \in X$. Thus we've shown that there exists a unique point $P \in X$ such that $(P)$ is linearly equivalent to $D + (\infty)$.
Taking stock of this result, what we've found is that there is a bijection between the set of points on $X$ and the divisor classes of degree zero divisors on $X$. (A "divisor class" is an equivalence class of divisors, where the equivalence relation is linear equivalence.) Under this bijection, a point $P$ on $X$ maps to the divisor class of the degree zero divisor $(P) - (\infty)$.
$$ \{ \text{points on } X \} \longleftrightarrow \{ \text{degree zero divisor classes on $X$} \} $$ $$ P \mapsto [(P) - (\infty)]$$
But the set of degree zero divisor classes on $X$ is an abelian group! The group operation is simply the addition operation on divisors.
Therefore, the set of points on $X$ inherits a group operation from the set of degree zero divisor classes, via our bijective correspondence.
Let's spell out how the group operation works. Let $P$ and $Q$ be two points on $X$. Via our bijective correspondence, $P$ and $Q$ map to the divisor classes of the divisors $(P) - (\infty)$ and $(Q) - (\infty)$ respectively.
If $P + Q$ is the sum of $P$ and $Q$ under the group operation on points, then $P + Q$ must be the unique point on $X$ that maps to sum of the divisor classes of $(P) - (\infty)$ and $(Q) - (\infty)$ via our bijective correspondence. In other words, $P + Q$ is the unique point on $X$ such that $(P + Q) - (\infty)$ is linearly equivalent to $(P) - (\infty) + (Q) - (\infty)$, i.e. $P + Q$ is the unique point on $X$ such that $(P) + (Q) - (P + Q) - (\infty)$ is a principal divisor.
Edit: As per the discussion in the comments below, the OP is most used to thinking of an elliptic curve as a smooth cubic in $\mathbb P^2$. $$ X = \{ [x:y:z] \in \mathbb P^2 : zy^2 = x(x-z)(x-cz) \}$$
The group law that OP is most familiar with is characterised by the condition that $P + Q + R$ is the zero point if and only if $P$, $ Q$ and $R$ are collinear. This zero point is the point $ \infty = [0:1:0]$.
I feel like for completeness' sake, we ought to explain why the divisor-based group law described in Terence Tao's notes is equivalent to the group law that the OP is familiar with.
Definition: Given any line $\{ [x:y:z] \in \mathbb P^2 : a_1x + a_2y + a_3z = 0\}$, let $P$, $Q$ and $R$ be the three points where this line intersects the elliptic curve $X$. We define the line divisor $H_{(a_1, a_2, a_3)}$ to be the divisor $(P) + (Q) + (R)$. (Note that $P$, $Q$ and $R$ don't have to be distinct: intersections between the line and the elliptic curve should be counted with multiplicity.)
Lemma: Any two line divisors $H_{(a_1, a_2, a_3)}$ and $H_{(b_1, b_2, b_3)}$ are linearly equivalent.
Proof: $H_{(a_1, a_2, a_3)} - H_{(b_1, b_2, b_3)}$ is the principal divisor associated to the meromorphic function $f = (a_1x + a_2y + a_3z)/(b_1x + b_2y + b_3z)$.
Lemma: Any line divisor $H_{(a_1, a_2, a_3)}$ is linearly equivalent to $3(\infty)$.
Proof: The line divisor $H_{(0, 0, 1)}$ is equal to $3(\infty)$. The claim then follows from the previous lemma.
Proposition: The zero point with respect to the divisor-based group law is the point $\infty$.
Proof: Via our bijective correspondence, the point $\infty$ maps to the divisor class of the divisor $(\infty) - (\infty)$, which is clearly the zero divisor.
Proposition: $P$, $Q$ and $R$ are collinear if and only if $P + Q + R$ is the zero element. (Here, the $+$ symbol represents the group operation for the divisor-based group law.)
Proof of $\implies$: If $P$, $Q$ and $R$ are collinear, then there exists a line divisor $H_{(a_1, a_2, a_3)}$ such that $H_{(a_1, a_2, a_3)} = (P) + (Q) + (R)$. By our lemma, we know that $H_{(a_1, a_2, a_3)}$ is linearly equivalent to $3(\infty)$. So $(P) + (Q) + (R)$ is linearly equivalent to $3(\infty)$, i.e. $(P) - (\infty) + (Q) - (\infty) + (R) - (\infty)$ is linearly equivalent to the zero divisor. Via our bijective correspondence, the point $P + Q + R$ maps to the divisor class of the divisor $(P) - (\infty) + (Q) - (\infty) + (R) - (\infty)$, and we've just shown that this is the zero divisor class. Hence $P + Q + R$ must be the zero point.
Proof of $\impliedby$: Given $P$ and $Q$, there is a unique $R$ such that $P + Q + R$ is equal to the zero point. By our proof of $\implies$, this $R$ must be the third point of intersection between the elliptic curve $X$ and the line through $P$ and $Q$.