"symmetry" in the sense of Abel-Jacobi map of hyperelliptic curves I met such a very interesting problem in algebraic curves (from Griffiths' Introduction to algebraic curves):
Denote by $u: Div(C)\rightarrow Jac(C)$ the Abel-Jacobi map of the compact Riemann surface $C$. Identify $C$ with the subset $\{1\cdot p ~|~p\in C\}$ of $Div(C)$. Then $u(C)=-u(C)$ iff $C$ is hyperelliptic.
I'm a beginner and I only know the basic definition of a hyperelliptic surface, namely admitting a 2-fold holomorphic mapping onto $\mathbb{P}^1$. Is there any further properties of hyperelliptic surfaces that I can use to tackle this problem? Thanks in advance for any suggestions!
 A: As I mentioned in the comment, the statement is based on some elementary observations in Calculus:
$$\textit{The antiderivative of an even function is an odd function},$$
when picking the critical point as the base point, and the other direction is more or less:
$$\textit{If an antiderivative of a function}~f~\textit{is odd, then}~f~ \textit{is even}.$$
We should apply the "odd function" to the Abel-Jacobi map $u:X\to JX$, and the "even function" to the canonical mapping $u:X\to \mathbb P^{g-1}$ (or maybe its dual space) by noting that the canonical mapping derived from the derivative of the Abel-Jacobi map $$du:TX\to \mathbb PT(JX)\cong JX\times \mathbb P^{g-1}\to \mathbb P^{g-1},$$
by sending $x$ to $[\omega_1(x):\omega_2(x):\cdots:\omega_g(x)]$ which vanish at the tangent line at $x$.
Here is the proof. For sufficiency: Suppose $X$ is a hyperelliptic curve, then it is a two-sheeted cover of $\mathbb P^1$ along $2g+2$ points $p_1,...p_{2g+2}$. We draw slits between $p_i$ and $p_{i+1}$ for $i$ odd, then $X$ can be thought as two copy of $\mathbb C$ glued along the $g+1$ segments (and with an $\infty$ point added). Let's say $p_1=0$ and $p_2=1$. We choose $0$ to be our base point.
Then a point on the "lower sheet" can be joined to $0$ by a path $l$ which does not cross any of the slit, while a point on the "upper sheet" can be joined to $0$ by a path $l'$ which cross the slit $[0,1]$.
In particular, if the point on the lower sheet $x$ and the point on upper sheet $x'$ have the same "shadow" (same image under the canonical map), $$l+l'+\gamma=0,$$ where $\gamma$ is a loop homologous to a small circle encompassing $[0,1]$ and represents a nonzero homology class in $H_1(X,\mathbb Z)$. In other words, the Abel-Jacobi image of $x$ and $x'$ differ by a sign:
$$u(x)=-u(x').$$
For necessity, since $u(X)=-u(X)$, for any $x\in X$, the inverse map on the ambient torus defines a holomorphic map $\tau:X\to X$. Since it is an involution,
$$x+\tau(x)=0,$$
and the equation is complex differentiable with respect to the parameter $x$. Taking the differential gives
$$dx+d\tau(x)=0.$$
This says that as the holomorphic differential, $dx=-d\tau(x)$, so in the projective space $(\mathbb P^{g-1})^*$, $[dx]=[d\tau(x)]$. Therefore, the canonical mapping of $X$ is not injective, thus $X$ has to be Hyperelliptic.
