I am learning about hyperelliptic curves and hyperelliptic integrals. I encountered some problems when reading the book by Gesztesy and Holden (F. Gesztesy, H. Holden, Soliton Equations and Their Algebro-Geometric Solutions, Volume 1: (1+1)-Dimensional Continuous Models, Cambridge University Press, Cambridge, 2003, pp. 360–363).
Consider the hyperelliptic curve
$$
y^2 = R_{2n+1}(z) = \prod^{2n}_{j=0}(z – E_j), \ \ E_0<E_1<\dots <E_{2n}.
$$
It is known that this curve can be compactified by adding a point $P_{\infty}$, which becomes a two-sheeted compact Riemann surface of genus $n$.
A basis of holomorphic differentials on the curve is given by
$$\eta_j = \dfrac{z^{j-1}{\mathrm d}z}{y}, j=1,\dots, n.$$
The homology basis $\{a_1, \dots, a_n, b_1, \dots, b_n\}$ of the curve is chosen as follows. The $a_j$ cycle encircles the interval $[E_{2j-2}, E_{2j-1}], j=1,\dots, n$ clockwise on the upper sheet. The $b_j$ cycle starts at a point in $(E_{2j-2}, E_{2j-1}), j=1,\dots,n$, on the upper sheet, proceeds clockwise to intersect $a_j$, and then continues clockwise on the upper sheet until it hits a point on the cut $[E_{2n}, \infty)$. Then $b_j$ returns clockwise on the lower sheet to its original starting points. As shown below
The authors wrote in the book
$$
\int_{a_k}\eta_j = 2\int^{E_{2k-1}}_{E_{2k-2}}\dfrac{z^{j-1}{\mathrm d}z}{y},
$$
and
$$
\int_{b_k}\eta_j = 2\sum^n_{l=k}\int^{E_{2l}}_{E_{2l-1}}\dfrac{z^{j-1}{\mathrm d}z}{y}.
$$
I don't understand why the integration path changed from the cycle $a_k$ to $[E_{2k-2}, E_{2k-1}]$ (as well as $b_k$).
Can someone give me some tips?
Best Answer
In McRae' PhD thesis (W. D. McRae, Riemann theta functions on degenerate surfaces, Phd thesis, 1997, pp. 43), the homology basis $\{\tilde a_j, \tilde b_j\}$ is given as follows. Let $\alpha_j$ be a line segment that goes from $E_{2j-2}$ to $E_{2j-1}$ on the top sheet for $j=1,\dots,n$. Let $\beta_j$ be a line segment that goes from $E_{2j-1}$ to $E_{2j}$ on the top sheet for $1,\dots,n$. Let $\Gamma$ be the anti-holomorphic involution: $$ \Gamma: (z, y) \mapsto (z, -y). $$ Then the $a_j$- and $b_j$-cycles are defined by $$ \tilde a_j = \alpha_j - \Gamma \alpha_j, $$ $$ \tilde b_j = \sum^n_{k=j}(\beta_k- \Gamma\beta_k). $$ In this case, it is easy to see that $$ \int_{\tilde a_k}\eta_j = \int_{\alpha_k}\eta_j - \int_{\Gamma\alpha_k}\eta_j = 2 \int_{\alpha_k}\eta_j = 2\int^{E_{2k-1}}_{E_{2k-2}}\eta_j. $$ Similarly, $$ \int_{\tilde b_k}\eta_j = 2\sum^n_{l=k}\int^{E_{2l}}_{E_{2l-1}}\eta_j. $$ Note that $a_j$ is homologous to $\tilde a_j$ and $b_j$ is homologous to $\tilde b_j$, one infers from Stoke's theorem that $$\int_{a_j}\eta_k = \int_{\tilde a_j}\eta_k,\ \ \ \int_{b_j}\eta_k = \int_{\tilde b_j}\eta_k.$$