I am currently studying compact Riemann surfaces (more specifically hyperelliptic surface), and have a problem understanding the definition of the normalization of differentials of the second or third kind. This normalization seems dependent on the specific path defined for the $a$- and $b$-cycles, instead of just their homotopy-class, which I would expect. Can anyone explain to me either why this dependence on the exact $a$ and $b$-cycles is not a problem, or what I am doing wrong?

I provided an extensive description below.

Let a Riemann surface be defined by $$\mu = \prod_{j=1}^{2n+2}(\lambda-\lambda_j),$$ where the $\lambda_j$ are all different complex numbers.

The usual procedure (see for instance this text, page 25) is then to define $2N$ closed paths on the Riemann surface, $N$ $a$-cycles, and $N$ $b$-cycles. These cycles are constrained by their intersections being strictly defined:

  • The $a$-cycles don't cross each other, the $b$-cycles don't cross each other either.
  • Labeling the different cycles, an $a$-cycle and $b$-cycle only cross if they have the same label, so $a_i$ crosses $b_j$ iff $i=j$. Moreover these cross only once.

We can now define holomorphic differentials (or Abelian differentials of the first kind) on this surface, as the differentials that in any local map are represented by $$\omega = h(z)dz,$$ with $h$ a holomorphic function. For a hyperelliptic surface a basis of $N$ vectors of these holomorphic differentials is given by $$\omega_j = \frac{\lambda^{j-1}d\lambda}{\mu}.$$ The integrals of these differentials do not depend on the exact path, but just on the homotopy class of the closed path (roughly, what zeros of $\mu(\lambda)$the path encloses)

The differentials of the second and third kind are also known as meromorphic differentials and are defined similarly, except the function $h$ is meromorphic. If all poles of $h$ have residue 0, it is known as a differential of the second kind. If $h$ has poles with residue unequal to 0, the differential is a differential of the third kind.

All texts then continue to 'normalize' these differentials of the second and third kind by subtracting holomorphic differentials such that the integral of $h$ over any $a$-cycle is 0, and I don't understand how this is well defined.

Take for example (from here, page 34) the differential of the third kind $$\hat{\Omega}_{RQ} = (\frac{\mu+\mu_R}{\lambda-\lambda_R}-\frac{\mu+\mu_Q}{\lambda-\lambda_Q})\frac{d\lambda}{2\mu}.$$

With $\lambda_R = \lambda_Q$ and $\mu_R = -\mu_Q$, this becomes $$\hat{\Omega}_{RQ} = (\frac{d\lambda}{\lambda-\lambda_R})$$.

If I were to compute the integral of this over any closed path on the Riemann-surface, I would only look if the path encloses\lambda_R. If it does the integral is $\pm 1$. If it does not, the integral is $0$. So $$\int_{a_j} \hat{\Omega}_{RQ} $$ depends on whether $\lambda_R$ is enclosed in $a_j$ or not. But I can change $a_j$ to include this point or not, freely, since the definition of $a_j$ doesn't care about this. This means that the normalization of the meromorphic differentials depends on the exact path chosen for the $a$- and $b$-cycles, which seems incorrect to me.


1 Answer 1


I think the main issue is that the differentials of the third kind are only closed away from the poles, so they only define cohomology classes on the punctured Riemann sphere. Since they do not define a cohomology class on the surface without punctures there is no reason why their integrals should only depend on the homology class of the integration cycle.

The punctured surface includes new cycles that wrap the punctures. So the different paths that you are referring to are actually in different homology classes, whose difference is precisely the cycle that wraps the puncture. Once you choose a basis for the homology, taking these extra cycles into account, the dual basis of cohomology is well defined and everything follows the usual rules.


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