# Riemann bilinear relations and meromorphic abelian differentials

I am getting quite confused with Riemann bilinear relations.

Let $$\Sigma$$ be a compact Riemann surface of genus $$g$$, with a canonical homology basis $$a_1,b_1,\dots,a_g,b_g$$, with associated normalised holomorphic and anti-holomorphic one-forms $$\omega_I,\bar \omega_I$$, $$I=1\dots,g$$, such that $$\oint_{a_I} \omega_J = \delta_{IJ}$$.

Let further $$\tau_{P,Q}$$ be an abelian differential on $$\Sigma$$ with poles at $$P,Q$$ of residues $$\pm1$$, normalised so that its $$a$$-cycles periods are zero : $$\oint_{a_I} \tau_{PQ}=0.$$ It is called an abelian differential of the third kind and the normalisation guarantees that it is unique. Besides, it is closed on $$\Sigma-\{P,Q\}$$. For more details, see Bobenko or textbooks on Riemann surfaces like Farkas & Kra.

I've read in a few places the following statement: $$\iint_\Sigma \tau_{PQ} \wedge \bar \omega_J \propto \int_Q^P \Im \omega_J$$ up to some factors of $$2,\pi,i$$ and sign.

I can't get where the imaginary part comes from. Calling $$f=\overline{\int^z\omega_J}$$, it seems to me that the standard proof of the Riemann identities based on Stokes theorem, $$\int_{\partial \Sigma} f \tau_{PQ} = \iint \bar \omega_J \wedge \tau_{PQ}$$ will hold ($$\partial \Sigma$$ means the union of $$a,b,a^{-1}$$ and $$b^{-1}$$ cycles that constitute a canonical dissection of the surface; the standard 4g-gon) and because $$\tau_{PQ}$$ has zero a-periods this is evaluated to $$\iint_\Sigma \tau_{PQ} \wedge \bar \omega_J = \sum_{I=1}^g \oint_{a_I} \bar \omega_J \oint_{b_I} \omega_{P,Q},$$which, using some reciprocity formula like $$\oint_{b_i} \omega_{P,Q} = 2i\pi \int_Q^P \omega_i$$ will yield $$\iint_\Sigma \tau_{PQ} \wedge \bar \omega_J = 2i\pi \int_Q^P \omega_J$$ without the imaginary part ?

• It's too hard to figure out what's going on from what little you've put here. For example, what are the $1$-forms $\omega_{P,Q}$? Do you have a link to a complete text? – Ted Shifrin Jan 5 at 0:31
• I've updated a bit the text... Don't know if this helps. – picop Jan 5 at 8:35