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Let $R$ be a Riemann surface of genus g and $p\in R$ a point. I'm searching for a way to compute linearly independent differential 1-forms on $R$ which:

  • are closed and not exact, not holomorphic, not anti-holomorphic
  • are zero on $p$

Is there a standard way to do so?

Thank you

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  • $\begingroup$ If you specify the degree of a smooth curve in $\mathbb{PC}^2$ then its genus is already determined by the degree-genus formula for plane curves (e.g. for degree 3 the genus is 1). Moreover, by that formula it is impossible that a smooth curve in $\mathbb{PC}^2$ have genus 2. So what you're asking for doesn't exist. Do you have a slightly different question in mind? $\endgroup$ – john Aug 29 '16 at 5:17
  • $\begingroup$ you're right, I fixed it $\endgroup$ – user00169 Aug 29 '16 at 9:43
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The hyperelliptic curve of genus $g$ defined by the affine equation $y^2 = f(x)$ with $f$ of degree $2g + 1$ has for basis of its space of holomorphic differential forms $y^{-1}x^k\,dx$, $k = 0, 1, \ldots, g - 1$ (you may check that each of these have no pole at a Weierstrass point, including the one at infinity).

Does this answer your question?

No$\ldots$ the forms can't be holomorphic and they have be zero on a prescribed Weierstrass point $p$.

You are right; I was a bit too hasty and I did not carefully read the question. Here is I hope an answer that is more to the point.

Suppose we are given a compact connected Riemann surface $R$ of genus $g$ and a basis $\omega_1, \ldots, \omega_g$ of its space of holomorphic differentials. Denote by $\omega_{g + 1}$ the complex conjugate of $\omega_i$ so that the cohomology classes $[\omega_1], \ldots, [\omega_{2g}]$ make up a basis of $H^1(R, \mathbb{C})$. (Counting down the days until Jacobin throws a party to celebrate HRC's win!)

Now assume also given a point $p$ of $R$ (in your case: $g = 2$ and $p$ is a Weierstrass point). Integrate $\omega_i$ in a neighborhood of $p$ to a complex valued function. Multiply this function with a differentiable function that is constant $1$ in a neighborhood of $p$ and zero outside a neighborhood of $p$ in the domain of the integral. This is gives a complex valued differentiable function $f_i$ on $R$ with the property that $df_i$ coincides with $\omega_i$ in a neighborhood of $p$. So if we put $\eta_i$ $:=$ $\omega_i$ $-$ $df_i$, then $[\eta_i]$ $=$ $[\omega_i]$ and $\eta_i$ is zero near $p$. Hence the $\eta_i$'s are as desired.

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  • $\begingroup$ no... the forms can't be holomorphic and they have be zero on a prescribed weierstrass point $p$.. $\endgroup$ – user00169 Sep 10 '16 at 16:56

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