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Let $R$ be a Riemann surface of genus $g\ge 2$ and $p\in R$ a point. I need to find a way to produce a closed 1-form $\omega$ on $R$ which satisfies the following conditions:

  • $\omega$ is not exact, not holomorphic and not anti-holomorphic
  • $\omega$ is zero in $p$
  • $\int_R\omega\wedge \overline{\omega}>0$

Can you help me?

The answer of this question Constructing one-forms on a Riemann surface using the uniformization theorem suggests to use the uniformization theorem, but they do it for meromorphic quadratic forms and I don't know how to apply it to my case.

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Here is what I suggest. First, start with a holomorphic (1,0)-form $\alpha$ which is not exact. We know it exists, since the Dolbeault group $H^{(1,0)}(R)$ is $g$-dimensional. $\alpha$ is automatically closed since $R$ has complex dimension $1$. Moreover, we immediately get that

$i \int_R \alpha \wedge \bar{\alpha} > 0$.

Also, since the Euler characteristic of $R$ is non-zero (it is $2-2g < 0$), it follows that $\alpha$ vanishes somewhere in $R$. If it vanishes precisely at $p$, then this is great. Otherwise, pull back $\alpha$ using a smooth diffeomorphism $f: R \to R$ which maps $p$ to a point where $\alpha$ vanishes (does there exist such a diffeomorphism?). $f^*(\alpha)$ is automatically closed, and not exact, and also vanishes at $p$. Also the positivity condition of the integral still holds if $f$ is assumed to be orientation-preserving.

If $f^*(\alpha)$ happens to be either holomorphic or anti-holomorphic, then just add to it $\epsilon$ times the complex conjugate of $f^*(\alpha)$, where $\epsilon$ is some small enough positive real number, so as to preserve positivity of the integral.

Thus I have solved your problem, provided the answer to the following is yes: given any two points $p_1$ and $p_2$ in $R$, does there exists an orientation-preserving diffeomorphism which maps $p_1$ to $p_2$?

Edit: I am surprised I wrote the answer above. There is no $\omega$ satisfying all the requirements in the original post. First of all, the integral in the inequality is pure imaginary, but I suppose that it is just because of a typo, and the author forgot to multiply by $i$. However, $i$ times the integral is negative, by the Gauss-Bonnet theorem, since $g \geq 2$. But let us assume that the author had asked the same questions, but with the inequality being $ i \int_R \omega \wedge \bar{\omega} < 0$ which is actually automatically satisfied, by the Gauss-Bonnet theorem.

We start with a holomorphic $(1,0)$-form $\alpha$ which is not exact, as I did in the beginning of my answer. If at $p$ we have that $i \alpha \wedge \bar{\alpha} \neq 0$, then we just add $df$ to $\alpha$, where $f$ is a smooth complex-valued function, chosen so that $i df \wedge d\bar{f}$ is minus $i \alpha \wedge \bar{\alpha}$ at $p$. This will make sure that $\omega = \alpha + df$ is still closed, and vanishes at $p$. It remains to destroy holomorphicity of $\omega$, in case $\omega$ happens to be holomorphic, but this can be done by replacing $\omega$ with a suitable linear combination of $\omega$ and $\bar{\omega}$. This should fix my wrong answer above.

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  • $\begingroup$ I understand your reasoning, but to me your last hypothesis seems a little too strong to be verified for every couple of points on every Riemann surface.. $\endgroup$ – user00169 Aug 27 '16 at 22:17
  • $\begingroup$ @user00169: see mathoverflow.net/questions/104104/… $\endgroup$ – Malkoun Aug 27 '16 at 22:40
  • $\begingroup$ in particular, I quote Liviu Nicolaescu's comment: The group of diffeomorphisms of a smooth connected manifold acts transitively on that manifold. $\endgroup$ – Malkoun Aug 27 '16 at 22:42
  • $\begingroup$ See the answer here $\endgroup$ – user99914 Aug 28 '16 at 4:18
  • $\begingroup$ @Arctic Char: thank you! It is a nice proof via the isotopy extension theorem. $\endgroup$ – Malkoun Aug 28 '16 at 4:24

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