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.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.