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Disclaimer: My knowledge of the theory of Riemann surfaces is limited to Miranda's book, where in particular the integration theory is not very detailed and moreover currents are not explained. So please forgive me if the following question is too silly.


Let $X$ be a compact Riemann surface and let $f$ be a meromorphic function on $X$ which induces a principal divisor $(f)=\sum_x a_x[x]$. Then the Poincare-Lelong formula says that the following equality of currents holds:

$$\frac{i}{\pi}\partial\bar\partial\log|f|=\sum_x a_x[x]\quad\quad(\ast)$$

I have two questions about this equation:

  1. I'd like to know if the following explicit interpretation of $(\ast)$ is correct. $$\text{for any $g\in C^\infty(X):\quad$} \int_X\frac{i}{\pi}g\,\partial\bar\partial\log|f|=\sum_x a_xg(x)$$
  2. Let's assume that $1.$ is correct, then let's look at the integral $\int_X \omega$ where $ \omega=g\,\partial\bar\partial\log|f|$. Let $U$ be the Zariski open set of $X$ given by the points where $f$ is well defined and holomorphic, then: $$\omega\in\mathcal E^{1,1}(U)$$ that is $\omega$ is a smooth $(1,1)$-form on $U$. So what is the meaning of writing $\int_X \omega$? Shouldn't be $\int_U\omega$? The integration of $(1,1)$-form is defined only when they are smooth! So the equation $(\ast)$ should be written as $$\text{for any $g\in C^\infty(X):\quad$} \int_U\frac{i}{\pi}g\,\partial\bar\partial\log|f|=\sum_x a_xg(x)$$

Thank you.

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Miranda doesn't explain what currents are? That's odd.

Anyway, a current is basically a differential form with distribution coefficients (instead of coefficients that are smooth functions). There's some good discussion of currents and their use in complex geometry in Demailly's book 1 and notes 2 (see singular Hermitian metrics on line bundles, of which your situation is an example).

Your interpretation (1) is correct. If we look at the local picture and consider a meromorphic function $f$ on an open set $U$, the current $\frac i\pi \partial \bar\partial \log |f|$ is the current of integration over the sets of poles and zeros of $f$. In the case of a curve, that becomes a set of points, and the current associated to a point is the Dirac delta at that point.

For your point (2), here we run into the fact that your $\omega$ is not a smooth form, but a current. In fact, the integral over any subset of $X$ where $f$ is holomorphic and non-zero is zero, because there we find $$ \partial \bar\partial \log |f|^2 = \partial \biggl( \frac{f \bar\partial \overline f}{f \overline f} \biggr) = \partial \biggl( \frac{\bar\partial \overline f}{\overline f} \biggr) = 0. $$ Thus the only non-zero contributions to the integral come from the set of zeros and poles of $f$, where $\omega$ is really a current.

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  • $\begingroup$ Thank you for you helpful answer! By the way, if you forget for a moment about the theory of currents and look at the integral $\int_X\omega$, this is a honest integral of a $(1,1)$-form. But $\omega$ is not well defined at the poles of $f$, therefore we are dealing with a sort of "improper integral"... The Poincare-Lelong formula basically says also that this integral converges. Is it correct? $\endgroup$ – Dubious Oct 20 '16 at 15:56
  • $\begingroup$ More or less, yes. The Poincaré-Lelong formula connects the very functional-analytic object $\partial \bar\partial \log |f|$ to geometry, by letting us interpret it as weighted integration over the set of zeros and poles of $f$. $\endgroup$ – Gunnar Þór Magnússon Oct 20 '16 at 16:03
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    $\begingroup$ Also, $\int_X \omega$ is really not an honest integral of a $(1,1)$-form. As we've seen, $\omega$ is zero almost everywhere, so if it were a form this integral should be zero. But $\omega$ is a current, so we get a nonzero integral in the end. This is very similar to what happens with the Dirac delta and is worth thinking about a bit. $\endgroup$ – Gunnar Þór Magnússon Oct 20 '16 at 16:06
  • $\begingroup$ $\partial \overline{\partial} = \frac{\partial}{\partial x^2}+\frac{\partial}{\partial y^2}$ so that if $h(z)$ is harmonic on $U$ then $\partial \overline{\partial} h(z) = 0$ on $U$. Now $\log |f(z)|$ is harmonic everywhere except at $\rho$ a zero of order $k$ of $f$ (or a pole of order $-k$) where $\log |f(z)| = k\log |z-\rho| + h(z)$ with $h(z)$ harmonic. Thus (around $z= \rho$) $\partial \overline{\partial} \log |f(z)| =k\, (\frac{\partial}{\partial x^2}+\frac{\partial}{\partial y^2})\log |z-\rho| = k \, \delta(z-\rho)$ where $\delta$ is the "$(1,1)$-form" Dirac delta $\endgroup$ – reuns Oct 20 '16 at 16:11
  • $\begingroup$ @user1952009 I think you're missing a couple of $\log$, but yes, this is what happens. The fun is really in showing that $i\partial\bar\partial \log |z|^2$ is (a constant multiple of) the Dirac delta. $\endgroup$ – Gunnar Þór Magnússon Oct 20 '16 at 16:14

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