Is exterior differentiation of forms well defined

Question

This is from 9.13. Exterior Differentiation of Forms, Otto Forster's Lectures on Riemann Surfaces. In the book, for $\omega=\sum f_k \,dg_k$, define $d'\omega=\sum d'f_k\land dg_k$. Is this well defined? See one example, $$d'(z\,d(z\bar{z}))=dz\land (z\,d\bar{z}+\bar{z}\,dz)=z\,dz\land d\bar{z}$$ $$d'(z^2\,d\bar{z}+z\bar{z} \,dz)=2z\,dz\land d\bar{z}$$

Answer 1

First of all, you should say what $d'$ means. Most of us write that as $\partial$, writing $d=\partial+\bar\partial$. So we should have (on a Riemann surface with local coordinate $z$) $$\partial (f\,dz) = \partial f\wedge dz = 0 \quad\text{and}\quad \partial(f\,d\bar z) = \partial f\wedge d\bar z = \frac{\partial f}{\partial z}\,dz\wedge d\bar z.$$ Since his function $g$ is just assumed to be smooth, then he is defining $d'(f\,dg) = \partial f\wedge dg$. But $\partial(f\,dg) = \partial f\wedge dg + f \partial(dg) = \partial f\wedge dg + f \partial\bar\partial g$, so you need $g$ to be harmonic in order for the second term to vanish. This is exactly what went wrong in your example. If $g$ is either holomorphic or anti-holomorphic, it works fine.