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Let $R,S$ be Riemann surfaces and $f: R \to S$ an orientation preserving diffeomorphism. Then $f$ determines what is called a Beltrami differential denoted by $\mu \frac{d\bar{z}}{dz}$.

Local expression of $\mu$ is clear, however some texts mention that $\mu$ is a $(-1,1)$ form on $R$. What does that mean? More specifically is $\mu$ the section of a vector bundle, which one? Is there a coordinate free definition for $\mu$?

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  • $\begingroup$ It's a section of $\omega^{*} \otimes \omega$ where $\omega$ is the canonical bundle. It's determines the deformation of complex structures. $\endgroup$
    – user40276
    Aug 25, 2015 at 7:12
  • $\begingroup$ More precisely, the Kodaira-Spencer map of a first order deformation around a given Riemann surface (i.e, a deformation of an $X$, the special fiber, over $\text{Specan} (\mathbb{C}[x]/(x^2))$) is the multiplication by the Beltrami differential. So this gives a coordinate free definition. See for instance, the beginning of amazon.com/Advances-Moduli-Translations-Mathematical-Monographs/… $\endgroup$
    – user40276
    Aug 25, 2015 at 7:36
  • $\begingroup$ @user40276 Do you mean it is a section of $\omega^* \otimes \bar{\omega}$ where $\omega$ is the canonical bundle. In that case $\omega^*$ is just the tangent bundle. That does make sense. $\endgroup$ Aug 25, 2015 at 8:44
  • $\begingroup$ Yes, I've forgotten the over line. $\endgroup$
    – user40276
    Aug 25, 2015 at 9:54
  • $\begingroup$ $(-1,1)$ form is a ratio of $(0,1)$ form and a $(1,0)$ form. The definition of Beltrami differential is similar to definition of any other differential. "To each local coordinate it puts a function into correspondence, which transforms in such and such way when we change the local coordinate". $\endgroup$ Aug 25, 2015 at 12:14

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