I am learning about Riemann surfaces as covering, and was interested to know how Riemann first thought of them. Looking at his collected papers, I found the following definition in his lecture called “Foundations for a general theory of functions of a complex variable” (p. 3-4 in “Collected Papers. Bernhard Riemann” of Kendrick Press):

“For the following treatment we permit $x, y$ to vary only over a finite region. The position of the point $0$ is no longer considered as being in the plane $A$, but in a surface $T$ spread out over the plane. We choose this wording since it is inoffensive to speak of one surface lying on another, to leave open the possibility that the position of $0$ can extend more than once over a given part of the plane. However, in such a case we suppose that the portions of surface lying upon one another do not connect along a line. Thus a folding [Umfaltung] of the surface, or a splitting of the surface into superimposed parts, does not occur.”

Here $A$ is the complex plane $\mathbb{C}$ or the sphere, i.e. $\mathbb{C} \cup \{ \infty \} $. After this definition Riemann defines the branching point of such a cover, where, if $f: T → A$ is the cover, then $f(z)$ is the branch point if it is locally of the form $f(z) = z^m$. However, what I didn’t understand is what is actually this “folding”. There is later (p. 24-25) a definition, where it seems that Riemann means by that, that $df/dz$ vanishes along a line, but I am not sure that I understood it correctly…. Are there more possibilities that this “folding” would appear, acc. to Riemann? I also must admit, that I didn’t find any other references of this singularity in his writings, or maybe it was renamed...

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    $\begingroup$ just fold a sheet of paper along a line and place it on a desk - you see two layers of the paper connected along a line. It is this (I think) that Riemann had in mind. Such a thing cannot appear for Riemann surfaces (for various reasons, e.g. it reverses the orientation). $\endgroup$ – user8268 Sep 8 '16 at 19:50
  • $\begingroup$ I tend to agree, but what is the mathematical formulation that Riemann had in mind? also, what do you mean by "reverses the orientation"? $\endgroup$ – David Sep 8 '16 at 19:58

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