It may be worth adding another variation on the viewpoints/explanations already given: "ramification" is a property of a map $f:X\rightarrow Y$ of one (compact connected) Riemann surface $X$ to another, $Y$. Namely, one proves that a non-constant $f$ is a covering map away from a finite collection of (ramified) points, where the map is locally of the form $z\rightarrow z^n$. Very often the "bottom" Riemann surface $Y$ is just the plane or the Riemann sphere $\mathbb P^1$.

Then phrases like "the Riemann surface attached to $\sqrt{z}$" and more complicated versions can be construed as asking for $f:X\rightarrow \mathbb P^1$ on which the not-single-valued $\sqrt{z}$ function defined only *locally* on $\mathbb P^1$ really exists as a *global* function. That is, there is a function $F$ on $X$ such that $F(x)=\sqrt{f(x)}$ *locally*.

The common discussion of "slitting the plane" is one way of examining global topological constructions to existence of global functions. "Making a slit" is a way of killing off a part of $\pi_1(\mathbb P^1 - R)$ where $R$ is a finite set of points where there is *local* problem with existence of the algebraic function like $\sqrt{z}$ (or solutions of ODEs and such). There are no "slits", really, it's just a way of talking about killing homotopy (or homology) classes. We have to respect Riemann for coming up with a way to talk about such things!

Yes, as in other answers, copies of "slit planes" can be stuck together to give part of an "atlas" of coordinate charts, if desired.

(The Riemann-Hurwitz theorem does not need to refer to "slits" for its application to computing genus: it only needs the degree of the cover and the ramification. One style of *proof* of Riemann-Hurwitz does use slits to make comparable triangulations of the cover and the base surface, to apply an Euler characteristic argument to compute the genus of the cover.)

Edit: @SimonRose gave the genus computation in another answer. To *make* an n-sheeted cover of $\mathbb P^1$ with specified ramified points, and with some specified "monodromy behavior" (=requiring that analytic continuation along certain paths returns to the same sheet), taking an $n$-th root may suffice. The zeros and poles of $f(x)$ in $y^n=f(x)$ must be just the ramified points. The specific trick in the two papers mentioned in the question uses $y^n=\prod_j (x-x_i)/(x-x'_i)$. The effect of this is that analytic continuation of $y$ along a loop encircling *both* $x_i$ and $x_i'$ returns to the same sheet.

This post imported from StackExchange MathOverflow at 2015-04-13 10:49 (UTC), posted by SE-user paul garrett