# What is a branched Riemann surface with cuts?

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Edit: Let me restate the main claim being made in these two papers,

• Consider the "branched" Riemann surface which has "n" sheets stuck along the intervals, $[z_i, z_{i+1}]$ for $i=1,..,2N$ then it is of genus $(n-1)(N-1)$ and is represented by the equation, $y^n = \prod_{i=1...N} \frac{z-z_{2i-1} }{z-z_{2i}}$

• If the same intervals are labelled as, $[u_i,v_i]$ and $i=1,..,N$ then this is represented by the curve, $y^n = \prod_{i=1}^{N-1}(z-u_k)(z-v_k)^{n-1}(z-u_N)$ and has genus $(n-1)(N-1)$

(..and $v_N$ has somehow been sent to infinity using conformal maps..)

I would like to understand why the above claims are true and their derivation and why these are the same things. (...to begin with I can't see how a "genus" can even be defined for such an object - it isn't something compact and orientable!..)

I can't see how I can use any of the standard Riemann surface theory (like in the books by Griffiths or Jurgen Jost) on this weird structure!

================================================================================== The two paper references where this concept is mentioned which I would like to understand,

I guess both of them are constructing the same "branched Riemann surface" and writing down an algebraic curve for it and calculating its genus. I can't understand this construction and how the Riemann-Hurwitz formula is being used here.

If I go back to my usual references of the books by Jurgen Jost or Phillip Griffiths then I don't see anything there called "n sheeted branched Riemann surface with N cuts" and how one calculates it genus or its algebraic curve. (or am I missing something and hence not able to recognize the concept?)

This post imported from StackExchange MathOverflow at 2015-04-13 10:48 (UTC), posted by SE-user user6818
asked Jul 5, 2013
retagged Apr 13, 2015

## 6 Answers

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There are two different things which are called "Riemann surface" in the literature.

1. The modern notion (introduced by Hermann Weyl): complex 1-dimensional manifold. In older literature this is sometimes called "Abstract Riemann surface".

2. "Riemann surface spread over the plane" (or over the sphere, or over some other surface). Surface de Riemann etalee in French, Uberlagerungsflache in German. This is what many old authors (and physicists) call a Riemann surface.

In the work of Riemann you can see both notions, but they are not clearly distinguished.

The relation between these two things is the following. A Riemann surface spread over the sphere is a pair $(F,p)$ where $F$ is an abstract surface, and $p:F\to S$ is a holomorphic map to the Riemann sphere. Critical points of $p$ are called "branch points" of the surface $(F,p)$, and so on. If you have an analytic germ, and perform an analytic continuation over all paths on which continuation is possible, you obtain a Riemann surface spread over the sphere $(F,p)$ whose $F$-part is an abstract Riemann surface. To visualize a Riemann surface spread over the sphere, you make cuts and paste it from sheets. The cuts and sheets are arbitrary to some extent (they are not intrinsically connected with $(F,p)$; the fact that physicists frequently ignore).

Most of the modern literature uses the first meaning. Some modern mathematical works (besides the elementary complex variables textbooks) which deal with surfaces spread over the plane or sphere are the papers of J. Ecalle on resurgent functions, or my survey on Geometric theory of meromorphic functions, which can be found in http://www.math.purdue.edu/~eremenko/surveys.html, or the work of H. Stahl, arXiv:1205.3811.

For many people with modern education (who think that a Riemann surface is a "complex 1-dimensional manifold"), the expression "the Riemann surface of $\sqrt{z}$" is meaningless, because this is just the same as the sphere.

This post imported from StackExchange MathOverflow at 2015-04-13 10:49 (UTC), posted by SE-user Alexandre Eremenko
answered Jul 5, 2013 by (150 points)
@Alexendre Eremenko Thanks for your efforts. I am taking a look at your review. I have made my question more precise in the new edit at the top. It would be great if you can help derive these equations.

This post imported from StackExchange MathOverflow at 2015-04-13 10:49 (UTC), posted by SE-user user6818
The answer of Simon Rose below tells you how to compute the genus, using the Riemann-Hurwitz formula. What else are you asking?

This post imported from StackExchange MathOverflow at 2015-04-13 10:49 (UTC), posted by SE-user Alexandre Eremenko
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@user6818: To reiterate earlier replies, the Riemann surface constructed from slit planes is smooth and compact. Your authors' equation should be viewed as defining a locus in $S^2 \times S^2$, viewed as the componentwise compactification of $\mathbb{C}^2$. In this setting:

1. As Alexandre Eremenko and Paul Garrett noted, the "Riemann surface of $\sqrt{z}$" is the sphere $S^2$, equipped with the squaring mapping $f(w) = w^2$, namely, the (compact, smooth) locus of $z = w^2$ in $S^2 \times S^2$. The branch points (images of critical points of $f$) are $z = 0$ and $\infty$. Since $f$ is two-to-one, the classical construction is to take two copies of $\mathbb{C}$ (with coordinate $z$), make a slit between the branch points (e.g., by removing the negative real axis), and to "cross-glue" the edges of the slits. Here, the result is another copy of $\mathbb{C}$ (with coordinate $w$). Adding the branch point $z = \infty$ compactifies the $w$ plane to $S^2$.

2. Igor Khavine's example of the "(complex) circle" $w^2 + z^2 = 1$ (again, a compact, smooth locus in $S^2 \times S^2$) may be viewed as "the Riemann surface of $\sqrt{1 - z^2}$". The picture is identical to the squaring map (rotate the "$z$ sphere" to bring $0$ and $\infty$ to $-1$ and $-1$, say); the branch points are $\pm 1$, and the Riemann surface can be constructed by slitting two copies of $\mathbb{C}$ along the closed interval $[-1, 1]$ and "cross-gluing" the edges. In the diagram below, the slits are the upper and lower halves of the circle in the vertical plane above the real axis; the two "sheets" ($z$ planes) of the surface are the two "tilted planes". The locus is smooth even at the branch points $z = \pm 1$, i.e., $w = 0$. (The vertical axis is the real part of $w$. At risk of posting gratuitous eye candy, there's an animation loop at http://mathcs.holycross.edu/~ahwang/epix/misc/complex/quadric.gif that shows the result of rotating the $w$ plane while keeping $z$ fixed.)

Andy

This post imported from StackExchange MathOverflow at 2015-04-13 10:49 (UTC), posted by SE-user Andrew D. Hwang
answered Jul 9, 2013 by (40 points)
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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
answered Jul 5, 2013 by (120 points)
Can you kindly explain how in the referenced papers do the authors come up with that algebraic curve equation to represent that "branched Riemann surface"? How does one see that a "n sheeted Riemann surface stuck at N cuts" is of genus $(n-1)(N-1)$? Can you kindly sketch these two calculations?

This post imported from StackExchange MathOverflow at 2015-04-13 10:49 (UTC), posted by SE-user user6818
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Since no one has answered the genus computation, here goes:

The Riemann-Hurwitz formula states that for a map of Riemann surfaces $f : C_1 \to C_2$, that we have

$$\chi(C_1) = n\chi(C_2) - \deg R$$

where $n$ is the degree of the map, and $R$ the ramification divisor.

In the case you have, since we are covering $\mathbb{C}$, we are secretly covering $\mathbb{P}^1$, so the Euler characteristic of the base is 2. The degree is clearly $n$ (the number of sheets), and it is hopefully clear that the map is only ramified at the $2N$ points listed, with ramification index $(n - 1)$. Thus the Riemann-Hurwitz formula now reads

$$2 - 2g(C) = 2n - 2N(n - 1)$$

and solving for the genus gives you the desired answer.

As for why the genus is well-defined---well, here is the slight-of-hand that I used to say that it's secretly a map to $\mathbb{P}^1$. If you take the $n$ given sheets and glue them together you get a non-compact Riemann surface. To make it compact, you just have to add in a few points. The easiest way to do it is to replace the copies of $\mathbb{C}$ that you have with copies of $\mathbb{P}^1$. You can still perform the cutting-and-gluing construction to build a surface (this time with a map to $\mathbb{P}^1$). In this case, you will find that you have a compact surface. It is orientable, as at no point are we using anything but holomorphic transition functions.

This post imported from StackExchange MathOverflow at 2015-04-13 10:49 (UTC), posted by SE-user Simon Rose
answered Jul 9, 2013 by (70 points)
Regarding Simon's comment "it is hopefully clear that the map is only ramified at the $2N$ points listed, with ramification index $(n−1)$", this point represents either a hidden hypothesis or a hidden consequence of the precise description of how the $n$ "sheets" are glued together to form the covering space. One then uses this to show that the number of points one needs to add to compactify the non-compact Riemann surface is exactly $2N$, and that the noncompact covering map over $P^1 - \{u_i,v_u\}$ extends continuously to the "branched covering map" over $P^1$ itself.

This post imported from StackExchange MathOverflow at 2015-04-13 10:49 (UTC), posted by SE-user Lee Mosher
... the missing "precise description" should be that for each branch point $u_i$ or $v_i$, as one goes around the point in counterclockwise order, the sheets can be cyclically ordered from $1$ to $n$ so that sheet 1 glues to sheet 2 which glues to sheet 3 ... which glues to sheet $n$ which glues to sheet 1.

This post imported from StackExchange MathOverflow at 2015-04-13 10:49 (UTC), posted by SE-user Lee Mosher
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You are looking at the wrong references. Any introductory complex analysis textbook (e.g., Ahlfors) will define a Riemann surface by a branched covering of the complex plane, with different sheets meeting along cuts.

This post imported from StackExchange MathOverflow at 2015-04-13 10:48 (UTC), posted by SE-user Igor Rivin
answered Jul 5, 2013 by (40 points)
@But defined in that way it isn't a 2-manifold anymore. right? Now how all my technology of Riemann-Hurwitz apply when this basic tenent fails!?

This post imported from StackExchange MathOverflow at 2015-04-13 10:48 (UTC), posted by SE-user user6818
(...so are you saying that there is a totally different theory of Riemann surfaces than what I learnt from Griffiths' and Jurgen Jost's book?...because these two things don't seem to match at all!..)

This post imported from StackExchange MathOverflow at 2015-04-13 10:48 (UTC), posted by SE-user user6818
@user6818, you can see the branched covering with cuts as essentially an incomplete collection of charts on the smooth manifold you have in mind. Consider the simpler example of $S^1$ ($x^2+y^2=1$) and the charts given by the projection $S^1\to\mathbb{R}$ ($(x,y)\mapsto x$). You get charts by removing $x=\pm1$ (branch points). There are two of them (you have two branches). To complete the chart atlas, you need charts near the branch points. But these are easy to get from the alternative projection $S^1\to\mathbb{R}$ ($(x,y)\mapsto y$). The latter step is implicit in the "cut-branched" picture.

This post imported from StackExchange MathOverflow at 2015-04-13 10:48 (UTC), posted by SE-user Igor Khavkine
@IgorKhavkine Thanks for the explanations. Can you kindly explain how in the referenced papers do the authors come up with that algebraic curve equation to represent that "branched Riemann surface"? How does one see that a "n sheeted Riemann surface stuck at N cuts" is of genus $(n-1)(N-1)$? Can you kindly sketch these two calculations?

This post imported from StackExchange MathOverflow at 2015-04-13 10:48 (UTC), posted by SE-user user6818
@user6818, you need to follow the references in the article and in these answers to find the details. Eventually you should be able to follow this: encyclopediaofmath.org/index.php/Riemann-Hurwitz_formula. Also, the compactness comes from compactifying the $\mathbb{C}$ plane by adding a point at infinity.

This post imported from StackExchange MathOverflow at 2015-04-13 10:48 (UTC), posted by SE-user Igor Khavkine
@IgorKhavkine I am of course familiar with the Riemann-Hurwitz formula but I can't see how that can be used here! What is the definition of "genus" here for such non-compact branched/singluar surfaces? And how does one derive the polynomial equations given here?

This post imported from StackExchange MathOverflow at 2015-04-13 10:48 (UTC), posted by SE-user user6818
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For convenience I've added some figures from those two papers:

Especially since you are talking about Entanglement Entropy, I thought you were asking what are Riemann surfaces in a more philosophical sense.

This post imported from StackExchange MathOverflow at 2015-04-13 10:49 (UTC), posted by SE-user john mangual
answered Jul 9, 2013 by (310 points)

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