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Using Riemann Surfaces

  1. Aug 15, 2013 #1
    I'll preface this by saying that this just isn't getting through to me. I know the material, but my brain feels like it doesn't, for lack of a better word, "fit."

    Looking back at Riemann surfaces in complex analysis after familiarizing myself with some differential geometry really makes them more confusing. Or, at least, my understanding clashes with what the internet seems to think.

    One of the uses of Riemann surfaces is to be able to define multivalued functions between them. I'm having a lot of trouble understanding the visual aspect of them. Many of the sources I go to seem to think that a Riemann surface can be visualized as a singular embedded surface in ##\mathbb{R}^3##. Shouldn't it instead be 2 surfaces, or a surface in ##\mathbb{R}^4##?

    Also, there is the problem of defining functions between them. This is my primary problem. Let's take, for example, the complex logarithm. Wolfram Alpha gives 2 surfaces, which I imagine correspond with the real and imaginary parts of ##\ln(z)##. If we imagine it as a single valued function using a Riemann surface ##S##, are we defining a map ##\ln_S: \mathbb{C}\to S##, or a map ##\ln_S: S\to\mathbb{C}##? If it's the first one, are the surfaces shown by Wolfram just the real and imaginary parts of the image of ##\mathbb{C}## under ##\ln_S##? If it's the second one...what the heck is ##S##?

    I don't understand, and I've managed to give myself something of a headache. Yay, math? :tongue::cry:

    Any help is greatly appreciated.
    Last edited: Aug 15, 2013
  2. jcsd
  3. Aug 15, 2013 #2


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    Which sources are you talking about? Some Riemann surfaces can be embedded in ##\mathbb{R}^3##, but not all. You can certainly visualize them as a surface in ##\mathbb{R}^3##, but then you must allow self-intersections, so it's an immersion instead of an embedding.

    We use log for the complex logarithm, not ln.

    In the wolframalpha link they are clearly defining a surface associated with the Real part and a surface associated with the imaginary part. None of these is the actual Riemann surface associated with the function. But it's the only way to visualize them in 3D (without self-intersections), since we can't visualize 4D.

    The logarithm defines a map ##\log: S\rightarrow \mathbb{C}##.
  4. Aug 15, 2013 #3
    That makes a lot more sense. I remember reading in a couple pages about "embedding" the surface into ##\mathbb{R}^3##, which I found confusing.

    My sources came from my standard habit of opening up every result Google throws at me for the first two pages in separate tabs. :tongue:

    By what convention? I've managed to get away with using ##\ln## for the complex logarithm (##\operatorname{Ln}## for the principle value) for a while.

    Alright. This, I think, is the trickier part.

    Now that I have a somewhat tangible understanding of manifolds, I can understand somewhat what a complex 1-manifold is. But, I guess my problem is imagining, for lack of less concrete wording, "what the mapping would look like."

    Maybe this would clear my head a little better: Say we had a chart ##(U,\phi)## on some subset ##U\subseteq S##, where ##S## is the aforementioned Riemann surface with ##\log(z)##, with ##\phi: U\to\mathbb{C}##. For a point ##p\in U##, what would ##\phi(p)\in\mathbb{C}## be? I recognize this might be a really general question, but I hope for a general answer to (hopefully) help me understand.
  5. Aug 15, 2013 #4
    I'd like to try this at the risk of upsetting those who know this better than me:

    I do not feel the complex log funtion is a good place to start understanding Riemann surfaces. Rather, I suggest starting with the lowly [itex]\sqrt{z}[/itex]. Riemann's goal was to define a new coordinate system over which multi-valued functions such as these are single-valued holomorphic functions. Once done, the entire foundation of complex analysis for single-valued holomorphic functions could be applied to multi-valued functions such as Cauchy's Theorem and integral formula, the Residue Theorem, Laurent's Expansion Theorem, Argument Principle, Gauss' Mean Value Theorem, and the others.

    That's not at all difficult to do with [itex]\sqrt{z}[/itex]: take two Riemann spheres. At each z-point of one sphere, map one determination of [itex]\sqrt{z}[/itex] so that we have places on the sphere, [itex](z,\sqrt{z})[/itex]. Do the same with the other sphere using the other determination of [itex]\sqrt{z}[/itex]. Now, make a slit in each sphere running from the points [itex](0,0)[/itex] to [itex](\infty,\infty)[/itex]. Connect the two spheres along the slit (appropriately) and deform the resulting surface into a new double-covered sphere. That sphere now has all the values of [itex]\sqrt{z}[/itex] and each point is distinct from the others. For example, we have the point [itex](4,2)[/itex] and the point [itex](4,-2)[/itex]. Label the coordinates over this new sphere [itex](z,w)[/itex] and define single-valued functions of these points. For example, I could define the funtion [itex]f(z,w)=w[/itex] which is identical to the square root function.

    Since f(z,w) is single-valued and holomorphic, I can apply the residue theorem (over the square-root coordinate system which I've now mapped onto this new surface:

    [tex]\mathop\oint\limits_{|z|=1} f(z,w)dz=2\pi i \text{Res}f(z,w)[/tex]

    Can you see that a closed contour [itex]|z|=1[/itex] over this new coordinate system around the point [itex](0,0)[/itex] is actually a loop with winding number=2 over the z-plane?

    In this particular case the integral is zero since the residue of [itex]\sqrt{z}[/itex] is zero. That is, the fractional power expansion of [itex]\sqrt{z}[/itex] is simply [itex]\sqrt{z}=z^{1/2}[/itex]. That is, it has no 1/z terms.

    Now, the real and imaginary surfaces often described as "Riemann" surfaces are really just the real and imaginary surfaces of a multi-valued function mapped not over a Riemann coordinate system, such as the double-covered sphere above, but rather just mapped over the z-plane multiple times, one for each determination and that is why we see the figures wrapping around in various twisted conformations. However, those figures, including the one for [itex]\log(z)[/itex] are really not the surfaces Riemann intended to create that we now call Riemann surfaces. Rather surfaces like the double-covered sphere above is what Riemann really had in mind.
    Last edited: Aug 15, 2013
  6. Aug 19, 2013 #5


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    the standard example of a (compact) Riemann surface, is a curve in the complex plane CP^2, defined by an irreducible homogeneous polynomial whose partials have no common zeroes.

    the standard example of a holomorphic function defined on it is the projection map onto a projective line CP^1 in that plane.

    All the confusing baloney about sheets and winding and so on is merely a poor way to express the fact that the projection map is usually not injective.

    e.g. consider the curve xz = y^2, which is a compactification of the affine curve x=y^2 in C^2. projection on the y = 0 axis, takes (x,y,z) to (x,0,z), a compactification of the function on x = y^2 taking (x,y) to x, i.e. taking = (y^2,y) to y^2.

    This is a double over of the x axis, and the graph of x = y^2 admits a function which equals sqrt(x), namely the function y. therefore if we want a good domain for the function sqrt(x), it should not be the x axis, but the curve x = y^2. On this new domain, the function sqrt(x) is represented just by y.

    Note there are two points of this new curve x = y^2 over each point of the x axis, and this allows sqrt(x) to have two values for each x. I.e. for each x, there are two points (x,y) on the curve x = y^2, so we can have sqrt(x) take one of its values on one point and the other value on the other point.

    That's all there is to it.

    if you have a more complicated function like sqrt(x^3-X) and you want it to be well defined, just consider the plane curve y^2 = x^3-x, and again use the function y on this curve. the projective version is just zy^2 =x^3 - xz^2.
    Last edited: Aug 19, 2013
  7. Aug 20, 2013 #6
    Suppose I'm mainly looking at learning about holomorphic/meromorphic functions. What do you think might be boring about a compact Riemann surface when looking at these things (assuming it's connected)? :tongue:

    I'm really interested in manifolds, and I love complex analysis. It's a rich and beautiful combination, and I'm looking forward to reading through more of the various lecture notes that I found on the subject. I feel like I'm understanding the material much more thoroughly now. :biggrin:
  8. Aug 20, 2013 #7


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    well the holomorphic functions are somewhat "boring" since they are all constant, but the field of meromorphic functions actually determines the riemann surface up to isomorphism, giving a correspondence between riemann surfaces, and extension fields of transcendence degree one of the complex numbers.

    this also shows why there are no non constant holomorphic maps between most pairs of riemann surfaces, since such a map induces an inclusion of their meromorphic functions fields, and given two such fields usually neither is contained in the other.
  9. Aug 21, 2013 #8


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