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Is there a relation between Möbius strip and the complex square root multifunction? watch

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    Let's say you start at a point O somewhere on the Möbius strip, but not on the edge. Then moving parallel to the edges of the strip, you will end up going to the other side and reaching the point O' that is the point on the other side of O. If you keep on moving, you go back to point O eventually. (Horrible explanation, but you should get the gist if you know how the Möbius strip works).

    This, to me, mimics the behavior of f(z)=z^{1/2} as z loops around the origin.

    So is there a relation of some kind between those two things?
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    (Original post by gagafacea1)
    Let's say you start at a point O somewhere on the Möbius strip, but not on the edge. Then moving parallel to the edges of the strip, you will end up going to the other side and reaching the point O' that is the point on the other side of O. If you keep on moving, you go back to point O eventually. (Horrible explanation, but you should get the gist if you know how the Möbius strip works).

    This, to me, mimics the behavior of f(z)=z^{1/2} as z loops around the origin.

    So is there a relation of some kind between those two things?
    A Moebius strip is the right kind of idea, but I believe that it doesn't work for technical reasons.

    As you have noticed, \sqrt{z} has a problem: it produces two different values for each unique point on the complex plane, and the values depend upon which value of the polar argument that you use. If you imagine a circle in the complex plane, centre the origin, then you get \sqrt{z} for 0 \le \theta < 2\pi and -\sqrt{z} for 2\pi \le \theta < 4\pi and this pattern repeats for higher multiples of 2\pi.

    So on the complex plane, this function is multivalued. This is messy (just like drawing both branches of y=\sqrt{x} is messy for the real numbers). To fix this up, mathematicians define a new domain for \sqrt{z} on which the function is single valued at each point. This is called a Riemann surface for the function. For \sqrt{z}, the Riemann surface is made by:

    1. removing, say, \mathbb{R}^+ from \mathbb{C}
    2. copying this to create a new version of \mathbb{C} \setminus \mathbb{R}^+
    3. imagining one of these cut planes sitting above the other
    4. assigning +\sqrt{z} to the lower plane and -\sqrt{z} to the upper plane
    5. joining the upper plane to the lower plane along the cuts in such a way that if you traverse a circle about the origin, you move from the lower plane to the upper once you reach \theta = 2\pi, and from the upper plane back to the lower plane once you reach \theta = 4\pi

    The Riemann surface is essentially multiple copies of \mathbb{C}, stitched together in such a way as to make the function single valued. Google is your friend for a pictorial representation of this.

    On this new more complex 2d structure, the Riemann surface, \sqrt{z} is single valued since you are always either on the upper plane or the lower plane, and you no longer work with \sqrt{z} : \mathbb{C} \rightarrow \mathbb{C} but with \sqrt{z} : \mathcal{R} \rightarrow \mathbb{C} where  \mathcal{R} , the new domain of the function, is the Riemann surface.

    Note that every function will have its own Riemann surface, whose structure will depend on where its discontinuities lie (the way I've described it, \sqrt{z} is discontinuous at each point on \mathbb{R}^+; it jumps from + to - as you cross that line) i.e. there isn't a single thing called "the" Riemann surface - you get a different Riemann surface for each function.

    To make this rigorous rather than pictorial, we have to define a Riemann surface using some mathematically well-defined object. The appropriate object here is what's called a 2-dimensional manifold; that is something that looks locally like \mathbb{R}^2, but globally can have a more complicated structure (by "looks like", I mean that if you take a point on a manifold, then nearby that point, you can produce an invertible mapping onto some region of \mathbb{R}^2 i.e. you can give local x-y coords to that bit of the manifold). For example, the surface of a sphere is a manifold; on a small scale (local to some point on the surface), it looks flat like a plane, but globally we can see that it's more complicated than a plane.

    You usually need multiple different mappings to map every point of a manifold to \mathbb{R}^2, and some points of the manifold can be mapped using more than one of these mappings (i.e. the mappings overlap - Google is your friend for pictures again).

    In the case of a Riemann surface, though, we also need to be able to do complex operations with functions on it e.g. we need to be able to complex differentiation and integration. To make sure this works, the Riemann surface is defined as a structure that can mapped locally onto \mathbb{C} rather than onto \mathbb{R}^2 (via a set of different mappings to make sure all of the surface is mapped), and such that where the mappings overlap, the overlaps allow complex differentiation to work properly. (They preserve something called the holomorphic structure)

    However, it turns out that a Moebius strip cannot represent a Riemann surface since a Riemann surface is also an orientable manifold - it is possible to define a unique direction for "a clockwise curve" everywhere. (See http://www.open.edu/openlearn/scienc...nt-section-3.2 for some nice pictures). This can't be done on a Moebius strip, but as far as I understand, orientability follows immediately for a Riemann surface from the fact that the overlap of the mappings allow complex differentiation - I don't know how it works in any detail though.

    So the upshot of that waffle is: I don't think that a Moebius strip is a good structure as an analogy for \sqrt{z}, but you are thinking along the right lines.
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    (Original post by atsruser)
    A Moebius strip is the right kind of idea, but I believe that it doesn't work for technical reasons....
    That was absolutely brilliant!!
    I can't believe you just made me understand what a Riemann Surface is, what a manifold is, and why the Möbius strip isn't a good analogy for the square root, all in one post! That was EXTREMELY helpful, thank you very much!
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    (Original post by atsruser)
    A Moebius strip is the right kind of idea, but I believe that it doesn't work for technical reasons.

    As you have noticed, \sqrt{z} has a problem: it produces two different values for each unique point on the complex plane, and the values depend upon which value of the polar argument that you use. If you imagine a circle in the complex plane, centre the origin, then you get \sqrt{z} for 0 \le \theta < 2\pi and -\sqrt{z} for 2\pi \le \theta < 4\pi and this pattern repeats for higher multiples of 2\pi.

    So on the complex plane, this function is multivalued. This is messy (just like drawing both branches of y=\sqrt{x} is messy for the real numbers). To fix this up, mathematicians define a new domain for \sqrt{z} on which the function is single valued at each point. This is called a Riemann surface for the function. For \sqrt{z}, the Riemann surface is made by:

    1. removing, say, \mathbb{R}^+ from \mathbb{C}
    2. copying this to create a new version of \mathbb{C} \setminus \mathbb{R}^+
    3. imagining one of these cut planes sitting above the other
    4. assigning +\sqrt{z} to the lower plane and -\sqrt{z} to the upper plane
    5. joining the upper plane to the lower plane along the cuts in such a way that if you traverse a circle about the origin, you move from the lower plane to the upper once you reach \theta = 2\pi, and from the upper plane back to the lower plane once you reach \theta = 4\pi

    The Riemann surface is essentially multiple copies of \mathbb{C}, stitched together in such a way as to make the function single valued. Google is your friend for a pictorial representation of this.

    On this new more complex 2d structure, the Riemann surface, \sqrt{z} is single valued since you are always either on the upper plane or the lower plane, and you no longer work with \sqrt{z} : \mathbb{C} \rightarrow \mathbb{C} but with \sqrt{z} : \mathcal{R} \rightarrow \mathbb{C} where  \mathcal{R} , the new domain of the function, is the Riemann surface.

    Note that every function will have its own Riemann surface, whose structure will depend on where its discontinuities lie (the way I've described it, \sqrt{z} is discontinuous at each point on \mathbb{R}^+; it jumps from + to - as you cross that line) i.e. there isn't a single thing called "the" Riemann surface - you get a different Riemann surface for each function.

    To make this rigorous rather than pictorial, we have to define a Riemann surface using some mathematically well-defined object. The appropriate object here is what's called a 2-dimensional manifold; that is something that looks locally like \mathbb{R}^2, but globally can have a more complicated structure (by "looks like", I mean that if you take a point on a manifold, then nearby that point, you can produce an invertible mapping onto some region of \mathbb{R}^2 i.e. you can give local x-y coords to that bit of the manifold). For example, the surface of a sphere is a manifold; on a small scale (local to some point on the surface), it looks flat like a plane, but globally we can see that it's more complicated than a plane.

    You usually need multiple different mappings to map every point of a manifold to \mathbb{R}^2, and some points of the manifold can be mapped using more than one of these mappings (i.e. the mappings overlap - Google is your friend for pictures again).

    In the case of a Riemann surface, though, we also need to be able to do complex operations with functions on it e.g. we need to be able to complex differentiation and integration. To make sure this works, the Riemann surface is defined as a structure that can mapped locally onto \mathbb{C} rather than onto \mathbb{R}^2 (via a set of different mappings to make sure all of the surface is mapped), and such that where the mappings overlap, the overlaps allow complex differentiation to work properly. (They preserve something called the holomorphic structure)

    However, it turns out that a Moebius strip cannot represent a Riemann surface since a Riemann surface is also an orientable manifold - it is possible to define a unique direction for "a clockwise curve" everywhere. (See http://www.open.edu/openlearn/scienc...nt-section-3.2 for some nice pictures). This can't be done on a Moebius strip, but as far as I understand, orientability follows immediately for a Riemann surface from the fact that the overlap of the mappings allow complex differentiation - I don't know how it works in any detail though.

    So the upshot of that waffle is: I don't think that a Moebius strip is a good structure as an analogy for \sqrt{z}, but you are thinking along the right lines.
    I shall now solve the millenium problems, all of them.


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