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Metric on torus induced by identification of points on plane

  1. May 25, 2013 #1
    Hi all,

    Perhaps I'm asking the wrong question but I am wondering about the relationship between different definitions of, for the sake of argument, the torus.

    We can define it parametrically (or as a single constraint) and from there work out the induced metric as with any surface.

    But we can also define it by considering the plane and identifying opposite sides. Is it possible to use this definition, equip the plane with Cartesian coordinates and then deduce the metric on the torus caused by the identification?

    I hope I've explained myself and apologise if it's an ill-thought-out question.

    Thanks,
    I
     
  2. jcsd
  3. May 25, 2013 #2

    quasar987

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    Yes, this is one way to put a metric on the torus. More generally, if you have a riemannian manifold and a group G acting by isometries such that the orbit space M/G is a smooth manifold, then the metric on M descends naturally to a metric on M/G. Your construction is an example of this with M=R² and G=Z².
     
  4. May 25, 2013 #3

    lavinia

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    You can realize this metric by mapping the plane into R^4 by the equation

    1/[itex]\sqrt{2}[/itex](cosx,sinx,cosy,siny)

    The image is a flat torus.

    To add to Quasar's comment, one can put a flat metric on the Klein bottle because it is the quotient of a group of isometries of the plane. Just add to the standard lattice, the isometry

    (x,y) -> (x+1/2,-y)
     
  5. May 25, 2013 #4
    Hi - thank you so much for the replies :-D

    So in the two cases above (assuming that the plane was initially a piece of ℝ2) are you saying that the two methods lead to different induced metrics? In the case of the pullback onto a surface we end up with
    ds2 = (1 + a.cos(v))2du2 + a2dv2
    in terms of parameters u and v (from http://mathworld.wolfram.com/Torus.html :-s).

    Do I understand that you mean that if we do things by identification we have the torus with a flat metric
    ds2 = du2 + dv2?

    Sorry if I've misunderstood - topology isn't my strong point! If I have could you explain what the induced metric would be in the latter case and how it is worked out?

    Thanks again,
    I
     
  6. May 25, 2013 #5

    lavinia

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    Sorry. It is my fault for being vague. If one identifies the edges of a square in the plane to make a torus then this torus inherits the metric ds^2 = dx^2 + dy^2 from the plane.

    The embedding of the torus in R^4 parametrizes the torus with this metric. One can not realize the torus with this metric in R^3 since in R^3 no closed surface without boundary can have a completely flat metric.

    However a torus can have many metrics and each one comes from a different metric on the square. The only requirement is that the metric on the square lines up on its edges. This lining up is what Quasar was talking about.
     
    Last edited: May 25, 2013
  7. May 26, 2013 #6
    Ok great, thank you very much :-)

    Two minor questions...
    Presumably finding geodesics is non trivial if you use the plane plus identifications....or do you extend the fundamental domain and place image points for your end point and find the geodesics for those, then wrapping them back to the fundamental domain?

    And so if I wanted the metric I wrote above (u and v) I guess I must first parameterise the plane with u and v satisfying the identification, place that metric on the plane and then make the torus by revolution?

    Actually also I was thinking that since we often think of it also as S1xS1 I would have thought it natural to use two angles, say [itex]\theta, \phi[/itex] and use
    [itex]ds^{2}= a^2d\theta^{2} + b(\theta)^{2}d\phi^{2}[/itex]
    but I just realised whilst typing that of course b is precisely the factor 1 + a.cos(\theta) in my first post telling us how far away from the centre of the second circle we are given how far around the first one (I've fixed some radial parameter to 1 here, sorry)!!

    Ok, think I'm good - thanks very much indeed.
    James
     
  8. May 26, 2013 #7

    lavinia

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    In general finding geodesics involves solving a differential equation. If the torus inherits its metric from the plane then its geodesics are the projections - "wrap arounds" - of geodesics in the plane. For the flat metric they are the projections of straight lines.


    A good parametized torus in 3 space is the sterographic projection of the flat torus in 4 space parameterized above. This torus looks like a curled up slinky. Since stereographic projection is conformal - it preserves infinitesimal angles - this torus is conformally flat. Try to figure out the geodesics on it.
     
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