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Mapping 4D torus onto 4-sphere non-trivially

  1. May 19, 2013 #1
    Hello,

    I'm trying to construct an explicit map that takes the 4D torus to the 4-sphere such that the wrapping is non-trivial (i.e. homotopically, i.e. you can't shrink it continuously to zero). More concretely, I'm looking for
    [itex]\phi: T^4 \to S^4: (\alpha,\beta,\gamma,\delta) \mapsto ( x(\alpha,\beta,\cdots),y(\cdots),z(\cdots),t(\cdots),w(\cdots) )[/itex]
    where [itex]\alpha^2 + \beta^2 + \cdots + w^2 = 1[/itex].

    I was thinking of embedding the 4D torus in 5D (i.e. it's natural representation) and then dividing each vector out by its norm, but I'm not sure how to ensure that it wraps around the sphere non-trivially.
     
  2. jcsd
  3. May 20, 2013 #2

    Bacle2

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    I'm not sure of this, but I think if the induced map on homology is non-trivial, so is the map.
     
  4. May 20, 2013 #3
    Sadly I know nearly nothing of homology theory. Could you give a suggestion on how to make a concrete mapping?

    I was thinking of first embedding the torus in 5D positioned such that it goes "around" the origin, and then dividing each vector out by its norm, as to get a 4D sphere embedded in 5D. Not sure how to verify the torus goes around the origin though.
     
  5. May 20, 2013 #4
    Never mind, I succeeded :) Thanks for the help though.
     
  6. May 20, 2013 #5

    Bacle2

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    I'm curious; could you give me an outline of what you did?
     
  7. May 20, 2013 #6
    Sure,

    So I embedded the 4-torus into 5D, basically the generalization of
    [itex]x = (r_1 + r_2 \cos \beta) \cos \alpha \qquad y = (r_1 + r_2 \cos \beta) \sin \alpha \qquad z = r_2 \sin \beta[/itex]
    which is the case of embedding the 2-torus into 3D.

    Then I add a constant to x such that the torus encloses the origin. More concretely, the torus will be such that you won't be able to shrink it to zero if you are not allowed to pass through the origin. In the case above you add [itex]r_1[/itex] to x. What took me a while to figure out was that for the 4D-torus what I had to add to x was not [itex]r_1[/itex] but [itex]r_1+r_2+r_3[/itex].

    Then I divide out by the norm, i.e. [itex]x = \frac{\textrm{as above}}{\sqrt{x^2 + y^2 + \cdots}}[/itex] (note that you can check the norm of my mapping is never zero so all is well-defined).

    This determines a map of the 5D embedding of a 4-torus to something which lies on the 5D embedding of a 4-sphere. The wrapping around the origin gives the intuition that this should have non-zero homology. This claim I then checked by numerically calculating the 2nd Chern number, which gave 1 :)
     
  8. May 20, 2013 #7

    Bacle2

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    Thanks; nice job.
     
  9. May 20, 2013 #8
    Thanks! It's a bit of a low-brow way of doing it, but I'm a physicist after all :P
     
  10. May 22, 2013 #9

    Bacle2

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    Just another idea, just for the fun of it: maybe you can use the degree of a map to determine if the
    image is nullhomotopic; the degree of a map is a homotopy invariant; the degree of a constant map
    (i.e., contractible image) is zero. So if your map has degree non-zero, it should be non-trivial.
     
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