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Little Spivak-Winding Number

  1. Jun 10, 2013 #1
    So I'm working on Little Spivak and I found a solution to problem 4-24 at http://www.ms.uky.edu/~ken/ma570/homework/hw18/html/ch4c.htm. My question is, is the integral defined there the integral of a differential form over a chain, or something else?

    I only ask because the book did not reach the point of defining integration of forms over chains until 1 page later.

    Thanks
     
    Last edited: Jun 10, 2013
  2. jcsd
  3. Jun 10, 2013 #2
    Do you mean #24?
     
  4. Jun 10, 2013 #3
    Yes, sorry about that.
     
  5. Jun 10, 2013 #4

    mathwonk

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    my solution used the polar coordinate map, and lifted the chain through the convex domain space of that map. no integration needed. do you know about covering maps?
     
  6. Jun 11, 2013 #5
    I've seen a bit of covering maps, but I don't know too much about them. I took a course that studied the fundamental group (but oddly skipped covering maps). Is your solution online?

    As an aside, is the c* in the link I posted the pullback operator, or something else (once I know that I can rest and check out your solution).
     
    Last edited: Jun 11, 2013
  7. Jun 11, 2013 #6

    mathwonk

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    the point is the "polar coordinate" map P taking (r,t) to (e^rcos(t), e^rsin(t)), has the "lifting property".. I.e. any map f from a closed interval into the complement of (0,0) in the plane, factors through P.

    I.e. given f:I-->R^2 - (0,0), there is a map g:I-->R^2, such that f = Pog. That lets you work in R^2 instead of R^2 - (0,0). and in R^2 you have convexity to let you fill things in.
     
  8. Jun 11, 2013 #7
    Hmmm, I'll have to review covering spaces. So, then you would just fill the closed curve in and use that to get the 2-cube with boundary the difference of the two curves?
     
  9. Jun 12, 2013 #8
    Also, in the link I posted at the beginning of the thread, can anyone explain why n in problem 4-24 is necessarily an integer?
     
  10. Jun 12, 2013 #9

    lavinia

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    the form d[itex]\theta[/itex] integrates to 2[itex]\pi[/itex] over the circle. The formul in the excercise is just a change of variables.
     
  11. Jun 12, 2013 #10

    lavinia

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    Yes it is an integral over a chain. The author is asking you to verifiy that any closed curve along the circle integrates to 2[itex]\pi[/itex]n for some integer n. intuitively it wraps around the curcle n times.
     
  12. Jun 12, 2013 #11
    Maybe I'm just not used to the computations with pullbacks. I get ∫c*d[itex]\theta[/itex]=∫f1(c)det(c')+f2(c)det(c'), where f1 and f2 are the component functions of d[itex]\theta[/itex]. Is this right? Why does this get you an integer (once you normalize it by 2∏)

    And for it to be change of variables, don't we have to assume det(c') is positive?
     
    Last edited: Jun 12, 2013
  13. Jun 13, 2013 #12

    lavinia

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    I spoke a little loosely. Because of the change of variables theorem, the integral of a form over a smooth simplex can be defined as the integral of any pull back. Two different pull backs are connected by a change of variables.

    The form d[itex]\theta[/itex] is dual to the tangent unit length vector field of the circle. Its integral is just the length of the circle. As Mathwonk pointed out its integral is the same as integrating dx over an interval of length 2[itex]\pi[/itex].
     
  14. Jun 13, 2013 #13
    I think I sort of understand, but I'm not sure how to see that it must integrate to 2[itex]\pi[/itex]n, for some n, explicitly. I'm on a pretty basic level, since I just started learning about this stuff.
     
  15. Jun 14, 2013 #14

    lavinia

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    The point of the exercise is to show that every integral of a closed curve around the circle is the same same as the integral of a constant speed curve that winds around the circle n times. n can be any integer, positive, negative, or zero. That is why the integral is always an integer.
     
  16. Jun 14, 2013 #15
    I understand (or maybe not), but the solution that I posted asks that you first take as given that ∫c*dθ is n/2∏. I don't see this (especially if the goal of the entire problem is essentially to prove this).

    Am I completely missing the point?
     
  17. Jun 16, 2013 #16
    So I might have worked it out (or at least thought about it a bit more), so I have another question (answers to the above are also very welcome). It is possible to find a 2-cube which has boundary = c1,n-c1,m if n≠m, if you allow the 2-cube to map to (0,0)? Or, does no 2-cube exist with such boundary (here c1,n(t)=(cos(2∏nt),sin(2∏nt))).

    If I'm thinking about it correctly, one should exist since the fundamental group of the plane is trivial (so we can just linearly interpolate). But, I wasn't sure.
     
    Last edited: Jun 16, 2013
  18. Jul 6, 2013 #17

    lavinia

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    Maybe this will be helpful.

    In polar coordinates in the plane dθ is a closed 1 form in the plane minus the origin.

    By Stokes theorem its integral over any closed curve that bounds a smooth rectangle is zero. Note that such a rectangle does not to contain the origin and this means that its bounding curve does not wind around the origin.

    Note also that this rectangle does not seen to be a geometric rectangle but only the smooth image of one, a smooth singular 2 cube. This follows from the change of variables theorem and the relation

    fc* = f*d for any smooth function,f.

    Now suppose you have a curve that does wind around the origin. If you retrace it exactly backwards, then the one followed by its reverse no longer winds around the origin. And it is clear that the integral of c*dθ over the combined curve is zero since the first half is negated by the second.


    But you do not need to exactly retrace the curve backwards to unwind it. If it winds around the origin n times, then any curve that winds around -n times will undo it. For instance the curves

    R(cos-nθ,sin-nθ) will unwind a curve that winds around the origin +n times. In your problem you are asked to show this.

    The method is to show that the original curve followed by one the the R(cos-nθ,sin-nθ)'s is the boundary of a smooth singular 2 cube then apply Stokes theorem. Since ddθ = 0 the integral is zero so the integral over the boundary is zero. But the integral over the boundary is the sum of the integrals of dθ over the original curve and R(cos-nθ,sin-nθ) so their integrals are negatives of each other.
     
    Last edited: Jul 6, 2013
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