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Forms on the circle

  1. Dec 2, 2009 #1
    I read a problem a while ago which was to find a differential form on the circle which is not the differential of any function. Being a hapless physicist, this puzzled me for a while. I've found an answer in Spivak's Calculus on Manifolds, but I need a little help in following his reasoning.

    He argues that the form [tex]d\theta[/tex] is such a form, and shows that if it is the differential of a function [tex]f[/tex] then [tex]f = \theta + constant[/tex]. I am OK up to this point, but I fail to see how such an [tex]f[/tex] can't exist, like he argues.
  2. jcsd
  3. Dec 2, 2009 #2

    Ben Niehoff

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    The function

    [tex]f(\theta) = \theta[/tex]

    is not single-valued on the circle.

    Note that every closed form can be written as d of something locally; that is, over some finite region of the manifold. But only exact forms can be written as d of something globally. Here is an example on the circle of a form that is closed but not exact.

    Another example that you might look up is the Dirac monopole. It can be written as [itex]\vec B = \vec \nabla \times \vec A[/itex] over some region of space, but not globally. There is always a topological defect (the Dirac string) on which [itex]\vec A[/itex] is not defined.
  4. Jan 6, 2010 #3
    dtheta is dual to the unit length tangent vector to the circle. It's integral over the circle is therefore the length of the circle. If if were the differential of a function then its integral over the circle would be zero. Spivak's argument is correct but you need to see that the integral of dx on a line segment is just the integral of the 1 form that is dual to the unit tangent vector to the segment.

    The angle function ,theta, is discontinuous and so does not have a derivative everywhere. Where it is continuous though, its differential equals dtheta.
  5. Jan 6, 2010 #4


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    [itex]\theta= 0[/itex] and [itex]\theta= 2\pi[/itex] refer to the same point on the circle. You cannot have a function that assigns, to the same point 0 and [itex]2\pi[/itex] (not to mention [itex]2n\pi[/itex] for every integer n.
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