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Topology of Aharonov Bohm Effect - Lewis Ryder's QFT book.

  1. Apr 18, 2012 #1
    Hi,

    I am reading through Section 3.4 of Lewis Ryder's QFT book, where he makes the statement,

    This makes some sense intuitively, but can someone please explain this direct product equivalence to me as I do not have a firm background in topology (unfortunately, I need some of it for a project so I am reading it as I go along -- a bad idea, but I have no time :().

    Specifically, how does this direct product come about?

    Thanks in advance!
     
  2. jcsd
  3. Apr 18, 2012 #2

    Physics Monkey

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    Think about the plane in polar coordinates: [itex] ds^2 = dr^2 + r^2 d\theta^2 [/itex]. If you remove the point r=0 then you can cover the the remaining space with the coordinates [itex] -\infty < u < \infty [/itex] ([itex] r = e^u [/itex]) and [itex] \theta [/itex]. The metric is then [itex] ds^2 = e^{2u} (du^2 + d\theta^2 ) [/itex]. The appearance of this overall prefactor in the metric means that the plane with a point removed is not only topologically equivalent but also conformally equivalent to a cylinder [itex] \mathbb{R}_u \times S^1_\theta [/itex]. In plain terms, by locally deforming the punctured plane you can endow it with the metric of a flat cylinder which should make the topological equivalence clear.
     
  4. Apr 18, 2012 #3
    Thanks Physics Monkey, that small substitution step helped quite a bit. Can you suggest some good references where I could learn more about these things without getting too caught up in formalism (which I have appreciation for, but just no time :frown:). Ryder makes all these statements which I am sure make sense if one thinks about them the way you suggest, but that does require some background in topology which I lack (I am concurrently reading topology but I am way behind all this.)
     
  5. Apr 19, 2012 #4

    tom.stoer

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    Try Nakahara: Geometry, Topology and Physics
     
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