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Differentiation on orbifolds

  1. Dec 18, 2008 #1
    I'm trying to understand how Randall and Sundrum go from Eq. (9) to Eq. (10) in their RS1 paper:

    http://arxiv.org/abs/hep-ph/9905221

    I understand that since the extra dimension [itex]\phi[/itex] is periodic, we must have

    [itex]\frac{d^2}{d\phi^2}|\phi|\propto \delta(0) - \delta(\phi - \pi)[/itex].

    However, I'm not entirely sure why the proportionality constant is 2, i.e, why

    [itex]\frac{d^2}{d\phi^2}|\phi|= 2[\delta(0) - \delta(\phi - \pi)][/itex].

    I'm assuming that it's related to the [itex]\mathbb{Z}_2[/itex] symmetry of the [itex]S^1/\mathbb{Z}_2[/itex] orbifold, but I'm not sure how to show his.

    Thanks.
     
  2. jcsd
  3. Dec 18, 2008 #2

    Hurkyl

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    Well, if we lift the absolute value function back to be a function on R, we get a triangle wave; specifically, the function defined on [itex][-\pi, \pi][/itex] by [itex]f(x) = |x|[/itex], and extended periodically by [itex]f(x + 2\pi) = f(x)[/itex].

    The first derivative of this function is a square wave (defined almost everywhere), alternating between 1 and -1. The second (distributional) derivative is therefore

    [tex]f''(x) = \sum_{k = -\infty}^{+\infty} (-1)^k 2 \delta(x - k \pi)[/tex]

    This distribution has the appropriate symmetry to be a distribution on your orbifold, being even and periodic, so I would think it ought to be the correct result for the derivative.

    (Disclaimer: I haven't actually worked with orbifolds before)
     
  4. Dec 19, 2008 #3
    Yes, but where did the factor of 2 come from??

    Edit: Oh I get it. Thanks

    [itex]|z|'' = sgn(z)' = [2\theta(z)-1]' = 2 \delta(z)[/itex].
     
    Last edited: Dec 19, 2008
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