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Some questions about isometry

  1. Oct 3, 2014 #1
    Isometry is the symmetry s.t.
    [tex]
    g^\prime_{\mu\nu}(x)=g_{\mu\nu}(x)
    [/tex]
    under the transformation [itex]x^\mu\to x^{\prime\mu}(x)[/itex]. This means under infinitesimal transformation
    [tex]
    x^\mu\to x^\mu+\epsilon \xi^\mu(x)
    [/tex]
    where [itex]\epsilon[/itex] is any infinitesimal constant, the vector field [itex]\xi^\mu(x)[/itex] satisfies Killing equation:
    [tex]
    \nabla_\mu\xi_\nu(x)+\nabla_\nu\xi_\mu(x)=0.
    [/tex]

    I have questions about above description.
    1. Why the infinitesimal parameter [itex]\epsilon[/itex] is constant and [itex]\xi^\mu[/itex] is depend on [itex]x[/itex]? The most general form seems to be [itex]\epsilon(x)\xi^\mu(x)[/itex].

    2.Is this transformation global or local? It's global symmetry because [itex]\epsilon[/itex] is constant?

    3.I have seen the description like "isometry means the metric is independent on the coordinate of the isometry direction." That is, if the isometry transformation is [itex]x^0\to x^0+\epsilon\xi(x)[/itex], [itex]g_{\mu\nu}=g_{\mu\nu}(x^1,\dots)[/itex] or [itex]\partial_0g_{\mu\nu}=0[/itex]. But I think this is true only when the Killing equation have a constant solution. The statement is true in more general?
     
  2. jcsd
  3. Oct 3, 2014 #2

    Ben Niehoff

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    Science Advisor
    Gold Member

    Not quite. If you have a diffeomorphism ##\phi : M \to M##, then ##\phi## is an isometry when

    $$(\phi^* g'_{\mu\nu}) (\phi^{-1}(x')) = g_{\mu\nu}(x).$$
    That is, you must pull back ##g'_{\mu\nu}## along ##\phi##.

    ##\xi^\mu(x)## is already an arbitrary function, so there is no need to make the infinitesimal parameter also a function.

    Most mathematical texts are referring to global properties unless they say otherwise. That means if your manifold cannot be covered by a single chart, you need to concern yourself with transition functions to move between charts.

    Often a Killing vector will vanish at some points on the manifold, so that although you do have an isometry, it does not necessarily "move" every point. For example, if you rotate a sphere, the north and south poles do not change.

    That's a bit sloppy.

    It is true that if you have a continuous isometry ##\xi^\mu##, it is possible to choose coordinates such that ##\xi = \partial_\varphi## for some coordinate ##\varphi##. However, if you have multiple isometries, it is not necessarily true that you can do this for all of them simultaneously. Again, the sphere is a good example. The sphere has three rotational isometries, but you can only "coordinatize" one of them at a time.

    In general, you can choose adapted coordinates for any set of isometries that mutually commute (I believe this comes from the Frobenius theorem).
     
  4. Oct 4, 2014 #3
    Thank you, BenNiehoff

    Could you show the statement,
    I interpreted this means the Killing equation have a constant solution like [itex]\xi^\mu=(1,0,0,\dots)[/itex] if the manifold have a single continuous isometry. Is it true?

    Can we consider "gauged" isometries? If we make naively [itex]\epsilon[/itex] to be a function [itex]\epsilon(x)[/itex], there is no change by absorbing [itex]x[/itex]-dependence to [itex]\xi^\mu(x)[/itex].
     
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