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Diffeomorphism invariance of the Polyakov action

  1. Sep 7, 2014 #1
    [SOLVED] Diffeomorphism invariance of the Polyakov action


    I'm struggling with something that is quite elementary. I know that the Polyakov action is diffeomorphism invariant and Weyl invariant. Denoting the world-sheet coordinates [itex]\sigma^0 = \sigma[/itex] and [itex]\sigma^1 = t[/itex] and the independent world-sheet metric by [itex]h_{\alpha\beta}(\sigma)[/itex], the Polyakov action can be written as

    [tex]S_{P} = -\frac{T}{2}\int d^{2}\sigma \, \sqrt{-h}h^{\alpha\beta}\partial_\alpha X^\mu \partial_\beta X_\mu[/tex]

    where T is the string tension and [itex]\alpha, \beta[/itex] run over 0 and 1 (labeling [itex]\sigma[/itex] and [itex]t[/itex]).

    Now, if I define

    [tex]\delta_{E}h_{\alpha\beta}(\sigma) = h_{\alpha\beta}'(\sigma) - h_{\alpha\beta}(\sigma)[/tex]

    as the Einstein variation of the independent world-sheet metric, where the primed coordinates denote the transformed world-sheet coordinates under a diffeomorphism defined by

    [tex]{\sigma'}^{\gamma} = \sigma^\gamma - \xi^\gamma(\sigma)[/tex]

    then I would like to show that

    [tex]\delta_E h_{\alpha\beta} = \partial_\alpha \xi^\gamma h_{\gamma\beta} + \partial_\beta \xi^\gamma h_{\alpha\gamma} + \xi^\gamma \partial_\gamma h_{\alpha\beta}[/tex]

    This is supposed to follow from chain rule, but I do not quite get all the terms.

    My working

    [tex]h'_{\gamma\delta}(\sigma') = h_{\alpha\beta}(\sigma)\frac{d\sigma^\alpha}{d{\sigma'}^\gamma}\frac{d{\sigma}^\beta}{d{\sigma'}^\delta}[/tex]

    Now by definition,

    [tex]\sigma^\alpha = {\sigma'}^\alpha + \xi^\alpha(\sigma)[/tex]


    [tex]\frac{d\sigma^\alpha}{d{\sigma'}^\beta} = \delta^\alpha_\beta + \frac{d\sigma^\delta}{d{\sigma'}^\beta}\frac{d{\xi}^\alpha(\sigma)}{d {\sigma}^{\delta}}[/tex]

    But this seems to give only the first two terms of [itex]\delta_E h[/itex] and not the third one, which is the usual derivative (change along gradient) term that I'd expect to get.

    Any suggestions will be greatly appreciated.

    Last edited: Sep 7, 2014
  2. jcsd
  3. Sep 7, 2014 #2
    Solved. Thanks anyway!
  4. Sep 7, 2014 #3


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    Expand the left-hand-side as well
    [tex]h^{'}_{ \mu \nu } ( \sigma^{'} ) = h^{'}_{ \mu \nu } ( \sigma - \xi ) \approx h^{'}_{ \mu \nu } ( \sigma ) - \xi^{ \rho } \partial_{ \rho } h^{'}_{ \mu \nu } ( \sigma ) .[/tex]
    But infinitesimally
    [tex]\xi^{ \rho } \partial_{ \rho } h^{'}_{ \mu \nu } ( \sigma ) \approx \xi^{ \rho } \partial_{ \rho } h_{ \mu \nu } ( \sigma ) .[/tex]
  5. Sep 8, 2014 #4
    Thanks samalkhaiat, yes, I got that.

    I would also like to show that

    (a) [itex]\delta_E(h^{\alpha\beta}\partial_\alpha X^\mu \partial_\beta X_\mu) = \xi^\rho \partial_\rho(h^{\alpha\beta}\partial_\alpha X^\mu \partial_\beta X_\mu)[/itex] because "[itex]h^{\alpha\beta}\partial_\alpha X^\mu \partial_\beta X_\mu[/itex] is an Einstein scalar". What is an Einstein scalar?

    My guess is that [itex]h^{\alpha\beta}\partial_\alpha X^\mu \partial_\beta X_\mu[/itex] transforms as a scalar function.

    (b) [itex]\delta_E(\sqrt{-h}) = \partial_\rho(\xi^\rho \sqrt{-h})[/itex]

    This one worries me a little because of the placement of parenthesis...any ideas?

    How does one prove these relations?
    Last edited: Sep 8, 2014
  6. Sep 8, 2014 #5


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    That is correct.

    Use the following fact. For any operator [itex]\hat{ \delta }[/itex], satisfying the Leibnz rule such as [itex]( \partial , \delta )[/itex], the following can be proven
    [tex]\frac{ \hat{ \delta }( \sqrt{ - g } ) }{ \sqrt{ - g } } = \frac{1}{2} g^{ \mu \nu } \hat{ \delta } g_{ \mu \nu } .[/tex]

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