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Transformations for the non-linear sigma action

  1. Oct 8, 2014 #1
    For the non-linear sigma action,
    [tex]
    S_G=\frac{1}{4\pi\alpha^\prime}\int d^2\sigma\sqrt{-\gamma(\sigma)}\gamma^{\mu\nu}(\sigma)G_{ij}(X)\partial_\mu(\sigma) X^i\partial_\nu X^j(\sigma),
    [/tex]
    Let us consider an infinitesimal target space transformation [itex]X^\mu\to X^{\prime\mu}(X)=X+\epsilon\xi^\mu(X)[/itex]. The variation of the action under this transformation corresponds to the Lie derivative of the target space metric?:
    [tex]
    \delta_\xi S_G=\frac{\epsilon}{4\pi\alpha^\prime}\int d^2\sigma\sqrt{-\gamma}\gamma^{\mu\nu}\left(\mathcal{L}_\xi G_{ij}\right)\partial_\mu X^i\partial_\nu X^j
    [/tex]

    Indeed, it seems to be true by a straightforward calculation :
    [tex]
    \delta_\xi S_G=S_G[X+\epsilon\xi]-S_G[X].
    [/tex]
    But I don't know how to understand this is same to the Lie derivative.

    And how about the NS-NS 2-form term?
    [tex]
    S_B=\frac{1}{4\pi\alpha^\prime}\int d^2\sigma\varepsilon^{\mu\nu}B_{ij}\partial_\mu X^i\partial_j X^j=\frac{1}{4\pi\alpha^\prime}\int B_{ij}dX^i\wedge dX^j
    [/tex]

    The variation of this term may be [itex]\delta_\xi S_B=\frac{1}{4\pi\alpha^\prime}\int\left(\mathcal{L}_\xi B_{ij}\right)dX^i\wedge dX^j[/itex]. But I couldn't verify by the straightforward calculation.

    Please teach me the validity that the transformation of the action corresponds to the Lie derivative for the background field [itex]G, B[/itex].
     
  2. jcsd
  3. Oct 9, 2014 #2

    samalkhaiat

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    Consider how the metric changes under infinitesimal transformation
    [tex]G_{ i j } ( X ) \rightarrow \bar{ G }_{ i j } ( X + \epsilon \zeta ) = \frac{ \partial X^{ n } }{ \partial \bar{ X }^{ i } } \frac{ \partial X^{ m } }{ \partial \bar{ X }^{ j } } G_{ n m } ( X ) .[/tex]
    Expand both sides and keep terms linear in [itex]\epsilon[/itex]:
    [tex]\bar{ G }_{ i j } ( X ) - G_{ i j } ( X ) = - \epsilon \left( \zeta^{ k } \ \partial_{ k } G_{ i j } + G_{ i n } \ \partial_{ j } \zeta^{ n } + G_{ n j } \ \partial_{ i } \zeta^{ n } \right) .[/tex]
    So,
    [tex]\delta G_{ i j } \equiv \bar{ G }_{ i j } ( X ) - G_{ i j } ( X ) = - \epsilon \ \mathcal{ L }_{ \zeta } \left(G_{ i j } ( X ) \right) .[/tex]
     
  4. Oct 9, 2014 #3
    Thank you samalkhaiat.

    Your calculation seems to be just a definition of the Lie derivative, the variation at the same "coordinate point".
    My question is why the Lie derivative appears in the [itex]\sigma[/itex]-action in the form of [itex]\delta G_{ij}\partial_\mu X^i\partial_\nu X^j[/itex] when acting [itex]X\to X+\epsilon\xi[/itex] on the [itex]\sigma[/itex]-action.
     
  5. Oct 10, 2014 #4

    samalkhaiat

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    Okay, since I brought this on myself, I will experience the pain and show you how to do it. One can prove it in many different ways, but I will choose the simplest two.
    In both methods, my infinitesimal transformation is
    [tex]\bar{ X }^{ a } = X^{ a } + \epsilon f^{ a } ( X ) . \ \ \ \ (1)[/tex]
    Also, to save time, I will consider the part of the Lagrangian which changes under (1), i.e., [itex]G_{ a b } \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b }[/itex]
    Method 1:
    To first order in [itex]\epsilon[/itex], we can expand
    [tex]G_{ a b } ( \bar{ X } ) \partial_{ \mu } \bar{ X }^{ a } \partial_{ \nu } \bar{ X }^{ b } = G_{ a b } ( X ) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } + \epsilon f^{ d } \partial_{ d } G_{ a b } \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } + 2 \epsilon G_{ a b } \partial_{ \mu } X^{ a } \partial_{ c } f^{ b } \partial_{ \nu } X^{ c } .[/tex]
    In the 3rd terms, we make [itex]b \leftrightarrow c[/itex], and get
    [tex]\delta \left( G_{ a b } ( X ) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } \right) = \epsilon \left( f^{ d } \partial_{ d } G_{ a b } + 2 G_{ a c } \partial_{ b } f^{ c } \right) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } . \ \ (2)[/tex]
    Since (in the action) the world-sheet indices [itex]( \mu , \nu )[/itex] are contracted with the symmetric world-sheet metric, we can always make [itex]\mu \leftrightarrow \nu[/itex]. This allows us to make the RHS of (2) symmetric with respect to the field indices [itex]( a , b )[/itex]. So,
    [tex]\delta \left( G_{ a b } ( X ) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } \right) = \epsilon \left( f^{ d } \partial_{ d } G_{ ( a b ) } + 2 G_{ c ( a } \partial_{ b ) } f^{ c } \right) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } . \ \ \ (3)[/tex]
    We can also add the following zero to (3):
    [tex]\left( \partial_{ ( b } G_{ a ) d } - \partial_{ ( a } G_{ b ) d } \right) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } = 0 .[/tex]
    So, (3) becomes
    [tex]\delta \left( G_{ a b } ( X ) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } \right) = 2 \epsilon G_{ c ( a } \partial_{ b ) } f^{ c } \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } + \epsilon f^{ d } \left( \partial_{ d } G_{ ( a b ) } + \partial_{ ( b } G_{ a ) d } - \partial_{ ( a } G_{ b ) d } \right) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } .[/tex]
    Introducing the Christoffel symbols in the second term on the RHS, we get
    [tex]\delta \left( G_{ a b } ( X ) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } \right) = 2 \epsilon G_{ c ( a } \partial_{ b ) } f^{ c } \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } + 2 \epsilon f^{ d } \Gamma_{ ( a b ) d } \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } .[/tex]
    Or
    [tex]\delta \left( G_{ a b } ( X ) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } \right) = 2 \epsilon \left[ G_{ c ( a } \partial_{ b ) } f^{ c } + G_{ c ( a } \Gamma^{ c }_{ b ) d } f^{ d } \right] \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } = 2 \epsilon \left[ G_{ c ( a } \nabla_{ b ) } f^{ c } \right] \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } .[/tex]
    Since [itex]\nabla_{ a } G_{ b c } = 0[/itex], we arrive at
    [tex]\delta \left( G_{ a b } ( X ) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } \right) = 2 \epsilon \nabla_{ ( b } f_{ a ) } \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } = - 2 \left[ \mathcal{ L }_{ f } G_{ ab } ( X ) \right] \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } .[/tex]
    So, by integrating this over the world-sheet variables, we find
    [tex]\delta S[ X ] \sim \frac{ - \epsilon }{ 2 } \int d^{ 2 } \sigma \ \sqrt{ | h | } h^{ \mu \nu } ( \sigma ) \ \partial_{ \mu } X^{ a }( \sigma ) \ \partial_{ \nu } X^{ b } ( \sigma ) \ \mathcal{ L }_{ f } G_{ a b } ( X ) .[/tex]
    qed.
    Ok, I am now too tired to do the second method. So I will leave it for next time.
     
  6. Oct 11, 2014 #5
    Thank you ! It was very helpful to understand.
    It seems that for the [itex]\sigma[/itex]-action, the variation accidentally coincides with the Lie derivative of the metric.
    I have a feeling that this coincidence is necessary in view of the geometry of the target space.
    Can I understand in more geometrical way?
     
  7. Oct 11, 2014 #6

    samalkhaiat

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    There is no “accident”; on the contrary it is a general result in nature. Isometries of the metric (i.e. [itex]\mathcal{ L } g_{ a b } = 0[/itex]) are symmetries of the action. So it is natural to find [itex]\delta S \propto \mathcal{ L } g_{ a b }[/itex].
    May be the following second proof will help you gain more understanding.

    Method (2)
    First, I would like to show that [itex]\partial_{ \mu } X^{ a }[/itex] is a contravariant vector on the target space manifold (the X-manifold). For that we need the finite form of the transformation
    [tex]\bar{ X }^{ a } = \bar{ X }^{ a } ( X ; \epsilon ) .[/tex]
    “As a side remark; notice that the first two terms of the Taylor expansion of RHS give us back our infinitesimal (general coordinate) transformations:
    [tex]
    \bar{ X }^{ a } = \bar{ X }^{ a } ( X ; 0 ) + \epsilon \frac{ d }{ d \epsilon } \bar{ X }^{ a } ( X ; \epsilon ) |_{ \epsilon = 0 } + \cdots ,
    [/tex]
    with [itex]X^{ a } = \bar{ X }^{ a } ( X ; 0 )[/itex] and the vector field [itex]f^{ a } ( X )[/itex] is given by [itex]f^{ a } ( X ) = \frac{ d }{ d \epsilon } \bar{ X }^{ a } ( X ; \epsilon ) |_{ \epsilon = 0 }[/itex]”.
    So, using the chain rule, we see that
    [tex]\frac{ \partial \bar{ X }^{ a } }{ \partial \sigma^{ \mu } } = \frac{ \partial \bar{ X }^{ a } }{ \partial X^{ b } } \frac{ \partial X^{ b } }{ \partial \sigma^{ \mu } } .[/tex]
    Thus, [itex]\partial_{ \mu } X^{ a }[/itex] is a contravariant vector on the target space manifold.
    The transformation laws of covariant and contravariant vectors (under general coordinate transformations) can be used to define the following infinitesimal variation operator
    [tex]
    \delta^{ g } V_{ a } ( X ) \equiv \bar{ V }_{ a } ( \bar{ X } ) - V_{ a } ( X ) = - \partial_{ a } f^{ b } V_{ b } ,
    [/tex]
    [tex]
    \delta^{ g } V^{ a } ( X ) \equiv \bar{ V }^{ a } ( \bar{ X } ) - V^{ a } ( X ) = + \partial_{ b } f^{ a } V^{ b } .
    [/tex]
    Notice (i) the sign difference, and (ii) how the position of the free index changes. The above laws can be generalized to tensors of arbitrary rank. Also notice that a true scalar on the manifold gets killed by [itex]\delta^{ g }[/itex], i.e. remains invariant under general coordinate transformations.
    Infinitesimal “dragging” on the manifold defines another infinitesimal variation operator: infinitesimal shift of the X-coordinates, i.e., without reorienting the direction of a vector, induces the following infinitesimal variation
    [tex]
    \delta^{ D } V_{ a } ( X ) \equiv \bar{ V }_{ a } ( \bar{ X } ) - \bar{ V }_{ a } ( X ) = f^{ c } \partial_{ c } V_{ a } ,
    [/tex]
    [tex]
    \delta^{ D } V^{ a } ( X ) \equiv \bar{ V }^{ a } ( \bar{ X } ) - \bar{ V }^{ a } ( X ) = f^{ c } \partial_{ c } V^{ a } .
    [/tex]
    That is, the drag operator, [itex]\delta^{ D }[/itex], has the same action on all tensors (including scalars) on the manifold.
    Now, giving that the Lie derivative along the vector field [itex]f^{ a }[/itex] is defined by
    [tex]\mathcal{ L }_{ f } ( V_{ a } ) \equiv \bar{ V }_{ a } ( X ) - V_{ a } ( x ) ,[/tex]
    we can write the following operator identity
    [tex]
    \delta^{ g }_{ f } ( \cdot ) = \mathcal{ L }_{ f } ( \cdot ) + \delta^{ D }_{ f } ( \cdot ) . \ \ \ (1)
    [/tex]
    In the calculus of variation, equation (1) is (probably) the most important identity.
    Ok, let us apply these ideas on the non-linear sigma model action
    [tex]
    S[ X ] = \alpha \int d^{ 2 } \sigma \ \sqrt{ | h | } h^{ \mu \nu } ( \sigma ) \ L_{ \mu \nu } ( X ) ,
    [/tex]
    where, the object that concerns us is given by,
    [tex]L_{ \mu \nu } ( x ) = G_{ a b } ( X ) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } .[/tex]
    Since the action is a scalar on the X-manifold, it is invariant under [itex]\delta^{ g }_{ f }[/itex] for any choice of [itex]f^{ a } ( X )[/itex]. This means that changes in the covariant tensor [itex]G_{ a b }[/itex] cancel against changes in the contravariant vectors in [itex]\partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b }[/itex]:
    [tex]
    \left( \delta^{ g }_{ f } G_{ a b } \right) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } = - G_{ a b } \ \delta^{ g }_{ f } \left( \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } \right) . \ \ \ (2)
    [/tex]
    Moreover, [itex]\delta^{ D }_{ f }[/itex] kills the contravariant vector [itex]\partial_{ \mu } X^{ a }[/itex]:
    [tex]
    \delta^{ D }_{ f } ( \partial_{ \mu } X^{ a } ) = f^{ c } \partial_{ c } (\partial_{ \mu } X^{ a } ) = f^{ c } \partial_{ \mu } ( \partial_{ c } X^{ a } ) = f^{ c } \partial_{ \mu } ( \delta^{ a }_{ c } ) = 0 .
    [/tex]
    This means, from (1), that [itex]\delta^{ g }_{ f }[/itex] and [itex]\mathcal{ L }_{ f }[/itex] have identical actions on the contravariant vector [itex]\partial_{ \mu } X^{ a }[/itex].
    Thus, the infinitesimal variation of [itex]L_{ \mu \nu } ( X )[/itex] (and therefore the action) induced by an arbitrary variation in the coordinates is of the form
    [tex]
    \delta_{ f } L_{ \mu \nu } ( X ) = \left( \delta^{ D }_{ f } G_{ a b } \right) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } + G_{ a b } \ \delta^{ g }_{ f } \left( \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } \right) . \ \ \ (3)
    [/tex]
    Using (2), we rewrite (3) as
    [tex]
    \delta_{ f } L_{ \mu \nu } ( X ) = \left[ \left( \delta^{ D }_{ f } - \delta^{ g }_{ f } \right) G_{ a b } \right] \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } .
    [/tex]
    Finally, using the operator identity (1), we find
    [tex]
    \delta_{ f } ( G_{ a b } \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } ) = ( - \mathcal{ L }_{ f } G_{ a b } ) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } .
    [/tex]
    qed.
     
  8. Oct 14, 2014 #7
    [itex]\frac{\partial}{\partial X^a}[/itex] commutes with [itex]\frac{\partial}{\partial\sigma^\mu}[/itex]? Is it trivial?

    I couldn't follow this equation. Could you explain in more detail?
     
    Last edited: Oct 14, 2014
  9. Oct 15, 2014 #8

    samalkhaiat

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    Yes they commute, if they act on X or functions of X. Remember, the set [itex]\{ d X^{ a } \}[/itex] is one-form basis on the X-manifold and [itex]\partial_{ \mu } X^{ a }[/itex] are the components of the pull-back of such one-form: [itex]\chi^{ * } ( d X^{ a } ) = \partial_{ \mu } X^{ a } d \sigma^{ \mu }[/itex], with [itex]\chi[/itex] is the map from the world-sheet to the X-manifold.
    We assume that an arbitrary variation [itex]\delta[/itex] acts as derivation with non-zero contributions from both terms:
    [tex]\delta L = ( \delta G_{ a b } ) \ \partial X^{ a } \partial X^{ b } + G_{ a b } \delta ( \partial X^{ a } \partial X^{ b } ) .[/tex]
    Since we only have [itex]\delta^{ D }[/itex] and [itex]\delta^{ g }[/itex] at our disposal, the only equation that gives non-zero contributions from both terms is the one I wrote down. That is [itex]\delta G \equiv \delta^{ D } G[/itex] and [itex]\delta ( \partial X \partial X ) \equiv \delta^{ g } ( \partial X \partial X )[/itex].
     
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