Transformations for the non-linear sigma action

[synoe]

For the non-linear sigma action,
<br /> 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),<br />
Let us consider an infinitesimal target space transformation X^\mu\to X^{\prime\mu}(X)=X+\epsilon\xi^\mu(X). The variation of the action under this transformation corresponds to the Lie derivative of the target space metric?:
<br /> \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<br />

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

And how about the NS-NS 2-form term?
<br /> 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<br />

The variation of this term may be \delta_\xi S_B=\frac{1}{4\pi\alpha^\prime}\int\left(\mathcal{L}_\xi B_{ij}\right)dX^i\wedge dX^j. 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 G, B.

 

[samalkhaiat]

Consider how the metric changes under infinitesimal transformation
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 ) .
Expand both sides and keep terms linear in \epsilon:
\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) .
So,
\delta G_{ i j } \equiv \bar{ G }_{ i j } ( X ) - G_{ i j } ( X ) = - \epsilon \ \mathcal{ L }_{ \zeta } \left(G_{ i j } ( X ) \right) .

 

[synoe]

Thank you samalkhaiat.

Consider how the metric changes under infinitesimal transformation
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 ) .
Expand both sides and keep terms linear in \epsilon:
\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) .
So,
\delta G_{ i j } \equiv \bar{ G }_{ i j } ( X ) - G_{ i j } ( X ) = - \epsilon \ \mathcal{ L }_{ \zeta } \left(G_{ i j } ( X ) \right) .

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 \sigma-action in the form of \delta G_{ij}\partial_\mu X^i\partial_\nu X^j when acting X\to X+\epsilon\xi on the \sigma-action.

 

[samalkhaiat]

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 \sigma-action in the form of \delta G_{ij}\partial_\mu X^i\partial_\nu X^j when acting X\to X+\epsilon\xi on the \sigma-action.

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
\bar{ X }^{ a } = X^{ a } + \epsilon f^{ a } ( X ) . \ \ \ \ (1)
Also, to save time, I will consider the part of the Lagrangian which changes under (1), i.e., G_{ a b } \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b }
Method 1:
To first order in \epsilon, we can expand
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 } .
In the 3rd terms, we make b \leftrightarrow c, and get
\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)
Since (in the action) the world-sheet indices ( \mu , \nu ) are contracted with the symmetric world-sheet metric, we can always make \mu \leftrightarrow \nu. This allows us to make the RHS of (2) symmetric with respect to the field indices ( a , b ). So,
\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)
We can also add the following zero to (3):
\left( \partial_{ ( b } G_{ a ) d } - \partial_{ ( a } G_{ b ) d } \right) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } = 0 .
So, (3) becomes
\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 } .
Introducing the Christoffel symbols in the second term on the RHS, we get
\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 } .
Or
\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 } .
Since \nabla_{ a } G_{ b c } = 0, we arrive at
\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 } .
So, by integrating this over the world-sheet variables, we find
\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 ) .
qed.
Ok, I am now too tired to do the second method. So I will leave it for next time.

 

[synoe]

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
\bar{ X }^{ a } = X^{ a } + \epsilon f^{ a } ( X ) . \ \ \ \ (1)
Also, to save time, I will consider the part of the Lagrangian which changes under (1), i.e., G_{ a b } \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b }
Method 1:
To first order in \epsilon, we can expand
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 } .
In the 3rd terms, we make b \leftrightarrow c, and get
\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)
Since (in the action) the world-sheet indices ( \mu , \nu ) are contracted with the symmetric world-sheet metric, we can always make \mu \leftrightarrow \nu. This allows us to make the RHS of (2) symmetric with respect to the field indices ( a , b ). So,
\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)
We can also add the following zero to (3):
\left( \partial_{ ( b } G_{ a ) d } - \partial_{ ( a } G_{ b ) d } \right) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } = 0 .
So, (3) becomes
\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 } .
Introducing the Christoffel symbols in the second term on the RHS, we get
\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 } .
Or
\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 } .
Since \nabla_{ a } G_{ b c } = 0, we arrive at
\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 } .
So, by integrating this over the world-sheet variables, we find
\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 ) .
qed.
Ok, I am now too tired to do the second method. So I will leave it for next time.

Thank you ! It was very helpful to understand.
It seems that for the \sigma-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?

 

[samalkhaiat]

Thank you ! It was very helpful to understand.
It seems that for the \sigma-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?

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

Method (2)
First, I would like to show that \partial_{ \mu } X^{ a } is a contravariant vector on the target space manifold (the X-manifold). For that we need the finite form of the transformation
\bar{ X }^{ a } = \bar{ X }^{ a } ( X ; \epsilon ) .
“As a side remark; notice that the first two terms of the Taylor expansion of RHS give us back our infinitesimal (general coordinate) transformations:
<br /> \bar{ X }^{ a } = \bar{ X }^{ a } ( X ; 0 ) + \epsilon \frac{ d }{ d \epsilon } \bar{ X }^{ a } ( X ; \epsilon ) |_{ \epsilon = 0 } + \cdots ,<br />
with X^{ a } = \bar{ X }^{ a } ( X ; 0 ) and the vector field f^{ a } ( X ) is given by f^{ a } ( X ) = \frac{ d }{ d \epsilon } \bar{ X }^{ a } ( X ; \epsilon ) |_{ \epsilon = 0 }”.
So, using the chain rule, we see that
\frac{ \partial \bar{ X }^{ a } }{ \partial \sigma^{ \mu } } = \frac{ \partial \bar{ X }^{ a } }{ \partial X^{ b } } \frac{ \partial X^{ b } }{ \partial \sigma^{ \mu } } .
Thus, \partial_{ \mu } X^{ a } 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
<br /> \delta^{ g } V_{ a } ( X ) \equiv \bar{ V }_{ a } ( \bar{ X } ) - V_{ a } ( X ) = - \partial_{ a } f^{ b } V_{ b } ,<br />
<br /> \delta^{ g } V^{ a } ( X ) \equiv \bar{ V }^{ a } ( \bar{ X } ) - V^{ a } ( X ) = + \partial_{ b } f^{ a } V^{ b } .<br />
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 \delta^{ g }, 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
<br /> \delta^{ D } V_{ a } ( X ) \equiv \bar{ V }_{ a } ( \bar{ X } ) - \bar{ V }_{ a } ( X ) = f^{ c } \partial_{ c } V_{ a } ,<br />
<br /> \delta^{ D } V^{ a } ( X ) \equiv \bar{ V }^{ a } ( \bar{ X } ) - \bar{ V }^{ a } ( X ) = f^{ c } \partial_{ c } V^{ a } .<br />
That is, the drag operator, \delta^{ D }, has the same action on all tensors (including scalars) on the manifold.
Now, giving that the Lie derivative along the vector field f^{ a } is defined by
\mathcal{ L }_{ f } ( V_{ a } ) \equiv \bar{ V }_{ a } ( X ) - V_{ a } ( x ) ,
we can write the following operator identity
<br /> \delta^{ g }_{ f } ( \cdot ) = \mathcal{ L }_{ f } ( \cdot ) + \delta^{ D }_{ f } ( \cdot ) . \ \ \ (1)<br />
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
<br /> S[ X ] = \alpha \int d^{ 2 } \sigma \ \sqrt{ | h | } h^{ \mu \nu } ( \sigma ) \ L_{ \mu \nu } ( X ) ,<br />
where, the object that concerns us is given by,
L_{ \mu \nu } ( x ) = G_{ a b } ( X ) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } .
Since the action is a scalar on the X-manifold, it is invariant under \delta^{ g }_{ f } for any choice of f^{ a } ( X ). This means that changes in the covariant tensor G_{ a b } cancel against changes in the contravariant vectors in \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b }:
<br /> \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)<br />
Moreover, \delta^{ D }_{ f } kills the contravariant vector \partial_{ \mu } X^{ a }:
<br /> \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 .<br />
This means, from (1), that \delta^{ g }_{ f } and \mathcal{ L }_{ f } have identical actions on the contravariant vector \partial_{ \mu } X^{ a }.
Thus, the infinitesimal variation of L_{ \mu \nu } ( X ) (and therefore the action) induced by an arbitrary variation in the coordinates is of the form
<br /> \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)<br />
Using (2), we rewrite (3) as
<br /> \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 } .<br />
Finally, using the operator identity (1), we find
<br /> \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 } .<br />
qed.

 

[synoe]

Moreover, \delta^{ D }_{ f } kills the contravariant vector \partial_{ \mu } X^{ a }:
<br /> \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 .<br />

\frac{\partial}{\partial X^a} commutes with \frac{\partial}{\partial\sigma^\mu}? Is it trivial?

Thus, the infinitesimal variation of L_{ \mu \nu } ( X ) (and therefore the action) induced by an arbitrary variation in the coordinates is of the form
<br /> \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)<br />

I couldn't follow this equation. Could you explain in more detail?

 

[samalkhaiat]

\frac{\partial}{\partial X^a} commutes with \frac{\partial}{\partial\sigma^\mu}? Is it trivial?

Yes they commute, if they act on X or functions of X. Remember, the set \{ d X^{ a } \} is one-form basis on the X-manifold and \partial_{ \mu } X^{ a } are the components of the pull-back of such one-form: \chi^{ * } ( d X^{ a } ) = \partial_{ \mu } X^{ a } d \sigma^{ \mu }, with \chi is the map from the world-sheet to the X-manifold.

I couldn't follow this equation. Could you explain in more detail?

We assume that an arbitrary variation \delta acts as derivation with non-zero contributions from both terms:
\delta L = ( \delta G_{ a b } ) \ \partial X^{ a } \partial X^{ b } + G_{ a b } \delta ( \partial X^{ a } \partial X^{ b } ) .
Since we only have \delta^{ D } and \delta^{ g } at our disposal, the only equation that gives non-zero contributions from both terms is the one I wrote down. That is \delta G \equiv \delta^{ D } G and \delta ( \partial X \partial X ) \equiv \delta^{ g } ( \partial X \partial X ).