# Transformations for the non-linear sigma action

1. Oct 8, 2014

### synoe

For the non-linear sigma action,
$$S_G=\frac{1}{4\pi\alpha^\prime}\int d^2\sigma\sqrt{-\gamma(\sigma)}\gamma^{\mu\nu}(\sigma)G_{ij}(X)\partial_\mu(\sigma) X^i\partial_\nu X^j(\sigma),$$
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?:
$$\delta_\xi S_G=\frac{\epsilon}{4\pi\alpha^\prime}\int d^2\sigma\sqrt{-\gamma}\gamma^{\mu\nu}\left(\mathcal{L}_\xi G_{ij}\right)\partial_\mu X^i\partial_\nu X^j$$

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

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

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 $G, B$.

2. Oct 9, 2014

### 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,
$$\delta G_{ i j } \equiv \bar{ G }_{ i j } ( X ) - G_{ i j } ( X ) = - \epsilon \ \mathcal{ L }_{ \zeta } \left(G_{ i j } ( X ) \right) .$$

3. Oct 9, 2014

### synoe

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.

4. Oct 10, 2014

### samalkhaiat

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.

5. Oct 11, 2014

### synoe

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?

6. Oct 11, 2014

### samalkhaiat

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:
$$\bar{ X }^{ a } = \bar{ X }^{ a } ( X ; 0 ) + \epsilon \frac{ d }{ d \epsilon } \bar{ X }^{ a } ( X ; \epsilon ) |_{ \epsilon = 0 } + \cdots ,$$
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
$$\delta^{ g } V_{ a } ( X ) \equiv \bar{ V }_{ a } ( \bar{ X } ) - V_{ a } ( X ) = - \partial_{ a } f^{ b } V_{ b } ,$$
$$\delta^{ g } V^{ a } ( X ) \equiv \bar{ V }^{ a } ( \bar{ X } ) - V^{ a } ( X ) = + \partial_{ b } f^{ a } V^{ b } .$$
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
$$\delta^{ D } V_{ a } ( X ) \equiv \bar{ V }_{ a } ( \bar{ X } ) - \bar{ V }_{ a } ( X ) = f^{ c } \partial_{ c } V_{ a } ,$$
$$\delta^{ D } V^{ a } ( X ) \equiv \bar{ V }^{ a } ( \bar{ X } ) - \bar{ V }^{ a } ( X ) = f^{ c } \partial_{ c } V^{ a } .$$
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
$$\delta^{ g }_{ f } ( \cdot ) = \mathcal{ L }_{ f } ( \cdot ) + \delta^{ D }_{ f } ( \cdot ) . \ \ \ (1)$$
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
$$S[ X ] = \alpha \int d^{ 2 } \sigma \ \sqrt{ | h | } h^{ \mu \nu } ( \sigma ) \ L_{ \mu \nu } ( X ) ,$$
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 $\partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b }$:
$$\left( \delta^{ g }_{ f } G_{ a b } \right) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } = - G_{ a b } \ \delta^{ g }_{ f } \left( \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } \right) . \ \ \ (2)$$
Moreover, $\delta^{ D }_{ f }$ kills the contravariant vector $\partial_{ \mu } X^{ a }$:
$$\delta^{ D }_{ f } ( \partial_{ \mu } X^{ a } ) = f^{ c } \partial_{ c } (\partial_{ \mu } X^{ a } ) = f^{ c } \partial_{ \mu } ( \partial_{ c } X^{ a } ) = f^{ c } \partial_{ \mu } ( \delta^{ a }_{ c } ) = 0 .$$
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
$$\delta_{ f } L_{ \mu \nu } ( X ) = \left( \delta^{ D }_{ f } G_{ a b } \right) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } + G_{ a b } \ \delta^{ g }_{ f } \left( \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } \right) . \ \ \ (3)$$
Using (2), we rewrite (3) as
$$\delta_{ f } L_{ \mu \nu } ( X ) = \left[ \left( \delta^{ D }_{ f } - \delta^{ g }_{ f } \right) G_{ a b } \right] \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } .$$
Finally, using the operator identity (1), we find
$$\delta_{ f } ( G_{ a b } \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } ) = ( - \mathcal{ L }_{ f } G_{ a b } ) \partial_{ \mu } X^{ a } \partial_{ \nu } X^{ b } .$$
qed.

7. Oct 14, 2014

### synoe

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

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

Last edited: Oct 14, 2014
8. Oct 15, 2014

### samalkhaiat

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.
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 $\delta ( \partial X \partial X ) \equiv \delta^{ g } ( \partial X \partial X )$.