# Diffeomorphism invariance of the Polyakov action

1. Sep 7, 2014

### maverick280857

[SOLVED] Diffeomorphism invariance of the Polyakov action

Hi,

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 $\sigma^0 = \sigma$ and $\sigma^1 = t$ and the independent world-sheet metric by $h_{\alpha\beta}(\sigma)$, the Polyakov action can be written as

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

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

Now, if I define

$$\delta_{E}h_{\alpha\beta}(\sigma) = h_{\alpha\beta}'(\sigma) - h_{\alpha\beta}(\sigma)$$

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

$${\sigma'}^{\gamma} = \sigma^\gamma - \xi^\gamma(\sigma)$$

then I would like to show that

$$\delta_E h_{\alpha\beta} = \partial_\alpha \xi^\gamma h_{\gamma\beta} + \partial_\beta \xi^\gamma h_{\alpha\gamma} + \xi^\gamma \partial_\gamma h_{\alpha\beta}$$

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

My working

$$h'_{\gamma\delta}(\sigma') = h_{\alpha\beta}(\sigma)\frac{d\sigma^\alpha}{d{\sigma'}^\gamma}\frac{d{\sigma}^\beta}{d{\sigma'}^\delta}$$

Now by definition,

$$\sigma^\alpha = {\sigma'}^\alpha + \xi^\alpha(\sigma)$$

So,

$$\frac{d\sigma^\alpha}{d{\sigma'}^\beta} = \delta^\alpha_\beta + \frac{d\sigma^\delta}{d{\sigma'}^\beta}\frac{d{\xi}^\alpha(\sigma)}{d {\sigma}^{\delta}}$$

But this seems to give only the first two terms of $\delta_E h$ 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.

Thanks!

Last edited: Sep 7, 2014
2. Sep 7, 2014

### maverick280857

Solved. Thanks anyway!

3. Sep 7, 2014

### samalkhaiat

Expand the left-hand-side as well
$$h^{'}_{ \mu \nu } ( \sigma^{'} ) = h^{'}_{ \mu \nu } ( \sigma - \xi ) \approx h^{'}_{ \mu \nu } ( \sigma ) - \xi^{ \rho } \partial_{ \rho } h^{'}_{ \mu \nu } ( \sigma ) .$$
But infinitesimally
$$\xi^{ \rho } \partial_{ \rho } h^{'}_{ \mu \nu } ( \sigma ) \approx \xi^{ \rho } \partial_{ \rho } h_{ \mu \nu } ( \sigma ) .$$

4. Sep 8, 2014

### maverick280857

Thanks samalkhaiat, yes, I got that.

I would also like to show that

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

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

(b) $\delta_E(\sqrt{-h}) = \partial_\rho(\xi^\rho \sqrt{-h})$

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
5. Sep 8, 2014

### samalkhaiat

That is correct.

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

Sam