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In the Gauge fixing of Polyakov action we do general coordinate transformation where we take the transformation stated below

$$h_{\alpha\beta} = e^{\phi(\sigma)}\eta_{\alpha\beta}.$$

But here the left side has three free parameters (one less in the 2x2 h metric as it is symmetric in the indices) while the right side only involves one parameter $\phi$ ; taking the $\eta$ metric to be constant as

$$\eta{_\alpha}{_\beta} = diag(-1,1).$$

So how can we put an equality if there are not equal free parameters on both sides? What could be the underlying reason?

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It is important to also keep track of the local symmetries present in each case. In the first case the dynamical metric $h_{\alpha\beta}$ has three gauge symmetries - two diffeomorphisms and a Weyl symmetry.

It is possible to choose the parametrisation in a particular way such that $h_{\alpha\beta} = e^{\phi(\sigma)}\eta_{\alpha\beta}$, where one drops the diffeomorphism invariance by choosing a particular parametrisation. This means that initially one has three degrees of freedom and two local symetries (diffeomorphisms), and at the end one has only a single degree of freedom and no symmetries left. It is the number of DoF - local symmetries that actually matters, as these are the physical DoF.

One can take this a step further and use Weyl symmetry to set the metric locally to $h_{\alpha\beta} = \eta_{\alpha\beta}$. This step can be even extended globally if the worldsheet has a certain euler characteristic.

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  • $\begingroup$ Thank you very much. $\endgroup$ – MRITYUNJAY NATH Sep 8 at 20:27

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