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?


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 '19 at 20:27

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