Why can't the dynamical metric in the Polyakov Action be the induced metric?

Question

The Polyakov action is given by $$ S_{P} = -\frac{T}{2} \int d^2\sigma \sqrt{h} h^{\mu \nu} \gamma_{\mu \nu} \tag{1}$$

where $h_{\mu \nu}$ is the dynamical metric and $\gamma_{\mu \nu}$ is the induced metric. It is well known that this is equivalent to the Nambu-Goto Action

$$ S_{NG} = -T \int d^2\sigma \sqrt{\gamma}.\tag{2} $$

Theoretically, I don't see any reason why the dynamical metric cannot be the same as the induced metric. However, if the dynamical metric is set equal to the induced metric in $S_P$, the two actions differ by a factor of $\frac{1}{2}$. Am I correct in that this means that the dynamical metric cannot be the induced metric? If so, why?

Answer 1

Your factors of $2$ are wrong. You do not get the $1/2$ difference because

$$ \gamma^{\mu\nu}\gamma_{\mu\nu}=\delta_{\mu}^{\mu}=2 $$

So replacing $h^{\mu\nu}$ by $\gamma^{\mu\nu} $ in eq. 1 gives eq. 2.

Note that you can also replace $h^{\mu\nu}$ by $e^{-2\Omega(\sigma)}\gamma^{\mu\nu}$ in eq. 1 and still get eq. 2. This is so because $\sqrt{h}= e^{+2\Omega(\sigma)}\sqrt{\gamma}$ if $h^{\mu\nu}=e^{-2\Omega(\sigma)}\gamma^{\mu\nu}$. This is the Weyl symmetry of the Polyakov action, i.e. $h_{\mu\nu} $ is only defined up to a scaling factor.