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

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?

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.