ADM formulation of GR derivative on the 3-metric In the ADM formalism where the projector is given by ${P^\mu}_\alpha={\delta^\mu}_\alpha+n^\mu{n}_\alpha$ and $n^\alpha$ is a future pointing normal vector to the constant time hypersurface $\Sigma$. The 3 derivative can be defined using the projection operator $\mathcal{D}_\alpha={P^\mu}_\alpha\nabla_\mu$
I am trying to proof the claim that:
 $\mathcal{D}_\gamma(^{(3)}g_{\alpha\beta})=0$ where $^{(3)}g_{\alpha\beta}=g_{\alpha\beta}+n_\alpha{n}_\beta$.
Obviously $\nabla_\mu{g_{\alpha\beta}}=0$ but I am not seeing the argument why the term $\nabla_\mu(n_\alpha{n}_\beta)$ would vanish generally. Or is it the case that I can not proof this generally and need to substitute in the explicit values and Christopher symbols from the ADM metric? I might just not understanding the properties of the normal vector clearly, for example I can't see either why they vanish by parallel transport i.e $n^\mu\nabla_\nu{n_\mu}=0.$
 A: Actually, the derivative operator you defined is not the usual definition of the "3-derivative" on the spatial hypersurface.  The intrinsic covariant derivative is defined to project all indices tangential to the surface after computing the covariant derivative, i.e. for $V^\alpha$ a tangent vector, you can write
$$D_\beta V^\alpha = P^\mu_\beta P^\alpha_\nu \nabla_\mu V^\nu.$$
This differs from your definition $\mathcal{D}_\beta = P^\mu_\beta\nabla_\mu$ by terms involving the extrinsic curvature of the surface.  Also, $D_\beta$ is the unique derivative operator compatible with the the spatial 3-metric, so it is correct that $\mathcal{D}_\beta$ does not annihilate the three metric.  Instead, you should be able to show that 
$$
D_\alpha {}^{(3)}g_{\beta\gamma} = P^\mu_\alpha P^\nu_\beta P^\rho_\gamma\nabla_\mu(g_{\nu\rho} + n_\nu n_\rho) = 0.
$$
Showing this involves the identity you mention $n^\mu \nabla_\alpha n_\mu = 0$.  Note that this equation doesn't have anything to do with parallel transport (which would be related to the other contraction $n^\mu \nabla_\mu n_\alpha$), but instead is just a consequence of $n^\alpha$ being normalized, $n^\mu \nabla_\alpha n_\mu = \frac12 \nabla_\alpha(n^\mu n_\mu) = \frac12\nabla_\alpha(-1) = 0$.
A: $P^{b}{}_{c}n^{a}\nabla_{a}n_{b}$ also is zero.  To prove this, you need to see that the vector $n_{a}$ is the level set of the function $\tau$ used to generate the three-surfaces, so you have $n_{a} = \alpha \nabla_{a}\tau$, where $\alpha$ is the lapse function, and $\tau$ is the function that defines your embedding of spacelike three-surfaces into the four-space.
Using this fact, and the fact that, for any torsion-free connection, $\nabla_{a}\nabla_{b}\tau = \nabla_{b}\nabla_{a}\tau$, and the insight from the previous answer, you can derive your result.
