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There is an exercise 2.8 Baumgarte's Numerical Relativity (p. 32):

Show that 3-dimensional covariant derivative is compatible with the spatial metric $\gamma_{ab}$, that is, show that $$ D_a \gamma_{bc} = 0. $$

Where $$ D_a f \equiv \gamma_a^{\; b} \nabla_b f $$ and more generally $$ D_a T^b_{\; c} \equiv \gamma_a^{\; d} \gamma_e^{\; b} \gamma_c^{\; f} \nabla_d T^e_{\; f}. $$

The spatial metric is defined as following $$\label{eq:spatial-metric} \gamma_{ab} = g_{ab} + n_a n_b, $$ where $$ n^a \equiv -g^{ab} \omega_b, \\ \omega_a \equiv \alpha \Omega_a, \\ ||\Omega||^2 = g^{ab} \nabla_a t \nabla_b t \equiv -\frac{1}{\alpha^2}. $$


My attempt involved using (general) metric compatibility: $$ \nabla_a g_{bc} = 0 $$ in order to reduce the equation from the question to $$ D_a \gamma_{bc} = \gamma^d_{\; a} \gamma^e_{\; b} \gamma^f_{\; c} \nabla_d \left( n_b n_c \right). $$

Now I'm completely lost on what I could do next with it. Resolving the covariant derivative and using Christoffel symbol doesn't seem to lead anywhere and the calculation gets quite formidable.

What's the best way to make progress here?

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Your expression for the $3$-dimensional covariant derivative of the spatial metric has typo. It should read $$ D_a \gamma_{bc} = \gamma^d_{\; a} \gamma^e_{\; b} \gamma^f_{\; c} \nabla_d \left( n_e n_f \right). $$ Now, you can easily show that $\gamma^e_{\; b} n_e = 0$ either by using the relations you have for $n^a$, or intuitively by considering that $\gamma^e_{\; b} n_e$ is the projection of the unit normal vector on the spatial hypersurface which vanishes.

So you need to use the Leibniz's rule to show the required relation, i.e., $$ \gamma^e_{\; b} \gamma^f_{\; c} \nabla_d \left( n_e n_f \right) = \gamma^e_{\; b} \, (\nabla_d n_e) \, (\underbrace{\gamma^f_{\; c} n_f}_{=0}) + \gamma^f_{\; c} \, (\nabla_d n_f) \, (\underbrace{\gamma^e_{\; b} n_e}_{=0}) = 0 . $$

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