Vanishing of rotation of timelike unit vector in numerical relativity I'm a bit confused. I don't think it is that difficult, but still don't manage :-(
So the question is why the rotation of the timelike unit-vector in numerical relativity
$$n_\mu=-\alpha \nabla_\mu t$$
vanishes. $\alpha=\left(-g^{\mu\nu} \nabla_\mu \nabla_\nu t\right)^{-\frac{1}{2}}$ is a normalization constant and $t$ the coordinate time.
The rotation then is $$\nabla_\mu n_\nu - \nabla_\nu n_\mu$$ as far as I expect, which becomes
$$-\alpha \nabla_\mu \nabla_\nu t - (\nabla_\mu \alpha)(\nabla_\nu t) + \alpha \nabla_\nu \nabla_\mu t + (\nabla_\nu \alpha)(\nabla_\mu t) = (\nabla_\nu \alpha)(\nabla_\mu t) - (\nabla_\mu \alpha)(\nabla_\nu t)$$
because $$\nabla_{[\mu}\nabla_{\nu]}t=0 \, .$$ But I think $\alpha$ can depend on the coordinates so it doesn't seem clear why the entire expression should vanish, right?
edit: So to add further context this is basically a question from "Shapiro - Numerical Relativity" (p.27, Eq.2.23) or see also exercise 2.10 where it is to show that the twist $$\omega_{ab} = \gamma_a^c \gamma_b^d \nabla_{[c}n_{d]}$$ vanishes ($\gamma_a^b = \delta_a^b + n_a n^b$ is the projector onto the hypersurface). Shapiro also defines
$$ \omega_a = \alpha \nabla_a t = -n_a$$ and claims $$\omega_{[a}\nabla_b \omega_{c]}=0$$
so it doesn't really matter if I write $n$ or $\omega$. What matters is that either vector is normal to the $t=$const hypersurfaces. I also think the above bracketing must contain a typo, because $[abc]$ doesn't make sense. For the specific case of the SS-metric in isotropic coordinates we have
$$\alpha=\frac{1-M/2r}{1+M/2r}$$
so it does depend on the coordinates.

Hint: 
$$n_{[i} \nabla_j n_{k]} = n_i \nabla_j n_k + n_j \nabla_k n_i + n_k \nabla_i n_j - n_i \nabla_k n_j - n_j \nabla_i n_k - n_k \nabla_j n_i $$
