Proof about covariant derivative I'm asked to prove that if $v$ is a vector orthogonal to a family of hypersurfaces, then $v_{[a}\nabla_bv_{c]}=0$, being $\nabla$ a covarian derivative, and the $[,]$ meaning the three indices are antisymmetrised, and where $v_a=g_{ab}v^b$.
I've tried writing $v_ae^a$ as $\mathbf v=\alpha df$ for some functions $\alpha$ and $f$ and plugging it in $v_{[a}\nabla_bv_{c]}$, but I can't get anywhere.
 A: Your starting point is great. If you're familiar with the language of differential forms, this is quite easy after noting that $\nabla_{[i} v_{j]} = \partial_{[i} v_{j]} =: (dv)_{ij}$, which is easy to see from the torsion-free condition $\Gamma^i_{[jk]} = 0$. We have $v = \alpha\ df$, so since $d^2 = 0$ we have $dv = d\alpha \wedge df$ and thus $v \wedge dv = \alpha \ df \wedge d\alpha \wedge df = 0$ by antisymmetry of the wedge product on one-forms.
In coordinates this proof looks something like this: $\partial_{[i} v_{j]} = \partial_{[i} (\alpha\ \partial_{j]}f) = (\partial_{[i} \alpha)(\partial_{j]}f) $ since $\partial_{[i} \partial_{j]}f=0$ by the symmetry of partial derivatives. Thus $v_{[k}\partial_{i}v_{j]}=\alpha \partial_{[k} f \partial_i \alpha \partial_{j]} f$ which is zero because it is both symmetric and antisymmetric in $j,k$.
A: Tacitly the OP represents the family of hypersurfaces as the level sets $\Sigma^t :=\{ f=t \}$ of a smooth function $f$. The differential $\mathrm{d}f$ is a $1$-form that annihilates the tangent vectors to all these hypersurfraces. Assuming certain regularity conditions we can see that any $1$-form that annihilates tangent to $\Sigma^t$-s vectors is proportional to $\mathrm{d}f$.
In abstract indices the differential $\mathrm{d}f$ is written as $\nabla_a f$. Thus, as the OP does, we write $V_a = \alpha \nabla_a f$.
If we are in a Riemannian manifold $(M,g)$, we can use the metric $g_{a b}$ and its inverse $g^{a b}$ to juggle the indices, so a normal to all the hypersurfaces vector $V^a$ will be given by $\alpha (\mathrm{d}f)^\sharp$ using the musical isomorphism $\sharp$. In abstract indices this is done easier: $V^a = g^{a b} V_b$.
Now, the calculation:
$$
\begin{align}
V_a \nabla_b V_c & = (\alpha \nabla_a f) \nabla_b (\alpha \nabla_c f) \\
& = (\alpha \nabla_a f) \Big( (\nabla_b \alpha ) \nabla_c f + \alpha \nabla_b \nabla_c f \Big) \\
& = \alpha (\nabla_a f) (\nabla_b \alpha ) \nabla_c f + \alpha^2 (\nabla_a f) \nabla_b \nabla_c f
\end{align}
$$
Antisymmetrizing the indices $a,b,c$ in the last line in the above display we see that the first term vanishes because the tensor $\alpha (\nabla_a f) (\nabla_b \alpha ) \nabla_c f$ is symmetric in the indices $a$ and $c$, and the second term vanishes because it is symmetric in the indices $b$ and $c$ due to the torsion freeness of the Levi-Civita connection $\nabla_a$ corresponding to the metric $g_{a b}$, which is equivalent to $\nabla_{[b} \nabla_{c]} f = 0$ for any smooth function $f$.
