Symmetry of extrinsic curvature tensor I am trying to solve following problem:

In a spacetime of signature (+, −, −, −), let
  $$
u^au_a = 1, \quad A_{ab} = \nabla_cu_dh^c_{\; a}h^d_{\; b}, \quad h_{ab} = g_{ab} - u_au_b
$$
  Show that $A_{ab} = A_{(ab)}$ (extrinsic curvature tensor) if and only if $u^a$ is hypersurface orthogonal.

I assume that I need to prove that
$
\nabla_cu_d = \nabla_{(c}u_{d)}
$
but I don't really understand why that should hold. Can someone explain it to me please?
 A: No that is not what you must prove. It is not true that if $u^a$ is hypersurface orthogonal then $\nabla_a u_b = \nabla_{(a}u_{b)}$. In fact this is only true if $u^a$ is geodesic. 
If $u^a$ is hypersurface orthogonal then, by definition, $u_{[c}\nabla_b u_{a]} = 0$. Writing this out we have 
$$
u_c \nabla_b u_a - u_b \nabla_c u_a + u_a \nabla_c u_b -u_c \nabla_a u_b + u_b \nabla_a u_c -u_a \nabla_b u_c =0 
$$  
Contract both sides of the above with $u^c$ and re-express the result appropriately to show that $A_{[ab]} = 0$. 
It's a little bit more work to show that $A_{[ab]} = 0$ implies $u^a$ is hypersurface orthogonal. As a hint, show that $A_{[ab]} = 0 \Leftrightarrow \epsilon^{abcd}u_b \nabla_c u_d = 0$.  
EDIT: Here is a little bit more detail on proving that $A_{[ab]} = 0$ implies hypersurface orthogonality. To start with, if $\epsilon^{abcd}u_b \nabla_c u_d = 0$ then $\epsilon_{aefg}\epsilon^{abcd}u_b \nabla_c u_d = -3!u_{[e}\nabla_f u_{g]} = 0$ so $u^a$ is hypersurface orthogonal. 
Therefore all you have to show is $A_{[ab]} = 0 \Rightarrow \epsilon^{abcd}u_b \nabla_c u_d = 0$.      
