Extrinsic curvature components I'm trying to understand how to derive the extrinsic curvature (in order to understand some calculation on fluid/gravity dynamics). But I hit a wall in my progress. I stuck at trying to verify the extrinsic curvature of a slowing-rotating Kerr solution ( page 94, E.Poisson: A relativistic Toolkit: The mathematics of black holes ):
$$ds_+^2=g_{ab}\;dx^a dx^b=-f dt^2+f^{-1}dr^2+r^2 d\Omega^2-\frac{4Ma}{r}\;\sin^2(\theta)\;dr\;d\phi$$
where $f=1-2M/r$ and cut of $r=R$ which defines the hypersurface $\Sigma$ on which we'll work on. 
I derive the induced metric (with $\psi=\phi-\Omega t$ and $\Omega=\frac{2Ma}{R^3}$) :
\begin{equation}
h_{ab}\;dy^ady^b=-f dt^2+R^2(d\theta^2+\sin^2\theta d\psi^2)
\end{equation}
But from here on, I stuck. So my stucking points are:

  
*
  
*How can I derive the form of the unit normal vector $n_a$ here and in not so obvious metrics ?(The answer for this problem is $n_a=f^{-1/2}\partial_a\;r$, this means that $n_a=f^{-1/2}\delta^r_a$) ?
  
*This defines the extrinsic curvature as 
  
  
  $$K_{AB}=n_{a;b}\frac{\partial x^a}{\partial y^A}\frac{\partial x^b}{\partial y^B} \;\;\;\;\;\;\; (1)$$
where $x^a=x^a(y^A)$. My question is, the covarian derivative on the normal and the Christoffel symbols are calculated with the induced metric ($h_{ab}$) or the original ($g_{ab}$)?

  
*Finally, I want to ask for the different definitions of the extrinsic curvature that I have found on the literature, because this got very confused: Is the definition $(1)$ equivalent with :
  
  
  $$K_{ab}=-\vec{e}_a\cdot\nabla_b \vec{n}\;\;\;\;\;\;\; (2) $$
  where $\vec{n}$ is the normal unit vector and $\vec{e}_a$ basis unit vectors of $\Sigma$. How can we work with $(2)$, it's very unclear to me?

Sorry for the long post, but I really had to understand all this in order to procceed to the more compliacated metrics of fluid/gravity. 
 A: 1)
On first glance it seems to be correct. If your hypersurface is given by $r=R$, then it is the level set of the coordinate function $r$, thus $\mathrm{d}r\sim n$.
2)
The covariant derivative is calculated with respect to the original metric. The connection induced on $\Sigma$ by the induced metric is technically a connection that only works on fields that are intrinsically defined on $\Sigma$. The normal vector is not such a a field.
3)
First of all, let us note that in component-index notation, you can define the extrinsic curvature with spacetime indices, or with 3-indices.
The (2) definition seems awfully confused, with this regard. It is clear that $a$ has to be a 3-index, because $a$ numbers the coordinate-frame vector fields on $\Sigma$, and you have 3 of those, however the $b$ index labels $\nabla$, so it must be a 4-index.
The formula is either wrong, or by what they mean $\nabla_b$ is actually $e^\mu_{\ b}\nabla_\mu$, the restriction of the spacetime connection to directions that are purely tangential to $\Sigma$.
If the latter is the case, then the formula is (aside from a sign) consistent with your (1) definition, since writing the inner product out in index notation nets $$ e_a\cdot\nabla_bn=e^\mu_{\ a}\nabla_bn_\mu=e^\mu_{\ a}e^\nu_{\ b}\nabla_\nu n_\mu, $$ and since $e^\mu_{\ a}=\partial x^\mu/\partial y^a$, this is exactly your formula.
Aside from a sign that is, but some authors define the extrinsic curvature with a different sign.
