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I am having trouble showing a relation in Carroll's GR book in his appendices. He defines the extrinsic curvature tensor as $$K_{\mu\nu}\equiv\frac{1}{2}\mathcal{L}_nP_{\,u\nu},$$ where $$\qquad P_{\mu\nu}\equiv g_{\mu\nu}-\sigma n_\mu n_\nu,$$ and $\mathcal{L}_n$ is the Lie derivative in the direction of the vector $n$ that is normal to the hypersurface. He says that the extrinsic curvature tensor obeys \begin{equation}P^\alpha_{\;\mu} P^\beta_{\;\nu}\nabla_{(\alpha}n_{\beta)} = \nabla_\mu n_\nu-\sigma n_\mu a_\nu,\tag{1}\label{a}\end{equation}

where $$\sigma = n_\mu n^\mu, \qquad a_\mu=n^\nu\nabla_\nu n_\mu$$

I want to go from left to right in \ref{a}. The vector $n$ is hypersurface orthogonal, and the contraction $n_\mu n^\mu$ is constant, so we can say that, because we have a metric compatible connection, that $n^\mu\nabla_\nu n_\mu=0$. I have gotten to the point where I find that the left hand side is equal to $$\nabla_{(\mu}n_{\nu)}-\frac{1}{2}\sigma n_\nu n^\beta(\nabla_\beta n_\mu)-\frac{1}{2}\sigma n_\mu n^\alpha(\nabla_\alpha n_\nu),$$ but I am stuck here. Any tips would be great!

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I think you're pretty much almost there. I'd just expand out the symmetrization of $\nabla_{(a}n_{b)}$ and collect terms into your $P$'s. One trick to realize is that, since you have a foliation of 3-surfaces that can be labeled by some function $\tau$, you have:

$$n_{a} = \alpha \nabla_{a}\tau$$

where $\alpha = \frac{1}{\sqrt{|\nabla_{a}\tau\nabla^{a}\tau|}}$, which will let you work out identities involving $P^{b}_{c}P^{a}_{d}\nabla_{[a}n_{b]}$ and $P^{b}_{c}P^{a}_{d}\nabla_{(a}n_{b)}$

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Ok I think I figured it out. First, you show that $$K_{\mu\nu} = \nabla_{(\mu}n_{\nu)}-\frac{1}{2}\sigma n^\beta\nabla_\beta (n_\mu n_\nu)=\nabla_{(\mu}n_{\nu)}-\sigma n_{(\mu}a_{\nu)}.$$ Then, you can split this into the part without specified symmetry and the antisymmetric part,i.e $$K_{\mu\nu} = \nabla_{\mu}n_{\nu}-\sigma n_{\mu}a_{\nu}-\nabla_{[\mu}n_{\nu]}-\sigma n_{[\mu}a_{\nu]}.$$ If one then considers only the antisymmetric part, one sees that it equals zero when acting on vectors parallel to $n$. Then, one considers the action of this part on vectors normal to $n$, and using the fact that $n$ being hypersurface orthogonal implies that $$\nabla_{[\mu}n_{\nu]}V^\mu W^\nu=0$$ for vectors normal to $n$, you can conclude that the antisymmetric part is zero.

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