# Expression for extrinsic curvature

In Padmanabhan's book Gravitation Foundations and Frontiers, the following equation can regarding the extrinsic curvature of a hypersurface can be found in section 12.2 (see just above equation 12.19 in that book),

\begin{align} K_{\alpha\beta}=-\nabla_\alpha n_\beta=-N\Gamma^0_{\alpha\beta}. \end{align}

According to the book's convention greek indices run for spatial coordinates ($$\alpha=1,2,3$$) and latin indices run for space-time coordinates ($$a=0,1,2,3$$). Thus the above equation gives an expression for the spatial components of the extrinsic curvature, $$K_{\alpha\beta}$$. Here, $$n^a$$ is the vector field normal to the hypersurface and $$N$$ is the lapse function. Now the book claims if we expand the Christoffel symbol, we shall get the following expression (see equation 12.19 in the book),

$$K_{\alpha\beta}=\frac{1}{2N}\left(D_\alpha N_{\beta}+D_\beta N_{\alpha}-\partial_0 h_{\alpha\beta}\right)$$

Here, $$N^\alpha$$ is the shift vector, $$h_{\alpha\beta}$$ is the induced spatial metric on the hypersurface, and $$D_m$$ is the intrinsic covariant derivative on the hypersurface with its action on the purely spatial vectors $$X_s$$, which satisfies a constraint like $$X_sn^s=0$$, defined as

$$D_mX_s=h^a_mh^b_s\nabla_aX_b,$$

where, $$h^a_b=\delta^a_b+n^an_b$$ are the projection tensor on the hypersurface, and $$\nabla_a$$ is the usual covariant derivative for the spacetime.

I have failed to derive the equation 12.19 giving the expression for $$K_{\alpha\beta}$$. Below I show, how I tried to do it. The Christoffel symbol can be expanded as, \begin{align} \Gamma^0_{\alpha\beta}&=\frac{1}{2}g^{0a}\left(\partial_\alpha g_{\beta a}+\partial_\beta g_{\alpha a}-\partial_a g_{\alpha\beta}\right)\nonumber\\ &=\frac{1}{2}g^{00}\left(\partial_\alpha g_{\beta 0}+\partial_\beta g_{\alpha 0}-\partial_0 g_{\alpha\beta}\right)+\frac{1}{2}g^{0\gamma}\left(\partial_\alpha g_{\beta \gamma}+\partial_\beta g_{\alpha \gamma}-\partial_\gamma g_{\alpha\beta}\right)\nonumber\\ &=\frac{-1}{2}N^{-2}\left(\partial_\alpha N_{\beta}+\partial_\beta N_{\alpha}-\partial_0 h_{\alpha\beta}\right)+\frac{1}{2}N^{-2}N^{\gamma}\left(\partial_\alpha h_{\beta \gamma}+\partial_\beta h_{\alpha \gamma}-\partial_\gamma h_{\alpha\beta}\right)\nonumber\\ &=\frac{-1}{2}N^{-2}\left(D_\alpha N_{\beta}+D_\beta N_{\alpha}+2\Gamma^\gamma_{\alpha\beta}N_{\gamma}-\partial_0 h_{\alpha\beta}\right)+\frac{1}{2}N^{-2}N^{\gamma}\left(\partial_\alpha h_{\beta \gamma}+\partial_\beta h_{\alpha \gamma}-\partial_\gamma h_{\alpha\beta}\right)\nonumber\\ &=\frac{-1}{2}N^{-2}\left(D_\alpha N_{\beta}+D_\beta N_{\alpha}-\partial_0 h_{\alpha\beta}\right)-N^{-2}\Gamma^\gamma_{\alpha\beta}N_{\gamma}+\frac{1}{2}N^{-2}N^{\gamma}\left(\partial_\alpha h_{\beta \gamma}+\partial_\beta h_{\alpha \gamma}-\partial_\gamma h_{\alpha\beta}\right) \end{align} In the above, I have used the facts that, $$n_0=-N,\quad n_\alpha=0,$$ $$D_\alpha N_\beta=h^a_\alpha h^b_\beta\nabla_a N_b=\partial_\alpha N_\beta-\Gamma^\gamma_{\alpha\beta}N_\gamma,$$ $$h_{00}=N^\gamma N_\gamma,\quad h_{0\alpha}=N_\alpha,\quad h_{\alpha\beta}=g_{\alpha\beta}$$

The OP's calculation seems alright. If we proceed along that line the required expression can be achieved rather easily. First, I note that, $$D_\alpha N_\beta=\partial_\alpha N_\beta-{}^{(3)}\Gamma^\gamma_{\alpha\beta}N_\gamma\neq \partial_\alpha N_\beta-{}^{(4)}\Gamma^\gamma_{\alpha\beta}N_\gamma=\partial_\alpha N_\beta-\Gamma^\gamma_{\alpha\beta}N_\gamma,$$ Perhaps this substitution is what was confusing in OP's calculation. If we correct that then it follows, \begin{align} &\frac{1}{2}g^{0a}\left(\partial_\alpha g_{\beta a}+\partial_\beta g_{\alpha a}-\partial_a g_{\alpha\beta}\right)\nonumber\\ &=-\frac{1}{2}N^{-2}\left(D_\alpha N_{\beta}+D_\beta N_{\alpha}+2{}^{(3)}\Gamma^\gamma_{\alpha\beta}N_{\gamma}-\partial_0 h_{\alpha\beta}\right)+\frac{1}{2}N^{-2}N^{\gamma}\left(\partial_\alpha h_{\beta \gamma}+\partial_\beta h_{\alpha \gamma}-\partial_\gamma h_{\alpha\beta}\right)\nonumber\\ &=-\frac{1}{2}N^{-2}\left(D_\alpha N_{\beta}+D_\beta N_{\alpha}-\partial_0 h_{\alpha\beta}\right)-N^{-2}{}^{(3)}\Gamma^\gamma_{\alpha\beta}N_{\gamma}+\frac{1}{2}N^{-2}N^{\gamma}\left(\partial_\alpha h_{\beta \gamma}+\partial_\beta h_{\alpha \gamma}-\partial_\gamma h_{\alpha\beta}\right)\nonumber\\ &=-\frac{1}{2}N^{-2}\left(D_\alpha N_{\beta}+D_\beta N_{\alpha}-\partial_0 h_{\alpha\beta}\right)-N^{-2}{}^{(3)}\Gamma^\gamma_{\alpha\beta}N_{\gamma}\nonumber\\ &\qquad+\frac{1}{2}N^{-2}N_{\sigma}h^{\gamma\sigma}\left(\partial_\alpha h_{\beta \gamma}+\partial_\beta h_{\alpha \gamma}-\partial_\gamma h_{\alpha\beta}\right)\nonumber\\ &=-\frac{1}{2}N^{-2}\left(D_\alpha N_{\beta}+D_\beta N_{\alpha}-\partial_0 h_{\alpha\beta}\right)-N^{-2}{}^{(3)}\Gamma^\gamma_{\alpha\beta}N_{\gamma}+N^{-2}N_{\sigma}{}^{(3)}\Gamma^{\sigma}_{\alpha\beta}\nonumber\\ &=-\frac{1}{2}N^{-2}\left(D_\alpha N_{\beta}+D_\beta N_{\alpha}-\partial_0 h_{\alpha\beta}\right) \end{align} Therefore, $$K_{\alpha\beta}=-N\Gamma^0_{\alpha\beta}=\frac{1}{2N}\left[D_\alpha N_{\beta}+D_\beta N_{\alpha}-\partial_0 h_{\alpha\beta}\right].$$
1. The extrinsic curvature is defined in the ambient spacetime (rather than on the hypersurface) as $$n_a$$: $$K_{ab} = -P_\perp{}^c{}_{a}P_\perp{}^d{}_b \nabla_c n_d,$$ with $$P_\perp$$ the projection tensor on the hypersurface. Notice that by construction the extrinsic curvature is spatial and symmetric in its two indices.
2. Use the symmetry to write $$K_{ab}$$ as a Lie derivative:$$K_{ab} ={-\scriptsize\frac{1}{2}} P_\perp{}^c{}_{a}P_\perp{}^d{}_b \mathcal{L}_n \,g_{cd}.$$
3. Use the orthogonal decomposition of the metric and the adapted coordinate system $$t^a = Nn^a + N^a$$ for the lapse function and shift vector to arrive at $$K_{ab} = {\scriptsize\frac{1}{2}}N^{-1}\mathcal{L}_{(N-t)}h_{ab}.$$
• But you haven't reduced the expression in the form $K_{\alpha\beta}=\frac{1}{2N}\left(D_{\alpha} N_{\beta}+D_{\beta} N_{\alpha}-\partial_0 h_{\alpha\beta}\right)$. I'm not interested in the general derivation of the expression, I'm only interested in expanding $\Gamma^0_{\alpha\beta}$ to get the expression in question. Thank you. – Faber Bosch Aug 19 '20 at 5:12
• The Lie derivative is equal to your expression. Compare the spatial derivatives of $h_{\alpha\beta}$ to the spatial Christoffel symbols. – haelewiin Aug 19 '20 at 8:52