Extrinsic Curvature expression (ADM Formalism)

Question

I'm reading The ADM Formalism chapter of Baez's book Gauge Fields, Knots and Gravity and on page 429 we have the expression $$ K_{ij}=\frac{1}{2}N^{-1}(\dot{q}_{ij}- {}^3\nabla_i N_j - {}^3\nabla_j N_i).$$

Context

We are working on a global hyperbolic manifold. Then we can find some diffeomorphism $\phi:M\rightarrow \mathbb{R}\times \Sigma$ where $\Sigma$ denotes an space-like surface whose metric is denoted by $^3g$.

Thus, we can define the extrinsic curvature $K\rightarrow$ $K(u,v)=-g(\nabla_u v,n)$ or equivalently by $K(u,v)=g(\nabla_u n,v)$

We also define the Levi-Civita connection associated with $^3g$ as being $^3\nabla$ defined by $$^3\nabla_u v=\nabla_u v+g(\nabla_u v,n)n $$

Finally, we are using the local coordinates $\partial_0=\partial_\tau$ (with $\partial_\tau=N\;\vec n+\vec N$ defined using the normal vector to $\Sigma$, $\vec n$) and $\partial_i$ (with $\partial_i$ tangent to $\Sigma$) and we define that $q_{ij}:=\;^3g_{ij}$.

I'm trying to use those definitions to find the expression for $K_{ij}$ but I'm having trouble doing it.

I'd appreciate any hint.

Answer 1

From $\partial_t = N \mathbf{n} + \mathbf{N}$ and $g(\mathbf{n},\mathbf{n})=-1$ we get \begin{align*} g(\partial_t,\partial_t) &= -N^2 + N^i N_i, \\ g(\partial_t,\partial_i) &= N_i. \end{align*} where $N_i \equiv q_{ij} N^j$. Then the metric matrices are \begin{equation} g_{\mu\nu} = \begin{pmatrix} -N^2 + N_k N^k & N_j \\ N_i & q_{ij} \end{pmatrix}, \quad g^{\mu\nu} = \begin{pmatrix} -1/N^2 & N^j/N^2 \\ N^i/N^2 & q^{ij} - N^i N^j/N^2 \end{pmatrix}, \end{equation} The components of $\mathbf{n} = (\partial_t - \mathbf{N})/N$ is \begin{equation} n^\mu = \left(\frac{1}{N}, -\frac{N^i}{N}\right). \end{equation} Lowering the index, we get \begin{equation} n_\mu = g_{\mu\nu} n^\nu = (-N, \mathbf{0}). \end{equation} Then \begin{equation} \begin{split} K_{ij} &= K(\partial_i, \partial_j) \\ &= -g(\nabla_{\partial_i} \partial_j, \mathbf{n}) \\ &= -g(\Gamma_{ij}^\mu \partial_\mu, n^\nu \partial_\nu) \\ &= -\Gamma_{ij}^\mu\, n^\nu \, g(\partial_\mu, \partial_\nu) \\ &= -\Gamma_{ij}^\mu\, n^\nu \, g_{\mu\nu} \\ &= -\Gamma_{ij}^\mu\, n_\mu \\ &= -\Gamma_{ij}^0 n_0 \\ &= N \,\Gamma_{ij}^0. \end{split} \end{equation} The Christoffel symbol $\Gamma_{ij}^0$ can be easily derived from the definition, \begin{equation*} \begin{split} \Gamma^0_{ij} &= \frac{1}{2} g^{0\mu}(\partial_i g_{\mu j} + \partial_j g_{i \mu} - \partial_\mu g_{ij}) \\ &= \frac{1}{2}\left[ g^{00}(\partial_i g_{0j} + \partial_j g_{i0} - \partial_0 g_{ij}) + g^{0k}(\partial_i g_{kj} + \partial_j g_{ik} - \partial_k g_{ij}) \right] \\ &= \frac{1}{2}\left[ -\frac{1}{N^2} (\partial_i N_j + \partial_j N_i - \partial_t q_{ij}) + \frac{N^k}{N^2} (\partial_i q_{kj} + \partial_j q_{ik} - \partial_k q_{ij}) \right] \\ &= \frac{1}{2N^2}\left[ (\partial_t q_{ij} - \partial_i N_j - \partial_j N_i) + N_l q^{lk} (\partial_i q_{kj} + \partial_j q_{ik} - \partial_k q_{ij}) \right] \\ &= \frac{1}{2N^2}\left[ (\partial_t q_{ij} - \partial_i N_j - \partial_j N_i) + 2 N_l {\bar\Gamma}^l_{ij} \right] \\ &= \frac{1}{2N^2} \left(\partial_t q_{ij} - {^3\nabla}_i N_j - {^3\nabla}_j N_i\right), \end{split} \end{equation*} where \begin{equation} {\bar\Gamma}^l_{ij} = \frac12 q^{lk} (\partial_i q_{kj} + \partial_j q_{ik} - \partial_k q_{ij}) \end{equation} represents the Levi-Civita connection coefficients of the covariant derivative $^3\nabla$ on $\Sigma$. Then we finally obtain \begin{equation} K_{ij} = N\, \Gamma_{ij}^0 = \frac{1}{2N} \left(\partial_t q_{ij} - {^3\nabla}_i N_j - {^3\nabla}_j N_i \right). \end{equation}