Relation between the determinants of metric tensors

Recently I have started to study the classical theory of gravity. In Landau, Classical Theory of Field, paragraph 84 ("Distances and time intervals") , it is written

We also state that the determinanats $g$ and $\gamma$, formed respectively from the quantities $g_{ik}$ and $\gamma_{\alpha\beta}$ are related to one another by $$-g=g_{00}\gamma$$ I don't understand how this relation between the determinants of the metric tensors can be obtained. Could someone explain, or make some hint, or give a direction?

In this formulas $g_{ik}$ is the metric tensor of the four-dimensional space-time and $\gamma_{\alpha\beta}$ is the corresponding three-dimensional metric tensor of the space. These tensors are related to one another by the following formulas $$\gamma_{\alpha\beta}=(-g_{\alpha\beta}+\frac{g_{0\alpha}g_{0\beta}}{g_{00}})$$ $$\gamma^{\alpha\beta}=-g^{\alpha\beta}$$

Thanks a lot.

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1. Consider the $4\times 4$ matrix $g_{\mu\nu}$ with zeroth row $g_{0\nu}$.

2. Now for $i=1,2,3$, add to the $i$'th row the zeroth row times $-g_{i0}/g_{00}$.

3. This produces the following matrix $$\begin{bmatrix} g_{00} & g_{01} & g_{02}& g_{03} \\ 0 & -\gamma_{11} & -\gamma_{12}& -\gamma_{13} \\ 0 & -\gamma_{21} & -\gamma_{22}& -\gamma_{23} \\ 0 & -\gamma_{31} & -\gamma_{32}& -\gamma_{33} \end{bmatrix}.$$

4. Such row manipulations do not change the determinant. So it is still $g=\det(g_{\mu\nu})$.

5. On the other hand the determinant can be expanded in the zeroth column to yield $g_{00}\times \det(-\gamma_{ij})$.

6. Hence we obtain the result $g=-g_{00}\det(\gamma_{ij})$.

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In my opinion it is better to work in an explicit covariant form. In my answer I will use two different definitions, the Greek indexes always run from $0$ to $3$ and Latin indexes from $1$ to $3$ and the metric $g_{\mu\nu}$ has signature $(-1,1,1,1)$.

To translate the expressions to a explicit covariant form we define some timelike vector field $v^\mu$. We can define an adapted coordinate system such that $v^\mu = \delta^\mu{}_0$ and, therefore, $$g_{00} = g_{\mu\nu}v^\mu{}v^\nu.$$ However, it is better (in my opinion) to not work in such coordinate system, in this case we always use expression like the right hand side of the above, i.e., $v^2 \equiv g_{\mu\nu}v^\mu{}v^\nu$.

The projected metric from your equation can be expressed as (where I reworked the signs to conform to my definitions) $$\gamma_{ij} = g_{ij} - \frac{g_{0i}g_{0j}}{g_{00}} = g_{ij} - \frac{v^\mu v^\nu g_{\mu i}g_{\nu j}}{v^2} = g_{ij} + \frac{v_i v_j}{\sqrt{-v^2}\sqrt{-v^2}} = g_{ij} + n_i n_j,$$ where we defined, naturally, $v_\mu = g_{\mu\nu}v^\nu$ and the normalized version of $v^\mu$, i.e., $n^\mu \equiv v^\mu/\sqrt{-v^2}$. Note that here we introduced the rest of the coordinate basis given by the vector fields $e_i{}^\mu$ such that $g_{ij} = e_i{}^\mu{}e_j{}^\nu{}g_{\mu\nu}$, $v_i = v_\mu{}e_i{}^\mu$, etc. Then, to avoid the introduction of coordinates we define the projector $$\gamma_{\mu\nu} = g_{\mu\nu} + n_\mu{}n_\nu,$$ which reduces to the spatial metric in the adapted coordinate system and in general works as a projector $\gamma_{\mu\nu}n^\nu = 0$.

One can introduce the determinant in a covariant format using the space of totally antisymmetric tensors of type $(4,0)$, e.g. $\epsilon^{\mu\nu\alpha\beta}$ which is antisymmetric in any two adjacent indexes. We define $\epsilon^{\mu\nu\alpha\beta}$ using the expression $$\epsilon^{\mu\nu\alpha\beta}\epsilon_{\mu\nu\alpha\beta} = -4!,$$ where the indexes were lowered using the metric $g_{\mu\nu}$, this defines $\epsilon^{\mu\nu\alpha\beta}$ up to a signal since the space of totally antisymmetry tensors is one dimensional. Using the definition of determinant in terms of the Levi-Civita symbol it is easy to show (you can take a look at Appendix B of Wald 1984) that $$v^\mu{}e_1{}^\nu{}e_2{}^\alpha{}e_3{}^\beta\epsilon_{\mu\nu\alpha\beta} = \epsilon_{0123} = \sqrt{-g},$$ where $g$ is the determinant of $g_{\mu\nu}$ calculated in the coordinate basis formed by $(v^\mu,e_i{}^\mu)$.

Finally, $\epsilon^{\mu\nu\alpha\beta}\epsilon_{\mu\nu\alpha\beta}$ is expressed as follows, \begin{align} \epsilon_{\mu\nu\alpha\beta}\epsilon_{\lambda\sigma\phi\psi}g^{\mu\lambda}g^{\nu\sigma}g^{\alpha\phi}g^{\beta\psi} &= \epsilon_{\mu\nu\alpha\beta}\epsilon_{\lambda\sigma\phi\psi}(\gamma^{\mu\lambda}-n^{\mu}n^{\lambda})(\gamma^{\nu\sigma}-n^{\nu}n^{\sigma})(\gamma^{\alpha\phi}-n^{\alpha}n^{\phi})(\gamma^{\beta\psi}-n^{\beta}n^{\psi}), \\ &= -4n^{\mu}n^{\lambda}\epsilon_{\mu\nu\alpha\beta}\epsilon_{\lambda\sigma\phi\psi}\gamma^{\nu\sigma}\gamma^{\alpha\phi}\gamma^{\beta\psi}, \end{align} where we used that any contraction of two $n^\mu$ with $\epsilon_{\mu\nu\alpha\beta}$ is null. Now, using that $\epsilon^{\mu\nu\alpha\beta}\epsilon_{\mu\nu\alpha\beta} = -4!$, we obtain $$n^{\mu}n^{\lambda}\epsilon_{\mu\nu\alpha\beta}\epsilon_{\lambda\sigma\phi\psi}\gamma^{\nu\sigma}\gamma^{\alpha\phi}\gamma^{\beta\psi} = -v^{-2}\epsilon_{0ijk}\epsilon_{0lmn}\gamma^{il}\gamma^{jm}\gamma^{kn} = 3!,$$ using again the formula of the determinant through the antisymmetric symbol we obtain $$v^2\gamma = -(\epsilon_{0123})^2 = g \qquad \Rightarrow\qquad -g = -g_{00}\gamma,$$ where the difference of signs comes from the different signature definition.

In terms of volume forms this result is equivalent to $$\tilde{\epsilon} = -4\tilde{n}\wedge{}^3\tilde{\epsilon},$$ where $\tilde{\epsilon}$ is just the four volume form with its components given by $\epsilon_{\mu\nu\alpha\beta}$ and ${}^3\tilde{\epsilon}$ is the three dimensional induced volume form with its components given by $n^\mu\epsilon_{\mu\nu\alpha\beta}$.

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About the last equation: shouldn't it be $4! \tilde{\epsilon} = -3! \tilde{n}\wedge {}^3\tilde{\epsilon}$ by the definition of the wedge product? –  auxsvr Jun 30 '14 at 9:18

You have to be extremely careful about what conventions you are using for defining your 4-metric and your 3-metric and your time vector.

In particular, if you are using coordinates where your metric has off-diagonal components, be aware about what the value for your unit normal is, and I would stronlgy advice choosing your slicing condition as one of your four coordinates, so that you're choosing surfaces of $\tau =$constant.

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