# Relation between det(spacetime metric) and det(spatial metric)

## Main Question or Discussion Point

I have a metric g on spacetime and a spatial metric $\gamma$ such that the components of g can be written in matrix form as

$$g_ {\alpha, \beta} = \begin{pmatrix} g_{00} & g_{0 j} \\ g_{i 0} & \gamma_{ij} \end{pmatrix}$$

where $i,j = 1,2,3$ and $\alpha = 0,1,2,3$. Now I want to find a relation between the determinant of $g$ and the determinant of $\gamma$ expressed in terms of the components of $g$. Using Cramer's rule I get

$$det(g_{\alpha, \beta}) \equiv g = \frac{C_{00}}{g^{00}}$$

where $C_{00}$ is the cofactor of the (0,0)-element of the matrix above; i.e. it is $det(\gamma_{ij}) = \gamma$. But in order to find the full relation I will still need to find the (0,0) element of the inverse. What is the best way to go about doing this? Do I really have to solve the full set of equations $g^{\alpha \lambda}g_{\lambda \beta} = \delta^{\alpha}_{\ \ \ \beta}$? Or is there a better way?

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I have a metric g on spacetime and a spatial metric $\gamma$ such that the components of g can be written in matrix form as

$$g_ {\alpha, \beta} = \begin{pmatrix} g_{00} & g_{0 j} \\ g_{i 0} & \gamma_{ij} \end{pmatrix}$$

where $i,j = 1,2,3$ and $\alpha = 0,1,2,3$.
I'm assuming you mean $\displaystyle \mathrm{g}_{\alpha,\beta}=\begin{bmatrix} g_{0,0} & g_{0,1} & g_{0,2} & g_{0,3} \\ g_{1,0} & \gamma_{1,1} & \gamma_{1,2} & \gamma_{1,3} \\ g_{2,0} & \gamma_{2,1} & \gamma_{2,2} & \gamma_{2,3} \\ g_{3,0} & \gamma_{3,1} & \gamma_{3,2} & \gamma_{3,3}\end{bmatrix}$, right?

Now I want to find a relation between the determinant of $g$ and the determinant of $\gamma$ expressed in terms of the components of $g$. Using Cramer's rule I get

$$det(g_{\alpha, \beta}) \equiv g = \frac{C_{00}}{g^{00}}$$

where $C_{00}$ is the cofactor of the (0,0)-element of the matrix above; i.e. it is $det(\gamma_{ij}) = \gamma$.
I'm probably misinterpreting something here, but I'm not seeing how you're applying Cramer's Rule.

If we were talking about finding the determinant of the given matrix, you'd get $$\det(\mathrm{g}_{\alpha,\beta})=g_{0,0}~\det\begin{bmatrix}\gamma_{1,1} & \gamma_{1,2} & \gamma_{1,3} \\ \gamma_{2,1} & \gamma_{2,2} & \gamma_{2,3} \\ \gamma_{3,1} & \gamma_{3,2} & \gamma_{3,3}\end{bmatrix}-g_{1,0}~\det\begin{bmatrix} g_{0,1} & g_{0,2} & g_{0,3} \\ \gamma_{2,1} & \gamma_{2,2} & \gamma_{2,3} \\ \gamma_{3,1} & \gamma_{3,2} & \gamma_{3,3}\end{bmatrix}+g_{2,0}~\det\begin{bmatrix} g_{0,1} & g_{0,2} & g_{0,3} \\ \gamma_{1,1} & \gamma_{1,2} & \gamma_{1,3} \\ \gamma_{3,1} & \gamma_{3,2} & \gamma_{3,3}\end{bmatrix}-g_{3,0}~\det\begin{bmatrix}g_{0,1} & g_{0,2} & g_{0,3} \\ \gamma_{1,1} & \gamma_{1,2} & \gamma_{1,3} \\ \gamma_{2,1} & \gamma_{2,2} & \gamma_{2,3}\end{bmatrix}$$ by Laplace Expansion.

Given, I'm not really knowledgeable in general relativity (or differential geometry for that matter :tongue:) so I could be mistaking this determinant for something else, or misunderstanding your notation.