# Determinant g = g_{00} det |g_{ij}|

1. Jun 27, 2006

### wandering.the.cosmos

If we let latin alphabets {i,j,k,...} denote the spatial indices, and the greek ones run from 0 to 3, then I've seen the following

$$det| g_{\mu \nu} | = g_{00} det | g_{ij} |$$

both in Landau's "Classical theory of fields", as well as ADM's paper on the initial value formulation of GR ("Dynamics of General Relativity", Arnowitt et al.) I don't quite understand it though, or am I reading it right?

Specifically, what about the terms that involve $g_{01}, g_{02}, g_{03}$ when we do the co-factor expansion? Do they somehow vanish by symmetry?

In (2+1) dimensions I got the determinant to be

$$det| g_{\mu \nu} | = g_{00} det | g_{ij} | - g_{10}(g_{01} g_{22} - g_{02} g_{21}) + g_{20} (g_{01} g_{12} - g_{11} g_{02})$$

How could the last four terms cancel out?

2. Jun 27, 2006

### pervect

Staff Emeritus
$g_{\mu\nu}$ can be any symmetric matrix.

If we try for an example

\left[ \begin {array}{cccc} m&1&2&3\\\noalign{\medskip}1&4&5&6\\\noalign{\medskip}2&5&7&8\\\noalign{\medskip}3&6&8&9\end {array} \right]

we find the determinant is -m-2, which is not a multiple of m.

So I don't think your statement can be correct as written. Start reading the fine print :-).

(Is the ADM paper online anywhere?)

3. Jun 28, 2006

### wandering.the.cosmos

It is equation 3.12 of "The Dynamics of General Relativity", Arnowitt, Deser and Misner:

http://arxiv.org/abs/gr-qc/0405109

Perhaps someone could explain to me if I am mis-reading it.

4. Jun 28, 2006

### HallsofIvy

Staff Emeritus
That is correct assuming that the metric tensor is of the form
\left[ \begin {array}{cccc} g_{00}&0&0&0\\\noalign{\medskip}0&g_{11}&g_{12}&g_{13}\\\noalign{\medskip}0&g_{21}&g_{22}&g{23}\\\noalign{\medskip}0&g_{31}&g_{32}&g_{33}\end {array} \right], the "three plus one dimensional decomposition of the Einstein field" referred to in the text.

5. Jun 28, 2006

### George Jones

Staff Emeritus
When a metric is static, coordinates can be chosen such that the components of the metric take the form given by HallsofIvy.

I don't think equations (3.10) and (3.12) of the ADM paper give this.

6. Jul 1, 2006

### wandering.the.cosmos

$$g^{00} = \frac{\textrm{det} |g_{ij}|}{\textrm{det} |g_{\mu \nu}|}$$
where $\textrm{det} |g_{ij}|$ is the determinant of the space-space part of the metric and $\textrm{det} |g_{\mu \nu}|$ is the determinant of the whole metric tensor.