Calculating the determinant of a metric tensor Suppose the line element is
$$ds^2 = -A(t,r)^2dt^2+B^2(t,r)dr^2+C^2(t,r)d\theta^2+C^2\sin^2\theta d\phi^2.$$
Since the metric is diagonal, to find the determinant I can multiply the diagonal entries,
$$\det g_{ab} = g = -A^2B^2C^4 \sin^2\theta.$$
I have a few questions about this.


*

*First off, why do we call the metric determinant $g$?

*Why isn't it true that $g = g_{ab} g^{ab} = 4$? Isn't that how $g$ is defined?

*When will it be true that $g = 1$?

 A: 
  
*
  
*First off, why do we call the metric determinant $g$?
  

Because we define the notation that way.


  
*Why isn't it true that $g = g_{ab} g^{ab} = 4$? Isn't that how $g$ is defined?
  

If the determinant of the metric could be written using abstract index notation, without resorting to non-tensorial objects like the Levi-Civita tensor, then it would be an observable quantity that was a property of space at a particular point. There are such quantities: they're measures of curvature or related quantities involving derivatives of the curvature. But the metric itself (as opposed to its derivatives) doesn't let you compute a curvature, and therefore any such quantity that doesn't involve a derivative cannot be an observable.
What this tells you is that the determinant of the metric isn't a property of space, it's a property of the coordinates you've chosen. For example, if you use coordinates in which the basis vectors aren't orthogonal, then the determinant of the metric will be smaller.


  
*When will it be true that $g = 1$?
  

Never, in 3+1 dimensions, because $g$ will be negative. If you want to know when $g=-1$, then the answer is whenever you choose coordinates that make that true.
A: I will add something to 2.
First of all,
$$g \neq g_{ab}g^{ab}$$
but
$$ \delta g = gg^{ab} \delta g_{ba} $$
which you can prove from
$$ \partial_\mu(\det(A)) = \det(A)tr(A^{-1}\partial_\mu A) $$
which you can prove from
$$ \det \left( e^{B} \right) = e^{\mathrm{tr}(B)}$$
