Covariance and contravariance of a singular metric tensor

Question

I ran across this problem on an exam a couple days ago:

Consider the transformation to the new coordinates $(r,u,z,\bar{z})$ $$r=\sqrt{(x^1)^2+(x^2)^2+(x^3)^2} \qquad u = x^0-r \qquad z=\frac{x^1+ix^2}{x^3+r} \qquad \bar{z}=\frac{x^1-ix^2}{x^3+r}$$ Find the metric tensor $g_{\mu \nu}$

At first glance, it seamed easier to just solve for the contravariant metric and invert it later $$g^{\mu\nu} = \epsilon^{\nu} \epsilon^\mu = \frac{\partial q^\nu}{\partial x^j} \hat{e}^j \frac{\partial q^\mu}{\partial x^i} \hat{e}^i = \frac{\partial q^\nu}{\partial x^j} \frac{\partial q^\mu}{\partial x^i} \eta^{ij} $$

However, when I computed the elements, I found it to be singular, with only three entries non-zero.

When I tried the other way around and computed $x^i = f(q^j)$ to get the covariant metric directly $$g_{\mu\nu} = \epsilon_{\nu} \epsilon_\mu = \frac{\partial x^i}{\partial q^\nu} \hat{e}_i \frac{\partial x^j}{\partial q^\mu} \hat{e}_j = \frac{\partial x^i}{\partial q^\nu} \frac{\partial x^j}{\partial q^\mu} \eta_{ij} $$

I got a non-singular metric with clearly non-zero elements.

I've read a few posts to understand the significance of a singular metric tensor ($det(g)=0$), but I'm yet unclear as to why one representation (covariant) of the metric was non-singular, while the other (contra) was singular when they should be describing the same thing.

Aren't they related by $G_{co}= \frac{1}{G_{contra}}$?

How is this possible if only one of them is singular?

One last note: this paper(On Degenerate Metrics and Electromagnetism, T.P Searight) seemed to acknowledge this problem, but I got lost in the heavy algebra because I didn't take a GR class yet

Answer 1

Due to the invariance of the line element $ds^2$ we can use the following relation: (here we assume that the coordinates $x^\mu = (x^0, x^1, x^2, x^3)$ are from a cartesian Minkowski metric $\eta_{\mu\nu}$):

$$ g_{ij} dq^i dq^j =ds^2= dx^\mu dx^\nu \eta_{\mu\nu} = \frac{\partial x^\mu}{dq^i} dq^i \frac{\partial x^\nu}{dq^j} dq^j \eta_{\mu\nu} \equiv \frac{\partial x^\mu}{dq^i} \frac{\partial x^\nu}{dq^j}\eta_{\mu\nu} dq^i dq^j$$

Einsteinian summation convention is assumed. From this it should be plausible that one has the following transformation law:

$$g_{ij} = \frac{\partial x^\mu}{dq^i} \frac{\partial x^\nu}{dq^j}\eta_{\mu\nu} $$

The advantage of this particular case is that one only needs to compute the products of differential quotients for $(\mu,\nu) = (0,0;\, 1,1;\, 2,2;\, 3,3)$. This should allow to compute the components of the metric tensor $g_{ij}$ as it was required. I did not check on the singular character of the matrix $(g_{ij})$, but I guess as it was an exam question that the result should be kind of reasonable.

EDIT: Contravariant and covariant metric tensor components are related by

$$ g^{ij}g_{jk} = \delta^i_k$$

i.e. indeed there are reciprocal to each other. By the way, for particular values of the coordinates the metric tensor can form a singular matrix (i.e. without inverse) or even not be defined, as it is the case for instance for the Schwarzschild metric:

$$ds^2 = \left(1-\frac{GM}{r c^2}\right) dt^2 - \left(1-\frac{GM}{r c^2}\right)^{-1} dr^2 -r^2 d\Omega^2$$

which is not defined for $r_s=GM/c^2$, the Schwarzschild radius. Then in this area $r_s$ other coordinates have to be chosen, for instance Kruskal-coordinates.