Why is metric determinant negative for Lorentzian metrics?

Question

In Sean Carroll's GR book, the Lorentzian metric is defined as a metric $g_{\mu\nu}$ that when put in its canonical form,

$$g_{\mu\nu}=\text{diag}(-1,..,-1,+1,...,+1,0,...,0)$$ has no zeros and only a single minus.

It was also said that the metric determinant $g$ is always negative for a Lorentzian metric. I can see that this is true if the metric $g_{\mu\nu}$ is put into the canonical form as shown above. However, under an arbitrary coordinate transformation $x\rightarrow x'$, how can we be sure that the determinant $g'$ is still negative?

Answer 1

The metric is a rank 2 tensor, so it transforms like

$$\tilde g_{\mu\nu}=g_{\alpha\beta}T^\alpha_\mu T^\beta_\nu$$

that is, each of the indices gets contracted with one transformation matrix $T^\alpha_\beta=\frac{\partial \tilde x^\alpha}{\partial x^\beta}$. So, for the determinant we get:

$$\det(\tilde g)=\det(g)\det(T)^2$$

As you can see, no matter the sign of $\det(T)$, it will vanish due to the square and thus the metric tensor will keep its determinant's sign after the coordinate transformation.