A semi-riemannian manifold $\zeta^{3,1}:=\zeta^{1,1}\times \zeta^{1,0} \times \zeta^{1,0}.$ I calculated the non degenerate product metric of it: $ds^2=\frac{drdt}{rt}+\frac{du}{u}+\frac{dv}{v}.$ Previously I had studied $\zeta^{1,1}$ and came to understand that it is diffeomorphic to the Minkowski plane, $\Bbb M^{1,1}.$
So with the correspondence $\zeta^{1,1}\simeq \Bbb M^{1,1}$ I wondered if $\zeta^{3,1}$ is diffeomorphic to $\Bbb M^{3,1}.$
Is $\zeta^{3,1}\simeq \Bbb M^{3,1}?$ Why or why not?
To answer this I took each term in the product metric $ds^2$ (above) and performed a change in coordinates on each.
What I'm getting after the coordinate change is $ds^2=drdt+du+dv$. So it suffices to show whether this metric is related to Lorentz metric $ds^2=dx^2+dy^2-dt^2$ by some euclidean transformation $-$ a rotation possibly?
Here's how I tried to calculate the metric of $\zeta^{3,1}:$
I first calculated the metric of $\zeta^{1,1}$ by pulling back the Lorentz metric $ds^2=dxdy.$
Let $u=\frac{dr}{r}$ and $f=\frac{dt}{t}.$
Then I tried to use the same argument for the 1-dimensional manifolds $\zeta^{1,0}$ but as per the comments maybe I'm misunderstanding something about this being a proper non-degenerate metric: $ds^2=\frac{drdt}{rt}+\frac{du}{u}+\frac{dv}{v}.$
All of this makes me think I have not properly calculated the non-degenerate metric of $\zeta^{3,1}$ properly.
I'm pretty stuck so any help would be great.