Two surfaces embedded in different spaces

Question

Consider two punctured Euclidean spaces $X={\Bbb R}^4\setminus\{0\}$ where one is equipped with the following $(4,0)$ Riemannian metric (in polar coordinates): $$ g= dr^2 + r^2 d\Theta^2 $$ while the other with the $(1,3)$ pseudo-Riemannian metric (where the only difference is the sign): $$ g= dr^2 - r^2 d\Theta^2 $$ At a constant radius $r=r_o$ we obtain two embedded 3-surfaces. The first one obviously is the 3-sphere. My question is if the second surface is the same or different. For example, if these two 3-surfaces are viewed as 3-spaces with an intrinsic curvature (without considering the embedding), would there be any topological [EDIT: not topological, but rather geometric] difference between them? I would appreciate any thoughts even remotely relevant to this question.

Answer 1

There's certainly no topological difference, but I don't think this is what you meant - anything to do with a metric is geometric, not topological.

Since the submanifold you're considering is the slice $r=r_0$, the induced metric is very easy to calculate: we just drop the $dr$ terms and set $r=r_0$. Thus in the first case, we get the metric $r_0^2 d \Theta^2$, while in the second we get $-r_0^2 d \Theta^2.$

Thus we have two different induced metrics on the three-sphere: one is signature $(3,0)$, while the other is $(0,3)$. Of course, since they are related by a sign change, the intrinsic geometry will all be essentially the same - you just have to make sure you keep track of the sign. (For example, I think the scalar and secitonal curvatures will also have the opposite sign.)