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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.

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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.)

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  • $\begingroup$ Thank you for the insight! Could you please give an example of where a curvature would have the opposite sign? To explain my confusion, normally a positive curvature ("sphere-like") refers to a closed space while a negative curvature ("saddle-like") refers to an infinite open space. In this case both seem closed, so I wonder, in what sense their curvature would have the opposite sign. $\endgroup$
    – safesphere
    Nov 29 '17 at 4:14
  • $\begingroup$ @safesphere: If you have a look at the coordinate formula for the scalar curvature, you'll see that if you change $g \to -g$ then the overall sign will change. The usual interpretations (in the Riemannian case) of positive/negative curvature as being like spheres/saddles are always assuming that the metric is positive-definite; so don't bestow too much meaning upon this sign change - it really is just an artifact of using a negative-signature metric. $\endgroup$ Nov 29 '17 at 4:36
  • $\begingroup$ Still trying to comprehend the details, hope you could help with a couple quick points on the metric sign. (1) As long as $r=r_o$, is any test possible inside the 3-sphere to reveal the sign of the metric? (I'm thinking, no, just want to confirm.) (2) Is there any difference between the (1,3) space in my question and the (3,1) space obtained by the reversal of the metric sign? (This way the 3-sphere metric at $r=r_o$ would be positive.) Thanks so much! :) $\endgroup$
    – safesphere
    Nov 29 '17 at 14:19
  • $\begingroup$ (1) Measure the squared length of any non-zero vector: if it's positive you're in (3,0), if it's negative you're in (0,3). (2) There's the same kind of difference as between the (3,0) and (0,3): squared lengths, scalar curvatures have their signs flipped. Timelike and spacelike are interchanged. $\endgroup$ Nov 29 '17 at 21:44
  • $\begingroup$ Thanks again Antony! I don't want to bother you too much, but if you don't mind me asking for one more clarification... As an example, consider the Minkowski metric that people write both ways $dt^2-d\mathbf{r}^2$ and $d\mathbf{r}^2-dt^2$. I don't think flips spacelike and spacelike intervals in special relativity, does it? So, in a 3D slice there at $t=t_o$, there's no way to figure out the sign in the metric. The same way I'm thinking that there must be no way to figure the sign in $-r_0^2 d \Theta^2$, because there inside all vectors may be spacelike regardless. Where is my thinking wrong? $\endgroup$
    – safesphere
    Nov 29 '17 at 23:37

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