- #26

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So could we say that points of ##\mathbb{R}^2 \times \mathbb{S}^2##, seen as a independently topological space, will have rectangular coordinates ##(t, r, \theta, \phi)##. Now if we embed this topological space into ##\mathbb{R}^4## and let the metric be that of ##\mathbb{R}^4## and if the points on ##\mathbb{R}^4## are labeled ##(t,x,y,z)##, we haveRemember here that "coordinates" just means "sets of numbers that label points"; there is no metric since we're just talking about topological spaces.

$$x = x(r, \theta, \phi) \\ y = y (r, \theta, \phi) \\ z = z(r, \theta, \phi)$$ with the usual expressions on the RHS of these functions.

However, as a manifold is locally Euclidean, it is on each point like ##\mathbb{R}^4## and so we would not need to make that change of coordinates, we would just continue with our natural coordinates ##(t,r,\theta,\phi)##. Of course, as you said in previous post, we would not get the maximal extension of a solution if we do that way. So the conclusion would be that the most smart choice is to make that transformations above for ##x, y, z##?