Exploring the Physicality of Metric Solutions in Einstein's Field Equation

[TrickyDicky]

So now you know that there are intrinsically curved manifolds where you only need one chart. The topology doesn't fully specify the metric; there are infinitely many possibilities to bestow a given topological manifold with a metric.

I might have misinterpreted this from some notes on diff.geometry:
"Note finally that any manifold M admits a finite atlas consisting of dimM +1 (not
connected) charts. This is a consequence of topological dimension theory [Nagata,
1965], a proof for manifolds may be found in [Greub-Halperin-Vanstone, Vol. I]."
I took this to meanthat any manifold with n-dimensions needs at least n+1 charts to cover all its points.

We didn't understand each other saying "trivial" :)
...
"Trivial" is a word you need to use with caution; you have physical singularities (r=0) and coordinate singularities, and from the coordinate singularities you have ignorable ones (θ=0) and non-ignorable ones (r=2M).

Ok, just one thing, are the terms "trivial" and "ignorable" referred to singularities as you use them standard mathematical notions? I had never heard about ignorable and non-ignorable singularities, but then I'm just a layperson.

 

[George Jones]

I might have misinterpreted this from some notes on diff.geometry:
"Note finally that any manifold M admits a finite atlas consisting of dimM +1 (not
connected) charts. This is a consequence of topological dimension theory [Nagata,
1965], a proof for manifolds may be found in [Greub-Halperin-Vanstone, Vol. I]."
I took this to meanthat any manifold with n-dimensions needs at least n+1 charts to cover all its points..

I am not familiar with this result, but here is my guess. dimM + 1 refers to the "dimensionality" of the charts, not to the actual number of charts. Suppose M has dimension n. Standard charts on M are maps from subsets of M into R^n. A dimM + 1 chart is something like a homeomorphism between an open subset U of M and its image in R^(n+1).

 

[grey_earl]

With mathematical theorems you must be careful in intepretation, they assure only and exactly only what they state. For each manifold of dimension n (so in my toy example n=2) there is an atlas containing n+1 charts, so you don't need more. Of course you can take more if you wish, and there are examples where you can take less. Minkowski space (of dimension n) is such an example, it is a manifold of dimension n where you only need one chart to cover it completely (but of course you can find n+1 charts, or more). It is an interesting theorem, though; I can see why for some manifolds you need at least n+1 charts, but not why this is sufficient. Let's look up this Greub-Halperin-Vanstone!

"Ignorable" is not a standard mathematical term, it's physics parlor. Trivial is a standard mathematical term, but as in "that's trivial to show", not for coordinate singularities. Non-ignorable coordinate singularities define a horizon (which may be one of: event horizon, black hole horizon, null horizon, Killing horizon, ...), and that's a standard mathematical term you can look up.

 

[TrickyDicky]

With mathematical theorems you must be careful in intepretation, they assure only and exactly only what they state. For each manifold of dimension n (so in my toy example n=2) there is an atlas containing n+1 charts, so you don't need more. Of course you can take more if you wish, and there are examples where you can take less. Minkowski space (of dimension n) is such an example, it is a manifold of dimension n where you only need one chart to cover it completely (but of course you can find n+1 charts, or more). It is an interesting theorem, though; I can see why for some manifolds you need at least n+1 charts, but not why this is sufficient. Let's look up this Greub-Halperin-Vanstone!

"Ignorable" is not a standard mathematical term, it's physics parlor. Trivial is a standard mathematical term, but as in "that's trivial to show", not for coordinate singularities. Non-ignorable coordinate singularities define a horizon (which may be one of: event horizon, black hole horizon, null horizon, Killing horizon, ...), and that's a standard mathematical term you can look up.

Thanks for the info. You are right that the theorem doesn't mean what I thought.
I know now that there are curved manifolds that can be specified with jst one chart, you may want to look up the thread on manifold charts I started in the Topology subforum.

Which point is not covered by the coordinate system I gave you?

Hmm... How about point phi=pi?, I believe the maps have to be bijective, and a single chart can't have both phi=0 and phi=pi.

 

[grey_earl]

Hmm... How about point phi=pi?, I believe the maps have to be bijective, and a single chart can't have both phi=0 and phi=pi.

True, but that's not a point on the manifold (sure you meant φ=2π, no?). My φ coordinate is in the range [0, 2π), so it includes 0 but excludes 2π. So everything ok there.
(if you want strictest mathematical rigor: φ ∊ ℝ/{2πℤ})

 

[George Jones]

What I wrote was silly, but I guessed (wrongly) that the number was too small, e.g., that there probably exist 3-dimensional manifolds not coverable by four charts.

[EDIT]

It is an interesting theorem, though; I can see why for some manifolds you need at least n+1 charts, but not why this is sufficient. Let's look up this Greub-Halperin-Vanstone!

[/EDIT]

Not available to me.

 

[TrickyDicky]

True, but that's not a point on the manifold (sure you meant φ=2π, no?). My φ coordinate is in the range [0, 2π), so it includes 0 but excludes 2π. So everything ok there.
(if you want strictest mathematical rigor: φ ∊ ℝ/{2πℤ})

No, I actually meant two opposite points in the S^1 part of the your manifold (if you picture it as a one-sheet hyperboloid, think of a point at the front and the diametrically opposite point in the back side of the hyperboloid)
I know for sure that S^1 needs two charts to specify all its points (proof in Tensor calculus by D. Kay, page 244) due to the bijective (one-one) character of the manifold maps and so I figured that if the slices of constant x have circular shape in the hyperboloid, it seems logical that this manifold also needs more than one chart. Only one chart would for some applications map the 2 opposite points in the circumference of a slice of constant x to the same real number.