Ignoring boundary term in the action

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  • Thread starter davidge
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  • #26
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Remember here that "coordinates" just means "sets of numbers that label points"; there is no metric since we're just talking about topological spaces.
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 have
$$x = x(r, \theta, \phi) \\ y = y (r, \theta, \phi) \\ z = z(r, \theta, \phi)$$ 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##?
 
  • #27
PeterDonis
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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)##.
You can't say the coordinates are "rectangular" because you don't have a metric. (Note that those coordinates aren't usually called "rectangular" for the standard metrics they are usually used for.)

The rest of your post just builds on this mistake.
 
  • #28
PeterDonis
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Of course, as you said in previous post, we would not get the maximal extension of a solution if we do that way.
What I said in that previous post only applies if you have a metric; the concept of a maximal analytic extension only has meaning if you have a metric. You don't have a metric in this example; we're just talking about a topological space.
 
  • #29
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You can't say the coordinates are "rectangular" because you don't have a metric.

The rest of your post just builds on this mistake.
So even if I were not considering a metric in my post, those things about the coordinates are wrong?
 
  • #30
PeterDonis
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even if I were not considering a metric in my post, those things about the coordinates are wrong?
If there is no metric, the things you are saying about the coordinates make no sense. Read your own post again:

rectangular coordinates
let the metric be that of ##\mathbb{R}^4##
 
  • #31
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What I mean is, "let the natural coordinates --just the label of the points of ##\mathbb{R}^2 \times \mathbb{S}^2##-- be ##(t,r, \theta, \phi)##. Now, let's consider a metric space, namely ##\mathbb{R}^4## with its usual metric."
 
  • #32
PeterDonis
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What I mean is, let the natural coordinates --just label of the points of ##\mathbb{R}^2 \times \mathbb{S}^2##-- be ##(t,r, \theta, \phi)##. Now, let's consider a metric space, namely ##\mathbb{R}^4## with its usual metric.
Why do you want to do this? It makes no sense.

Also, you still haven't understood: you need to understand the properties of a topological space without using any embedding of that space in another space.

I'm sorry, but I don't see the point of continuing to repeat the same advice and have a discussion if you continue to ignore that advice.
 
  • #33
PeterDonis
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The OP question in this thread has been answered, and the thread is closed. @davidge if you want to ask questions about the topological concept of a manifold with boundary, you can do that in the appropriate math forum; but please take the time to work through a textbook first.
 

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