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How does a hemisphere differ from a disk? 
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#1
Feb209, 12:35 PM

P: 582

Can the surface of a hemisphere be distinguished from a disk without introducing a metric?
What is the minimum amount of info we need to know to identify one space as a disk and the other as the surface of hemisphere? Both can have the same coordinate chart. For example, distance from the center r + position [tex]\theta[/tex] on the circle passing through r. 


#2
Feb209, 01:16 PM

P: 159

Well, they are homeomorphic (for example, via (x,y,z)<>(x,y)), if that's what you mean.



#3
Feb209, 04:19 PM

P: 582

I know that the difference in curvature (one has constant nonzero curvature, one has zero curvature) can be identified through the Riemann curvature tensor once we assign a metric, but is there any difference between the two prior to the metric? 


#4
Feb209, 05:36 PM

P: 159

How does a hemisphere differ from a disk?



#5
Feb309, 04:51 AM

P: 491

Well, if you look at the surface in question as an abstract Riemannian manifold, they are actually the same as well (look at the pullback metric of the projection that gives the homeomorphism). This is the same object. So to distinguish them, we need to add something extra, like giving the hemisphere the induced metric from R3.



#6
Feb309, 11:27 AM

P: 582

If so, that is not right. I can illustrate if you wish. 


#7
Feb309, 01:23 PM

P: 491




#8
Feb309, 01:29 PM

P: 491

To answer your original question, unless you specify something else, they can't be distinguished. They are diffeomorphic for starters, so any invariant would require additional structure. The invariant you seem to be trying to find is that the curvature of the hemisphere with the induced metric is different than that of the disk with the induced metric. This requires not only a metric, but the metric induced by the specific embedding in Euclidean space.



#9
Feb309, 01:53 PM

P: 582

Thank you, zhentil. I had to look up "induced metric."
I think what we arriving at is that by a Reimannian "surface of a hemisphere" we mean precisely : That 2d manifold whose metric is the induced metric of a "hemisphere" embedded in 3D euclidean space where "hemisphere" in the definition means a certain locus of points as one would define it in basic geometry. And so the identification "hemisphere" (as a topological manifold) has no other meaning apart from this. Right? If so, the answer to my question is, "No, there is no difference between them prior to defining a metric." The terms "surface of a hemisphere" and "disk" are meaningless except as applied to metric spaces. 


#10
Feb309, 05:25 PM

P: 491

Precisely. Except in the last line, the term "disk" is used invariably to refer to anything that is homeomorphic to the standard Euclidean disk.



#11
Feb309, 09:41 PM

P: 582

Gotcha. Thanks!



#12
Feb509, 08:05 AM

P: 707

What you have said is true but in my opinion overlooks an important idea in the mathematical concept of equivalence. It is true that any homeomorph of the 2 dimensional disk may be given a metric that makes it isometric to the standard 2d hemisphere. But before one can have a Riemannian metric on a topological manifold it must first be given a differentiable structure. Two manifolds that are isometric must have equivalent differentiable structures. What if it turned out that the 2 disk had more than one differentiable structure? Then one of these structures could never be made isometric to the standard hemisphere no matter what metric it had. So there is a theorem here that you are using implicitly. That is that there is only one differentiable structure on the 2 disk. This allows one to ignore the differentiable category and view any manifold that is homeomorphic to the standard hemisphere as the 2 disk. In higher dimensions this may not work. When one removes the Riemannian metric from a manifold one is left with a differentiable manifold and its diffeomorphs not a topological manifold and its homeomorphs. 


#13
Feb709, 02:39 AM

P: 491

It's more a matter of convention. When we refer to S^7 in the smooth category, we're referring to it with its standard differentiable structure. The same with R^4. Hence the term "exotic," implying that there is a "nonexotic."



#14
Feb709, 07:00 AM

P: 707




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