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## How does a hemisphere differ from a disk?

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 $$\theta$$ on the circle passing through r.

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

Quote by Preno 					 				 				Well, they are homeomorphic (for example, via (x,y,z)<->(x,y)), if that's what you mean.
They are homeomorphic, yes, but I'm not sure if you mean the same thing as I do. I just mean the 2-D surface of the hemisphere, not the hemisphere itself. The surface is just a stretched disk.

I know that the difference in curvature (one has constant non-zero 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?

## How does a hemisphere differ from a disk?

Quote by pellman 					 				 				They are homeomorphic, yes, but I'm not sure if you mean the same thing as I do.  I just mean the 2-D surface of the hemisphere, not the hemisphere itself.  The surface is just a stretched disk.
Well, yes. The disk is obviously not homeomorphic to the whole "hemi-ball".

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.

Quote by zhentil 					 				 				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.
Are you saying that the 2D metrics on the two surfaces are not enough to distinguish them and that we have to embed them in a 3D space?

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

Quote by pellman 					 				 				Are you saying that the 2D metrics on the two surfaces are not enough to distinguish them and that we have to embed them in a 3D space?    If so, that is not right.  I can illustrate if you wish.
A priori, an induced (or pullback) tensor requires a map. Once the metric is constructed, it's not dependent on the ambient space, but I would be hard-pressed to come up with a canonical induced metric on a manifold.

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.

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.

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

Gotcha.  Thanks!

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

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.

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 "non-exotic."

Quote by zhentil 					 				 				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 "non-exotic."
I see what you are saying but don't agree. The question had to do with what happens when you ignore the metric. this does not give you the topological category. It gives you the differentiable category. In the topological category you may not be able to get the metric back again.

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