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April 6

3d surface

Is a hollow sphere i.e a spherical shell with almost zero thickness a 3 d surface or a 2 d one??? Does there exist a 3 d surface by the way?? —Preceding unsigned comment added by 119.235.54.67 (talk) 06:53, 6 April 2010 (UTC)[reply]

The hollow sphere has two 2d surfaces, one inner and one outer. The boundary of a 4d sphere is a 3d surface. Bo Jacoby (talk) 07:25, 6 April 2010 (UTC).[reply]

Strictly speaking, I think a surface has to be locally two-dimensional. The generalisation of a surface to different numbers of dimensions is a manifold - however, I expect most mathematicians would understand that a "3d surface" informally means a 3-manifold. In your example, because you said the spherical shell has "almost zero thickness", each point in the shell needs three co-ordinates to determine its location. Therefore the shell is a 3 dimensional space. We can go further - because the space around each point in the interior of the shell is toplogically the same as 3d Euclidean space, we can say that the shell is a 3-manifold. However, you cannot travel as far as you like in any direction within the shell, because you may hit the inner surface or the outer surface. These surfaces are boundaries, so the shell is a "3-manifold with boundaries". The boundaries themselves are 2-manifolds, which we can also call surfaces. Note that if you had said that the shell had zero thickness, then it only has two dimensions (each point on the shell now only needs two co-ordinates to locate it) and so a zero-thickness shell would be a 2-manifold or surface. Gandalf61 (talk) 08:33, 6 April 2010 (UTC)[reply]

that all makes sense, except don't you always need three coordinates to locate a point, at least since its a sphere which can only exist in a 3 dimensional space? 68.171.233.51 (talk) 09:05, 6 April 2010 (UTC)[reply]

I think I answered my own question, since you could arbitrarily draw perpindicular axes within the sphere, identify one of the intersections as the origin and then go from there in the same way as if it were a standard 2d cartesian system; so it identifies the point internal to the sphere without referencing a 3rd dimension. Is that right?68.171.233.51 (talk) 09:09, 6 April 2010 (UTC)[reply]

To locate a point that is on the surface of a sphere you only need, at a minimum, two co-ordinates - to locate a point on the surface of the Earth, for example, you only need to know its longitude and its latitude. You could use a co-ordinate system with more dimensions - you could, for example, use Cartesian co-ordinates relative to the centre of the Earth - but you are then using more co-ordinates than you need. The fact that the surface happens to be embedded in a 3d space is irrelevant to toplogists, who focus on properties that are intrinsic to the surface itself. There are some surfaces, such as the Klein bottle, that cannot be embedded in 3d space without intersecting themselves - but they are still legitimate 2d surfaces. Gandalf61 (talk) 09:39, 6 April 2010 (UTC)[reply]

See also Spherical coordinate system - you can describe any point on a sphere with only

\theta

and

\phi

. -- Meni Rosenfeld (talk) 11:39, 6 April 2010 (UTC)[reply]

I think the OP really had zero thickness in mind, but didn't know that was the right term for it. -- Meni Rosenfeld (talk) 11:39, 6 April 2010 (UTC)[reply]