# Is there a net of a sphere?

• December 13th, 2013, 06:24 PM
anticorncob28
Is there a net of a sphere?
The answer is NO, as far as I know. vsauce told me that Carl Friedrich Gauss was the first to prove that it is impossible to make a flat map of the earth that does not distort shape or size. If nobody could prove it until Gauss came around, I'd assume it must be a pretty complicated proof. But I recently came up with something:
Suppose that there exists a way to cut a sphere such that it can be laid flat on a plane, and that all straight lines remain straight lines, and all areas, lengths, and angles remain the same. We are also going to
assume that the flattened map can be folded back up into a sphere also. Then, on the flattened sphere, draw a triangle. What do it's angles add up to? 180 degrees. Now fold it back into the sphere. All the line
segments remain line segments, but the angles cannot remain the same measure since they now add up to strictly more than 180 degrees.
In fact, if it's true that all the angles in an equilateral triangle are congruent on a sphere, then there's a way of proving this without using the angles: in the net, draw six equilateral triangles with a common
vertex. If you fold it into a sphere, the result cannot be six equilateral triangles with a common vertex, because the angles must strictly be greater than 60 degrees, and six of them add up to more than 360
degrees and it cannot fit around a point. Therefore, these figures can no longer be equilateral triangles.
Any bending and folding of paper will keep straight lines straight, angles the same, and distances and areas, no?
Is there something wrong with this? Or is it simply that the geometry of a sphere is strictly different from the geometry of a plane?
• December 14th, 2013, 03:51 AM
Markus Hanke
Quote:

Any bending and folding of paper will keep straight lines straight, angles the same, and distances and areas, no?
That depends what you mean by "bending". But your main idea is correct - the intrinsic geometry of a sphere's surface is different from the Euclidean geometry of a flat plane, so there isn't any way to deform a plane into a sphere while keeping all lengths, angles and areas the same. It is impossible.

Quote:

If nobody could prove it until Gauss came around, I'd assume it must be a pretty complicated proof.
I don't know what Gauss's original proof was; I would simply parallel-transport a tangent vector around a small enough, closed path on the surface in question - if the transported vector is the same as the original vector after going around the loop, then the surface is flat on that patch, and is hence isomorphic to a flat plane. If not, then not. To make it even simpler than that, I would ask myself whether or not the covariant derivatives of a tangent vector taken in each direction commute or not; the "proof" then essentially becomes a one-liner :)
• December 14th, 2013, 04:13 AM
KJW
Quote:

Originally Posted by Markus Hanke
I don't know what Gauss's original proof was; I would simply parallel-transport a tangent vector around a small enough, closed path on the surface in question - if the transported vector is the same as the original vector after going around the loop, then the surface is flat on that patch, and is hence isomorphic to a flat plane. If not, then not. To make it even simpler than that, I would ask myself whether or not the covariant derivatives of a tangent vector taken in each direction commute or not; the "proof" then essentially becomes a one-liner :)

Gauss was the originator of differential geometry, and therefore tensor calculus originated with Gauss. Gauss showed that the curvature of a surface could be derived from the intrinsic properties of the surface without reference to the external three-dimensional space.
• December 14th, 2013, 05:41 AM
Sealeaf
"Suppose that there exists a way to cut a sphere such that it can be laid flat on a plane, and that all straight lines remain straight lines, and all areas, lengths, and angles remain the same. We are also going to
assume that the flattened map can be folded back up into a sphere also" That is a pretty exact statement of what can't be done.
• December 14th, 2013, 04:46 PM
GiantEvil
Quote:

Originally Posted by KJW
Quote:

Originally Posted by Markus Hanke
I don't know what Gauss's original proof was; I would simply parallel-transport a tangent vector around a small enough, closed path on the surface in question - if the transported vector is the same as the original vector after going around the loop, then the surface is flat on that patch, and is hence isomorphic to a flat plane. If not, then not. To make it even simpler than that, I would ask myself whether or not the covariant derivatives of a tangent vector taken in each direction commute or not; the "proof" then essentially becomes a one-liner :)

Gauss was the originator of differential geometry, and therefore tensor calculus originated with Gauss. Gauss showed that the curvature of a surface could be derived from the intrinsic properties of the surface without reference to the external three-dimensional space.

If we assume an infinite S^2 surface, then any arbitrary finite interval of coordinates would show a flat topology?
• December 14th, 2013, 05:38 PM
KJW
Quote:

Originally Posted by GiantEvil
Quote:

Originally Posted by KJW
Gauss was the originator of differential geometry, and therefore tensor calculus originated with Gauss. Gauss showed that the curvature of a surface could be derived from the intrinsic properties of the surface without reference to the external three-dimensional space.

If we assume an infinite S^2 surface, then any arbitrary finite interval of coordinates would show a flat topology?

I should remark at the outset that Gauss's Theorema Egregium (which this thread is about) is not the same as the Gauss-Bonnet theorem (which pertains to topology). When it comes to the surface integral of the curvature, which is invariant to continuous deformations of the interior region, and therefore a topological invariant, what happens at the boundary is an important consideration. Thus, if we have a topologically flat surface, for which the integral of the curvature over the entire surface is zero, the integral over a finite area of the surface need not be zero. It should be clear that if we remove a small area from a sphere, the integral of the curvature over the remaining part of the sphere will not be zero, and therefore that part of the sphere cannot be continuously deformed to a flat surface. The obstruction to this is the neighbourhood of the boundary of the removed region.
• December 14th, 2013, 09:50 PM
RedPanda
Quote:

Originally Posted by anticorncob28
Suppose that there exists a way to cut a sphere such that it can be laid flat on a plane, and that all straight lines remain straight lines, and all areas, lengths, and angles remain the same.

And if that sphere was a knitted kitten, it would make a very cosy pet.