B How to know if a Euclidean space is not a 3-sphere?

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You said in Post #1"
Then obviously it's a 3 sphere and verifying the metric is euclidean is not so hard.

Since the metric of this 3 sphere is euclidean..."

1) The manifold is not a 3 sphere.
2) A 3 sphere cannot have a Euclidean metric.
I supposed the definition of the sphere was : "the locus of points of a surface at equal distance from the center", then the 3-sphere was just that the points of a higher dimensional space at equal distance from a center ?
 

Orodruin

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I supposed the definition of the sphere was : "the locus of points of a surface at equal distance from the center", then the 3-sphere was just that the points of a higher dimensional space at equal distance from a center ?
The 3-sphere is the set of points in 4-dimensional Euclidean space that is at an equal distance from the center. The set of points in 8-dimensional space that is at an equal distance from the center is a 7-sphere. What you are presenting here is a 3-dimensional sub-manifold of a 7-sphere. This does not need to be a 3-sphere.
 

lavinia

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I supposed the definition of the sphere was : "the locus of points of a surface at equal distance from the center", then the 3-sphere was just that the points of a higher dimensional space at equal distance from a center ?
The set of points in Euclidean space that are equidistant from a center is a sphere. The dimension of the sphere as a manifold is one less than the dimension of the Euclidean space. So the sphere in Euclidean three space is a two dimensional manifold. The sphere in euclidean four space is a three dimensional manifold. ( This is the three sphere.)The sphere in eight dimensions is a seven dimensional manifold.

Your example is a three dimensional manifold in eight dimensional space whose points are equidistant from the origin. It is a submanifold of the seven dimensional sphere. But it is not a three sphere.

There are an infinite number of different three dimensional manifolds that are submanifolds of the seven dimensional sphere. The three sphere is only one possibility. In fact, any closed smooth three dimensional manifold is homeomorphic to a submanifold of the seven dimensional sphere. Your example seems to be a three dimensional torus.

Your manifold - if it is flat which I think it is - cannot be a sphere. No sphere in any dimension can be flat.

Comment: There certainly are examples of Lorentz manifolds that can be foliated by three dimensional closed flat submanifolds all of whose tangent vectors are space like. I have no idea if such manifolds have any use in Physics but mathematically they exist. For instance, the Cartesian product of the three dimensional torus with the real line ##T^3×R^1## can be given a Lorentz metric in which the coordinate tori are flat.
 
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Orodruin

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Comment: There certainly are examples of Lorentz manifolds that can be foliated by three dimensional closed flat submanifolds all of whose tangent vectors are space like. I have no idea if such manifolds have any use in Physics but mathematically they exist. For instance, the Cartesian product of the three dimensional torus with the real line T3×R1T3×R1T^3×R^1 can be given a Lorentz metric in which the coordinate tori are flat.
They certainly are of interest, in particular in terms of cosmology. However, a basic assumption in cosmology is usually that the submanifolds be homogeneous and isotropic, which only allows Euclidean space as a flat option (the torus is not isotropic).
 
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The usual torus in three dimensions has a euclidean metric, but the points are not at equal distance from a center, and it has a sum inside.

But is the previous parametrization not the tensor product of 2 circles ?

How does it come that in 4D they can be at equal distance from a center ?
 
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WWGD

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You said in Post #1"
Then obviously it's a 3 sphere and verifying the metric is euclidean is not so hard.

Since the metric of this 3 sphere is euclidean..."

1) The manifold is not a 3 sphere.
2) A 3 sphere cannot have a Euclidean metric.
Is this last related to the fact that Euclidean d
I supposed the definition of the sphere was : "the locus of points of a surface at equal distance from the center", then the 3-sphere was just that the points of a higher dimensional space at equal distance from a center ?
EDIT: 3-dimensional subspace of all whose points are at equal distance from the origin, or a fixed point ( the center)

EDIT2: Not to be pretentious here, but this is true up to homeomorphism. Start with a "standard" 3-sphere { (x,y,z): ||(x,y,z)||=1 } and apply any homeomorphism. You can get something as nasty as Alexander's horned sphere is for the 2-sphere.
 
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WWGD

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The 3-sphere is the set of points in 4-dimensional Euclidean space that is at an equal distance from the center. The set of points in 8-dimensional space that is at an equal distance from the center is a 7-sphere. What you are presenting here is a 3-dimensional sub-manifold of a 7-sphere. This does not need to be a 3-sphere.
What if your points in 8D are of the form {(x,y,z,0,...,0)} at an equal distance from the origin, i.e., this set may be 3-dimensional ( assuming x,y,z are pairwise independent). But I assume you were discard this option, maybe we have different language here.

@jk22: The condition of equal distance from the origin is sufficient but not necessary. If you deform the sphere homeomorphically but in a way that does not preserve the distance from the origin, you still have a sphere
 

lavinia

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The usual torus in three dimensions has a euclidean metric, but the points are not at equal distance from a center, and it has a sum inside.
What do you mean by a sum inside?

No torus in three dimensions can have its points equidistant from a center. Any such manifold must lie on a 2 sphere and the only closed submanifolds of the 2 sphere are topological circles.

When you say that the usual torus in 3 dimensions has a euclidean metric, what exactly do you mean? If you mean that it is flat then this is not possible.No closed smoothly embedded submanifold of three space can be flat. A famous Theorem of Hilbert says that any smoothly embedded closed surface in Euclidean 3 space must have a point of positive Gauss curvature. The usual torus is smoothly embedded.

Aside: There are non-smoothly embedded tori in Euclidean 3 space that are flat. There is a famous example where the unit normal is nowhere differentiable so the torus does not have a shape operator at any point. It looks sort of like a fractal.

But is the previous parametrization not the tensor product of 2 circles ?
No. It is the Cartesian product of two circles. I don't know what it means to take the tensor product of manifolds.

How does it come that in 4D they can be at equal distance from a center ?
Using the inverse of stereographic projection, the usual torus in 3 space is sent to a torus in four dimensional space that lies completely on the unit sphere. The moral is that the 3 sphere is Euclidean 3 space with one point added. So any geometric figure in 3 space can also be realized in the 3 sphere.

BTW: A good excercise is to derive the equations for the torus in 3 space that is the stereographic image of a flat torus in the 3 sphere. From the equations one can also visualize it using a graphics program. This torus is not flat since stereographic projection is smooth so Hilbert's Theorem applies. However stereographic projection is conformal. It preserves angles between tangent vectors. So given two intersecting curves on the torus, the angle between them is the same as the angle of their pre-images on the flat torus in four space. Such manifolds are sometimes called conformally flat.
 
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Orodruin

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What if your points in 8D are of the form {(x,y,z,0,...,0)} at an equal distance from the origin, i.e., this set may be 3-dimensional ( assuming x,y,z are pairwise independent). But I assume you were discard this option, maybe we have different language here.
What I said was that a 3-dimensional sub-manifold of a 7-sphere does not need to be a 3-sphere, i.e., there are 3-dimensional sub-manifolds of a 7-sphere that are not 3-spheres. This does not imply that a sub-manifold of the 7-sphere cannot be a 3-sphere. However, in the case of the OP, it is not a 3-sphere.
 

lavinia

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What if your points in 8D are of the form {(x,y,z,0,...,0)} at an equal distance from the origin, i.e., this set may be 3-dimensional ( assuming x,y,z are pairwise independent). But I assume you were discard this option, maybe we have different language here.
These points form a 2 sphere. You need 1 more independent variable.
 

WWGD

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