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

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• jk22
In summary, the 8-dimensional manifold given by the parametrization $$a_1=cos(x)cos(y)cos(z)$$$$a_2=cos(x)cos(y)sin(z)$$$$a_3=cos(x)sin(y)cos(z)$$$$a_4=cos(x)sin(y)sin(z)$$$$a_5=sin(x)cos(y)cos(z)$$$$a_6=sin(x)cos(y)sin(z)$$$$a_7=sin(x)sin(y)cos(z)$$$$a_8=sin(x)sin(y)sin(z)$$ represents a 3-sphere.

#### jk22

If we suppose the following 8-dimensional manifold given by

$$a_1=cos(x)cos(y)cos(z)$$
$$a_2=cos(x)cos(y)sin(z)$$
$$a_3=cos(x)sin(y)cos(z)$$
$$a_4=cos(x)sin(y)sin(z)$$
$$a_5=sin(x)cos(y)cos(z)$$
$$a_6=sin(x)cos(y)sin(z)$$
$$a_7=sin(x)sin(y)cos(z)$$
$$a_8=sin(x)sin(y)sin(z)$$

Then obviously it's a 3 sphere and verifying the metric is euclidean is not so hard.

Since the metric of this 3sphere is euclidean there would be no way to detect for a R^3 lander if he lives in fact in such a high dimensional hypersphere ? (Or maybe by solving EFE in higher dimension and check gravitation experimentally ?)

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This depends on the size of a triangle we can measure. Since spheres are locally Euclidean, we need big triangles to measure.

jk22
Sorry if I am missing something obvious, but how do you know it is a sphere? I assume it is the parametrization, but I don't see how you can collapse 6 parameters into a single one.

lavinia
Spheres of dimension higher than two can contain submanifolds that are flat i.e. whose Riemann curvature tensor is identically zero. Your example seems to be a flat submanifold of an 8 dimensional sphere although I have not checked everything. If you have done it right, it is a flat 3 dimensional torus not a 3 dimensional sphere.

A perhaps simpler example is a flat 3 dimensional torus embedded in ##R^6## given by the mapping
(x,y,z)→(sin(x),cos(x),sin(y),cos(y),sin(z),cos(z)). This torus lies on a 5 dimensional sphere of radius ##√3##.

No sphere of any dimension can ever be a flat Riemannian manifold. Without going into a proof, there is a general structure theorem for closed flat manifolds that classifies them by their fundamental group. This group is always infinite. For instance, the fundamental group of the 3 dimensional torus is ##Z×Z×Z##, the triple direct product of the integers with itself. The fundamental group of the Klein bottle is a non-abelian semi-direct product of the integers with the integers ##0→Z→π_1(K)→Z→0##. However, the fundamental group of every sphere is trivial.

Note that if the curvature tensor is identically zero then the exponential map at any point is non-singular so one gets a non-singular smooth map of Euclidean space(the tangent space at the point) onto the flat manifold. In fact, this map is a covering projection and the covering group is the fundamental group. No sphere is covered by Euclidean space.

On a flat manifold there is no way to tell locally that it is not Euclidean space. Its local geometry is the same. The sum of the angles of a triangle is always 180°. Globally however one finds that straight lines - actually geodesics - that start at the same point may intersect at a different point. This tells you that the manifold is not Euclidean space. With most closed flat manifolds there are large closed paths along which parallel translation is not trivial. A vector may return to a different vector. For instance on a flat Klein bottle there are vectors that return to their negatives under parallel translation. This non-trivial holonomy also tells the lander that he is not in Euclidean space.

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Spinnor, Klystron, WWGD and 3 others
WWGD said:
Sorry if I am missing something obvious, but how do you know it is a sphere? I assume it is the parametrization, but I don't see how you can collapse 6 parameters into a single one.
What do you mean by parameters? If I add all three angles of a triangle, I can decide the curvature. All I need are big triangles and many of them.

fresh_42 said:
What do you mean by parameters? If I add all three angles of a triangle, I can decide the curvature. All I need are big triangles and many of them.
So you find it is of constant curvature. Is that enough?

fresh_42 said:
What do you mean by parameters? If I add all three angles of a triangle, I can decide the curvature. All I need are big triangles and many of them.
Bah, never mind, we have just x,y,z as variables. Duh!

Klystron
It's a 3-sphere I supposed from ##\sum_{i=1}^8 a_i^2=1##.

A usual sphere is not locally euclidean in the sense : the metric at a given point is not the identity matrix.

I think I understood, I made the mistake to induce "locally euclidean at each point", with "globally euclidean". The latter would be formalized by checking ##\langle \vec{e}_i(x_1)|\vec{e}_j(x_2)\rangle=\delta_{ij}##, so at different points, which is the case for cartesian coordinates.

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jk22 said:
It's a 3-sphere I supposed from ##\sum_{i=1}^8 a_i^2=1##.

A usual sphere is not locally euclidean in the sense : the metric at a given point is not the identity matrix.

I think I understood, I made the mistake to induce "locally euclidean at each point", with "globally euclidean". The latter would be formalized by checking ##\langle \vec{e}_i(x_1)|\vec{e}_j(x_2)\rangle=\delta_{ij}##, so at different points, which is the case for cartesian coordinates.

You can check directly whether your parameterization defines a local isometry from Euclidean 3 space onto the 3 manifold.

More simply, in the 2 dimensional version one gets a map from the Euclidean plane into ##R^4##.

##T(x,y) = (cos(x)cos(y),sin(x)cos(y),cos(x)sin(y),sin(x)sin(y))##

The tangent vectors ##∂T/∂x## and ##∂T/∂y## are an orthonormal basis for the tangent spaces to ##T##.

##∂T/∂x = (-sin(x)cos(y),cos(x)cos(y),-sin(x)sin(y),cos(x)sin(y))##

##∂T/∂y=(-cos(x)sin(y),-sin(x)sin(y),cos(x)cos(y),sin(x)cos(y))##

So for instance the dot product is

##∂T/∂x⋅∂T/∂y =sin(x)cos(y)cos(x)sin(y)-cos(x)cos(y)sin(x)sin(y)-sin(x)sin(y)cos(x)cos(y)+cos(x)sin(y)sin(x)cos(y) =0##
.

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Yes but we could check the angle between the basis vectors of two distinct points, or is this physically undoable since to measure an angle we need to transport the vectors at the same point ?

jk22 said:
Yes but we could check the angle between the basis vectors of two distinct points, or is this physically undoable since to measure an angle we need to transport the vectors at the same point ?
You need to define a connection to define operations between vectors based at different tangent spaces.

I don't know what the connection is. What would be the formula to compute the angle between vectors based at different points ?

A connection is essentially a choice of isomorphism between vector spaces ( of the same dimension, of course).

Does the transport have to be done along geodesics ? I do't understand the wikipedia article on affine connection, I think I remember that ##dA^\mu=-\Gamma^\mu_{\alpha\beta}A^\alpha dx^\beta## but the Christoffel symbol can be written in term of derivative of the metric and in the example given in the OP the metric is the identity so the connection vanishes ?

But should not the transport include an integral of the derivative of the vector field along a geodesic joining the two points ?

Erratum : ##g_{ij}(\vec{r}_1,\vec{r}_2)## should be rather called a nonlocal metric. Where as usual ##g_{ij}=\langle\vec{e}_i|\vec{e}_j\rangle## and ##\vec{e}_i(\vec{r})=\frac{\partial\vec{a}(\vec{r})}{\partial r_i}##, ##(r_1,r_2,r_3)=(x,y,z)##

jk22 said:
Yes but we could check the angle between the basis vectors of two distinct points, or is this physically undoable since to measure an angle we need to transport the vectors at the same point ?

In Euclidean space you could but why would you do it?

Post 9 was meant to show that the surface is a flat by showing that it is locally isometric to the Euclidean plane.

Yes it is locally flat, but still it hurts my common sense because the manifold is a sphere in 4 dimension, so in 4d all the points are at an equal distance from the origin. I don't understand that. is there an analogon between dimension 2 and 3 or 2 and 1 i thought of a cylinder.

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jk22 said:
Yes it is locally flat, but still it hurts my common sense because the manifold is a sphere in 4 dimension, so in 4d all the points are at an equal distance from the origin. I don't understand that. is there an analogon between dimension 2 and 3 or 2 and 1 i thought of a cylinder.

The image of the mapping is not a sphere. It is a torus. If you think of the torus as the Cartesian product of two circles (just as the Euclidean plane is the Cartesian product of two lines) then it is not hard to see the torus.

The manifold is a surface that lives inside the sphere in 4 dimensions.But you can see right away that it can not be the entire sphere because it is parameterized by 2 variables whereas the sphere in 4 dimensions is a 3 dimensional manifold.

By The Way: Locally flat in this case means that the curvature tensor is identically zero. It is a computation worth doing directly.

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Klystron and WWGD
I consider the OP in 8 dimension. Yes the curvature tensor can be expressed as derivatives of the metric which here vanish since the metric is euclidean i.e. equal to diag(1,1,1).

Btw: this OP manifold depends on 3 parameters and hence is three dimensional.

This was the point : since the curvature tensor vanishes how to decide if we don't live in such a 3-sphere ? (In fact it would be the classical space in vacuum neglecting general relativity).

jk22 said:
I consider the OP in 8 dimension. Yes the curvature tensor can be expressed as derivatives of the metric which here vanish since the metric is euclidean i.e. equal to diag(1,1,1).

Btw: this OP manifold depends on 3 parameters and hence is three dimensional.

This was the point : since the curvature tensor vanishes how to decide if we don't live in such a 3-sphere ? (In fact it would be the classical space in vacuum neglecting general relativity).

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 think this requires an isometric embedding into Euclidean space , and I think it is not too hard to show that none exists.

WWGD said:
I think this requires an isometric embedding into Euclidean space , and I think it is not too hard to show that none exists.

Can you explain what you mean?

lavinia said:
Can you explain what you mean?
IIRC ( it's been a while) Euclidean metric implies curvature vanishes so we end up with a Euclidean distance function, or an isometry. Isn't this correct, or the two equivalent Edit: to the existence of an isometric embedding? Maybe it's been too long.

WWGD said:
IIRC ( it's been a while) Euclidean metric implies curvature vanishes so we end up with a Euclidean distance function, or an isometry. Isn't this correct, or the two equivalent Edit: to the existence of an isometric embedding? Maybe it's been too long.

You mean in order for a manifold to be flat?

WWGD said:
IIRC ( it's been a while) Euclidean metric implies curvature vanishes so we end up with a Euclidean distance function, or an isometry. Isn't this correct, or the two equivalent Edit: to the existence of an isometric embedding? Maybe it's been too long.
Any closed Riemannian manifold can be isometrically embedded in a Euclidean space of sufficiently high dimension.

A closed flat Riemannian manifold does not have a global Euclidean distance function, only a local one.

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lavinia said:
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 ?

jk22 said:
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.

jk22 and lavinia
jk22 said:
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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mathwonk and jk22
lavinia said:
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).

lavinia
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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lavinia said:
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
jk22 said:
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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Orodruin said:
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

jk22 said:
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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mathwonk
WWGD said:
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.

WWGD said:
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.

Orodruin