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

• B
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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## Answers and Replies

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fresh_42
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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
WWGD
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Sorry if I am missing something obvious, but how do you know it is a sphere? I assume it is the parametrization, but I dont see how you can collapse 6 parameters into a single one.

lavinia
lavinia
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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
fresh_42
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Sorry if I am missing something obvious, but how do you know it is a sphere? I assume it is the parametrization, but I dont 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.

WWGD
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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?

WWGD
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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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lavinia
Gold Member
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 ?

WWGD
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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 ?
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 ?

WWGD
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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)##

lavinia
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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 ?
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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lavinia
Gold Member
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).

lavinia
Gold Member
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.

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

lavinia
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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?

WWGD
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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.

lavinia
Gold Member
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?

lavinia