I Round 3-sphere symmetries as subspace of 4D Euclidean space

[cianfa72]

TL;DR Summary

About the symmetries of round 3-sphere as subspace of the ambient 4d euclidean space

As follow up of this thread in Special and General Relativity subforum, I'd like to better investigate the following topic.

Consider the 4d euclidean space in which there are 10 $\mathbb R$-linear independent KVFs. Their span at each point is 4 dimensional (i.e. at any point they span the entire tangent space). Now take an hypersurface of dimension 3, for instance a round 3-sphere about the origin (0,0,0,0).

Since it is a 3 dimensional restriction of the "ambient" 4d euclidean space, the maximum number of symmetries that, let me say, it may "support" is 3*4/2 = 6. Therefore, from the full set of 10 "ambient symmetries", only up to 6 of them can apply to it.

In the case of round 3-sphere, which subset of the 10 KVFs ambient symmetries actually apply to it ? The round 3-sphere should be a maximally symmetric space, therefore I believe 6 of them should apply. However, the 4 ambient translational KVFs do not apply since they basically "exit" from the 3-sphere itself as submanifold.

What about the other 6 KVFs of the ambient 4d euclidean space ?

 

[Ibix]

They're just the rotations. Think of writing a 4×4 rotation matrix - how many ways are there to place the non-zero terms?

 

[cianfa72]

They're just the rotations. Think of writing a 4×4 rotation matrix - how many ways are there to place the non-zero terms?

I believe you're actually talking of the 4x4 infinitesimal matrix generators of rotations in 4D that are skew-symmetric (and not of the matrix representatives of the rotations themselves). Then there are 6 non-zero terms in each of them (i.e. 6 degree of freedom). Basically any rotation in 4D occurs as combination of rotations around pairs of orthogonal axes and not just around a 1D straight line as in 3D.

 

[Ibix]

and not of the matrix representatives of the rotations themselves

You're overcomplicating it. This is a Euclidean space. Just write down the rotation matrices and count how many distinct possibilities there are.

 

[cianfa72]

This is a Euclidean space. Just write down the rotation matrices and count how many distinct possibilities there are.

I found here the set of 6 basis rotating matrices for a generic rotation in 4D euclidean space. Each of those matrix subscripts represent a rotation on the plane being rotated (i.e. the orthogonal axes to those planes are left fixed by the rotation including their linear combination). And yes, any rotation in 4D can be given as combination of those 6, so it depends on 6 parameters $\theta$.

Therefore I believe each of those basis rotating matrices has a degenerate eigenvalue 1 (i.e. its eigenspace has dimension 2).

 

[Orodruin]

This is trivial. The Lie algebra of the rotation matrices in 4D is anti-symmetric real matrices. It is evident that they constitute a 6-dimensional vector space.

 

[cianfa72]

The Lie algebra of the rotation matrices in 4D is anti-symmetric real matrices. It is evident that they constitute a 6-dimensional vector space.

Ok. Since $\text{det} (e^A)= e^{\text{tr}(A)}$ then the exponential of a Lie algebra element has det 1 (as any element of $SO(4)$ must have). And yes, the set of 4x4 skew-symmetric matrices has a vector space structure of dimension 6.

Therefore the 6 non translation KVFs of 4D euclidean space, are the 6 rotation KVFs symmetries that apply to any 3-sphere.

 

[cianfa72]

Btw, in the general case of a 4D Riemann manifold, say I give you 6 KVFs w.r.t. the metric tensor $g$.

Suppose they form a Lie subalgebra (i.e. they form a vector subspace of the vector space of all manifold's KVFs and that close w.r.t. Lie bracket/commutator) therefore, by Frobenius's theorem, the distribution they define is integrable. So far so good.

Does the structure of such Lie subalgebra (i.e. its structure constants) determinate the rank of the distribution they define (i.e. the rank of the subbundle being defined) ?