How Many Parameters Are Needed to Parametrize the 3-Sphere?

[jk22]

The equation is $$\|\left(\begin{array} &a\\b\\c\\d\end{array}\right)\|^2=1$$

I was wondering if the number of parameters is 6 and not 3, since we can consider rotations in the differents planes : we choose 2 directions among 4 hence $$C^4_2=6$$ possibilities ?

 

[Orodruin]

The 3-sphere is not the same thing as the set of rotations in 4 dimensions.

 

[jk22]

indeed, the set that keep a direction fixed has to subtracted hence $$C^n_2-C^{n-1}_2=n-1$$ which seems more correct. Thanks.

 

[Orodruin]

The easy way to see that it must be $n-1$ is that it is a level surface of a function in $\mathbb R^n$.

 

[lavinia]

indeed, the set that keep a direction fixed has to subtracted hence $$C^n_2-C^{n-1}_2=n-1$$ which seems more correct. Thanks.

The rotations of $R^4$ that keep a point in the 3 sphere fixed are the subset of rotations of the orthogonal 3 dimensional plane (not 2 dimensional).

These rotations are three dimensional since the dimension of the rotations of 3 space is 3. The rotations of $R^4$ are six dimensional so the 3 sphere has dimension three by your argument. That is: the three sphere has dimension $6-3$.

For the 4 sphere the rotations that keep a point fixed are the subset of rotations of the 4 dimensional orthogonal plane.

So by your argument the dimension of $S^4$ is the dimension of the rotations of $R^5$ minus the dimension of the rotations of $R^4$ which is $10 - 6$.

In general the dimension of the rotation group of $R^n$ is $.5n(n-1)$

 

[mathwonk]

i.e. by these arguments, the dimension of the rotation group of n space = dim of n-1 sphere + dim of rotation group of n-1 space = (n-1) + (n-2) + ...+2+1 = n(n-1)/2, using induction. nice observation.

 

[WWGD]

The easy way to see that it must be $n-1$ is that it is a level surface of a function in $\mathbb R^n$.

This is also a nice, high-level way of showing it is a manifold.

 

[fresh_42]

I like $\mathbb{S}^n = \operatorname{SO}(n+1)/\operatorname{SO}(n)$ which sums it up.

 

[WWGD]

I like $\mathbb{S}^n = \operatorname{SO}(n+1)/\operatorname{SO}(n)$ which sums it up.

Guess you are more into Lie Theory/ Geometric group theory than I am.

 

[Orodruin]

Guess you are more into Lie Theory/ Geometric group theory than I am.

I don't think you need to be so much into Lie theory to appreciate it, it has a very nice visual interpretation. A point on the sphere $\mathbb S^n$ is picking one out of $n+1$ direction, whereas a rotation in $\mathbb R^{n+1}$ is fixing all $n+1$ directions. Since you do not care about what happens the other directions on $\mathbb S^n$, you can rotate at will around it and will still end up with your direction being the same, which is a $SO(n)$ freedom.

 

[lavinia]

Guess you are more into Lie Theory/ Geometric group theory than I am.

$\mathbb{S}^n = \operatorname{SO}(n+1)/\operatorname{SO}(n)$
also shows that the sphere is a manifold.

 

[lavinia]

i.e. by these arguments, the dimension of the rotation group of n space = dim of n-1 sphere + dim of rotation group of n-1 space = (n-1) + (n-2) + ...+2+1 = n(n-1)/2, using induction. nice observation.

Another way to get the induction is to observe first that $SO(3)$ acts transitively and without fixed points on the tangent circle bundle of the $2$ sphere. The action is the differential of the action on $S^2$. So $SO(3)$ is diffeomorphic to the tangent circle bundle of $S^2$ and has dimension $3$.

$SO(4)$ acts transitively on the tangent $2$ sphere bundle of $S^3$ but not without fixed points since the tangent sphere at a rotation axis point will itself have a rotation axis.

But its differential (this is the differential of the differential of the action of $SO(4)$ by rotations on $S^3$) does act transitively and without fixed points on the bundle of circles tangent to the fiber $2$ spheres. This bundle is a six manifold ($3 + 2 + 1$).

For $SO(5)$ one needs three differentials to get a transitive fixed point free action on the bundle of tangent fiber circles over the bundle of tangent fiber $2$ spheres over the bundle of tangent $3$ spheres over the $4$ sphere. So $SO(5)$ is $10$ dimensional ($4+3+2+1$).

In general $SO(n)$ acts transitively and without fixed points by repeated differentials on a circle bundle over a tower of sphere bundles over $S^{n-1}$. The fiber dimensions in the tower are $(n-2) + (n-3)+... +1$ so adding in the dimension of $S^{n-1}$ gives dimension $n(n-1)/2$.

 

[WWGD]

I take it back, I guess, I am way less into Lie Theory /Geometric Group theory than most.

 

[lavinia]

I take it back, I guess, I am way less into Lie Theory /Geometric Group theory than most.

Since you are interested in fiber bundles you might like to learn a little about the classical groups and their quotient spaces. Steenrod's Topology of Fiber Bundles goes into this.

 

[fresh_42]

I take it back, I guess, I am way less into Lie Theory /Geometric Group theory than most.

Since you are interested in fiber bundles you might like to learn a little about the classical groups and their quotient spaces. Steenrod's Topology of Fiber Bundles goes into this.

... or help me to figure out (by foot) why the Ehresmann connection is a connection and how it is different from Levi-Civita.

 

[WWGD]

Since you are interested in fiber bundles you might like to learn a little about the classical groups and their quotient spaces. Steenrod's Topology of Fiber Bundles goes into this.

Thanks, I am indeed, but I found his book a bit dry for my taste. Can you think of any other good books, links, videos on this topic?

 

[fresh_42]

Thanks, I am indeed, but I found his book a bit dry for my taste. Can you think of any other good books, links, videos on this topic?

I know a good one for the groups, but not so much for the bundles.

 

[lavinia]

Thanks, I am indeed, but I found his book a bit dry for my taste. Can you think of any other good books, links, videos on this topic?

Not off hand. I will look around.

 

[lavinia]

... or help me to figure out (by foot) why the Ehresmann connection is a connection and how it is different from Levi-Civita.

Sure. How shall we start?

 

[fresh_42]

Sure. How shall we start?

Thanks, really. However, I should start a new thread for this. I also need to give some preliminaries, as e.g. where to start, and what for. I'll let you know via a quote, but thanks again.