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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 ?

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- #1

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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 ?

- #2

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

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- #5

lavinia

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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)##

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- #6

mathwonk

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- #7

WWGD

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

- #8

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

- #9

WWGD

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Guess you are more into Lie Theory/ Geometric group theory than I am.I like ##\mathbb{S}^n = \operatorname{SO}(n+1)/\operatorname{SO}(n)## which sums it up.

- #10

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

- #11

lavinia

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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.

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- #12

lavinia

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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##.

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- #13

WWGD

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

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lavinia

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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.

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

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

- #16

WWGD

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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?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.

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

- #18

lavinia

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Not off hand. I will look around.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?

- #19

lavinia

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

- #20

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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.Sure. How shall we start?

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