- #1

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Smooth manifold --> Find a coordinate chart --> use the coordinatebasis in tangentspace --> we have three pairwise commuting generators.

Where does this break down?

Thanks in regards!

/Kontilera

- Thread starter Kontilera
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- #1

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Smooth manifold --> Find a coordinate chart --> use the coordinatebasis in tangentspace --> we have three pairwise commuting generators.

Where does this break down?

Thanks in regards!

/Kontilera

- #2

quasar987

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Smooth

The result in general will differ from the local coordinate frame you started with and so the Lie bracket won't be 0 anymore.

- #3

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In my GR course a coordinate basis is defined by satisfying:

[tex] [e_\mu, e_\nu] = 0. [/tex]

while in my mathematics literature the are defined by being directional derivatives along a set of coordinate axes (given by a chart on the manifold).

But these two statements are very different for Lie groups.

- #4

quasar987

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The two definitions of coordinate basis you mention are indeed equivalent. You can consult Lee Thm 18.6 for a proof. The key is that two vector fields Lie-commute (i.e. [X,Y] = 0) iff their flows commute for all times. Given a basis of Lie-commuting vector field, you can thus compose the flows to construct coordinates.

I don't see in what sense the statements are different "for Lie groups". The statements concern only smooth manifold and are independent of any additional structures the manifold might carry.

- #5

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- #6

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[tex][e_\mu, e_\nu] = 0[/tex]

but these coordinate axes will not be left invariant and therefore doesnt say anything about the bigger structure of the group or its Lie algebra.

- #7

quasar987

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Right.

- #8

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Lets take the 2-sphere as an example. The tangentspace can at every point be spanned by derivatives along logitude and latitude - so its 2-dimensional.

But making the identification that our derivative along the longitudinal axis is the L_z operator when we represent this algebra in QM, it seems that we should have a 3-dimensional algebra; L_z, L_y, L_x.

So we have a 3-dimensional vectorspace cosisting of the leftinvariant vectorfields but our manifold and tangentspace is two dimensional.

Is this correct?

(I am thinking of SO(3) as topologically equivalent to the 2-sphere, this is correct, right?)

- #9

quasar987

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In particular SO(3) is a 3-dimensional manifold. It has S³ as its universal (2-sheeted) cover.

- #10

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Yeah, just realized this, glad you verified it. :)

My misstake (which I commited the last time I dealt with SO(3) aswell) is that I reason like this:

"So, I want a group of rotations on R^3. Lets start with the vector (1,0,0), this can now be rotated to every vector on the 2-sphere by using all the elements of SO(3). However we can not rotate it to a vector not lying on the 2-sphere. So it should be a 1-1 correspondense between the elements of SO(3) and the 2-sphere."

Where does this logic fail? :/

EDIT: This logic fails on the problem I brought up earlier I guess.. Obviously the dimensions arent correct, starting from (1,0,0) we can only go along \phi or \theta, but SO(3) is generated by rotation around three diffrent axes.

My misstake (which I commited the last time I dealt with SO(3) aswell) is that I reason like this:

"So, I want a group of rotations on R^3. Lets start with the vector (1,0,0), this can now be rotated to every vector on the 2-sphere by using all the elements of SO(3). However we can not rotate it to a vector not lying on the 2-sphere. So it should be a 1-1 correspondense between the elements of SO(3) and the 2-sphere."

Where does this logic fail? :/

EDIT: This logic fails on the problem I brought up earlier I guess.. Obviously the dimensions arent correct, starting from (1,0,0) we can only go along \phi or \theta, but SO(3) is generated by rotation around three diffrent axes.

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- #11

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Because there isn't a"So, I want a group of rotations on R^3. Lets start with the vector (1,0,0), this can now be rotated to every vector on the 2-sphere by using all the elements of SO(3). However we can not rotate it to a vector not lying on the 2-sphere. So it should be a 1-1 correspondense between the elements of SO(3) and the 2-sphere."

Where does this logic fail? :/

The moral is that when you relate the topology of some manifold to a Lie group that acts transitively on it, you have to factor out the [STRIKE]isometry[/STRIKE] isotropy group. So, in this case, ##S^2 = SO(3)/SO(2)##. I don't know about real projective groups, but I suppose that must be equivalent to what quasar987 said.

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- #12

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As soon as I can find isometries, not including the identity element, I should be careful making my fast assuptions about topological equivalents. :)

I will continue my hiking down the road of GR.

Thanks for the help - both of you!

- #13

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Oops, sorry—meant to say "isotropy group", not "isometry group"

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