
#1
Jan1513, 04:49 AM

P: 101

Hello! Im currently trying to get things straight about Lie group from two different perspectives. I have encountered Lie groups before in math and QM, but now I´m reading GR where we are talking about coordinate and noncoordinate bases and it seems that we should be able to find commuting generators, to for example SU(2), by just:
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
Jan1513, 07:43 AM

Sci Advisor
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PF Gold
P: 4,768

You forgot a step:
Smooth Lie group > Find a coordinate chart > use the coordinate basis in tangent space to identity> extend to a leftinvariant vector field 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
Jan1513, 09:17 AM

P: 101

Ah okey, so my misunderstanding is a confusion about the difference between smooth manifolds and Lie groups. My line of thinking was that since every lie group is a smooth manifold (correct?) every coordinate chart will induce a tangentspace basis which will commute pairwise.
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
Jan1513, 10:02 AM

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PF Gold
P: 4,768

Confusion regarding Lie groups
Yes, I understood your line of thinking. Your error was in forgetting to keep track of how exactly is the lie algebra of a lie group G identified with the tangent space of G at the identity. Namely: if X and Y are two arbitrary vector fields on G, then they restrict to vectors on TeG (and hence induce elements of Lie(G) = {leftinvariant vector field on G}), but these induced elements X' and Y' are not going to be X and Y if X and Y were not leftinvariant to begin with, and so [X,Y] will not equal [X',Y'] in general.
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 Liecommute (i.e. [X,Y] = 0) iff their flows commute for all times. Given a basis of Liecommuting 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
Jan1513, 10:13 AM

P: 101

Thanks, it doesnt seem obvious right now but I will give it some time.. maybe I come back tomorrow. :)




#6
Jan1513, 10:27 AM

P: 101

Last question: So on SU(2) I can find a coordinate basis which will satisfy
[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. 



#8
Jan1813, 06:12 AM

P: 101

Its getting clearer and clearer for me but I dont really get the dimensions to add up.
Lets take the 2sphere as an example. The tangentspace can at every point be spanned by derivatives along logitude and latitude  so its 2dimensional. 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 3dimensional algebra; L_z, L_y, L_x. So we have a 3dimensional 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 2sphere, this is correct, right?) 



#9
Jan1813, 07:39 AM

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PF Gold
P: 4,768

mh, no topologically, SO(3) is real projective 3space RP³. See the very nice explanation of this on wiki: http://en.wikipedia.org/wiki/SO%283%29#Topology.
In particular SO(3) is a 3dimensional manifold. It has S³ as its universal (2sheeted) cover. 



#10
Jan1813, 07:50 AM

P: 101

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 2sphere by using all the elements of SO(3). However we can not rotate it to a vector not lying on the 2sphere. So it should be a 11 correspondense between the elements of SO(3) and the 2sphere." 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. 



#11
Jan1813, 10:09 AM

P: 718

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 



#12
Jan1813, 02:42 PM

P: 101

Right, I´m with you!
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
Jan1813, 04:57 PM

P: 718

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



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