Confusion regarding Lie groups


by Kontilera
Tags: confusion, groups
Kontilera
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#1
Jan15-13, 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 Im reading GR where we are talking about coordinate and non-coordinate 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
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quasar987
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#2
Jan15-13, 07:43 AM
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You forgot a step:

Smooth Lie group --> Find a coordinate chart --> use the coordinate basis in tangent space to identity--> extend to a left-invariant 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.
Kontilera
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#3
Jan15-13, 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.

quasar987
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#4
Jan15-13, 10:02 AM
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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) = {left-invariant 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 left-invariant 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 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.
Kontilera
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#5
Jan15-13, 10:13 AM
P: 101
Thanks, it doesnt seem obvious right now but I will give it some time.. maybe I come back tomorrow. :)
Kontilera
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#6
Jan15-13, 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.
quasar987
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#7
Jan15-13, 12:01 PM
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Right.
Kontilera
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#8
Jan18-13, 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 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?)
quasar987
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#9
Jan18-13, 07:39 AM
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mh, no topologically, SO(3) is real projective 3-space RP. See the very nice explanation of this on wiki: http://en.wikipedia.org/wiki/SO%283%29#Topology.

In particular SO(3) is a 3-dimensional manifold. It has S as its universal (2-sheeted) cover.
Kontilera
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#10
Jan18-13, 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 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.
LastOneStanding
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#11
Jan18-13, 10:09 AM
P: 718
Quote Quote by Kontilera View Post
"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? :/
Because there isn't a unique SO(3) rotation that yields a given point on the 2-sphere. If I take your (1,0,0) vector and rotate it so it's directed to a given point on ##S^2##, I can now take that vector as an SO(3) rotation axis. Clearly, a rotation by any angle around that vector doesn't move the vector to a new point on the 2-sphere. Thus, these are all distinct rotations in an SO(2) subgroup of SO(3) that all correspond to the same point on the 2-sphere.

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 isometry 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.
Kontilera
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#12
Jan18-13, 02:42 PM
P: 101
Right, Im 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!
LastOneStanding
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#13
Jan18-13, 04:57 PM
P: 718
Oops, sorry—meant to say "isotropy group", not "isometry group"


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