# Confusion regarding Lie groups

1. Jan 15, 2013

### Kontilera

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 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

2. Jan 15, 2013

### quasar987

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.

3. Jan 15, 2013

### Kontilera

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:
$$[e_\mu, e_\nu] = 0.$$
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. Jan 15, 2013

### quasar987

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.

5. Jan 15, 2013

### Kontilera

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

6. Jan 15, 2013

### Kontilera

Last question: So on SU(2) I can find a coordinate basis which will satisfy
$$[e_\mu, e_\nu] = 0$$
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. Jan 15, 2013

### quasar987

Right.

8. Jan 18, 2013

### Kontilera

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?)

9. Jan 18, 2013

### quasar987

mh, no topologically, SO(3) is real projective 3-space RP³. See the very nice explanation of this on wiki: http://en.wikipedia.org/wiki/SO(3)#Topology.

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

10. Jan 18, 2013

### Kontilera

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.

Last edited: Jan 18, 2013
11. Jan 18, 2013

### VantagePoint72

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 [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.

Last edited: Jan 18, 2013
12. Jan 18, 2013

### Kontilera

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. Jan 18, 2013

### VantagePoint72

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