# Lie Groups and Canonical Coordinates

1. Mar 25, 2012

### Sina

Hello. I have a question that has been on my mind for some time. I always see in mathematical physics books that they identify elements of the Lie algebra with group elements "sufficiently close" to the identity. I have never seen a real good proof of this so went on an gave a proof. Let Xi be the generators of the Lie algebra and Yi are their images in coordinates. Let β be the coordinate map around identity of e. (identifying that of Rn with its tangent space at 0) I have shown for analytic group operation mapping and for some small nbd around e,

[tYi,tYj] = β[exp(tYi)exp(tYj)] - β[exp(tYj)exp(tYi)] + o(t3)

The last order of correction basically comes from the exponential coordinates and other higher corrections. In writing the exponential coordinates, I think the correct expansion is:

β(exp(tYi)) = tYi + o(t2)

Now that is where most other books say that for small enough t we have

β(exp(tYi)) = tYi

In fact, Dubrovin, Novikov, Fomenko's book they directly say that in canonical coordinates of the first kind we have β(exp(tYi)) = tYi. And by throwing out that last order of t2, they show that if the lie algebra of a connected lie group is abelian then lie group is also abelian (which also follows from my estimate above if you disregard o(t2), but can you?).

So my questions are

1- What happens to that order of correction to the expansion :) Why do people treat it like "lets take the limit zero and it is no more important". I could not find a way to gauge away the lesser t orders out of the equation by taking derivatives so I really don't understand the methodology here.

2- In many books, when they build this kind of correspondance between Lie groups and Lie algebras, they assume the group operation is analytic and carry out the proof with Taylor expansions. But then they say that actually analyticity is not required but is harder to give proof without that assumption. I also used that assumption to have a taylor series expansion. But that is of course bothersome because generally theory of differentiable manifolds is built assuming the manifold is not analytic so that one can use hat functions. So my second question is what is the idea of passage to non-analytic coordinates. For instance can one build a non-analytic charts for a Lie group such that at least in some nbd of identity coordinates are analytic? When we give up real analyticity of the manifold, do we lose anything other than Taylor expansions?

2. Mar 25, 2012

### dextercioby

The Lie group is assumed by definition to be an analytic manifold. It's what singles them out from the greater set of topological groups.

3. Mar 25, 2012

### Sina

But in many books like Lee's where non-analytic manifolds are used, Lie groups are simply introduced as smooth manifolds with smooth group operation and inversion? I guess the all the concepts of vector fields etc there belong their construction with partition of unities.

If they are analytic how can one build vector fields etc by giving them some values in nbds? Shouldn't one proceed with a sheaf theoretic approach, through by germs etc.

I have found this on wikipedia

" Hilbert's fifth problem asked whether replacing differentiable manifolds with topological or analytic ones can yield new examples. The answer to this question turned out to be negative: in 1952, Gleason, Montgomery and Zippin showed that if G is a topological manifold with continuous group operations, then there exists exactly one analytic structure on G which turns it into a Lie group (see also Hilbert–Smith conjecture)."

But I don't really understand what this implies. Does this mean that if we try to build topological groups using non-analytic tools like partition of unity, then are we making a mistake?

4. Mar 25, 2012

### Alesak

Hmm, I´m reading a book on matrix Lie groups, so I might help with insights from this perspective. What is the precise statement of theorem you are trying to prove?

5. Mar 25, 2012

### Sina

I am not trying to proove anything, I gave a proof of what I wanted to proove in my own way :) But somethings in very well known books bother me and I am trying to understand those. (the questions are at the end of my previous message).

In matrix lie groups every thing is okay anyway. One of the things I am trying to understand the general case where some authors say that for sufficiently close to the identity, in coordinates exp(tX) = 0 + tX' (where 0 is the image of identity and X' is image of X in coordinates) and totally disregard second order in t without taking derivatives. I was wondering how that could be justified.

6. Mar 25, 2012

### Sina

I will put my first question more simply. Let X,Y be two elements of the lie algebra and x(t), y(t) be their 1-parameter subgroups. I have seen some well known books write (for t small enough) in coordinates,

[tX,tY] = x(t)y(t) - y(t)x(t)

without any order of correction. In my opinion the correct expression should be

[tX,tY] + o(t3) = x(t)y(t) - y(t)x(t)

however somebooks totally disregard that factor as it seems and calculates properties of the Lie algebra from the above equations for "t small enough" (like the commutator etc). I can not understand how one can make analytical calculations disregarding an order of factor.

7. Mar 27, 2012

### morphism

They're just being sloppy.

8. Mar 31, 2012

### Sina

I think I might have a question now from the matrix lie groups too. Suppose you have a matrix lie group like SL(2,R) and say you take its quotient with {-1,1}. This is also a lie group. Do you know how to calculate the induced maps on lie algebras? Lie algebra of SL(2,R) are traceless matrices. How does the pushforward map of the quotient map transform the lie algebra? Is there a general procedure for such things?

Thanks

9. Apr 4, 2012

### morphism

The pushforward of that map will be an isomorphism. This is easy to check using the differential geometric definition of the pushforward, for example. (The dimensions of the Lie algebras will be the same, so all you need to do is check that the pushforward is injective, but this is more or less obvious. You could also use the exponential mapping for this last bit.)

In general, a covering map of Lie groups (e.g. $\text{SL}(2,\mathbb R) \twoheadrightarrow \text{SL}(2,\mathbb R)/\{\pm 1\}$) will induce an isomorphism of Lie algebras.

10. Apr 4, 2012

### Sina

thanks for the suggestions, that is what I was hoping since it has discrete kernel and well it should just be identifying two tangent spaces at two points. I will refresh my memory on covering groups and look at it from that point of view