Question on definition of Lie groups

[mnb96]

Hello,

I have a doubt on the definition of Lie groups that I would like to clarify.
Let's have the set of functions $$G=\{ f:R^2 \rightarrow R^2 \; | \; < f(x),f(y)>=<x,y> \: \forall x,y \in R^2 \}$$, that is the set of all linear functions ℝ²→ℝ² that preserve the inner product. Let's associate the operation of composition to the elements of G and we obtain a group $(G,\circ)$.

Now in order to say that G is indeed a Lie group we must prove that G is also a smooth manifold. How can I do this if we don't specify a parametrization (e.g. a matrix representation) for the group G ?

And also in that case, wouldn't the definition of G being a smooth manifold depend on the parametrization?

 

[chiro]

Can you use continuity and smoothness properties of the inner product and also use some assumptions for f and g to ensure this?

 

[mnb96]

I don't know. To be honest I cannot come up with other different strategies than associating each function $f\in G$ with an element of a vector space where I could perform ordinary partial differentiation w.r.t. to the parameters. But this approach is then dependent on the parametrization that I give, which may or may not be differentiable.

Perhaps there might be some theorem that states that if we are able to find just one parametrization by which we manage to prove that the "parametrized G" is a differentiable manifold, then also our "non-parametrized G" must be a differentiable manifold?

 

[chiro]

I don't know. To be honest I cannot come up with other different strategies than associating each function $f\in G$ with an element of a vector space where I could perform ordinary partial differentiation w.r.t. to the parameters. But this approach is then dependent on the parametrization that I give, which may or may not be differentiable.

Perhaps there might be some theorem that states that if we are able to find just one parametrization by which we manage to prove that the "parametrized G" is a differentiable manifold, then also our "non-parametrized G" must be a differentiable manifold?

It sounds like you need some topological argument since you are talking about a general continuity property rather than some specific one (like a specific parameterization). I don't really know any topology though in depth so I'm not even aware of the major theorems.

R^2 I'm sure (as for R^n) has known topological results: can you assume that the topology is the same or at least something that keeps the smoothness properties in tact?

 

[Bacle2]

Sorry, I need to go now, but maybe try using the fact that the group you described

is usually called the orthogonal group--consisting of the elements that preserve

a given quadratic form Q* -- and it is a well-known Lie group. * Remember that your inner-product can be described thru a quadratic form.
The answer

to your question, tho, may have to see with using matrices in a representation of

the group. A group representation of a group G is just a homomorphism between

G and a group of matrices, so that each g in G is assigned a matrix. Once you have

a matrix, you can start talking about a topology . Just use the subspace topology

in R^{n^2} , where you assign to a matrix an n^2-ple (a11, a12,..., ann ).

See, e.g:

http://en.wikipedia.org/wiki/Orthogonal_group