# Test for Lie Group

1. Feb 24, 2012

### mnb96

Hello,

if I have a set of functions of the kind $\{ f_t | f_t:\mathbb{R}^2 \rightarrow \mathbb{R}^2 \; ,t\in \mathbb{R} \}$, where t is a real scalar parameter. The operation I consider is the composition of functions. What should I do in order to show that it forms a Lie Group?

Last edited: Feb 24, 2012
2. Feb 24, 2012

First check if you have a group. Then check if the map $t\mapsto f_t$ is one to one. Next steps will depend on the results of these two.

3. Feb 24, 2012

### mnb96

let's consider a toy example, in which we have that each ft is a linear function ℝ2→ℝ2 given by a counter-clockwise rotation of angle t around the origin, where t is in the range [0,2π). Each ft can effectively be represented as a 2x2 matrix, and the composition would be given by matrix multiplication.

Matrix multiplication is associative, and for any pair of rotation matrices the composition yields another rotation matrix in the set, so we have closure too. Moreover we have an identity element which is f0, and also an inverse which is f2π-t ===> so, we have a group.

We also consider the map $t \mapsto R(t)$, where R(t) is a 2x2 rotation matrix and t is the angle parameter. The map is clearly 1-to-1, provided t is in the range [0,2pi).

Now, what else do we need?

4. Feb 24, 2012

Now you want it to be a Lie group. So it needs to be a manifold. In this case you need first to convince yourself that the circle (because effectively you are dealing with a circle here) is naturally a manifold, according to the definition of a smooth manifold (Hint: in this case it will be one-dimensional).
Next step will come after you are done with this exercise.

5. Feb 24, 2012

### mnb96

I see...so I have to prove basically that I am dealing with a group, that is also a differentiable manifold. Unfortunately I feel I start walking on thin cracking ice.
The definitions of differentiable manifolds that I can find seem to be very abstract, and in order to truly understand them it seems to me that one has to recursively "unwrap" quite many concepts in topology.

Before giving up, I can make an intuitive non-rigorous attempt.
I would take one generic point in ℝ2, say x=(x0,y0), and then consider its orbit under the action of the rotation group. The orbit is a circle (we can prove it easily), and thanks to the bijective mapping t↦R(t) mentioned before, each point of the orbit can be identified with a specific value of t. That effectively means we have parametrized a 1-dimensional curve in ℝ2 with the parameter t. I would then simply try to prove analytically that by varying t, the curve is smooth.

Does it go along those lines?

6. Feb 24, 2012

In fact you want to give your group a parametrization. For this you do not have to consider orbits. You have your parametrization already: it is your t. Composition law in the group is effectively that of $t+t'$ modulo $2 \pi$. So, roughly speaking, all you need to do is checking that this composition law is differentiable (and also the inverse operation $t\mapsto 2\pi -t$). And addition of real numbers is differentiable in both variables. The only subtle point is "modulo $2 \pi$" part. It needs some mathematical precision, like covering the circle with two overlaping open segments - you can leave it aside for a while.

Last edited: Feb 24, 2012
7. Feb 24, 2012

### mnb96

ok, thanks!

so if my group is parametrized by say two parameters u, v, then I have to figure out how the group operation (composition of functions in our previous case) can be expressed in terms of the parameters u, v. In general, the composition will be $\phi:(U\times V)\times (U\times V) \rightarrow (U\times V)$, so basically I have to check that the 4-variable function:
$$\phi(u_1,v_1,u_2,v_2)$$
is differentiable with respect to all the four variables.
Am I right?
If so, how am I supposed to do the same test for the inverse operation? Do I simply consider the function
$$\phi(f(u_1),f(v_1),f(u_2),f(v_2))$$, where f denotes the inverse operation (in terms of parameters), and check that it is differentiable?

Last edited: Feb 24, 2012
8. Feb 24, 2012

In fact, checking the composition law is sufficient. See the attached image taken from P. M. Cohn, "Lie groups". ("Analytic" means that it can be expressed locally as a power series). So, it is more than just "differentiable". But usually, in applications, we are dealing with elementary functions that are analytic in this sense. Problems would arise with functions like "absolute value" which is not differentiable at 0.

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Last edited: Feb 24, 2012
9. Feb 24, 2012

### mnb96

thanks a lot for the book excerpt!
everything should be clearer now, although I would need to study some topology...