# Lie group

1. Dec 2, 2013

### LagrangeEuler

Is it $(\mathcal{R} without \{0\},\cdot)$ Lie group?

2. Dec 2, 2013

### R136a1

Yes, it is a Lie group.

3. Dec 2, 2013

### Office_Shredder

Staff Emeritus
Try \setminus in the future by the way.
$$\mathbb{R} \setminus \{0\}.$$

Was there anything about the definition of a Lie group which you thought it might not satisfy?

4. Dec 3, 2013

### LagrangeEuler

Tnx. I can not find generators of Lie algebra for this group $(\mathbb{R}\setminus\{0\},\cdot)$. What is Lie algebra for $(\mathbb{R}\setminus\{0\},\cdot)$?

Last edited: Dec 3, 2013
5. Dec 3, 2013

### George Jones

Staff Emeritus
Have you heard of the Lie groups $\mathrm{GL}\left( n, \mathbb{R} \right)$ with corresponding Lie algebras $\mathfrak{gl}\left( n, \mathbb{R} \right)$?

6. Dec 3, 2013

### LagrangeEuler

I think you want to say that $(\mathbb{R}\setminus\{0\},\cdot)$ is $GL(1,\mathbb{R})$. I never heard about algebra $gl(1,\mathbb{R})$.

7. Dec 3, 2013

### George Jones

Staff Emeritus
Right. $\mathfrak{gl}(1,\mathbb{R}) = \mathbb{R}$.

See 2.1 in

http://www.math.jhu.edu/~fspinu/423/8.pdf

Here, $G^0$ is the connected component of $G$.

8. Dec 3, 2013

### LagrangeEuler

I always thought that advantage of Lie algebra is finite number of generators, while group has infinity elements. But in this case, if I understand you well, you have infinite number of generators. So I do not understand concept of Lie algebra any more. :(

9. Dec 3, 2013

### George Jones

Staff Emeritus
A Lie algebra is a vector space. The number of independent generators is the dimension of the Lie algebra, i.e., the set of independent generators is a basis for the Lie algebra. The actual Lie algebra is the set of all possible linear combinations of the generators, and thus has an infinite number of elements.

In this case, $\mathbb{R}$ is a one-dimensional vector space, and thus there is one (independent) generator.

10. Dec 3, 2013

### George Jones

Staff Emeritus
For example, take the generator to be $X = 1$. Then the Lie algebra is $\left\{tX\right\}$, and $e^{tX}$ gives positive real numbers, i.e., elements of the connected component of the Lie group.

11. Dec 4, 2013

### LagrangeEuler

Thanks for the answer. When I write $(\mathbb{R}\setminus\{0\},\cdot)$ I think about all real numbers without $0$. When you take exponentials of real numbers you would not get negative numbers. So I'm not sure in this particular case how you will reproduce all members of group $(\mathbb{R},\setminus\{0\})$, starting from Lie algebre $\mathbb{R}$.

Except this I understand now concept better. Thank you. Generators of Lie algebra (vector space), are l. independent vectors. So $\mathbb{R}^n$ has $n$ generators of algebra (still finite number).

Last edited: Dec 4, 2013
12. Dec 4, 2013

### jgens

Depending on what you mean by "reproduce all members of a group" this is silly for a couple reasons:
1. The exponential map is not usually surjective. So unless you restrict your attention to compact connected groups, or something like that, then it is too much to expect the exponential to reproduce the entire group this way.
2. If you have a connected (possibly non-compact) Lie group, then the image of the exponential map gives you enough group members to generate the rest. This does not apply to R× since it is not connected.
What happens in this case (really every case) is that the image of the exponential map generates the identity component of the group. So I have no idea what the problem here is.

Last edited: Dec 4, 2013