# SL(2) group is a Lie group

1. Apr 16, 2013

### LayMuon

1. The problem statement, all variables and given/known data

Prove that in SL(2) group the matrix $\begin{pmatrix} -1 & \lambda \\ 0 & -1 \end{pmatrix}$ can not be presented as a single exponentail but instead as product of two exponentials of $sl(2)$ algebra. $\lambda \in \mathbb{R}$

2. Relevant equations

I don't understand how an element of Lie group cannot be presented by a single exponential. Does this mean that SL(2) is not a Lie group? or a Lie group that has some elements that cannot be cast into exponential form? So Lie group with no exponential form for some elements?

3. The attempt at a solution

stuck with definitions.

2. Apr 17, 2013

### fzero

The elements of a Lie group that can be written as $\exp(\alpha^a t_a)$, where $t_a$ are generators of the Lie algebra and $\alpha^a$ are some parameters, are said to be the "identity component" of the Lie group, or sometimes "connected component" (meaning component connected to the identity). This is because we smoothly recover the identity element ($\mathbb{1}$) as we take the $\alpha^a \rightarrow 0$. We can think of this adjustment of parameters as defining a path in the Lie group viewed as a topological space. Sometimes we can connect all elements of the group to the identity in this way, in which case the group is called connected. Often, however, the group is composed of disconnected components.

SL(2) is an example of a Lie group that is not connected as a topological space. The element you write down is one that is not connected to the identity element by a smooth deformation of the parameter $\lambda$.

Last edited: Apr 17, 2013
3. Apr 17, 2013

### Kreizhn

It is certainly not simply connected, but I think you mean just connected in this case. Simple connectivity is unrelated to surjectivity of the exponential.

4. Apr 17, 2013

### fzero

Yes, I didn't mean simply connected. Thanks.

5. Apr 17, 2013

### LayMuon

What does concern me is that this matrix can be presented as product of two exponentials and then with Campbell-Hausdorff relation merged into one, the commutators should yield something because Lie algebra is closed, so we would get one exponential.

Does this mean that the series of commutators does not converge or what? It seems to be a paradox, I am missing something here apparently.

6. Apr 17, 2013

### fzero

You can write the matrix down in the form $e^X e^Y$ with $Y$ in the Lie algebra sl(2), but $X$ is not in the Lie algebra. I could probably be more specific if you wrote down the expression yourself. But you should try to convince yourself of my claim in the meantime.

7. Apr 17, 2013

### Kreizhn

There is also some subtlety in the convergence of the BCH formula. It does not generally converge on a global scale. A quick glance at this paper should hopefully convince you of that.

I believe BCH always converges in connected, simply connected groups, though that does not apply in this case.