# Lie groups: Exponential map

1. Mar 3, 2013

### parton

Hi!

I was wondering why it is possible to write any proper orthochronous Lorentz transformation as an exponential of an element of its Lie-Algebra, i.e.,
$$\Lambda = \exp(X),$$
where $\Lambda \in SO^{+}(1,3)$ and X is an element of the Lie Algebra.

I know that in case for compact, connected Lie groups, every element of the group can be expressed by the exponential of an element of the corresponding Lie algebra. But this is not necessarily true for non-compact groups.

In case of $SO^{+}(1,3)$ we are dealing with the connected component of the full Lorentz group. But it is non-compact. So why is it nevertheless possible to express every element of $SO^{+}(1,3)$ by an exponential of an element of the Lie algebra?

I hope someone could help me understanding that.

2. Mar 4, 2013

### dextercioby

Usually the restricted Lorentz group is denoted by SO$\uparrow$+(1,3) and it's the component of the full Lorentz group connected to the identity.

Off the top of my head, the existence of the exponential mapping as a mean to link the Lie group to its corresponding Lie algebra is not determined by whether the group is compact or not, but rather if it's locally compact or not.

Last edited: Mar 4, 2013
3. Mar 4, 2013

### parton

I think the usual notation is $\mathcal{L}_{\uparrow}^{+}$ or $SO^{+}(1,3)$.

In the latter case, the arrow is not necessary, because the 'S' already expresses the fact that the determinant is 1.

The exponential map links the Lie group in a neighborhood of the identity to the Lie Algebra. So you can describe the a group element near the identity by the Lie Algebra.

But on the other hand you cannot reconstruct every element from the Lie Algebra in general.

If we have a compact connected group than every element of the group can be built by the Lie Algebra (via the exponential map). But it is clearly not sufficient to have a locally compact group.
By the way, every Lie group is locally compact.

So, I am curious why every element of the group $SO^{+}(1,3)$ which is clearly connected, but not compact, can be constructed from its Lie Algebra. Maybe anyone has an explanation.

Maybe, there is a way: if one determines the general structure of a matrix of $SO^{+}(1,3)$ one finds that it can be described by 6 real parameters. And this is a finite number which is not obvious because the group is not compact.

Maybe, one can argue that if the group can be described by a finite set of paramters, than every element can be described by its Lie Algebra (via exp).

4. Mar 4, 2013

### dextercioby

I stand corrected for the notation. There's been a while since college.

The exponential mapping and its existence for a general Lie group has nothing to do with the group manifold being compact or not. Likewise, if the group is non-compact, then it doesn't mean that the number of parameters describing the local charts is infinite. Mathematicians simply do not study infinite-dimensional manifolds (in particular Lie groups) as <text-book material>.

The group SO+(1,3) is not only connected, but also path-connected, this means that every element of it can be reached through curves (uni-parametric subgroups) passing through the origin. Again, this has nothing to do with its compactness or lack of.

5. Mar 4, 2013

### parton

Yes, your are right of course. But I did not say the opposite.

I was just saying that if a Lie group is compact and connected every element of the group (and not only locally, i.e., in a neighborhood of the identity) can be expressed as exp(X), where X is an element of
the Lie algebra. This is a typical textbook theorem.

But because SO+(1,3) is clearly not compact, I cannot use this theorem, so I need another reason to justify why every element of SO+(1,3) can be expressed as exp(X).
And this is the question.

Maybe you know a theorem or something which might help to justify that.

6. Mar 4, 2013

### samalkhaiat

For any Lie group $G$, $\exp$ gives a diffeomorphism from a neighbourhood of $0 \in \mathcal{ L }_{ G }$ to a neighbourhood of $e \in G$. Thus, using inverse function theorem, one can show that for any CONNECTED Lie group $G$, $G$ is generated by the subset $\exp ( \mathcal{ L }_{ G } ) \subset G$.

Sam

7. Mar 4, 2013

### dextercioby

Not quite so. The exp:g->G is surjective iff G is compact, connected and linear, see theorem VII of page 390, 2. Volume of Cornwell (Group Theory). Then a generalization is given by theorem VIII (proved in Appendix E, Section 2), when G is connected, non-compact and linear.

The connected component of the Lorentz group in 4D is linear, so theorem VIII applies. In particular, the number of factors is 3, see page 175 of Fonda and Ghirardi.

8. Mar 10, 2013

### wisvuze

Actually, what samalkhaiat is said is true. Any connected Lie group is generated by any neighbourhood around the identity ( that is, take an open neighbourhood U of the identity and consider all possible products and inverses ). This is because the subgroup generated by an open neighbourhood U will be both open and closed. Since the exp map is a local diffeomorphism, the image of an open nhood around 0 of the Lie algebra will be an open neighbourhood around the identity.

This is not the same as the exponential map being surjective. However, if your Lie group is both connected and COMPACT, then the exponential map is actually surjective. This comes from the fact that you can put a riemannian metric on your Lie group. Then, the exponential map will correspond to the usual exp map from differential geometry, and the 1-parameter subgroups and the geodesics. Then, by the Hopf-Rinow theorem, you can connect any two points with a geodesic, so the exp map is surjective.

Last edited: Mar 10, 2013