# Lie group=exp(Lie Algebra)?

1. Sep 9, 2010

### geodesic42

I feel like this has to have already been addressed somewhere, but apparently my search skills are lacking.

I rarely hear a mathematician talk about a Lie group and its generating algebra as related via the exponential map, though I have seen it somewhere. When I do see it, it requires more background development than I have immediate time for.

I ALWAYS see physicists talk about this relation (indeed, almost all physics-oriented math texts define the algebra by starting with the group and writing down the exponential). So I've heard many, many times: the exponentiation of the algebra gives the connected part of the group.

I don't believe this (or at least am confused by this). At least at an abstract level. Given how universal it is, I'd like to be convinced otherwise.

My first and most nit-picky gripe is: multiplication isn't defined on the algebra; a bracket is, which becomes a commutator when we find a rep and thus specifies only half of the multiplication operator (i.e. exp(algebra) makes no sense until we find a rep).

Second, su(2)=so(3) (the algebras, not the groups). exp is a 1-1 map. So we should find exp(su(2))=exp(so(3))...but this isn't the case. Choose $$\frac{1}{2}\{\sigma^i\}$$ as our generators for the 2-D rep, and we indeed get SU(2)'s multiplication table. Choose the canonical generators for the 3-D rep and we get SO(3), which obviously does not have SU(2)'s multiplication table. SU(2) and SO(3) are both completely connected (though SU(2) is simply connected), so the whole "it generates the connected piece" bit doesn't solve anything. So how can we possibly say Group=exp(Algebra) if different groups have the same algebra? I'm pretty sure the answer must be that the group we get out of exponentiating is rep-dependent since the other half of the algebra's multiplication table isn't specified until we choose one anyway.

tl;dr: How can we possibly say Group=exp(Algebra) if different groups have the same algebra?

Thanks!

2. Sep 9, 2010

But who says that it is so in general? In physics we usually deal with matrix groups. Exp(Algebra) gives you a neighborhood of identity. Finite products of such elements (from this neighborhood) will give you the connected component. For quite a number of applications that is all you need.

Details about different cases can be found in this paper:

"[URL [Broken] surjectivity question for the exponential function of real Lie groups:
A status report[/URL]
Dragomir Z. Dokovic and Karl H. Hofmann

Last edited by a moderator: May 4, 2017
3. Sep 9, 2010

### geodesic42

You said matrix groups. I.e. the answer is rep-dependent. So I was right?

4. Sep 9, 2010

It depends on what you mean by "answer". It depends on what kind of representations (continuity properties), and what kind of groups etc. we are talking about. Each problem may need a separate discussion.

Last edited: Sep 9, 2010
5. Sep 9, 2010

### Office_Shredder

Staff Emeritus
If two Lie groups have the same universal cover they have the same Lie algebra. But the Lie algebra uniquely determines the simply connected associated Lie group

6. Sep 18, 2010

### element4

You have to be careful here, a matrix (Lie) group is a closed subgroup of $$\text{GL}(n;\mathbb C)$$ (most groups used in physics is of this kind). This is, however, NOT the same as representations.

This is not true. Generally, Lie groups are manifolds equipped with a group product compatible with the underlying differential structure. The corresponding Lie algebra is isomorphic to the tangent space at the identity element, and it is possible to construct a map EXP from the Lie algebra to a neighborhood around the identity element.

7. Sep 19, 2010

### BruceG

I original learnt about all this from Cornwell "Group Theory in physics".
Chapter 10-11 does full justice to this topic.

The real Lie Algebra does not determine the structure of its corresponding Lie groups globally but only locally.

If G is a compact linear Lie group (such as SU(2) and SO(3)) every element of the connected subgroup of G can be expressed in the form exp(a) for some element a of the corresponding real Lie Algebra L. So if G is connected and compact (as SU(2) and SO(3) both are), every element of G has the form exp a for some a in L.

So you question is how come SU(2) and SO(3) are different?

This can occur because the exponential mapping from L -> G are not one-to-one maps.
For example, with SO(3), exp(theta.a) = exp ((theta+2pi).a).
So the maps exp:L->SU(2) and exp:L->SO(3) end up with different groups. In fact there is a 2-1 map from SU(2) to SO(3).

8. Sep 20, 2010

### Petr Mugver

You can say this if the group is simply connected. So if two simply connected groups have the same algebra, they are homeomorphic. If a group is not simply connected, a representation of the algebra is a representation of the group only "up to a phase", i.e. a projective representation. For example, SO(3) is not simply connected, in fact it is doubly connected. To use Weinberg's words, the phase in the projective representation of SO(3) is then just a sign, which is precisely the classic example you mentioned.

9. Sep 20, 2010

### George Jones

Staff Emeritus
Unfortunately, this isn't always true. For example, consider the simply connected group $SL \left( 2 , C \right)$, which is the universal cover of the restricted Lorentz group. There are elements of the simply connected $SL \left( 2 , C \right)$ that are not exponentials of the Lie algebra $sl \left( 2 , C \right)$, i.e.,

$$SL \left( 2 , C \right) \ne \exp \left( sl \left( 2 , C \right) \right).$$

arkajad, Office_Shredder, and BruceG have given various aspects of the situation.

It is true that every element of $SL \left( 2 , C \right)$ can written as the product of two exponentials of elements of $sl \left( 2 , C \right)$, so $\exp \left( sl \left( 2 , C \right) \right)$ does generate $SL \left( 2 , C \right)$. Note that $SL \left( 2 , C \right)$ is not compact.

Last edited: Sep 20, 2010
10. Oct 23, 2010

### vigvig

The exp map is not necessarily surjective. In some case it is, we call such lie groups, exponential lie groups. Some example of exponential lie groups are nilpotent groups, and extension of nilpotent lie groups with diagonal action of a real line. However, in general, your claim is untrue. For example extend a Nilpotent lie group with a diagonal action of a complex torus. Such lie group is not exponential.
Vignon S. Oussa

11. Oct 23, 2010

### mathwonk

I know less about lie groups and algebras than any other topic, hence I throw in my 2 cents. It seems to me part of the OP's problem is a tacit assumption that exp refers to a power series expansion, hence requires a multiplication. from my minimal experience, (reading the first 2-3 pages of a book on lie groups such as that by j.f.adams), exp is defined on a tangent vector by producing an integral curve in the manifold with that tangent vector at the identity and then running along it for time 1 in the manifold. Hence the surjectivity problem is equivalent to the coverage of a manifold by such integral curves, a problem in diff eq.

(of course it goes without saying that the primordial connection is that the lie algebra is the tangent space to the lie group at the identity. incredibly some books on lie algebras do not mention this fact.)
experts can no doubt correct the harm my ignorance does here.

12. Oct 23, 2010

I think that for finite-dimensional matrix groups there is nothing wrong with power series expansion. There are no multiplications, just powers of one matrix. Things nicely converge, you get a trajectory in the group. The only question is how far from the identity can you get this way.

13. Oct 23, 2010

### vigvig

You are correct, but to be more precise, the lie algebra is the vector space of left invariant tangent vectors based at the identity of the group.

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