This should be an elementary question in Lie group theory, although I'm pretty sure I'm mixing up concepts. Any help clarifying would be much appreciated.
Set up
Let $G$ be finite-dimensional real Lie group, and take it, for the sake of simplicity, to be connected. There is a three-fold correspondence between:
- tangent space at the identity element: $T_eG$,
- left-invariant vector fields: $\mathfrak{g}$,
- one-parameter subgroups of $G$: $\text{Hom}_{\text{LieGr}}(\mathbb{R},G)$,
which are interchangeably called the Lie algebra of $G$.
The exponential map of $G$ is defined as
$$
\exp_G:\mathfrak{g}\to G\,,\quad X\mapsto \exp_G(X):=\phi_X(1)\,,
$$
where $\phi_X$ is the one-parameter subgroup generated by the (left-)invariant vector field corresponding to the tangent vector $X\in T_eG$. In particular, we have for all $t\in\mathbb{R}$ that $\phi_X(t)=\exp_G(tX)$ and therefore the exponential map sends the one-dimensional vector subspace through $X$ to the whole one-parameter subgroup generated by $X$.
(a) The map $\exp_G$ is just a local diffeomorphism. Nevertheless, its image $\exp_G(\mathfrak{g})$ is a (connected) neighbourhood of the identity. Hence, the subgroup $\langle \exp_G(\mathfrak{g})\rangle$ algebraically generated by this image is the whole Lie group $G$ back again (since $G$ is assumed connected) Prop A4.25.
(b) On the other hand, $\exp_G$ is but a local diffeomorphism and therefore shouldn't be able to retrieve global information about $G$. That is, Lie groups with isomorphic Lie algebras are not necessarily isomorphic, and the issue here is the usual simply-connectedness. However, it seems that in (a) we are perfectly able to distinguish the different Lie groups with a given Lie algebra.
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Question(s)
Where is the topological information coming from in the generating process in (a)? Or, where is it lost in (b)? Or, if the questions don't make sense, where is the confusion?
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My thoughts is that in (a) we are really generating the Lie group $G$ by the image of one-parameter subgroups $\text{Hom}(\mathbb{R},G)$, which already contain enough topological information about $G$. In particular, by simple Yoneda if $\text{Hom}(\mathbb{R},G)\cong \text{Hom}(\mathbb{R},H)$ then $G\cong H$.[See comment.] However, the Hom is equally well isomorphic to the tangent space at the identity $T_eG$ and therefore it seems that we can always reproduce the whole group by this small bit of simply `vectorial' data.
In (b), when we worry about non-isomorphic Lie groups with isomorphic Lie algebras we are really completely forgetting about their underlying exponential maps, and just keeping the local diffeomorphism property. The Lie algebras are then purely algebraic objects with no a priori attached (global) Lie groups. The `corresponding' local Lie groups may be constructed via the BCH formula, which is defined entirely in terms of the Lie algebra structure.
Therefore, in (a) saying we reconstruct the whole Lie group from the exponential map may be seen as a tautology because the exponential map itself is defined globally with respect to a given Lie group. And, although Lie algebras may be isomorphic they are not the same, and the exponential map distinguishes just this.
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Please, (re)tag more appropriately and/or edit the question if need be.
Best Answer
Let's focus our attention on a straightforward special case: the quotient map $\mathbb{R} \to \mathbb{R} / \mathbb{Z}$, as a map of Lie groups, induces an isomorphism on Lie algebras, but it is not an isomorphism. So the Lie algebra cannot even tell whether a Lie group is compact or not (although it can get surprisingly close, as it turns out, modulo this example).
When you try to reconstruct a connected Lie group as the group generated by the image of the exponential map, the problem is that there are some relations between these generators which take place "far from the identity," and without more global information than just the Taylor expansion of the Lie group multiplication at the identity, you don't know which of these relations to impose or not. In the above example, one such relation is $\frac{1}{2} + \frac{1}{2} = 0$, which holds in $\mathbb{R} / \mathbb{Z}$ but not in $\mathbb{R}$.
(What's worse, you don't even know a priori that you can consistently impose these relations in such a way that you get a Lie group globally. That is, it's not at all obvious, although it is true, that every finite-dimensional Lie algebra is the Lie algebra of a Lie group.)