Reconstructing Lie group globally from the exponential map

Solution 1:

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.)

Solution 2:

I guess I convinced myself of the remarks following and motivated by Qiaochu's answer, and I'll post it here as an answer in case someone gets similarly confused.

So indeed, the issue here was just a question of interpretation. The best, and probably the right, way of thinking about the exponential map of a Lie group $G$ is as a gadget meant to organize its one-parameter subgroups: $$ \exp_G:\text{Hom}(\mathbb{R},G)\to G\,,\quad \phi\mapsto \phi(1)\,. $$

The question in (a) is akin to the question about the surjectivity of $\exp_G$. This asks about how whether or not the one-parameter subgroups of $G$ cover $G$. In the same vein, in (a), when we say that the image of $\exp_G$ generates the Lie group $G$ (when connected) is telling exactly how the one-parameter subgroups of $G$ sit inside $G$ itself. The answer to the question "Where is the topological information coming from in the generating process in (a)?" is nowhere, because this information is all already there in $G$ which we haven't forgotten. What this does provide is more refined information about how the Lie group $G$ `behaves' in terms of maps of $\mathbb{R}$ into $G$. In particular, via the isomorphism $\mathfrak{g}\cong T_eG\cong \text{Hom}(\mathbb{R},G)$ (as given with a full knowledge of $G$), the result $\langle \exp_G(\mathfrak{g})\rangle=G$ comes as not so tautological because it tells we may write any group element $g\in G$ as a finite product of exponentials $g=\exp_G(X_1)\cdots \exp_G(X_k)$, where $X_1,\ldots, X_k\in T_eG$. This may be extremely useful in reducing a proof about $G$ to a proof about $\text{Hom}(\mathbb{R},G)$.

When we look at things such as in (b), and use $\exp_G$ to conclude that isomorphic Lie algebras may come from non-isomorphic Lie groups, we're simply using the local diffeomorphism property of $\exp_G$ (i.e. $(d \exp)_e=\text{id}_{\mathfrak{g}}$), but otherwise forgetting about $\exp_G$. This answers the question "Where is it lost in (b)?" At this point we simply have an abstract Lie algebra $\mathfrak{g}$ together with the BCH-formula (which equivalently encodes the local properties, i.e. around the identity, of a forgotten Lie group). The Lie algebra $\mathfrak{g}$ alone determines at most a local Lie group, but no more.

Note that there is a priori no notion of exponentiation of $\mathfrak{g}$ in the sense above. However, by Ado's theorem, any finite-dimensional Lie algebra $\mathfrak{g}$ is a subalgebra of $\mathfrak{gl}(n,\mathbb{R})$ for some $n$, and hence we may exponentiate $\mathfrak{g}$ as a matrix Lie algebra via the usual power series. For $U$ a neighbourhood of $0$ in $\mathfrak{g}\subseteq \mathfrak{gl}(n,\mathbb{R})$, we may form $\exp(U)\subseteq \text{GL}(n,\mathbb{R})$ (as a submanifold). The Baker-Campbell-Hausdorff formula in $\text{GL}(n,\mathbb{R})$ then gives $\exp(U)$ the structure of a local Lie group which can in fact be extended to a global Lie group. A useful reference on this latter part are Terry's notes on local groups.