Lie's third theorem (global)

Question

(The image corresponds to the book "Lie Groups, Lie Algebras, and their Representations" by Varadarajan)

Hello. I have a question regarding Lie groups and Lie algebras. If I have a Lie algebra $\mathfrak{g}$ the theorem mentions that there is a simply connected Lie group $G$ such that its Lie algebra is isomorphic to $\mathfrak{g}$.

Question 1. Is such a group $G$ related to the exponential application? i.e. $G=\left\{\exp(X):X\in\mathcal{G}\right\}$?

Answer 1

(All Lie algebras are finite-dimensional here.) Generally:

• If $G$ is any Lie group with Lie algebra $\mathfrak{g}$ then there is an exponential map $\exp : \mathfrak{g} \to G$, which can be defined using left-invariant vector fields in general but in the case of matrix Lie groups is the matrix exponential.
• If $G$ is connected, it is generated by the image of the exponential map, because $\exp$ is always a diffeomorphism near the identity and any connected topological group is generated by an open neighborhood of the identity.

These remarks apply in particular to the unique simply connected Lie group with Lie algebra $\mathfrak{g}$. However, generally it's not true that the exponential map is surjective.

Edit: Maybe I've misunderstood your question. Maybe you were asking: can one construct the unique simply connected Lie group $G$ with Lie algebra $\mathfrak{g}$ using the exponential map somehow?

The answer is not obviously yes, since in the absence of a specific Lie group it's not clear what "the exponential map" ought to mean. We can do something like this by first using Ado's theorem to find a faithful representation $\mathfrak{g} \to \mathfrak{gl}(V)$. This means we can apply the ordinary matrix exponential to $\mathfrak{g} \subset \mathfrak{gl}(V)$ which will generate a subgroup of $GL(V)$. This group is not necessarily simply connected, but it does have Lie algebra $\mathfrak{g}$ (this is not obvious but is standard) and we can take its universal cover.

Answer 2

As the comments above already point out, if $G$ is any Lie group with Lie algebra $\mathfrak g$, then there is an exponential map $\text{exp}_G\colon \mathfrak g \to G$ whose image generates $G^0$, the connected component of $G$ containing the identity element of $G$. This map is in general neither injective nor surjective however.

The failure of injectivity is easy to see already for $S^1$ the circle, and it is not too hard to see that the exponential map $\text{exp}\colon \mathfrak{sl}_2(\mathbb R) \to \mathrm{SL}_2(\mathbb R)$ from the Lie algebra $\mathfrak{sl}_2(\mathbb R)= \{X\in \text{Mat}_2(\mathbb R): \text{tr}(X)=0 \}$) to the special linear group $\mathrm{SL}_2(\mathbb R) = \{A\in \text{Mat}_2(\mathbb R): \det(A)=1\}$ which is given by $\text{exp}(X) = \sum_{k=0}^\infty \frac{X^k}{k!}$ is not surjective.

None of the above involves the fundamental group of $G$ however. Where the issue of simple-connectedness does naturally arise in the context of Lie's theorems, is when one asks about homomorphisms: if $\phi\colon \mathfrak g\to \mathfrak h$ is a Lie algebra homomorphism, and $\mathfrak g = \text{Lie}(G)$, $\mathfrak h = \text{Lie}(H)$, then it is natural to ask if one can "integrate" $\phi$ to a homomorphism of Lie groups $\Phi\colon G \to H$, such that $$ \require{AMScd} \begin{CD} \mathfrak{g} @>{\phi}>> \mathfrak h\\ @V{\exp_G}VV @VV{\exp_H}V\\ G @>{\Phi ?}>> H \end{CD} $$ Now since $\exp\colon \mathfrak g\to G$ is locally a diffeomorphism from $0_{\mathfrak g}$ to $e_G$ and the exponential of a derivation is a homomorphism, one expects $\phi$ to "locally lift" to a homomorphism from $G$ to $H$ -- indeed it must be given by $\exp_H(\phi(\log_G(g))$ where $\log$ is a local inverse to $\exp_G$ at $e_G$ -- but in general the ability to lift locally will not ensure the existence of a globally defined homomorphism from $G$ to $H$. Perhaps the simplest example of this global failure is given by taking $G=S^1$ and $H= \mathbb R$ -- then $\mathfrak g = \mathrm{Lie}(G)=\mathrm{Lie}(H) = \mathfrak h$, with both being equal to $\mathfrak{gl}_1(\mathbb R)$, that is, the real numbers as a Lie algebra with the zero Lie bracket. It follows that we have an obvious homomorphism from $\phi\colon \mathfrak g \to \mathfrak h$ given by the identity map, but there is no corresponding homomorphism $\Phi\colon S^1\to \mathbb R$ for it to integrate to (the image of $S^1$ under a continuous map will be compact, but $(\mathbb R,+)$ has no compact subgroups other than the trivial subgroup, hence the only continuous homomorphism from $S^1$ to $\mathbb R$ is the constant map to $0_{\mathbb R}$. On the other hand, if we let $G$ and $H$ trade places, then the exponential map itself gives the lift of $\phi^{-1}$ to a homomorphism of groups from $\mathbb R$ to $S^1$.

The crucial difference between the two groups turns out to be the fact that $\mathbb R$ is simply connected: whenever $G$ is simply connected, using the fact that a path can then be lifted from $\mathfrak g$ to $G$ uniquely up to homotopy, one can show that any $\phi\colon \mathfrak g\to \mathfrak h$ lifts to a group homomorphism $\Phi\colon G \to H$. (Note that, as we saw in the example of $\mathbb R^1$ and $S^1$, the group $H$ is not required to be simply-connected here, only $G$.)