I'm trying to learn some Lie algebra theory with bare bones knowledge of differentiable manifolds, and little knowledge of Lie groups. I see why the exponential map $\exp: \mathfrak{g} \to G$ is surjective if $G$ is a Lie subgroup of $GL_n(\mathbf{C})$. However, my intuition is a little loose when it comes to isomorphisms of Lie groups. If $H$ is a Lie group isomorphic to some matrix Lie group, does this imply that the exponential map $\exp: \mathfrak{h} \to H$ is surjective. Furthermore, is there a nicer condition to guarantee that a Lie group is isomorphic to a matrix Lie group (it seems like almost all interesting examples of Lie groups are, except for the covering spaces of certain matrix Lie groups)?
Differential Geometry – Surjectivity of Exponential Map of a Lie Group
differential-geometrylie-algebraslie-groups
Related Solutions
Solution 1
The left translation maps have the form $L_a:\mathbb R^n \rightarrow \mathbb R^n: x\mapsto a+x$. So $D L_a(x)=\operatorname{id}$ for all $a,x\in\mathbb R^n$ ($D L_a$ is the total derivative of $L_a$). So the set of all left-invariant vector fields is the set of all constant vector fields $\mathfrak g=\{f:\mathbb R^n\rightarrow T\mathbb R^n: f = \operatorname{const}\}$ (which is isomorph to $\mathbb R^n$ because $\mathfrak g$ is $n$-dimensional).
Because $X\in \mathfrak g$ is constant, one has $[X,Y]=\mathcal L_X(Y)=0$ for any given vector field $Y$.
Solution 2
$(\mathbb R^n, +)$ can be embedded into $GL_n$ via $$f:\mathbb R^n \rightarrow GL_n: (x_1,x_2,\ldots,x_n) \mapsto \left(\begin{matrix} e^{x_1} & & & & \\ & e^{x_2} & & \\ & & \ddots & \\ & & & e^{x_n}\end{matrix}\right)$$ $f$ is well defined because the determinant $e^{x_1 + x_2 + \ldots + x_n}$ is always positive and because of $e^{a+b}=e^a\cdot e^b$ the function $f$ is a group homomorphism.
Via $f$ one can show that $(\mathbb R^n,+)$ is isomorph to the set $D^+$ of diagonal matrices with positive entries on the diagonal. The lie algebra $\mathfrak g$ of $D^+$ is given by $$\mathfrak g = \{\dot\gamma(0): \gamma:(-\epsilon,\epsilon)\rightarrow D^{+}\}$$ which is also $D^{+}$ (as one can easily show). Because $D^{+}$ is in the center of $GL_n$ the lie bracket is always zero.
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.)
Best Answer
First, it is not true that the exponential map is surjective if $G$ is a Lie subgroup of $\operatorname{GL}_n(\mathbb{C})$. One trivial situation in which it might fail to be surjective is if $G$ is not connected as can be already seen in the zero-dimensional case. Even if $G$ is connected, the exponential map might not be onto (see this example).
However, if $G$ is compact and connected then the exponential map is onto. In this case, $G$ is also isomorphic to a matrix lie group.