Surjectivity of Exponential Map of a Lie group

Question

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)?

Answer 1

For G=SL(2,R) the exponential is not surjective. To prove this, consider a 2x2 lmatrix with Zero trace. There are 3 possibilities:

1. The eigenvalues are 0 with multiplicity two
2. The eigenvalues are $\pm x$ with $x$ real.
3. The eigenvalues are $\pm i x$ with $x$ real.

In situación 1, the exponencial is Id+N with N nilpotent, so the trace is 2.

In situación 2, the exponential has trace $e^x+e^{-x}$ that is a possitive real number.

In situation 3, the trace of the exponencial is $2 cos(x)$.

In any case, the trace of an exponential of a real 2 by2 matrix with cero trace is grater or equal to -2.

For example, the diagonal matrix diag(-2, -1/2) is a element of SL(2,R) that is not in the image of the exponential.

Answer 2

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.