# Surjectivity of Exponential Map of a Lie group

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)?

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.

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.

In this blog post Terry Tao mentions a nice criterion:

For a (connected) Lie group $$G$$, if the exponential map $$\mathfrak{g} \rightarrow G$$ is surjective, then every element of $$G$$ is a square in $$G$$ (in fact, is an $$n^{\text{th}}$$ power in $$G$$ for every $$n\in\mathbb{Z}_{>0}$$).

Proof: Indeed, for $$g \in G$$, if there exists $$x \in \mathfrak{g}$$ with $$\mathrm{exp}(x) = g$$, then for each $$n\in\mathbb{Z}_{>0}$$ we have $$\exp(x/n)^n = \exp(x)= g$$, hence $$g$$ is the $$n^{\text{th}}$$ power of $$\mathrm{exp}(x/n) \in G$$.

As is mentioned (in a specific example) in the above blog post, one can show if $$a\in\mathbb{R} , \,\&\,a>0 ,\,\&\,a\neq1$$, then $$\bigl(\begin{smallmatrix} -a & 0 \\ 0 & -1/a \end{smallmatrix}\bigr) \in \mathrm{SL}_2(\mathbb{R})$$ is not a square in (the Lie group) $$\mathrm{SL}_2(\mathbb{R})$$, or even in $$\mathrm{GL}_{2}(\mathbb{R})$$ as proved below.

One proof: Let $$A := \bigl(\begin{smallmatrix} -a & 0 \\ 0 & -1/a \end{smallmatrix}\bigr)$$. Aiming for a contradiction, suppose there exists $$B \in \mathrm{Mat}_{2\times 2}(\mathbb{R})$$ with $$B^2 = A$$. Note $$A$$ has minimal polynomial $$(X+a)(X+1/a)$$ over $$\mathbb{R}$$ (indeed this polynomial does vanish on $$X=A$$; and $$A$$'s minimal polynomial cannot have degree 1 since $$A$$ is not a scalar multiple of the identity matrix, since $$a\neq\pm1$$). Therefore the monic minimal polynomial $$P_B \in \mathbb{R}[X]$$ of $$B$$ (over $$\mathbb{R}$$) divides $$(X^2 + a)(X^2+1/a)$$. Note $$X^2+a,\,X^2+1/a$$ are irreducible in $$\mathbb{R}[X]$$ (since $$a>0$$), and $$P_B$$ has degree $$\leq2$$ (since $$B$$ is 2-by-2). Therefore $$P_B \in \{X^2+a,X^2+1/a\}$$, and $$P_B$$ also equals the (monic) characteristic polynomial of $$B$$ (since e.g. the char poly has degree 2 since $$B$$ is 2-by-2, and the min poly $$P_B$$ already has degree 2 and must divide the char poly). The complex roots of $$P_B$$ (which are the complex eigvals of $$B$$) are then either $$\{\pm i\sqrt{a}\}$$ or $$\{\pm i\sqrt{1/a}\}$$, both with multiplicity 1; but then $$\mathrm{det}(B)$$, which equals the product w/ multiplicity of its complex eigenvalues, is either $$a$$ or $$1/a$$; so then $$\mathrm{det}(B)^2 \in \{a^2,1/a^2\}$$. However $$\mathrm{det}(B)^2 = \mathrm{det}(B^2) = \mathrm{det}(A) = 1$$, and $$1 \notin \{a^2,1/a^2\}$$ (since $$a\neq\pm1$$); this gives the contradiction.

For your question about a condition to guarantee that a Lie group is isomorphic to a matrix Lie group: this is true for compact Lie groups; see the following Stackexchange post: Are all Lie groups Matrix Lie groups?