# 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)?

## 3 Answers

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.

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.

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?