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)?


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?


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