I am reading a book on matrix Lie algebras (Brian Hall's). Corollary 2.30. says that if $G$ is a connected matrix Lie group, then every element $A$ of $G$ can be written in the form
$$A=e^{X_1}e^{X_2}\ldots{}e^{X_m}$$
for some $X_1$,$X_2$$\ldots$$X_m$ in the Lie algebra. Immeadiately after it is stressed that even if $G$ is connected, it is not true that any every element $A$ of $G$ canbe written
$$A=e^{X}$$
where $X$ is a Lie algebra element. My background is on physics, and I have many times seen Lie groups written using ony one exponential with absolute impunity (for example with $SU(2)$). Can anybody tell me when it is true that there is some $X$ Lie algebra element for every $A$ in a Lie group?
If $G$ is compact and connected, one can prove (by constructing a bi-invariant metric on $G$ and relating the metric and Lie group exponential maps) that the exponential map $\exp : {\frak g} \to G$ is surjective. This justifies the claim for, e.g., $SU(2)$. Using the Baker-Campbell-Hausdorff formula, one can show that $\exp$ is also surjective for Lie groups that are connected, simply connected, and nilpotent. (See this blog post of Terry Tao for more.)
There is a condition on Lie groups equivalent to surjectivity of $\exp$, namely, divisibility: A group is divisible iff for every $g \in G$ and every $k \in \Bbb Z_+$ there is some $h \in G$ such that $h^k = g$. (See the paper cited below.) Checking this condition is not necessarily easier than checking surjectivity directly, however.
We can use this criterion to show readily that $\exp$ is not surjective for all connected Lie groups: For example, we can check directly that $B := \pmatrix{-1&0\\0&-2}$ has no square root in $GL(2, \Bbb R)$ (say, by writing out the components of $X^2 = B$ and deriving a contradiction). So, $GL_+(2, \Bbb R) := \{A \in GL(2, \Bbb R) : \det A > 0\}$ is not divisible, and hence $\exp: {\frak gl}(2, \Bbb R) \to GL_+(2, \Bbb R)$ is not surjective.