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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?

Travis Willse
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PhoenixPerson
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    It's true for all compact groups (like $SU(2)$) and also $\mathbb{R}^n$ and products $G\times \mathbb{R}^n$ where $G$ is compact, but I don't know if it's true or not for anything else. Maybe someone else can chime in? (Later, I can supply a handwavy proof for compact groups.) – Jason DeVito - on hiatus Feb 04 '16 at 16:55
  • See here, here and many other discussions on this subject. – Dietrich Burde Feb 05 '16 at 08:21

1 Answers1

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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.

Hoffman, Lawson. Divisible Semisubgroups of Lie Groups. J. London Math. Soc. (1983) s2-27 (3): 427-434.

Travis Willse
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  • Thanks for the answer. Could you justify the case of the identity component of the Lorentz group? – PhoenixPerson Feb 04 '16 at 17:12
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    You're welcome. I don't know offhand whether it is true for $SO_0(1, 3)$. I suspect it's false, and that one might be able to find a counterexample using that $SO_0(1, 2)$ has the same universal cover as $SL(2, \Bbb R)$, for which we've seen that $\exp$ is not surjective. – Travis Willse Feb 04 '16 at 17:24
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    Apparently it is surjective. https://en.wikipedia.org/wiki/Representation_theory_of_the_Lorentz_group#Surjectiveness_of_exponential_mapping – Brian Klatt Feb 04 '16 at 21:18
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    @BrianKlatt Thanks for pointing this out---for anyone who doesn't bother to click through, one can show this using that (1) $SO_0(1, 3) \cong PGL(2, \Bbb C)$ and (2) $\exp: {\frak gl}(2, \Bbb C) \to GL(2, \Bbb C)$ is surjective. (I'd forgotten about the exceptional isomorphism in (1).) – Travis Willse Feb 05 '16 at 00:20