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When is it the case that all elements in the connected component of the identity of a matrix Lie group can be written as a single exponential of some element of the corresponding Lie algebra?

SWV
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  • The title question seems to be different from the one you are asking. You also have to specify what a "single exponential" means. Interpreted literally, a group has this property iff it is trivial. – Moishe Kohan Nov 04 '13 at 20:00
  • I changed the title to say matrix Lie groups. Aren't all one parameter subgroups generated by the exponential of some element in the lie algebra? By single, I mean not a product of exponentials. – SWV Nov 04 '13 at 20:07
  • What do you mean by "generatedby? – Moishe Kohan Nov 04 '13 at 20:09
  • A is generated by B if for some real t, A = exp(tB). – SWV Nov 04 '13 at 20:13
  • I see, what you mean is that each 1-parameter subgroup has the form $exp({\mathbb R}\cdot A)$ for some $A$ in the Lie algebra. That's correct, of course, just one does not use the terminology "generate" in this case. In any case, the title of your question seems to ask if the component $G_o$ of identity of a matrix Lie group $G$ is the image of the exponential map $exp: {\mathfrak g}\to G$, where ${\mathfrak g}$ is the Lie algebra of $G$. Equivalently, $G_o$ is the union of its 1-parameter subgroups. The answer to this would be negative. – Moishe Kohan Nov 04 '13 at 20:45
  • See http://math.stackexchange.com/questions/200743/ for the explicit computation in the $GL(2,R)$ case. – Moishe Kohan Nov 04 '13 at 20:48
  • I am asking if there are conditions in which it is the union? If I recall, elements of the Lorentz group can be written as exponentials. – SWV Nov 04 '13 at 20:54
  • Also, according to John Lee's "Introduction to Smooth Manifolds," you can say a "one-parameter subgroup is generated by X". – SWV Nov 04 '13 at 20:57

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The question you are actually asking is: "For which connected matrix Lie groups $G$ the exponential map is surjective?" I do not think there is a comprehensive answer beyond "sometimes yes, sometimes not". It is known that for compact connected Lie groups exponential map is surjective. For some rather general partial sufficient conditions in the noncompatc case see the paper The surjectivity of the exponential map for certain Lie groups by M.Moskowitz.

Moishe Kohan
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