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Jan 16, 2018 at 3:13 vote accept CommunityBot
Dec 18, 2017 at 18:39 answer added Torsten Schoeneberg timeline score: 5
Dec 2, 2017 at 20:10 answer added Jim Humphreys timeline score: 3
Dec 2, 2017 at 16:39 comment added YCor To be a little more precise: there is a compact open subring of the Lie algebra on which the exponential is defined, defines a homeomorphism onto a compact open subgroup of $G$, and on which the Baker-Campbell-Hausdorff formula converges and computes $\log(\exp(x)\exp(y))$.
Dec 2, 2017 at 12:57 history edited YCor
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Dec 2, 2017 at 12:54 comment added YCor Yes, at least locally (the exponential is defined on a compact open subgroup of the Lie algebra). This is developed in detail in Chapter 2 of Bourbaki, Lie algebras and Lie groups.
Dec 2, 2017 at 12:15 history asked user113393 CC BY-SA 3.0