Maybe the following is trivial or folklore, but I can't find any concrete proof of the theorem, that higher order derivatives of Lie groups don't give any new information above what is coded in its Lie algebra.
Can someone explain why this is true?
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1$\begingroup$ See this <a href="en.wikipedia.org/wiki/… formula</a>. But I'm voting to close. (Why is there no better way to indicate "not research-level" when voting to close than "too localized"?) $\endgroup$– Tom GoodwillieCommented Aug 21, 2011 at 15:37
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$\begingroup$ ok. maybe i have to reformulate my question a bit: In a sense the Lie algebra is the first Taylor approximation of the lie group, i.e. of all structure maps. If the higher approximations don't give anything new, why are there still non vanishing higher terms in the taylor polinomials? $\endgroup$– MircoCommented Aug 21, 2011 at 15:48
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$\begingroup$ It's not just "not research level": the question doesn't make sense (unless the OP specifies what he means by "higher order derivatives of a Lie group"). $\endgroup$– QfwfqCommented Aug 21, 2011 at 16:41
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$\begingroup$ I think you want to read about the Lie correspondence. The book by Wulf Rossmann gives a good account with a minimum of prerequisites. Having said that, the question is not appropriate for this site and I'm voting to close. $\endgroup$– José Figueroa-O'FarrillCommented Aug 21, 2011 at 17:06
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1$\begingroup$ I think of the Lie algebra as more of a 2nd order approximation to the Lie group, not 1st order -- 1st order is addition of vectors. $\endgroup$– Ryan BudneyCommented Aug 21, 2011 at 19:54
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The reason is the Campbell-Baker-Hausdorff equation, which proves that all higher derivatives of the multiplication map are expressed in exponential coordinates explicitly in terms of iterated Lie brackets. Once you know the Lie bracket operation, you can calculate the Taylor series expansion of the multiplication operation explicitly, order by order. The Taylor series has positive radius of convergence, so the multiplication is given, near the identity element, by this series expansion. (I shouldn't be giving this answer, as it is in all basic introductions to Lie groups, not a research level question.) The explicit formula is in Serre's book Lie Algebras and Lie Groups, chapter 4, p. 28.
answered Aug 21, 2011 at 19:37