MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

closed as too localized by Steve Huntsman, Tom Goodwillie, Deane Yang, José Figueroa-O'Farrill, Ryan Budney Aug 21 '11 at 19:55

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

See this <a href="… 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"?) – Tom Goodwillie Aug 21 '11 at 15:37
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? – Mirco Aug 21 '11 at 15:48
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"). – Qfwfq Aug 21 '11 at 16:41
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. – José Figueroa-O'Farrill Aug 21 '11 at 17:06
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. – Ryan Budney Aug 21 '11 at 19:54
up vote 5 down vote accepted

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.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.