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

This is an eccentric question: recall that a smooth Lie group structure on $\mathbb R^n$ is uniquely identified by a triple $(\mu,\iota,e)$ where $\mu:\mathbb R^n\times\mathbb R^n\to\mathbb R^n$ is the multiplication, $e\in\mathbb R^n$ is the neutral element, and $\iota:\mathbb R^n\to\mathbb R^n$ is the inversion. Since we can canonically identify the tangent space $T_e\mathbb R^n$ with $\mathbb R^n$, for every triple $(\mu,\iota,e)$ we obtain (by differentiation) a Lie bracket as a bilinear map $\beta:\mathbb R^n\times\mathbb R^n\to\mathbb R^n$. My question is the following: do there exist two triples $(\mu,\iota,e)$ and $(\mu',\iota',e)$ (note: same neutral element) such that $\mu\not=\mu'$ but $\beta=\beta'$?

I conjecture that using some nilpotent Lie groups one can construct an example but that's only mere speculation.

share|cite|improve this question

Well, literally, the answer is 'yes'. Just let $\phi:\mathbb{R}^n\to\mathbb{R}^n$ be a diffeomorphism other than the identity that fixes $e$ (which you can assume to be $0\in\mathbb{R}^n$) to some order at least $2$ and then define $$ \mu'(x,y) = \phi^{-1}\bigl(\mu\bigl(\phi(x),\phi(y)\bigr)\bigr) \qquad\text{and}\qquad \iota'(x) = \phi^{-1}\bigl(\iota\bigl(\phi(x)\bigr)\bigr). $$

For example, take $n=1$ with $\mu(x,y) = x+y$ and $\iota(x) = -x$ and set $\phi(x) = x + x^3$. Then $\beta = \beta' = 0$, but $\mu\not=\mu'$.

I suspect you meant to ask something else, though.

share|cite|improve this answer

Sure. You'll always get isomorphic Lie groups, but they'll be different actual multiplication maps $\mu$. Just pick any Lie group diffeomorphic to $\mathbb{R}^n$ sending the identity to 0, and let $\phi:\mathbb{R}^n\to \mathbb{R}^n$ be any diffeomorphism whose Taylor series at 0 coincides with that of the identity, but which is not the identity, and let $\mu'=\phi^{-1}\circ \mu\circ \phi$. All examples will involve doing non-analytic funny business, since an analytic multiplication on $\mathbb{R}^n$ is determined from the Lie bracket by Campbell-Baker-Hausdorff.

EDIT: as Terry points out below, this isn't quite the right way to say things. However, no matter how you look at it, the issue is just one of picking strange coordinates on a Lie group where morally the multiplication really is recoverable from the Lie algebra structure.

share|cite|improve this answer
Strictly speaking, the BCH formula only determines the multiplication law in exponential coordinates; if one uses a different (but still analytic) coordinate system instead, one can certainly get analytic multiplication laws that don't correspond directly to the BCH formula. – Terry Tao Apr 10 '12 at 20:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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