# Different Lie group structures on a vector space with the same Lie algebra structure

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.

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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.

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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.