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I'm new to Lie theory.

The cross product is the Lie bracket of the Lie algebra $\mathfrak{so}(3)$. But this Lie algebra also "belongs" to the unit quaternions $Sp(1)$. Moreover $\mathfrak{so}(3)$ might prefer $Sp(1)$ to $SO(3)$, because of the simple connectivity of the former. Is it therefore possible to construct $Sp(1)$ from $\mathfrak{so}(3)$ without knowing anything about $Sp(1)$ to start with? That would provide a way to start with the cross product, turn a crank, and get out the quaternions. Even better, it would provide a way to start with $SO(3)$, turn two cranks (the first being the one that takes the Lie algebra $\mathfrak{so}(3)$), and get out the quaternions.

I'm guessing the exponential map $\exp$ won't produce $Sp(1)$ from $\mathfrak{so}(3)$. Is the exponential map sensitive to which of the (arbitrary) Lie groups we associate a Lie algebra to?

wlad
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1 Answers1

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Yes. In particular, $\mathrm{Sp}(1)$ is the universal cover of $\mathrm{SO}(3)$, so we can simply take that as a definition, and define $\mathrm{Sp}(1)$ to be the universal cover (which always exists) of $\mathrm{SO}(3)$. Alternatively, you could use Lie's Third Theorem to do it directly: the usual proof of that theorem is a construction of the associated Lie group that gives exactly $\mathrm{Sp}(1)$ when given $\mathfrak{so}(3)$ as an input: specifically, it constructs the universal cover of the subgroup of $\mathrm{GL}_n(\mathbb{R})$ generated by $e^{\mathfrak{so}(3)}$, and that subgroup is $\mathrm{SO}(3)$.

user3482749
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