I'm introducing myself to Lie Groups theory, and I read Ado's Theorem says that every finite real Lie algebra is isomorphic to a Lie subalgebra of $\mathfrak{gl}(n,\mathbb{R})$.

I found the following problem as a consequence of Ado's Theorem: "Give infinite examples of non isomorphic Lie groups with Lie algebra $\mathfrak{so}(3)$."

Because of the following statement:

"If a Lie group has Lie algebra $\mathfrak{g}$, then it is isomorphic to $G/\Gamma$, where $G$ is the simply connected Lie group with Lie algebra $\mathfrak{g}$, and $\Gamma$ is a discrete subgroup of the center of $G$, $Z(G)$."

I think I must work with a group $G$ non simpy connected, but I don't achieve the examples. Could you help me?

  • 4
    $\begingroup$ Consider $SO(3)\times \Bbb Z_n$ for $n\in\Bbb N$. Note that these groups are not connected. $\endgroup$ – s.harp Nov 27 '20 at 17:09
  • 2
    $\begingroup$ Are you sure your exercice is exactly as you stated? Because there exists only a finite number of connected Lie groups with Lie algebra $\mathfrak{so}3$, and the statement following your sentence "becausse of the following.." refers to connected Lie groups. $\endgroup$ – Didier Nov 27 '20 at 18:04
  • $\begingroup$ "Give infinite examples" is poor English, I hope you didn't find this in an exercise set... it should be "Give infinitely many [non-isomorphic] examples". $\endgroup$ – YCor Nov 28 '20 at 12:52
  • $\begingroup$ By the way Ado's theorem seems irrelevant to the whole discussion. $\endgroup$ – YCor Nov 28 '20 at 12:53

One cannot find an infinite number of connected Lie groups with Lie algebra $\mathfrak{so}(3)$ because such a Lie group is either $\mathrm{SO}(3)$, either $\mathbb{S}^3$, the universal cover of $\mathrm{SO}(3)$. This is because $\mathbb{S}^3$ is a simply connected Lie group with Lie algebra $\mathfrak{so}(3)$ and has center $\{\pm 1\}$, thus the only discrete subgroups of its center are $\{1\}$ and $\{1,-1\}$, and the results follows from what you stated.

On the other hand, there are infinitely many non-connected Lie groups with Lie algebra $\mathfrak{so}(3)$. For example, $\mathbb{S}^3 \times \mathbb{Z}/n\mathbb{Z}$ for all $n \geqslant 2$.

  • $\begingroup$ Sorry, but why the lie algebra of that direct products are $\mathfrak{so}(3)$ ? I know lie algebra of the product is isomorphic to the direct sum of the lie algebras, but why is the lie algebra of the second factor null? Thank you. $\endgroup$ – Jotabeta Nov 28 '20 at 12:18
  • 1
    $\begingroup$ Because $\mathbb{Z}/n\mathbb{Z}$ is a $0$ dimensional Lie group, hence its Lie algebra is $\{0\}$. $\endgroup$ – Didier Nov 28 '20 at 12:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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