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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?

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    $\begingroup$ Consider $SO(3)\times \Bbb Z_n$ for $n\in\Bbb N$. Note that these groups are not connected. $\endgroup$
    – s.harp
    Commented Nov 27, 2020 at 17:09
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    $\begingroup$ 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
    Commented Nov 27, 2020 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
    Commented Nov 28, 2020 at 12:52
  • $\begingroup$ By the way Ado's theorem seems irrelevant to the whole discussion. $\endgroup$
    – YCor
    Commented Nov 28, 2020 at 12:53

1 Answer 1

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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 $n \geqslant 2$.

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  • $\begingroup$ 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
    Commented Nov 28, 2020 at 12:18
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    $\begingroup$ Because $\mathbb{Z}/n\mathbb{Z}$ is a $0$ dimensional Lie group, hence its Lie algebra is $\{0\}$. $\endgroup$
    – Didier
    Commented Nov 28, 2020 at 12:26

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