# Infinite examples of non isomorphic Lie groups with Lie algebra $\mathfrak{so}(3)$

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?

• Consider $SO(3)\times \Bbb Z_n$ for $n\in\Bbb N$. Note that these groups are not connected. – s.harp Nov 27 '20 at 17:09
• 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. – Didier Nov 27 '20 at 18:04
• "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". – YCor Nov 28 '20 at 12:52
• By the way Ado's theorem seems irrelevant to the whole discussion. – YCor Nov 28 '20 at 12:53

## 1 Answer

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 $$n \geqslant 2$$.

• 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. – Jotabeta Nov 28 '20 at 12:18
• Because $\mathbb{Z}/n\mathbb{Z}$ is a $0$ dimensional Lie group, hence its Lie algebra is $\{0\}$. – Didier Nov 28 '20 at 12:26