# Proving $\mathbb{R}^3 \textrm{\\} {0}$ not Lie group, without using homotopy equivalence?

From this question, I found that Lie group structure cannot be granted onto $$\mathbb{R}^n \textrm{\\} {0}$$ for odd n. I am specifically interested in the minimal case, which is n = 3.

The answer there is based on a short comment, which involves homotopy equivalence and Euler characteristic. I am fairly unfamiliar with these concepts, so I wonder if there is an easier way to achieve this proof. (In particular, I also dislike that it uses Euler characteristic, which seems to be involved concept for general manifolds)

There is also this question, but $$\mathbb{R}^3 \textrm{\\} {0}$$ part is not proved.

Is there more specific, perhaps painstaking way to prove that $$\mathbb{R}^3 \textrm{\\} {0}$$ can never be a Lie group?

• Also I am curious what happens if we cut out more "holes". Would it continue to be inadmissible to Lie group? Would R^n with holes stop being Lie group for even n as well? Commented Jun 14, 2022 at 8:10
• First, $\Bbb R^3\setminus 0$ is homotopy equivalent to the sphere $S^2$, see this post. This is not so difficult, and it is useful anyway (even necessary) that you learn this definition. Then we have elementary proofs at MSE, why this isn't a Lie group, see for example here. Commented Jun 14, 2022 at 8:40
• Definition of homotopy equivalence by itself does not seem difficult, but I cannot make the connection to Lie group (or the homeomorphism without fixpoint). Does auto-homeomorphsms on $\mathbb{R}^3$ \ $0$ translate to $S^2$? Commented Jun 14, 2022 at 8:56
• I understand. My idea was, to "view" $\Bbb R^3\setminus 0$ simply as $S^2$. Then things are easier. You can also use the results of this post, but this is also not elementary. Commented Jun 14, 2022 at 9:03
• You cannot beat something with nothing. You are asking an advanced question and, yet, refuse to accept advanced tools for answering it. Afaik, using $\pi_2$ is the simplest way to solve this. Alternatives involve structural results for Lie groups, which are harder. Commented Jun 14, 2022 at 15:01

You really cannot avoid learning some algebraic topology to answer this question: using "bare hands" I don't even know how to show that $$\mathbb{R}^3$$ (which is obviously a Lie group) is not diffeomorphic to $$\mathbb{R}^3 \setminus \{ 0 \}$$ (try it!), whereas with some algebraic topology they can be easily distinguished using $$\pi_2$$ or $$H_2$$ or the Euler characteristic.
The question of Lie group structures on $$\mathbb{R}^n \setminus \{ 0 \}$$ is also related to the classification of real division algebras (if an $$n$$-dimensional division algebra over $$\mathbb{R}$$ exists then its multiplication induces a Lie group structure on $$\mathbb{R}^n \setminus \{ 0 \}$$ and also on $$S^{n-1}$$; we know by the classification that this occurs only when $$n = 2$$ and $$n = 4$$, and $$S^1$$ and $$S^3$$ are the only positive-dimensional spheres with Lie group structures) which requires some unavoidable actual work to establish one way or another.
• I see, then it seems that this was harder than I thought. I guess $S^2$ with its always vanishing vector field is the most trivial case then. Commented Jun 14, 2022 at 21:58
If you don't like homotopy topology ), then you can use Lie group theory. The space under consideration may be diffeomorphic only to some three-dimensional simply connected Lie group. Its maximal compact subgroup must be diffeomorphic to $$S^2$$. But the two-dimensional sphere is not a Lie group (there are only two two-dimensional simply connected Lie groups, and both of them are solvable and diffeomorphic to $$R^2$$).
• What is the argument that ensures that the maximally compact subgroup must be $S^2$? Why not a torus, a point, a circle? Commented May 13, 2023 at 7:53