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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?

Abastro
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  • 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? – Abastro Jun 14 '22 at 08:10
  • 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. – Dietrich Burde Jun 14 '22 at 08:40
  • 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 $S^2$? – Abastro Jun 14 '22 at 08:56
  • 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. – Dietrich Burde Jun 14 '22 at 09:03
  • Using $\pi_2(M) \neq 0$ would be way over my head. How do I "view" $\mathbb{R}^3$ \ $0$ as $S^2$? Hairy ball theorem does not extend to this manifold, so I cannot apply the same logic. Sorry, I banged my head at this and am utterly lost.. – Abastro Jun 14 '22 at 12:25
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    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. – Moishe Kohan Jun 14 '22 at 15:01

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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.

Qiaochu Yuan
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  • 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. – Abastro Jun 14 '22 at 21:58
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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 $R^2$).

In short, the absence of a Lie group structure in the indicated space follows easily from the description of the structure of three-dimensional simply connected Lie groups. (In this case, the Euler characteristic is not used!)

Тyma Gaidash
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Vladimir47
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    What is the argument that ensures that the maximally compact subgroup must be $S^2$? Why not a torus, a point, a circle? – Didier May 13 '23 at 07:53