5

Let $X=\mathbb{R}^2-\{p_1, p_2,..., p_n\}$, then if $n=0$ then clearly $X=\mathbb{R}^2$ is a Lie group under addition. If $n=1$, then $X=\mathbb{R}^2-\{p_1\}$ is isomorphic to $\mathbb{C}\setminus\{0\}$ which has a Lie group structure under multiplication. But I don't know how to proceed in the case in which $n>1$. I have thought of using that the tangent bundle of a Lie group is trivial but I don't know how to do this. I would appreciate any help or suggestions. Thank you.

Josef E. Greilhuber
  • 1,662
user425181
  • 1,231
  • 1
    Your question seems more akin to finding the fundamental group of topological spaces. Removing points from a vector space definitely doesn't make it a Lie group. – Mnifldz Nov 20 '19 at 04:14
  • 1
    "$X = \mathbb R^2 - \lbrace p_1 \rbrace$ is isomorphic to $\mathbb S^1$" -- "isomorphic" in what category? They are homotopic, but not homeomorphic as topological spaces. So are you asking for which $n$ the underlying topological space of your $X$ is homotopic to a topological space which can be given a Lie group structure? Then again, the case of one point removed is homeomorphic to the underlying topological space of a Lie group, namely, $\mathbb R^\ast$, so maybe you want to ask in which cases $X$ is homeomorphic to the underlying topological space of a Lie group? – Torsten Schoeneberg Nov 20 '19 at 06:13

1 Answers1

7

Mnifldz has the right idea. The fundamental group of $\mathbb{R}^2$ with $n$ points removed is a free group on $n$ generators (I guess there is a MSE question explaining that if you look for it). In particular these groups are non-abelian for $n > 1$

There is however a really nice and surprising result involving the (non)-abelianness of fundamental groups of Lie groups. Namely that The fundamental group of a topological group is abelian.

So this gives your answer: $\mathbb{R}^2 - \{p_1, \ldots, p_n\}$ can only be homeomorphic to (the underlying topological space of) a topological group if $n \leq 1$.

Vincent
  • 11,280
  • 2
    Great answer, +1. One can strengthen the last statement by replacing "homeomorphic" with "homotopy equivalent", if indeed the OP asked about that, as I speculated in my comment. – Torsten Schoeneberg Nov 20 '19 at 19:21