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I want to find all maps $g: \mathbb R^n\setminus \{0\} \rightarrow GL_n(\mathbb R)$ which satisfy the properties

  • $g$ is differentiable and injective
  • $g(g(a)b) = g(a)g(b)$ for all $a,b\in\mathbb R^n\setminus \{0\}$
  • $g(e_1)=I_n$ whereby $e_1$ is the vector $(1,0,\ldots,0)^T$ and $I_n$ is the identity matrix of $GL_n(\mathbb R)$

Do you have any tip, how I can solve my problem? Can you recommend a book or article I shall read?

My Attempt:

If I define $\circ: \mathbb R^n\setminus \{0\} \times \mathbb R^n\setminus \{0\} \rightarrow \mathbb R^n\setminus \{0\}$ via $a\circ b = g(a)b$, then $(\mathbb R^n\setminus \{0\}, \circ)$ should be a lie group. $g$ should be a representation of $(\mathbb R^n\setminus \{0\}, \circ)$.

So I thought the lie group theory and lie group representation theory should help me. Unfortunately after scanning some books about those theories I still have no idea how to solve my problem. I have the feeling, that in the theory of lie groups and their representations the lie group structure is always known but in my case it is not. Did I miss something? Can I use those theories to find all maps $g$?

Stephan Kulla
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    The title of your question is misleading. –  Jul 20 '14 at 16:01
  • @VahidShirbisheh: Yes, unfortunately I did not find a better title. Which title would you choose? – Stephan Kulla Jul 20 '14 at 16:04
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    $\mathbb{R}^n \setminus { 0 }$ can't have a Lie group structure for $n$ odd and greater than $1$ because it's homotopy equivalent to $S^{n-1}$ and in particular has Euler characteristic $2$. But a connected Lie group of positive dimension has Euler characteristic either $0$ or $1$. – Qiaochu Yuan Jul 20 '14 at 16:47
  • @QiaochuYuan: Thanks a lot! Can you write your comment as an short answer, so that I can accept it? – Stephan Kulla Jul 20 '14 at 17:16
  • A group always has Euler characteristic 0. – jspecter Jul 20 '14 at 18:07
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    @jspecter: you mean a compact connected Lie group of positive dimension. This is false if any of those hypotheses are dropped: $\mathbb{R}$ has Euler characteristic $1$, $\mathbb{Z}_2$ has Euler characteristic $2$, there are topological groups without a well-defined Euler characteristic... – Qiaochu Yuan Jul 20 '14 at 20:06

2 Answers2

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Comment by Qiaochu Yuan: "$\mathbb R^n\setminus\{0\}$ can't have a Lie group structure for $n$ odd and greater than 1 because it's homotopy equivalent to $S^{n−1}$ and in particular has Euler characteristic 2. But a connected Lie group of positive dimension has Euler characteristic either 0 or 1."

Community
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Stephan Kulla
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Here is my brief answer for the question

"Finding a lie group structure on $R^n\setminus\{0\}$"

For $n$ odd - see above.

Now let $n$ be even.

$n=2$ this space is diffeomorphic to the Lie group $\mathbb{R} \times S^1$

$n=4$ this space is diffeomorphic to the Lie group $\mathbb{R} \times S^3 =\mathbb{R}\times SU(2)$

$n=2k>4$ - there is no structure of a Lie group, since the space of three-dimensional cohomology for non-Abelian compact Lie groups is always non-trivial (due to E.Cartan), while for the space under consideration (considering up to homotopy equivalence!) it is trivial for $n>4$.

user113988
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Vladimir47
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  • As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community May 13 '23 at 08:17
  • I don't really understand that downvote or prior two comments. This answer is no less terse than the accepted answer, and it provides a complete (and completely correct) answer to the Title question. – Jason DeVito - on hiatus May 13 '23 at 20:48