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I like to introduce (matrix) Lie groups without the notion of manifolds. (And I am happy to sacrify the "few" Lie groups which are not matrix groups in favor of a simpler definition.)

I was thinking of the following definition:

  • $G$ is a (matrix) Lie groups $:\Leftrightarrow$ $G$ is a closed subgroups of $GL(n,\mathbb{R})$.

(I do not care about Lie groups over finit fields either...)

This definition seems to be okay for my purpose but it requires to equip $GL(n,\mathbb{R})$ with a metric (to give closeness a meaning). My question:

Am I correct: If I do not want to use the notion of a manifold (or a non-standard replacement with similar complexity), I need to equip $GL(n,\mathbb{R}$ with a metric to sufficiently characterize matrix Lie groups?

B0rk4
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  • Why did you ask this again? You can edit your question. – Tobias Kildetoft Jan 17 '13 at 02:50
  • Hi Tobias, my previous question was slightly different. I deleted it since I just found the answer to it once I submitted it. Sorry for the confusion, I know how to edit ;-) – B0rk4 Jan 17 '13 at 02:53
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    Anyway, you only need a topology, not a metric to get the closed subsets. And there is certainly no way to avoid topological questions when dealing with Lie groups. – Tobias Kildetoft Jan 17 '13 at 02:55

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You can take the approach given in Hall's book on Lie groups where he defines a matrix Lie group to be a closed subgroup of $\text{GL}_n(\Bbb{C})$. Of course when we mean closed it is with respect to the usual Euclidean topology on $\Bbb{C}^{n^2}$.

You may choose not to equip $\text{GL}_n(\Bbb{C})$ with a metric but I think ultimately at the end of the day you want to equip it with at least some kind of topology. From my experience the usual topology on these groups is the one induced from the Euclidean metric.