Let's say I want to prove that a closed subgroup of GL(n,R) or GL(n,C) is a Lie group, with an atlas given by exponential of matrices (restricted to an appropriate subalgebra of gl(n)), without using any manifold or Lie theory. Can you provide the necessary argument? Maybe it's trivial, but I can't see it at the moment.
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asked Nov 3, 2009 at 18:41
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2$\begingroup$ This sort of thing is standard in many textbooks. What do you mean by "without using any manifold or Lie theory"? $\endgroup$– Kim MorrisonCommented Nov 3, 2009 at 19:11
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1$\begingroup$ The notion of Lie group requires the use of manifolds, so your constraint does not seem to make sense. In the absence of a clarifying edit, I'd suggest that this question be closed. $\endgroup$– S. Carnahan ♦Commented Nov 3, 2009 at 20:01
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$\begingroup$ There's a good answer below, so I'll leave it be. $\endgroup$– Kim MorrisonCommented Nov 3, 2009 at 22:04
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$\begingroup$ Sorry if my request wasn't so clear, but Fran Burstall's answer is exactly what I wanted. $\endgroup$– Gian Maria Dall'AraCommented Nov 3, 2009 at 22:22
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1 Answer
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Well, the crux of the matter is to cook up the linear subspace of $gl(n)$ that will eventually be the subalgebra.
If $H$ is the subgroup: set $h=\{X\in gl(n): (\forall t \in \mathbb{R}) \, exp(tX) \in H \}$.
You need to show two not completely obvious things:
$h$ is a linear subspace
$exp(h)$ is a nbhd of $1$ in $H$ (give $H$ the induced topology from $GL_n(\mathbb{R})$).
Adams has ingenious short arguments for these on pages 17-19 of his Lectures on Lie groups which require nothing but a little real analysis.
I remark that all this uses no manifold or Lie theory beyond the necessary definitions and the argument works completely without change for closed subgroups of any Lie group.
