My Math Forum Homomorphism of Lie groups as groups but not as manifolds

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 October 16th, 2010, 05:04 PM #1 Senior Member   Joined: Jan 2009 From: Russia Posts: 113 Thanks: 0 Homomorphism of Lie groups as groups but not as manifolds I was wondering if there exists an example of a homomorphism (or even better isomorphism) between Lie groups as groups, but not as manifolds. If anyone could give such an example, that would be of great value!
 October 18th, 2010, 02:18 AM #2 Senior Member   Joined: Sep 2008 Posts: 150 Thanks: 5 Re: Homomorphism of Lie groups as groups but not as manifold Hi lime, Well, what does the abstract group $(\IR,+)$ look like? It is just some infinite dimensional vector space over $\IQ$. Now the same is true for the vectorgoup $(\IR^2,+)$. It is easy to see, that both dimesions are the same as cardinal numbers, thus there is an isomophism as $\IQ$-vector spaces thus in particular one as groups. But of course they don't have the same dimension. It is a bit harder with compact Lie groups, but at least there should be many more automophisms of $\IR/\IZ$ as an abstract group then as a Lie group. You get lots of abstract morpisms for free realising that $\IR/\IZ\cong \IQ/\IZ\oplus V$, where V is again an infinite dimensional vector space. best Peter
 October 18th, 2010, 02:05 PM #3 Senior Member   Joined: Jan 2009 From: Russia Posts: 113 Thanks: 0 Re: Homomorphism of Lie groups as groups but not as manifold Peter thank you for your answer. That's exactly what I was looking for. Non-compact space example was absolutely clear to me, though I'm still a bit confused with your counterexample for compact Lie groups. Do you say that there must be a group isomorphism $\mathbb R / \mathbb Z \simeq \mathbb Q / \mathbb Z \oplus V$ ? If yes then how to construct it? Why the last one is compact? Why these are different manifolds? Would be very helpful if you elucidate here a bit.
 October 25th, 2010, 07:30 AM #4 Senior Member   Joined: Sep 2008 Posts: 150 Thanks: 5 Re: Homomorphism of Lie groups as groups but not as manifold Sorry for the late answer. The second example was not entirely written out, and I was meaning to say: $\mathbb{R}/\mathbb{Z}$ is has automorphisms as an abstract group, which are not continuous or even differentiable. I don't know any example of two compact Lie groups, which are abstractly isomorphic, but not as Lie groups. What you can do, is to show, that $\mathbb{R}/\mathbb{Z}$ is isomorphic to $\mathbb{R}/\mathbb{Z}\times \mathbb{R}$ as an abstract group but not as a Lie group. (Basicly you have to show $\mathbb{R}/\mathbb{Z}= \mathbb{Q}/\mathbb{Z} \times V$ as abstract groups for a infinite dimensional $\mathbb{Q}$ vector space V. I will think a bit, if I can come up with an example of two different compact Lie groups. rgds Peter

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group homomorphism between lie groups not continuous example

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