Re: Question on Lie Groups

From: Igor Khavkine (k_igor_k_at_lycos.com)
Date: 01/31/05 

Date: Mon, 31 Jan 2005 10:15:12 -0500

On Mon, 31 Jan 2005 07:58:12 +0100, Jannick Asmus wrote:
> On 31.01.2005 01:45, igor.kh@gmail.com wrote:

>> This
>> curve is homeomorphic to the unit interval [0,1] and hence is compact.
>> If U is a neighborhood of the identity, then cover this curve by left
>> (or right) translations of U by each group element lying on the curve.
>> Since the curve is compact, a finite subcover can be selected. Now
>> leapfrog between overlapping neighborhoods.
>>
> After a quick look at your argument the question came to my mind whether
> the image of this curve needs to be multiplicatively closed? E.g. if the
> curve is multiplicative, i.e. it enjoys the property of a group
> homomorphism.

I don't see why it should be. I'm not covering the curve by open segments
like (0,1), I'm covering the curve with copies of U. Once I find a finite
cover, I can forget that the curve was even there. Now all I need to get
from identity to the desired group element is multiplication by elements
of U, or perhaps UU^{-1}, but I'm sure those can be made from finitely
many elements of U.

Igor

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