Re: Question on Lie Groups

From: Jannick Asmus (jannick.news_at_web.de)
Date: 01/31/05 

Date: Mon, 31 Jan 2005 16:48:14 +0100
To:  igor.kh@gmail.com

On 31.01.2005 01:45, igor.kh@gmail.com wrote:
> Tony wrote:
>
>>Hi everyone,
>>
>>Doing some problems on my own outside of class, and can't seem to get
>
> the
>
>>following :
>>
>>Let G be a connected Lie group, and let U in G be any open
>
> neighborhood of
>
>>the identity. Show that every element of G can be written as a
>
> finite
>
>>product of elements of U.
>>
>>I can't seem to figure this out.
>
>
> How about this. Since G is connected, there is a continuous curve from
> the identity to any element. Strictly speaking, this requires path
> connectedness, but I think for manifolds they are equivalent. 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.

Igor,

I just went through your argument. And now I see that it goes straight
up the hill if U is shrunk to U\cap U^{-1} (without loss of generality,
of course).

I would like to add that I prefer the topological proof already given
somewhere in the thread, since the assertion is true for any topological
connected group.

Best,
J.
>
> Igor
>

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