Re: Which norm is compatible with Lie bracket?
- From: israel@xxxxxxxxxxx (Robert Israel)
- Date: 21 Sep 2005 12:54:37 -0400
In article <dgrghm$642$1@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Pavel Pokorny <Pavel.Pokorny@xxxxxxxxxxxxxxxxxxxxxxx> wrote:
>how can we define a norm for smooth maps
> f,g: R^n -> R^n
>so that the Lie bracket
> [f,g] = g'.f-f'.g
>satisfies
>|| [f,g] || <= M ||f|| ||g||
In particular, if f is constant [f,g] is the directional derivative
of g in the direction of f. Fix such a nonzero f, and let D_f be
the operator g -> [f,g]. Then D_f would have to be bounded. But
consider g(x) = exp(k f.x) f for constant k, which satisfies
D_f g = k (f.f) g. Thus ||D_f|| >= (f.f) |k|. Since k is arbitrary,
D_f must be unbounded. So there is no such norm.
Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.
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- From: Pavel Pokorny
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