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Re: Which norm is compatible with Lie bracket?
Posted:
Sep 23, 2005 3:00 PM
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In article <dh0u8n$1j1$1@news.ks.uiuc.edu>,
Pavel Pokorny <Pavel.Pokorny@vscht.RemoveMe.MeToo.cz> wrote:
>Robert Israel <israel@math.ubc.ca> wrote:
>> 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.
>That is sad. I need some estimate.
>Let me refine the question:
>
>How can we take a subset of smooth maps
>and define a norm for
> f,g: R^n -> R^n
>so that the Lie bracket
> [f,g] = g'.f-f'.g
>satisfies
>|| [f,g] || <= M ||f|| ||g||
>where M is a constant?
>
>E.g. if we take n=1 and polynomials of order not more than 2,
>f(x) = f0 + f1 x + f2 x^2
>g(x) = g0 + g1 x + g2 x^2
>then
>[f,g] = f0 g1 - f1 g0 + 2(f0 g2 - f2 g0)x + (f1 g2 - f2 g1)x^2
>(resembling a vector product in R^3)
>and with the Euclidean norm we have
>|| [f,g] || <= 2 ||f|| ||g||
>
>I would like to see some infinite dimensional subset, if possible.
For simplicity, let's work in the case n=1.
According to my example, there's no possibility of such a norm on
any subspace that includes 1 and exp(kx) for arbitrarily large k.
For polynomials, there's bad news too. Since
[x, x^m] = (m-1) x^m, there is no such norm on a subspace that
includes x and x^m for arbitrarily large m. Moreover,
if your subset is a Lie algebra (i.e closed under addition,
multiplication by scalars, and your Lie bracket operation) and
contains x^2 and x^p for some integer p > 2, then it must
contain x^m for all integers m >= p since
[x^2, x^m] = (m-2) x^(m+1).
However, if you avoid 1 and x the news is better. Consider
the space of polynomials with no constant term and no x term,
and use the norm
|| sum_{j=2}^d c_j x^j || = sum_{j=2}^d |c_j| exp(-j^2)
Then this will work with M = 1/(2 e^2) since for j, k >= 2
|| [x^j, x^k] || = || (k-j) x^(j+k-1) || = |k-j| exp(-(j+k-1)^2)
= |k-j| exp(-2(j-1)(k-1)+1) ||x^j|| ||x^k||
<= M ||x^j|| ||x^k||
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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