Hadamard three-lines theorem: Difference between revisions

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Applications: punctuation
Proof: This is already "displayed" so "displaystyle" is redundant.
 
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In [[differential calculus]], the '''domain-straightening theorem''' states that, given a [[vector field]] <math>X</math> on a [[manifold]], there exist local coordinates <math>y_1, \dots, y_n</math> such that <math>X = \partial / \partial y_1</math> in a neighborhood of a point where <math>X</math> is nonzero. The theorem is also known as '''straightening out of a vector field'''.


The [[Frobenius theorem (differential topology)|Frobenius theorem]] in differential geometry can be considered as a higher dimensional generalization of this theorem.
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== Proof ==
It is clear that we only have to find such coordinates at 0 in <math>\mathbb{R}^n</math>. First we write <math>X = \sum_j f_j(x) {\partial \over \partial x_j}</math> where <math>x</math> is some coordinate system at <math>0</math>. Let <math>f = (f_1, \dots, f_n)</math>. By linear change of coordinates, we can assume <math>f(0) = (1, 0, \dots, 0).</math> Let <math>\Phi(t, p)</math> be the solution of the initial value problem <math>\dot x = f(x), x(0) = p</math> and let
:<math>\psi(x_1, \dots, x_n) = \Phi(x_1, (0, x_2, \dots, x_n)).</math>
<math>\Phi</math> (and thus <math>\psi</math>) is smooth by smooth dependence on initial conditions in ordinary differential equations. It follows that
:<math>{\partial \over \partial x_1} \psi(x) = f(\psi(x))</math>,
and, since <math>\psi(0, x_2, \dots, x_n) = \Phi(0, (0, x_2, \dots, x_n)) = (0, x_2, \dots, x_n)</math>, the differential <math>d\psi</math> is the identity at <math>0</math>. Thus, <math>y = \psi^{-1}(x)</math> is a coordinate system at <math>0</math>. Finally, since <math>x = \psi(y)</math>, we have: <math>{\partial x_j \over \partial y_1} = f_j(\psi(y)) = f_j(x)</math> and so <math>{\partial \over \partial y_1} = X</math>
as required.
 
[[Category:Differential calculus]]

Latest revision as of 22:53, 2 January 2015

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