Talk:Killing vector field

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If the Ricci curvature is positive, then a Killing field must have a zero.

May I ask how come? I had some idea that the Hopf fibration defines a nonvanishing vector field on S3, and that this field acts by isometries, but I might be wrong... (too tired to do the calculations tonight :/ ) Someone here who can help me, please? (A google on "killing field" + Hopf gave me e.g. this paper...see section III) \Mikez 03:06, 21 Dec 2004 (UTC)

Hmm, that seems right to me. Writing the metric on S3 as

ds^2 = d\theta^2 + \sin^2\theta \;d\xi_1^2 + \cos^2\theta \;d\xi_2^2

one sees that the vector field

\frac{\partial}{\partial \xi_1} + \frac{\partial}{\partial \xi_2}

which generates the U(1) action in the Hopf fibration, is a Killing field which has norm 1 everywhere (and hence is nonvanishing). -- Fropuff 07:11, 2005 Jan 9 (UTC)

I have now removed that statement. \Mikez 18:58, 7 Feb 2005 (UTC)

Killing Field??? Maths is awesome!!! 130.195.2.100 03:50, 8 August 2006 (UTC)

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