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Killing Fields on R^n

  1. Nov 24, 2008 #1
    1. The problem statement, all variables and given/known data
    A linear vector field on R^n, defined by a matrix A, is a Killing field if and only if A is antisymmetric.


    2. Relevant equations



    3. The attempt at a solution
    I took R^2. Let A be the following matrix:
    0 -10
    10 0

    A is antisymmetric. However, how in the heck is A a Killing field? For example, let
    v = (1, 10), then Av = (-100, 10)
    u = (2, 3), then Au = (-30, 20)
    and <u, v> = 32 while <Au, Av> = 3200, so the dot product after we apply A is a multiple of the original dot product by the determinant of A (in this case it's 100).

    What exactly am I misreading? The only restriction on A is that it is antisymmetric. A Killing field is a vector field that preserves the metric, i.e. in particular it will preserve the dot products of vectors.
     
  2. jcsd
  3. Nov 24, 2008 #2

    Dick

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    Science Advisor
    Homework Helper

    Ok, I finally figured out what you mean here. On R^2, they are referring to a vector field of the form F(x*i+y*j)=(ax+by)*i+(cx+dy)*j, where i and j are the usual basis for R^2. Using the usual metric, what do Killing's equations look like? The covariant derivatives are just the ordinary partial derivatives.
     
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