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Independent Killing vector field

  1. Dec 13, 2015 #1
    Hello everyone
    How is it possible that a n-dimensional spacetime admits m> n INDEPENDENT Killing vectors where m=n(n+1)/2 if the space is maximally symmetric?
     
  2. jcsd
  3. Dec 13, 2015 #2

    bcrowell

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    Could you explain a little more what you're asking? Are you asking for a proof? Does it seem impossible to you that m>n? Have you googled to see what the 10 Killing vectors are for Minkowski space?
     
  4. Dec 13, 2015 #3
    I know that from Killing equation the Killing field takes the form (for flat space): ##V_{\alpha}= c_{\alpha} + A_{\alpha \beta} x^{\beta}##. I know this vector field can be expressed as a linear combination of 10 certain vectors. I am wondering how it is possible that those 10 vectors are linearly independent if we are in 4 dimensions.
     
  5. Dec 13, 2015 #4

    bcrowell

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    When we say "Killing vector," we mean "Killing vector field." The dimensionality of the space of vector fields is infinite. This is similar to the idea that the set of all real-valued functions is in some sense much bigger than the set of all real numbers.

    You might want to look at the Killing vectors of Minkowski space as an example. There are 10 of them, so at any given point, some of them are not independent of the others. However, they are independent if you consider them as functions defined on all of space.
     
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