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Satisfies Killing equation, but not a Killing field?

  1. Aug 22, 2010 #1

    bcrowell

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    A cone has all the same local geometrical properties as a plane, so if you take a piece of graph paper and form it into a cone, [itex]\partial_x[/itex] and [itex]\partial_y[/itex] still satisfy the Killing equation. On the other hand, the cone has intrinsic geometrical properties that are different from those of the plane, e.g., parallel transport around a loop enclosing the tip will cause a vector to rotate. This singles out the tip and gives it a special geometrical role, which is clearly not consistent with translational symmetry. Does this mean that we can have a field that satisfies the Killing equation without being a Killing vector, or is the Killing equation violated at the tip of the cone? Does it matter if you extend the cone to make a double cone, so that orbits of a Killing vector can pass smoothly through the tip?
     
  2. jcsd
  3. Aug 22, 2010 #2
    I think due to topological reasons, [tex]\partial_x[/tex] is no longer a global Killing field, i.e. it doesn't join up at the cut of the cone.

    The mapping between the plane and the cone fails to be a diffeomorphism at the tip because it's not smooth. The tip must be excluded from the cone for the cone to form a differentiable manifold.

    Feel free to correct me as I'm not entirely familiar with these issues.
     
  4. Aug 22, 2010 #3

    bcrowell

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    Thanks, petergreat, that's very helpful!
     
  5. Aug 23, 2010 #4

    George Jones

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    Due to topological reasons, not only is [tex]\partial_x[/tex] not a global Killing field, [tex]\partial_x[/tex] is not a smooth global vector field.
     
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