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Laplacian of the metric

  1. Jul 31, 2011 #1
    When reading about the ideas behind ricci flow, I've often read that the ricci tensor is proportional to the laplacian of the metric, but only in harmonic coordinates. Can someone explain this to me? What laplacian operator would one use to show this as there are many different laplacians in differential geometry? I think it is the Laplace beltrami operator, but I was under the impression that this operator can only be used on scalar functions, and that the laplacian of a metric is not itself a tensor, so how can the ricci tensor be composed of this? Any help would be much appreciated.
     
  2. jcsd
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