Laplacian of the metric

[unchained1978]

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.

 

[pat_connell]

The Laplace-Beltrami operator is indeed the correct operator to use when discussing the Ricci flow in harmonic coordinates. The Laplace-Beltrami operator is defined on scalar functions, but it can also be used to operate on tensor fields. For example, the Laplace-Beltrami operator applied to a metric tensor field will yield a new tensor field, which is the Ricci tensor. This is because the Ricci tensor is defined as the trace of the Hessian of the metric tensor field. In other words, the Ricci tensor is the (trace of the) second derivative of the metric tensor field with respect to the coordinates. The Laplace-Beltrami operator can be used to calculate this second derivative, which yields the Ricci tensor. So, to summarize, the Ricci tensor is proportional to the Laplace-Beltrami operator applied to the metric tensor field, and this equation holds true only in harmonic coordinates.