Is the Laplacian of a Function Simply the Trace of its Hessian Matrix?

[icurays1]

Stupid thing I noticed today:

\nabla^2 U=tr(H(U))

Or, in other words, the Laplacian of a function is just the trace of its Hessian matrix. Whoop-de-frickin do, right? Is this useful knowledge or should I forget it immediately?


N!

 

[Ben Niehoff]

This is not true in curvilinear coordinates.

 

[icurays1]

But then isn't the 'Hessian' a tensor with co/contravariant components? is the trace even defined for tensors like that? And how is a differential operator like \nabla^2 even defined in a nonlinear coordinate system? (Genuine questions, I'm just starting to learn tensors and differential geometry and whatnot...)

 

[Ben Niehoff]

Actually, it looks like I am wrong. One can define a Hessian tensor, and then the Laplacian is the trace of it. Read about it here:

http://en.wikipedia.org/wiki/Laplace–Beltrami_operator

 

[icurays1]

okay, cool. thanks!

would the trace of an arbitrary tensor be \sum_{i}{A_{ii...i}} i.e. summing over the elements of the tensor with identical indicies? For tensors rank>2 'diagonal' is sort of vague, or does 'diagonal' always mean 'elements with the same index'? I guess this doesn't have anything to do with the original question...thanks again Ben!

 

[Ben Niehoff]

Tensor traces are taken by contracting any two indices with the metric tensor. The result will be a tensor whose rank has been reduced by 2. For example, the Ricci tensor is the trace of the Riemann tensor on the 1st and 3rd indices:

R_{bd} = g^{ac} R_{abcd}

and then the scalar curvature is the trace of the Ricci tensor, or the double trace of the Riemann tensor:

R = g^{bd} R_{bd} = g^{ac} g^{bd} R_{abcd}.

In flat space, the metric tensor is just the identity matrix, so this reduces to the more familiar definition of the trace.