laplacian
If \(f\) is a function \(R^n \rightarrow R\), then the Laplacian is \(\Delta f = \sum_i \frac{\partial^2 f}{\partial x_i^2}\)
It can be shown using an argument from variational calculus that functions \(f\) such that \(\Delta f = 0\) are stationary points of the dirichlet energy functional.
1. fourier transform
The fourier transform diagonalizes the Laplacian. First, note that \(\sin\) and \(\cos\) are eigenfunctions of the laplacian. Second, remember that the fourier transform decomposes a function into sines and cosines. You can think of this decomposition as returning an infinite length vector, where the elements are the fourier transform coefficients. Then, applying the Laplacian to this decomposition just means that each element needs to be scaled. By analogy think about what happens for a diagonalizable matrix: you apply your unitary matrix \(U\) and then all you need to do is scale each component of the vector.
- see terence tao
2. isometry
- the laplacian is invariant under isometric deformations. This makes it useful for building algorithms that need pose invarianc. You can assume humans only undergo rigid deformation. But only mostly. There is some accounting that needs to be done for deviance from isometry.