Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question is motivated by this paper. There, they develop a stippling method which requires a kernel to be diagonal. Meaning a symmetric bilinear function

$K\colon \chi\times \chi\to \mathbb{R}$

from the Riemannian surface $\chi$ to the reals, and it diagonal that admits a Fourier expansion of the form

$K=\sum_l\lambda_l\psi_l(x)\overline{\psi_l(y)}$

where $\psi_l$ is a Fourier basis, eg $e^{2\pi i (n_1x_1+n_2x_2)}$. This Fourier expansion is diagonal in the sense that all the terms $\lambda_{l,k}\psi_l(x)\overline{\psi_k(y)}$ with $k\neq l$ are $0$.

In the case when $\chi$ is the Torus it is not hard to prove that if $K$ depends only on the distance, ie $K(x,y)=\varphi(\|x-y\|)$, then $K$ is diagonal.

Is it also true in the case when $\chi=\mathbb S^2$ and the Fourier basis is the spherical harmonics?

In which sense can this be generalized? Is there always a basis of $L^2(\chi)$ such that this happens?

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.