Let $M$ be compact manifold without boundary equipped with a smooth measure. Since continuous complex-valued functions are dense in $L^2(M)$, the Gramm-Schmidt process shows that $L^2(M)$ admits an orthonormal basis $(e_n)$ of continuous functions.
Question: Is it always possible to find $(e_n)$ as above so that each $e_n$ is nowhere-vanishing? If not, what obstructs us from doing this?
- If $M=G$ is a compact abelian Lie group with Haar measure, then this is possible since we may use the dual group $\widehat G$ as a basis. These are not only nowhere-vanishing, but actually circle-valued.
- If $M=S^2$, I suspect things don't work out. As with any Riemann manifold, we can consider the eigenfunctions of the Laplacian to get a basis of nice smooth functions. For $S^2$, these are the spherical harmonics. However, eigenspaces here have common zeros, apparently. One can learn more by googling "nodal sets".