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Consider the $n$-dimensional sphere $S^n$. I'm especially interested in the $n=4$ case. The Hilbert space $L^2(S^n)$ can be decomposed into a direct sum of eigenspaces of the Laplacian, which are finite dimensional. I'm looking for non-isometric conformal transformations

$$f: S^n \to S^n$$

s.t. for some $\lambda, \mu > 0$ if $\psi$ is an eigenvector of the Laplacian with eigenvalue $\alpha < \lambda$ then $f(\psi)$ is a sum of eigenvectors with eigenvalues $< \mu$.

Do such $f$ exist? If so, is it possibly to classify them?

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2 Answers 2

up vote 2 down vote accepted

The only possibility is the trivial one when $\lambda$ is so small that the only eigenfunctions with eigenvalue less than $\lambda$ are constants (eigenvalue zero). Otherwise the eigenfunctions with eigenvalue less than $\lambda$ span the space of polynomials of degree at most $d$ for some positive integer $d$, and then composition with a non-isometric conformal transformation takes it outside the space of polynomials, and thus outside the span of eigenfunctions of eigenvalue less than $\mu$ for any finite $\mu$.

P.S. In case you've not seen this yet: the group of conformal transformations is described by this theorem of Liouville, identifying it with the group of Möbius transformations.

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Let me add some remarks... The group of conformal transformations of $S^n$ is generated by isometries, scalings ( $x \to \lambda x$ conjugated by stereographical projection and it's inverse) and spherical inversions $x\to \frac{x}{\|x\|}$. As Noam Elkies indicated in his answer - the eigenfunctions of the Laplace equation (the so called spherical harmonics) are restrictions of harmonic polynomials on $\mathbb{R}^{n+1}$.

Also there's a modification of the Laplace-Beltrami operator by a scalar term which is called Yamabe operator. This Yamabe operator is conformally invariant when acting on densities of certain conformal weight. Maybe this operator will be more useful for theorems you want to prove...

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