7
$\begingroup$

The ordinary Laplacian on $\mathbb{R}^N$ behaves nicely under a stereographic projection onto $\mathbb{S}^N\setminus\{P\}$. (Here $P$ is either the north or south pole of the unit sphere $\mathbb{S}^N$.) Namely, letting $$ \Omega=\frac{2}{1+\lvert x \rvert^2},\qquad \left(x\in \mathbb{R}^N\right) $$ denote the conformal factor of the stereographic projection, one has the formula $$\tag{1} -\Delta_{\mathbb{R}^N}\phi = \Omega^{\frac{2+N}{2}} \left(-\Delta_{\mathbb{S}^N} + \frac{N(N-2)}{4}\right)\left(\Omega^{1-\frac{N}{2}}\phi\right). $$ My question is: does there exist an analogue of (1) for the fractional Laplacian $(-\Delta_{\mathbb{R}^N})^\alpha$? Actually I am interested in the cases $\alpha=\pm \frac{1}{2}$.

$\endgroup$
3
$\begingroup$

Unless I'm misreading, there exist such an analogue at least for some $\gamma$. See equation 2.5 and theorem 3.2 of Fractional Laplacian in Conformal Geometry by Sun-Yung Alice Chang and María del Mar Gonzáles.

To spell it out, the theorem 3.2 says that for $\gamma \in (0,\frac{n}{2})\setminus \mathbb{N}$ and a smooth function $f:\mathbb{R}^n\to \mathbb{R}$ one has $$ P_\gamma[g_\mathbb{H},|dx^2|] f = (-\Delta_x)^\gamma f, $$ where $P_\gamma[g^+,\widehat{g}]$ is the fractional power of the conformal Laplacian with power $\gamma$ associated to the ambient metric $g^+$ and the conformal class $[\widehat{g}]$. The metric $g_\mathbb{H}$ is the standard hyperbolic metric of the upper space and $|dx^2|$ is the standard Euclidean metric.

Now if we take the rescaled metric $\widehat{g}_v = v^{\frac{4}{n-2\gamma}}\widehat{g}$, equation 2.5 claims that $$ P_\gamma[g^+,\widehat{g}_v] \phi = v^{-\frac{n+2\gamma}{n-2\gamma}}P_\gamma[g^+,\widehat{g}] (v \phi), $$ for all smooth functions $\phi$.

The conformal fractional Laplacian $P_\gamma[g_\mathbb{H},v^{\frac{4}{n-2\gamma}}|dx^2|]$ plays the role of the spherical conformal Laplacian in the classical formula (1) of OP.

$\endgroup$
1
$\begingroup$

Just a short addendum: long ago Stefan Samko observed this connection between Euclidean and spherical fractional Laplacians, see item d) in page 360 of his conference paper from 2000.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.