Let $U = \mathbf{R}^n - \{ 0 \}$, $n > 2$ and consider for a function $f \in C^2(U)$ the Kelvin transform

$$f^\star(x) = r^{2-n} f\left(\frac{x}{r^2}\right),$$

where $r = \lvert x \rvert$. One can verify by explicit computation that,

$$r^2 \Delta(f^\star(x))= (r^2\Delta f)^{\star}(x),$$

where $\Delta$ denotes the usual Laplace operator. In particular the Kelvin transform maps harmonic functions to harmonic functions. Is there a way to see this last fact without resorting to explicit computation? I suspect it has to do with how the group of conformal transformations acts.

  • 3
    $\begingroup$ Yes, that's it. You can think of $\mathbb{R}^n \setminus \{0\}$ as being stereographically projected from the sphere. Then the Kelvin transform is just "flipping" the sphere. $\endgroup$
    – Ray Yang
    Jun 11, 2012 at 7:35

1 Answer 1


A conceptual view on the Kelvin transform is use the conformal invariance of the Yamabe operator : If $(M,g)$ is a Riemannian manifold of dimension $n$, the Yamabe operator is the differential operator $$L_g=\Delta+\frac{n-2}{4(n-1)} \mathrm{Scal}_g$$ It is a conformally invariant operator that is to say if $\widehat g=u^{\frac{4}{n-2}} g$ where $u$ is a smooth positive function then $$ L_{ g}(u\varphi)=u^{\frac{n+2}{n-2}}L_{\widehat g}(\varphi)$$ (see the formula (2.7) in the article of Lee and Parker : (The Yamabe problem , Bull. AMS, (17) n° 1 (1987) pp 37--91).

...Well this a consequence of the computation of the scalar curvature under conformal change and this is not really simpler than the computation involved in the proof of the Kelvin transform...

Now the Kelvin transform is a special case of this conformal transformation law for the Yamabe operator : Let $I\colon \mathbb{R}^{n}\setminus\{0\}\rightarrow \mathbb{R}^{n}\setminus\{0\}$ be the inversion : $$I(x)=\frac{x}{\|x\|^2}.$$ It is a conformal map and the pull back of the Euclidean metric (we called it $\mathrm{eucl}$) by $I$ is: $$I^*\mathrm{eucl}=\frac{1}{\|x\|^4} \mathrm{eucl}$$ Indeed in polar coordinate : $$I^*\mathrm{eucl}=\left(d\frac{1}{r}\right)^2+\left(\frac{1}{r}\right)^2 (d\sigma)^2=\frac{(dr)^2+r^2 (d\sigma)^2}{r^4}.$$ Call $g=\mathrm{eucl}$ and $\widehat g=I^*\mathrm{eucl}$, the conformal transformation for the Yamabe operator yields: $$L_{\widehat g}(\varphi)=u^{-\frac{n+2}{n-2}}L_{ g}(u \varphi)$$ where $u=r^{2-n}$. But $I\colon \left(\mathbb{R}^{n}\setminus\{0\}, \widehat g\right) \rightarrow \left(\mathbb{R}^{n}\setminus\{0\}, \mathrm{eucl}\right) $ is by definition an isometry hence the scalar curvature of the Riemannian metric $\widehat g$ is zero and $$L_{\widehat g}(\varphi\circ I)=\left(\Delta\varphi\right)\circ I $$ Eventually we get : $$\left(\Delta \varphi\right)\circ I=r^{n+2}L_{ g}(u \varphi\circ I)= r^{n+2} \Delta\left(r^{2-n}\, \varphi\circ I\right),$$ this formula is equivalent to the Kelvin transform.


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