# Can one understand the Kelvin transform conceptually?

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.

• 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. Jun 11, 2012 at 7:35

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).
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.