Kelvin transformation in fully nonlinear equation

Question

Let $g_\text{flat}$ denote the Euclidean metric on $\mathbb{R}^n$ and $A^u$ denote the $(1,1)$-Schouten tensor of $u^{\frac{4}{n-2}}g_\text{flat}$, $$ A^u = -\frac{2}{n-2}u^{-\frac{n+2}{n-2}}\nabla^{2}u+\frac{2n}{(n-2)^2} u^{-\frac{2n}{n-2}}\nabla u\otimes \nabla u - \frac{2}{(n-2)^2} u^{-\frac{2n}{n-2}}|\nabla u|^2 g_\text{flat}.$$

Consider the equation $$ \sigma_k(\lambda(A^{u}))=K(x) \quad \text {and}\quad \lambda(A^{ u}) \in \Gamma_k \text{ on } \, \mathbb{R}^n,$$where $\lambda(A^{ u})=(\lambda_1,\ldots,\lambda_n)$ denote the eigenvalues of $A^u$, $\sigma_k$ is the $k$-elementary symmetric function, i.e., $$\sigma_k(\lambda)=\sum\limits_{i_1 < \cdots < i_k} \lambda_{i_1} \cdots \lambda_{i_k},$$ and $\Gamma_k$ is the connected component of $\{\lambda \in \mathbb{R}^n: \sigma_k(\lambda) > 0\}$ which contains the positive cone $\{\lambda \in \mathbb{R}^n: \lambda_1, \ldots, \lambda_n > 0\}$ (to ensure the equation is elliptic.) $K(x)$ is a function defined on $\mathbb{R}^n$.

Let $\hat{u}(z)=|z|^{2-n} u\left(\frac{z}{|z|^2}\right)$, is it true $$\sigma_k(\lambda(A^{\hat{u}(z)}))=K\left(\frac{z}{|z|^2}\right) \quad \text{and} \quad \lambda(A^{ \hat{u}}) \in \Gamma_k \text{ on } \, \mathbb{R}^n?$$ This is well-known that $\hat{u}(z)$ is a Kelvin transformation of $u$, but seems not easy to check the above equation. Did I miss someting? Thanks for any help.

Answer 1

This follows easily from the fact that the Kelvin transform is a conformal diffeomorphism and the naturality of the Schouten tensor.

Let $\Phi(z) := \frac{z}{\lvert z\rvert^2}$ denote the Kelvin transform. It is easy to check that $\Phi$ is a conformal diffeomorphism. In fact, $$ \Phi^\ast g_0 = \lvert z\rvert^{-4} g_0 . $$ (As suggested in the comments, this is very easy to check in spherical coordinates, where $g_0 = dr^2 + r^2g_{S^{n-1}}$.) Therefore $$ \Phi^\ast g_u = \Phi^\ast (u^{\frac{4}{n-2}}g_0) = (u \circ \Phi)^{\frac{4}{n-2}}\lvert z\rvert^{-4} g_0 = \left( \lvert z\rvert^{2-n}(u\circ \Phi)\right)^{\frac{4}{n-2}}g_0 = \hat u^{\frac{4}{n-2}}g_0 . $$

The naturality of the Schouten tensor is the statement $\Psi^\ast (A^g) = A^{\Psi^\ast g}$ for any diffeomorphism $\Psi$. Therefore $\Psi^\ast (g^{-1}A^g) = (\Psi^\ast g)^{-1}A^{\Psi^\ast g}$ where, as usual, $g^{-1}$ is the musical operator that raises one index.

Finally, recall that $\sigma_k(\lambda(A^u)) = \sigma_k(g_u^{-1}A^u)$, as mentioned in my answer to your other question. Combining the above gives \begin{align*} K\left( \frac{z}{\lvert z\rvert^2} \right) & = (\Phi^\ast K)(z) \\ & = \left( \Phi^\ast \sigma_k(g_u^{-1}A^{u}) \right)(z) \\ & = \sigma_k(g_{\hat u}^{-1}A^{\hat u})(z) . \end{align*} Repeating this argument with different values of $k$ also shows that the positive cone is preserved by the Kelvin transform.