Let $n\geq 3$, for a metric $g$ on $\mathbb{S}^n$, the $\sigma_k$-curvature of $g$ is defined as follows. Let $Ric_{g}$, $R_{g}$ and $A_{g}$ denote respectively the Ricci curvature, the scalar curvature and the Schouten tensor of $g$: \begin{equation*} A_{g}=\frac{1}{n-2}\left(Ric_{g}-\frac{R_{g}}{2(n-1)}g\right). \end{equation*} Let $\lambda(A_g)$ denote the eigenvalues of $A_g$ with respect to $g$. For $1 \leq k \leq n$, the $\sigma_k$-curvature of $g$ is then the function $\sigma_k(\lambda(A_g))$ where $\sigma_k$ is the $k$-elementary symmetric function, $\sigma_k(\lambda)=\sum\limits_{i_{1}<\cdots<i_{k}}\lambda_{i_{1}}\cdots\lambda_{i_{k}}$. Our equation of interest is \begin{equation}\label{1} \sigma_k(\lambda(A_{g})) = K(x)\quad \text{ and }\quad \lambda(A_{g}) \in \Gamma_k \text{ on }\,\mathbb{S}^n. \end{equation} where $g$ is the unknown metric which is conformal to the standard metric $g_0$, $K$ is a function defined on $\mathbb{S}^n$, and $\Gamma_k$ is the connected component of $\{\lambda \in \mathbb{R}^n: \sigma_k(\lambda) > 0\}$ which contains the positive cone $\{\lambda \in \mathbb{R}^n: \lambda_1, \ldots, \lambda_n > 0\}$. This is a fully nonlinear equation when $2\leq k\leq n$.

Let $g_0$ be the standard metric on $\mathbb{S}^n$ and write the metric $g$ as $g_v = v^{\frac{4}{n-2}}g_0$ for some positive function $v$. Note that \begin{equation*} A_{g_{v}}=A_{g_0}-\frac{2}{n-2}v^{-1}\nabla^{2}_{g_0}v+\frac{2n}{(n-2)^{2}}v^{-2}dv\otimes d v-\frac{2}{(n-2)^{2}}v^{-2}|dv|^{2}_{g_0}g_0. \end{equation*} Now, we seek the solution $v$ to the following equation \begin{equation} \sigma_k(\lambda(A_{g_v})) = K(x)\quad \text{ and }\quad \lambda(A_{g}) \in \Gamma_k \text{ on }\,\mathbb{S}^n. \end{equation}

If $-\Delta u=f$ in $\mathbb{R}^n$, we know from Newtonian potential ($f$ suitably well), $$u(x)=\int_{\mathbb{R}^n} \frac{1}{|x-y|^{n-2}}f(y)\,\mathrm{d} y.$$ My question is whether the solution to the above fully nonlinear equation has similar representation formula?

  • $\begingroup$ When you typed reprentation, did you mean representation? $\endgroup$ Commented Apr 23, 2023 at 20:25
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    $\begingroup$ Yes, some typos, sorry. $\endgroup$ Commented Apr 24, 2023 at 1:29
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    $\begingroup$ The operator $u \mapsto \sigma_k(\lambda(A_{g_u}))$ is not linear, even after taking a power to make it homogeneous of degree $1$. For this reason, you do not have a linear representation formula. $\endgroup$ Commented Apr 24, 2023 at 11:21


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