Representation formula for solutions to fully nonlinear equations

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, $$\sigma_k(\lambda)=\sum\limits_{i_{1}<\cdots. Our equation of interest is $$$$\label{1} \sigma_k(\lambda(A_{g})) = K(x)\quad \text{ and }\quad \lambda(A_{g}) \in \Gamma_k \text{ on }\,\mathbb{S}^n.$$$$ where $$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 $$$$\sigma_k(\lambda(A_{g_v})) = K(x)\quad \text{ and }\quad \lambda(A_{g}) \in \Gamma_k \text{ on }\,\mathbb{S}^n.$$$$

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?

• When you typed reprentation, did you mean representation? Commented Apr 23, 2023 at 20:25
• Yes, some typos, sorry. Commented Apr 24, 2023 at 1:29
• 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. Commented Apr 24, 2023 at 11:21