The linearization problem of fully nonlinear equation on standard sphere

Question

For a metric $g$ on $\mathbb{S}^{n}$ $(n\geq 3)$, 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, $\sigma_k(\lambda)=\sum\limits_{i_{1}<\cdots<i_{k}}\lambda_{i_{1}}\cdots\lambda_{i_{k}}$. We want to find the unknown metric $g$ which is conformal to the standard metric $g_0$, such that \begin{align}\label{1} \sigma_k(\lambda(A_{g})) = 1\quad \text{ and } \lambda(A_{g}) \in \Gamma_k \text{ on }\,\mathbb{S}^{n},\qquad \qquad \qquad \qquad \qquad\qquad\qquad\qquad\qquad\qquad\qquad (1) \end{align} and $\Gamma_k$ is the connected component of $\{\lambda \in \mathbb{R}^n: \sigma_k(\lambda) > 0\}$ which contains the positive cone $\{\lambda \in \mathbb{R}^n: \lambda_1, \ldots, \lambda_n > 0\}$. (1) is also called the $\sigma_k$-Yamabe problem on $\mathbb{S}^n$.

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*} For $2 \leq k \leq n$, we need to solve a fully nonlinear elliptic equation for $v$.

Define the operator $$ F_{\mu}[v]:=\sigma_{k}(\lambda(A_{g_{v}}))-1. $$ and $$ \mathcal{S}_{0} =\Big\{v\in C^{4,\alpha}(\mathbb{S}^n):\int_{\mathbb{S}^n}x |v(x)|^{\frac{2n}{n-2}}\,dv_{g_0}(x) =0\Big\}. $$ We parametrize $C^{4,\alpha}(\mathbb{S}^n)$ as $\mathcal{S}_0 \times \mathbb{R}$ where the $\mathbb{R}$-factor takes into account the the action of the M"obius group on $\mathbb{S}^n$ on axisymmetric functions and where the element $1 \in \mathcal{S}_0$ corresponds to the so-called axisymmetric standard bubbles on $\mathbb{S}^n$. To this end, for $t \in \mathbb{R}$, let $\varphi_{t}$ be the M"obius transformation on $\mathbb{S}^n$ which, under stereographic projection with respect to the north pole, sends $y$ to $ty$. For function $v$ defined on $\mathbb{S}^n$, we let $$ T_{t}v:=v\circ\varphi_t|\det d\varphi_t|^{\frac{n-2}{2n}}, $$ where $d\varphi_t$ denotes the Jacobian of $\varphi_t$. In particular, the pull-back metric of $g_v = v^{\frac{4}{n-2}}g_0$ under $\varphi_t$ is given by $\varphi_t^* (g_v) = g_{T_t v}$.

The linearized operator of $F_\mu[\pi(\cdot,t)]$ at $\bar w \equiv 1$ is readily found to be $$ \mathcal{L} := D_w (F_\mu \circ \pi)(w,\xi)]\Big|_{w = \bar w} = - d_{n,k}(\Delta + n) \quad \text{ with } \quad d_{n,k} := \frac{2^{2-k}}{n-2} \Big(\begin{array}{c}n\\k\end{array}\Big) $$ and with domain $D(\mathcal{L})$ being the tangent plane to $\mathcal{S}_0$ at $w = \bar w$: $$ D(\mathcal{L}) := T_1(\mathcal{S}_0) = \Big\{\eta \in C^{4,\alpha}(\mathbb{S}^n): \int_{\mathbb{S}^n} x \eta(x)\,dv_{g_0}(x) = 0\Big\}. $$ I don't know how to obtain the expression of the linearized operator of $F_\mu[\pi(\cdot,t)]$ at $\bar w \equiv 1$ since the fully nonlinear term $\sigma_k(\lambda(A_{g_v}))$ is very hard to calculate, I have no idea to check the above results $\mathcal{L}$. Thanks for any help!

Answer 1

Here is my favorite way to do this computation.

Let $$ \delta_{i_1 \dotsm i_k}^{j_1 \dotsm j_k} = \begin{cases} \mathrm{sgn}\, \sigma, & \text{if $j_k = i_{\sigma(k)}$}, \\ 0, & \text{otherwise} \end{cases} $$ be the generalized Kronecker delta. Then $$ \sigma_k(\lambda(A)) = \frac{1}{k!}\delta_{i_1 \dotsm i_k}^{j_1 \dotsm j_k} A_{j_1}^{i_1} \dotsm A_{j_k}^{i_k} , $$ as can be checked, for example, by diagonalizing $A$. It immediately follows that $$ \left. \frac{d}{dt} \right|_{t=0} \sigma_k(\lambda(A + tB)) = \frac{1}{(k-1)!}\delta_{i_1 \dotsm i_k}^{j_1 \dotsm j_k} B_{j_1}^{i_1} A_{j_2}^{i_2} \dotsm A_{j_k}^{i_k} . $$ It is also easy to check that on an $n$-dimensional vector space, $$ \frac{1}{k!}\delta_{j_k}^{i_k}\delta_{i_1 \dotsm i_k}^{j_1 \dotsm j_k} = \frac{n-k+1}{k!}\delta_{i_1 \dotsm i_{k-1}}^{j_1 \dotsm j_{k-1}} . $$

Now let’s specialize to the sphere, where $A_{g_0} = \frac{1}{2}g_0$. Using your equation for $A_{g_v}$ for $g_v = (1+tv)^{\frac{4}{n-2}}g_0$, we see that $$ \left. \frac{\partial}{\partial t}\right|_{t=0} A_{g_{tv}} = -\frac{2}{n-2}\nabla^2v . $$ Therefore $$ \left. \frac{\partial}{\partial t}\right|_{t=0} (A_{g_{tv}})_i^j = -\frac{2}{n-2}v_i^j - \frac{2}{n-2}v\delta_i^j $$ (Recall that you need to raise an index to define $\sigma_k$, hence the second summand.) Putting everything together yields \begin{align*} \left. \frac{\partial}{\partial t}\right|_{t=0} \sigma_k(\lambda(g_{tv})) & = \frac{1}{(k-1)!}\delta_{i_1 \dotsm i_k}^{j_1 \dotsm j_k}\left( -\frac{2}{n-2}v_{j_1}^{i_1} - \frac{4}{n-2}v\delta_{j_1}^{i_1} \right)\left( \frac{1}{2}\delta_{j_2}^{i_2}\right) \dotsm \left( \frac{1}{2}\delta_{j_k}^{i_k} \right) \\ & = -\frac{2^{1+1-k}}{n-2}\frac{(n-k+1)(n-k+2)\dotsm(n-k+k-1)}{(k-1)!}\delta_i^j\left(v_j^i + v\delta_j^i \right) \\ & = -\frac{2^{2-k}}{n-2}\binom{n-1}{k-1}(\Delta v + n) . \end{align*} I’m not entirely sure why your expression has an extra factor of $\frac{n}{k}$.