Let $(\mathbb{S}^n,g_0)$ be the standard sphere, $n\geq 3$, consider the Nirenberg problem$$ k(n) \Delta_{g_0} u+R_0 u=R u^{\frac{n+2}{n2}}, \quad u>0\,\text{ on }\, \mathbb{S}^n, $$ where $k(n)=\frac{4(n1)}{n2}$ and $R_0=n(n1)$ is the scalar curvature of $g_0$. Let $t \in \mathbb{R}$, 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 $$ T_{t}v:=v\circ\varphi_t\det d\varphi_t^{\frac{n2}{2n}} $$ where $d\varphi_t$ denotes the Jacobian of $\varphi_t$. In particular, the pullback metric of $g_v = v^{\frac{4}{n2}}g_0$ under $\varphi_t$ is given by $\varphi_t^* (g_v) = g_{T_t v}$. Is it true that $$ k(n) \Delta_{g_0}\left(T_\phi u\right)+R_0\left(T_\phi u\right)=(R \circ \phi)\left(T_\phi u\right)^{\frac{n+2}{n2}}\quad \text{ on }\, \mathbb{S}^n? $$ I don't konw how to verify the above equation since the $ \det d\varphi_t$ term in $\Delta_{g_0}$ is hard to calculate, Thanks a lot for any help.
1 Answer
Yes, it is true. This is a consequence of the conformal invariance of the conformal Laplacian.
Let $(M^n,g)$ be a Riemannian manifold. The conformal Laplacian is $$ L_2^g = \Delta + \frac{n2}{4(n1)}R . $$ (I use the convention $\Delta \geq 0$.) We require two properties of this operator.
First, it is natural: If $\Phi \colon M \to M$ is a diffeomorphism, then $$ \label{1}\tag{1} \Phi^\ast (L_2^gu) = L_2^{\Phi^\ast g}(\Phi^\ast u) $$ for all $u \in C^\infty(M)$, where $\Phi^\ast f = f \circ \Phi$ is the pullback of a function and $\Phi^\ast g$ is the pullback of the initial metric.
Second, it is conformally covariant: If $\Upsilon \in C^\infty(M)$, then $$ \label{2}\tag{2} L_2^{e^{2\Upsilon}g}(u) = e^{\frac{n+2}{2}\Upsilon} L_2^g \left( e^{\frac{n2}{2}\Upsilon}u \right) $$ for all $u \in C^\infty(M)$.
Now let’s specialize to your setting of the sphere $(S^n,g)$ with its metric of constant sectional curvature one. Then $R=n(n1)$, so $L_2^g$ is (proportional to) the operator you wrote down. Let $\Phi \colon S^n \to S^n$ be a conformal diffeomorphism; i.e. $\Phi^\ast g = e^{2\Upsilon}g$ for some $\Upsilon \in C^\infty(M)$. (Your $\phi_t$ is one example of a conformal diffeomorphism.) Then the volume element $\mathrm{dv}$ transforms by $$ \label{3}\tag{3} \mathrm{dv}_{\Phi^\ast g} = e^{n\Upsilon}\mathrm{dv}_g , $$ and so $\lvert \det d\Phi \rvert = e^{n\Upsilon}$. Combining Equations \eqref{1}, \eqref{2}, and \eqref{3} yields \begin{align*} (L_2^gu) \circ \Phi & = L_2^{\Phi^\ast g}(\Phi^\ast u) \\ & = \lvert \det d\Phi \rvert^{\frac{n+2}{2n}} L_2^g \left( \lvert \det d\Phi \rvert^{\frac{n2}{2n}} (u\circ\Phi) \right) . \end{align*} In particular, if you set $T_\Phi u := \lvert \det d\Phi \rvert^{\frac{n2}{2n}} (u \circ \Phi)$ and assume $$ L_2^gu = \frac{n2}{4(n1)}fu^{\frac{n+2}{n2}} , $$ then you conclude that \begin{align*} \frac{n2}{4(n1)}(f \circ \Phi)(T_\Phi u)^{\frac{n+2}{n2}} & = \frac{n2}{4(n1)}\lvert \det d\Phi \rvert^{\frac{n+2}{2n}} (fu^{\frac{n+2}{n2}})\circ \Phi \\ & = \lvert \det d\Phi \rvert^{\frac{n+2}{2n}} (L_2^gu) \circ \Phi \\ & = L_2^g(T_\Phi u) . \end{align*} This is precisely your equation.


$\begingroup$ @ Davidi Cone If you find an answer to your question correct and useful, you should accept it by clicking on the "check " sign. I think your questions are interesting. Good Luck! $\endgroup$– MedoCommented Apr 25, 2023 at 15:45