# Nirenberg problem in conformal change

Let $$(\mathbb{S}^n,g_0)$$ be the standard sphere, $$n\geq 3$$, consider the Nirenberg problem$$-k(n) \Delta_{g_0} u+R_0 u=R u^{\frac{n+2}{n-2}}, \quad u>0\,\text{ on }\, \mathbb{S}^n,$$ where $$k(n)=\frac{4(n-1)}{n-2}$$ and $$R_0=n(n-1)$$ 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 $$T_{t}v:=v\circ\varphi_t|\det d\varphi_t|^{\frac{n-2}{2n}}$$ 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 $$\varphi_t^* (g_v) = g_{T_t v}$$. Is it true that $$-k(n) \Delta_{g_0}\left(T_\phi u\right)+R_0\left(T_\phi u\right)=(R \circ \phi)\left(T_\phi u\right)^{\frac{n+2}{n-2}}\quad \text{ on }\, \mathbb{S}^n?$$ 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.

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 $$L_2^g = -\Delta + \frac{n-2}{4(n-1)}R .$$ (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 $$\label{2}\tag{2} L_2^{e^{2\Upsilon}g}(u) = e^{-\frac{n+2}{2}\Upsilon} L_2^g \left( e^{\frac{n-2}{2}\Upsilon}u \right)$$ 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 $$R=n(n-1)$$, 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{n-2}{2n}} (u\circ\Phi) \right) . \end{align*} In particular, if you set $$T_\Phi u := \lvert \det d\Phi \rvert^{\frac{n-2}{2n}} (u \circ \Phi)$$ and assume $$L_2^gu = \frac{n-2}{4(n-1)}fu^{\frac{n+2}{n-2}} ,$$ then you conclude that \begin{align*} \frac{n-2}{4(n-1)}(f \circ \Phi)(T_\Phi u)^{\frac{n+2}{n-2}} & = \frac{n-2}{4(n-1)}\lvert \det d\Phi \rvert^{\frac{n+2}{2n}} (fu^{\frac{n+2}{n-2}})\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.

• Good job! Many thans! Commented Apr 15, 2023 at 13:04
• @ 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!
– Medo
Commented Apr 25, 2023 at 15:45