Hypersurface in hyperbolic space and conformal metric on Sphere

Suppose \Omega is a domain in \mathbb{S}^n, g=e^{2\rho}g_0, where g_0 is the standard metric on \mathbb{S}^n. One can construct a hypersurface in hyperboloid \mathbb{H}^{n+1}\subset \mathbb{L}^{n+2} by

\phi(x)=\frac{1}{2}(1+\sigma^2+|\nabla^0\sigma|^2)\psi(x)-\sigma(0,x)-(0,\nabla^0 \sigma)

where \sigma=e^{-\rho} and \psi=e^{\rho}(1,x)=\sigma^{-1}(1,x). What are the induced metric on \phi? One can calculate as the following,

Suppose u\in T_x\Omega is a unit eigenvector associated to the eigenvalue s of Hess(\sigma)_x. We need to calculate d\psi_x(u). Firstly it is easy to see

d\psi_x(u)=-\sigma^{-1}u(\sigma)\psi+\sigma^{-1}(0,u)

and

d(\nabla^0 \sigma)_x(u)=\ \nabla_u^0\nabla^0 \sigma+(d \nabla^0\sigma_{x}(u))^{\perp}=Hess(\sigma)_x(u)+\langle d \nabla^0\sigma_{x}(u),x\rangle x
= \ su-\langle\nabla^0\sigma_{x},u\rangle x= \ su-u(\sigma)x

Then

d\phi_x(u)=u(\sigma)\sigma \psi+\frac{1}{2}u(|\nabla \sigma|^2)\psi+\frac{1}{2}(1+\sigma^2+|\nabla \sigma|^2)d\psi_x(u)-\sigma(0,u)-(0,su)

So this gives the following formula

d\phi_x(u)=\ \left(\frac{1}{2}-\lambda\right)d\psi_x(u).

where \lambda=s\sigma+\frac{1}{2}(\sigma^2-|\nabla^0\sigma|^2). Suppose u_i is the eigenvector corresponding to s_i and u_i,i=1,\cdots, n are orthonormal basis, then

\ll d\phi_x(u_i),d\phi_x(u_j)\gg=\left(\frac{1}{2}-\lambda_{i}\right)\left(\frac{1}{2}-\lambda_{j}\right)\frac{1}{\sigma^{2}}\delta_{ij}

where \lambda_i=s_i\sigma+\frac{1}{2}(\sigma^2-|\nabla^0\sigma|^2). Recall that Schouten tensor has the following formula under conformal change,

Sch_g=Sch_0+d\rho\otimes d\rho-\nabla^{2}_0\rho-\frac{1}{2}|\nabla^0\rho|^2g_0

=\frac{1}{2}g_0+\frac{1}{\sigma}\nabla_0^2\sigma-\frac{1}{\sigma^2}|\nabla_0\sigma|^2

for g=e^{2\rho}g_0 and Sch_0=\frac{1}{2}g_0. Therefore the eigenvalues of Sch_g are \lambda_i.

\ll d\phi,d\phi\gg=\left(\frac{1}{2}g-Sch_g\right)^2{\sigma^{2}}\delta_{ij}

\textbf{Remark:} I am using the thesis of Dimas Percy Abanto Silva and some communication with him.

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