## Unit normal to a radial graph over sphere

Consider $\Omega\subset \mathbb{S}^n$ is a domain in the sphere. $S$ is a radial graph over $\Omega$.

$\boldsymbol{F}(x)=\{v(x)x:x\in \Omega\}$

What is the unit normal to this radial graph?

Suppose $\{e_1,\cdots,e_n\}$ is a smooth local frame on $\Omega$. Let $\nabla$ be the covariant derivative on $\mathbb{S}^n$. Tangent space of $S$ consists of $\{\nabla_{e_i}\boldsymbol{F}\}_{i=1}^n$ which are

$\nabla_{e_i}\boldsymbol{F}=v(x)e_i+e_i(v)\cdot x$

In order to get the unit normal, we need some simplification. Let us assume $\{e_i\}$ are orthonormal basis of the tangent space of $\Omega$ and $\nabla v=e_1(v)e_1$. Then

$\nabla_{e_1}\boldsymbol{F}=v(x)e_1+e_1(v)x, \quad \nabla_{e_i}\boldsymbol{F}=v(x)e_i, \quad i\geq 2$

Then we obtain an orthonormal basis of the tangent of $S$

$\{\frac{1}{\sqrt{v^2+|\nabla v|^2}}\nabla_{e_1}\boldsymbol{F},\nabla_{e_2}\boldsymbol{F},\cdots,\nabla_{e_n}\boldsymbol{F}\}$

We are able to get the normal by projecting $x$ to this subspace

$\nu=x-\frac{1}{v^2+|\nabla v|^2}\langle x,\nabla_{e_1}\boldsymbol{F}\rangle\nabla_{e_1}\boldsymbol{F}=\frac{v^2x-v\nabla v}{v^2+|\nabla v|^2}.$

After normalization, the (outer)unit normal can be written

$\frac{vx-\nabla v}{\sqrt{v^2+|\nabla v|^2}}$

Remark: Guan, Bo and  Spruck, Joel. Boundary-value Problems on \mathbb{S}^n for Surfaces of Constant Gauss Curvature.