Suppose $\tilde{M} \subset M$ is a hypersurface sitting inside a Riemannian manifold $(M,g)$. The second fundamental form of $M$ evaluated on $u,v \in T_pM$ is denoted $II(u,v)$ and defined as the normal component of $\tilde{\nabla}_UV$ for some extensions $U$ and $V$ of $u$ and $v$. Define $l(u,v)$ by the relation $II(u,v) = l(u,v)\nu$ for $\nu$ a fixed unit normal vector field of $M$ around $p$.
I recently came across the following exercice : The hypersurface $M$ is said to be totally umbilical if $l = fg$ for $f \in C^{\infty}(M)$ and $g$ the restriction of the metric to $M$. It is asked to show that if $\dim M \geq 2$, then $f$ must be a constant.
The book says it is a consequence of the Gauss-Codazzi equation $$\tilde{R}(x,y,u,\nu) = \nabla_xl(y,u) - \nabla_yl(x,u).$$ I tried playing around with this equation, taking traces hoping to find a Bianchi identity, but didn't have any success. Can anyone help me out?
(1) If $\widetilde{M}={\bf R}^{n+1}$, then from Codazzi, we have $$ x(f) g(y,z) - y(f) g(x,z) =0$$
If $y=z\perp x$, then $x(f)=0$ That is $f$ is constant
(2) If $\widetilde{M}$ is general, then we have an counter example : $\widetilde{M}=({\bf R}^2\times {\bf R}, G:=F(x_3)^2g_1(x_1,x_2)+ g_2(x_3)),\ M={\bf R}^2$ Here $g_i$ are canonical metric on each factor
If $E_i$ is coordinate vector field, then $\nu :=E_3 $ are orthonormal And $$ \Gamma_{13}^1 = \frac{1}{2} G^{11} \{ G_{11,3} + G_{13,1} - G_{13,1} \} = \frac{F_3 }{F} = \Gamma_{23}^2 $$ $$ \Gamma_{13}^2 = \frac{1}{2} G^{22} \{ G_{12,3} + G_{23,1} - G_{13,2}\} =0 $$ So $$ l(E_1,E_1)=(\nabla_1 E_3, E_1)=F^2\Gamma_{13}^1 =FF_3= l (E_2,E_2) $$ And $$ l (E_1,E_2)=(\nabla_1 E_3,E_2)= F^2\Gamma_{13}^2 = 0 $$ Hence $$ f:=\frac{F_3}{F} $$