I'm working on problem 8.3 of Lee's Riemannian Manifold textbook: let $\Omega \subset R^{n+1}$ be an open set, $F:\Omega\rightarrow R$ a smooth submersion, and $M=F^{-1}(0)$. Show the scalar second fundamental form of M with respect to the unit normal vector field $N=grad F/\|grad F\|$ is given by $h(V,V)=-\frac{\partial_i\partial_jFV^{i}V^j}{\|gradF\|}$, where $V=V^i\partial_i$ in the Euclidean coordinates on $R^{n+1}$. (the original book had an error and the corrected form is given here).
I made two attempts. I wonder why they differ from each other AND the true answer:
#1
The scalar second fundamental form is the inner product between the second fundamental form and the unit normal. The definition of the second fundamental form: $II(V,V)=\tilde \nabla_VV-\nabla _VV$, where $\tilde \nabla$ is the connection in the ambient euclidean space and $\nabla$ on the manifold). If I take the inner product with the unit normal on both sides, since $\langle\nabla_VV,N\rangle=0$, I get the scalar second fundamental form $h(V,V)=\langle II(V,V),N\rangle=\langle\tilde\nabla_VV,N\rangle=\langle V(V^i)\partial_i,(\partial_iF)\partial_i/|gradF| \rangle =V(V^i)(\partial_iF)$. This is clearly not right so what's wrong?? Is expanding $gradF$ into $\partial_iF\partial_i$ right? Is doing the inner product this way (only taking the coefficients of $\partial_i$ on both terms and multiply them) correct?
#2 The other way is to use the Weingarten equation: $h(V,V)=\langle N,II(V,V)\rangle=-\langle\tilde\nabla_VN,N\rangle=-\langle V(N^j)\partial_j,V^j\partial_j\rangle=V^i\partial_i(\partial_jF/|gradF|)V^j$.
Now from here, if I treat $|gradF|$ as constant and take it outside of the $\partial_i$, then I would recover the solution. But I can't see a reason why that is the case. If I continue the derivation, I get: $(\partial_i\partial_jF / |grad F| + \partial_jF\partial_iF/|gradF|^3)V^iV^j$. So what's wrong in my derivation??
I would really appreciate any help!! Thanks!!
Given $X,Y$ tangent to $M = F^{-1}(0)$, we have that $D_XY = \nabla_XY+h(X,Y)N$, where $D$ and $\nabla$ are the Levi-Civita connections of $\Bbb R^{n+1}$ and $M$, and $N = {\rm grad}\,F/\|{\rm grad}\,F\|$.
However, we know that $D_XY = {\rm d}Y(X)$ and that $h(X,Y) = \langle D_XY, N\rangle$, as $N$ is a unit vector. Noting that $\langle D_XY,N\rangle = -\langle Y, D_XN\rangle$ as a consequence of $\langle Y,N\rangle = 0$, we compute $$D_XN = {\rm d}N(X) = {\rm d}\left(\frac{{\rm grad}\,F}{\|{\rm grad}\,F\|}\right)(X) = \frac{\|{\rm grad}\,F\| {\rm d}({\rm grad}\,F)(X) - {\rm d}(\|{\rm grad}\,F\|)(X) {\rm grad}\,F}{\|{\rm grad}\,F\|^2}.$$We must take the inner product of the above quantity with $Y$, but the second term in the numerator is orthogonal to $Y$. Simplifying, we have $$h(X,Y) = -\langle Y, D_XN\rangle = -\left\langle Y, \frac{{\rm d}({\rm grad}\,F)(X)}{\|{\rm grad}\,F\|} \right\rangle = -\frac{\langle Y,{\rm d}({\rm grad}\,F)(X)\rangle}{\|{\rm grad}\,F\|}.$$The numerator $\langle Y,{\rm d}({\rm grad}\,F)(X)\rangle = \langle Y, D_X({\rm grad}\,F)\rangle$ is the Hessian of $F$ evaluated at $X$ and $Y$, so that $$h(X,Y) = -\frac{({\rm Hess}\,F)(X,Y)}{\|{\rm grad}\,F\|},$$as claimed by Lee.