Induced Second Fundamental Form for a Graph

Question

If one has a Riemannian $3$-manifold $(M,g)$ and a warped product space $(M \times\mathbb{R}, g + \phi^2 dt^2)$, the induced metric of a hypersurface $\Sigma$ given by a graph $t = f(x)$ is $g + \phi^2 df^2$, but in that case how does one define the induced second fundamental form of $\Sigma$?

Answer 1

Here's a sketch: further simplifications should be done as needed. You need to find a normal field to the hypersurface, as usual. Namely, if $$\Sigma = \{ (x,f(x)) \mid x \in M \},$$parametrize it with $\psi\colon M \to \Sigma \subseteq M \times_\phi \Bbb R$ given by $\psi(x) = (x,f(x))$, so $${\rm d}\psi_x(v) = v+ {\rm d}f_x(v)\partial_t$$ for all $v \in T_xM$, according to the decomposition $T_{(x,t)}(M\times_\phi \Bbb R) \cong T_xM \oplus \Bbb R\partial_t$. Let $N = N_0 + h\partial_t$, with $N_0$ tangent to $M$ and $h$ a smooth function on $M$, be a normal field. Being normal means that $$g_x(v, (N_0)_x) + \phi(x)^2 h(x){\rm d}f_x(v) = 0$$for all $v \in T_xM$. Rewrite this as $$g_x(v, (N_0)_x + \phi(x)^2 h(x)\,{\rm grad}_g f|_x) = 0,$$and use non-degeneracy of $g$ to see that a unit normal field along $\Sigma$ is $$N= \frac{-\phi^2{\rm grad}_gf + \partial_t}{\phi\sqrt{1+\phi^2\|{\rm d}f\|^2}}.$$The second fundamental form will be given by ${\rm II}(X,Y) = \langle A_N(X),Y\rangle N$, where $A_N(X) = - (\nabla_XN)^\top$ is the Weingarten operator associated to $N$. If $v = v+c\partial_t \in T_{(x,f(x))}\Sigma$, a straightforward computation gives that $$v^\top = v + (c-{\rm d}f_x(v))\frac{\phi^2\,{\rm grad}_gf}{1+\phi^2\|{\rm d}f\|^2} + \left(c+\frac{{\rm d}f_x(v)-c}{1+\phi^2\|{\rm d}f\|^2}\right)\partial_t.$$Writing $X = X_0+a\partial_t$, $Y = Y_0+b\partial_t$, with $a = X_0(f)$ and $b = Y_0(f)$, and $X_0, Y_0 \perp \partial_t$, applying the above formula for $v = -\nabla_XN$ gives $A_N(X)$, and then $\langle A_N(X),Y\rangle$ gives ${\rm II}$. One should expect ${\rm d}\phi\,{\rm d}f$ and ${\rm Hess}(f)$ in the final formula.