Trying to get the idea of what the second fundamental form of a hypersurface is Let $M \subset \mathbb{R}^n$ be a submanifold of codimension $1$. Suppose it is given by $M = f^{-1}(\{ 0\})$ for some $f: \mathbb{R}^n \to \mathbb{R}$. I was wondering if someone could give me an idea of what exactly the second fundamental form of $M$ measures, in particular what does it mean if the second fundamental form vanishes identically on $T_p M$ for some $p \in M$?
I vaguely understand that it measures curvature and I had the following idea which is possibly incorrect.
Suppose $\gamma: [-1, 1] \to M$ be a smooth function such that $\gamma(0) = p$.  Is it then true that:
the second fundamental form vanishes identically on $T_p M$ if and only if given any such $\gamma$ the second derivative of $f \circ \gamma(t)$ vanishes at $t =0$?
I would appreciate if someone could explain me the intuition behind the second fundamental form, and also if I my idea above is incorrect or not.. Any comments appreciated. Thank you!
 A: The second fundamental form of a hypersurface refers to different -equivalent- notions: sometimes it refers to a scalar bilinear form, sometimes, to an endomorphism, and sometimes, to a vector valued functions on the tangent space. These notions are equivalent but some authors chose to talk only about one or another. I will here present the idea behind the endomorphism notion.
Suppose $M \subset \mathbb{R}^n$ is an orientable hypersurface, and let $\nu$ be a normal vector field, that is a -smooth- function $\nu : p \in M \mapsto \nu(p) \in \mathbb{R}^n$ such that $\|\nu(p)\| = 1$ and $\nu(p) \perp T_pM$.
Suppose $M$ is just a linear -or affine- hyperplane. Then the function $\nu$ is constant, equal to the normal of the hyperplane. Thus, there is no local change in $\nu(p)$.
Suppose $M$ is the unit sphere. Then $\nu(p)=p$ because the normal of a point in the unit sphere is directed by this point. Locally, $p \mapsto \nu(p)$ has the same variation than $p$.
These two examples shows that the infinitesimal change in $\nu(p)$ gives some information on the local geometry of $M$: if $M$ is flat around $p$ (like an hyperplane), $\nu$ does not change around $p$, and if $M$ is convex (like a sphere) around $p$, $\nu$ changes.
The second fundamental form, defined in terms of endomorphisms and usually denoted by $S$, and also known as the shape operator or weingarten operator, is the derivative of the function $p\mapsto \nu(p)$. More precisely, if $\nabla$ denotes the covariant derivative -or Levi-Civita connexion- of $M$, and if $p\in M$, then $S(p)$ is defined to be the linear transform $v \in T_pM \mapsto \nabla_{v}\nu(p) \in T_pM$. In terms of derivation along a curve, if $\gamma : I \to M$ is a curve such that $\gamma(0) = p$ and $\gamma'(0) = v$, then
$$ S(p)v = {\nu(\gamma)'(0)}^{\perp}$$
where $\perp$ refers to the orthogonal projection onto $T_pM$ in $\mathbb{R}^n$.
Geometrically, $S(p)v$ measures how much the hypersurface $M$ is curved at $p$ in the direction of $v$. If it is vanishing, then $M$ looks flat at $p$ in the direction of $v$.
A property of $S$ is that it is a symmetric endomorphism of $T_pM$, that is $\langle S(p)v,w\rangle = \langle v,S(p)w\rangle$. Hence, $S(p)$ is a diagonalisable endomorphism with othonogonal eigenvectors: this says that there exist orthogonal eigen-directions in $T_pM$ for $S(p)$. They are the principal directions in which $M$ is curved at $p$. Moreover, if $S(p)$ is symmmetric -definite- positive, then $M$ is -strictly- convex at $p$.
The bilinear definition of the second fundamental form is the bilinear form associated to the symmetric endomorphism $S$, that is $II_p(u,v) = \langle S(p)u,v\rangle$. They have the exact same properties.
Note that $S$ is an extrinsic property of $M$: as an abstract manifold in itslef, $M$ does not have any notion of shape operator at all. It depends on the geometry of the ambiant space (here $\mathbb{R}^n$). One can remember this by remembering that $S$ is defined by a normal vector field on $M$ and then, refers to an ambiant space in which $M$ is embedded.
Edit There was an edit in the question stating that $M$ is defined to be the zero level-set of a smooth function $f : \mathbb{R}^n \to \mathbb{R}$, and asking if the second fundamental form $S$ is zero at $p\in M$ if and only if for every smooth path $\gamma : I \to M$ with $\gamma(0)=p$ one has $\left(f\circ \gamma\right)''(0)=0$. The answer is no for the simple reason that if $\gamma : I \to M$ is a smooth function, then $f\circ \gamma : I \to \mathbb{R}$ is always the null function (because $M = f^{-1}(\{0\})$!). Consequently this condition is always true.
For example, $\mathbb{S}^n \subset \mathbb{R}^{n+1}$ is defined as $f^{-1}(\{0\})$ with $f(x) = \|x\|^2 -1$. Any path $\gamma : I \to \mathbb{S}^n$ has $\|\gamma\|^2 = 1$ by definition, thus $f\circ\gamma$ is the constant null function.
