When majority of authors textbook of differential geometry (like a O'Neill, Boothby, and some PDFs from Google) deal with a shape operator of $M$, I have never seen that the shape operator is defined on a manifold and solely have seen the surface $M$ in $\mathbb{R}^3$.
To begin with, The shape operator $S$ in a surface on $\mathbb{R}^3$ is $S :T_pM \to T_pM , v \mapsto S(v):=-\bigtriangledown _{v}N $(where $N$ is a unit normal vector at a point $p\in M$. in this case, the basis of unit normal vector is only one whenever we pick a point $p\in M$.
However, my thought -- the shape operator is defined only on a sufrace on $\mathbb{R}^3 $ -- would be wrong when I see the generalization of the second fundamental form on a Riemannian manifold.
Example) One defines a second fundamental form on a surface $M$ in $\mathbb{R}^3$ like :,
$II(v,w):=\langle S(v),w\rangle = \langle-\bigtriangledown _{v}N ,w \rangle$ (where $u,v$ in $T_pM$, and $\langle\cdot,\cdot\rangle$ is inner product)
Next, from the second fundamental form on a "$n$-dimensional Riemannian manifold $M$", I recognized that shape operator is also well defined on any Riemannian manifold (or hypersurface).
Here is my question about shape operator on Rimannian manifold : how to select a unit vector $N$ at $p \in M$ when we extend shape operator into a n-dimensional Riemannian manifold? Unlike a surface in $\mathbb{R}^3$, the basis of a unit normal vector at a point $p \in T_pM$ is not unique. However, anybody does not seem to consider the choice of unit normal vector. Based on this, the choice of unit normal vector does not seem to depend on the the result of shape of operator. More explicitly, if $N, N'$ are unit normal vector which of each is independent, then $\bigtriangledown _{v}N \overset{?}{=} \bigtriangledown _{v}N'$ ?? Here is my trouble when I admit the shape operator is well-defined on a Riemannian manifold.
