The first fundamental form is enough to determine the geometry of a hypersurface?

Question

Question:

As we know, For two dimensional surfaces there are many examples for which their first fundamental forms are the same, but their second fundamental forms are not.

However, it seems that for hypersurfaces (dimension $\ge 3$) their first fundamental forms are enough to determine the geometry of the hypersurfaces completely, provided that $Rank(L) \ge 3$, where $L$ is the shape operator. I want to know whether this is true or not. If the answer is yes, why?

Definitions:

The first fundamental form $I$ of a surface element is just the restriction of the Euclidean inner product in $\mathbb R^n$ to all tangent hyperplanes $T_uf$, i.e., $$I(X,Y):=\langle X,Y \rangle$$ for any two tangent vectors $X,Y \in T_uf$ or for vectors $X,Y \in \mathbb R^n$ which are tangent to the surface element$.^1$

Shape operator of a surface is the minus derivative of the unit normal vectors on the surface. Formally speaking, let $f:U \to \mathbb {R}^3$ be a surface element with unit normal vector map $\nu$, $\nu: U \to S^2$ is defined by $$\nu (u_1,u_2):=\frac{\frac{\partial f}{\partial u_1} \times \frac{\partial f}{\partial u_2}}{\left \Vert \frac{\partial f}{\partial u_1} \times \frac{\partial f}{\partial u_2} \right \Vert},$$

then for every $u\in U$ we have the linear map $$D\nu|_u:T_uU \to T_uf,$$ where $T_uU=\{u\} \times \mathbb R^2$ and $T_uf=Df|_u\left(T_uU\right)$, and $$Df|_u:T_uU \to T_uf$$ is a linear isomorphism. Then the shape operator $L:=-D\nu \circ (Df)^{-1}$ is defined pointwise by $$L_u:=-\left(D\nu|_u \right) \circ \left(Df|_u\right)^{-1}:T_uf \to T_uf\,.^2$$ The above definition can be easily generalized to the general $\mathbb R^n$ space.

Let $f:U \to \mathbb R^3$ be given. Then for tangent vectors $X$ and $Y$, one defines:

the second fundamental form $I\!I$ of $f$ by $$I\!I(X,Y):=I(LX,Y),$$ where $L$ is the shape operator$.^3$

The above definition can be easily generalized to the general $\mathbb R^n$ space.

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[1], [2], [3] Wolfgang Kühnel, "Differential Geometry Curves-Surfaces-Manifolds", Second Edition, American Mathematical Society, 2006.

Answer 1

I should read my own references - the first theorem in Spivak Vol 5 Ch 12 is exactly what you're asking for: $\def\2{I\!I}$

Theorem. Let $M, \bar M$ be immersed hypersurfaces in $\mathbb R^{n+1}$, and let $\phi: M \to \bar M$ be an isometry. Suppose that the shape operator $L : T_p M \to T_p M$ has rank at least $3$. Then $(\phi^* \bar\2)_p=\pm\2_p$, and consequently $\phi$ is the restriction of some rigid motion of $\mathbb R^{n+1}$.

Sketch of Proof. This is a pointwise fact, so we can fix an orthonormal basis for $T_p M$ so that $\2$ and $L$ are the same $n \times n$ matrix, and transfer this basis to $T_{\phi p} \bar M$ via $D\phi$ so that we don't have to worry about the pullback. Gauss's equation $$R_{ijkl} = \2_{ik}\2_{jl} - \2_{il}\2_{jk}$$ along with the fact that $R = \bar R$ (since $\phi$ is an isometry) tells us that the determinants of corresponding $2 \times 2$ submatrices of $\2, \bar \2$ are equal. This means precisely that the induced endomorphisms $L_*, \bar L_* : \Lambda^2 T_p M \to \Lambda^2 T_p M$ are equal.

Given any unit vector $v$ with $Lv \ne 0$, the rank assumption along with the symmetry of $L$ means we can find a three-dimensional subspace $V \ni v$ such that $L|_V$ is a isomorphism of $V$. Since $V$ is three-dimensional and $L|_V$ is invertible, ${L|_V}_* = {\bar L|_V}_*$ implies $L|_V= \pm \bar L|_V$ (see e.g. this question) and thus $Lv = \pm \bar Lv.$

For $Lv = 0$ we must thus have $\bar L v = 0$, since the equality of ranks of $L_*, \bar L_*$ implies equality of ranks of $L, \bar L$. Thus we conclude $L = \pm \bar L$ and thus $\2 = \pm \bar \2$ as desired.

Spivak gives a different proof of most of this that is probably worth reading - this is just how I convinced myself.