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

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.

, ,  Wolfgang Kühnel, "Differential Geometry Curves-Surfaces-Manifolds", Second Edition, American Mathematical Society, 2006.

• According to the first page of this article, it follows from an old theorem of Allendoerfer that "general" hypersurfaces are determined up to ambient isometry in dimensions $n -1\ge 3.$ For the case of hypersurfaces this "generality" condition probably reduces to some condition on the rank of $D\nu$ - I haven't worked through the details. Have a look at Vol 5, Ch12 of Spivak if you can. – Anthony Carapetis Aug 28 '17 at 14:44
• The first fundamental form determines the geometry of the hypersurface but not how it is immersed in the ambient space. Consider a part of a plane and a part of a cylinder. Both are hypersurfaces with the same first fundamental form (they are isometric). But they have different second fundamental form. – mfl Aug 28 '17 at 14:45
• @mfl: this kind of example is why some non-triviality assumption on the second fundamental form is essential. – Anthony Carapetis Aug 28 '17 at 23:55
• @AnthonyCarapetis Thanks for your useful comments. – user87128 Oct 13 '17 at 11:08
• @mfl Thanks for your useful comment. – user87128 Oct 13 '17 at 11:08

## 1 Answer

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.

• Thanks for your nice answer. – user87128 Oct 13 '17 at 11:07