The second fundamental form and isometry.

Question

What is the effect on the second fundametal form of asurface of applying an isometry of $\Bbb R^3$ ? Or a dilation?

I posted its answer.

This answer is not understandable for me in general. Please is someone explains it, I'll be happy. Thank you.

Answer 1

Let $r=r(u,v)$ be a parametrization of a given surface and let $n=n(u,v)$ denote the normal vector. We use the notation

$$II:=Ldu^2 + 2Mdudv + Ndv^2$$ where $L = r_{uu}\cdot n$; $M = r_{uv}\cdot n$; $N = r_{vv}\cdot n$ for the second fundamental form.

In this answer I proved that the first fundamental form is invariant under $r\mapsto r':= Or+b$, where $O$ is a $3\times 3$ orthogonal matrix and $b$ real. Note that $r'=r'(u,v)$. and

$$r'_j=\sum_{k=1}^3 O_{jk}r_k+b_j, $$

for all $j=1,2,3$.

Then $r'_u:=\frac{\partial r'}{\partial u}=O\frac{\partial r}{\partial u}$ and $r'_v:=\frac{\partial r'}{\partial v}=O\frac{\partial r}{\partial u}$, as $O$ does not depend on $(u,v)$; similarly $r'_{uu}=r_{uu}$ and $r'_{vv}=r_{vv}$. The normal vector w.r.t. $r'$ is then

$$n'=r'_u \times r'_v=\pm n,$$

as proved in the original reference (prop. A.1.6). It follows that

$$L'= r'_{uu}\cdot n'=\pm r_{uu}\cdot n=\pm L,$$ $$M' = r'_{uv}\cdot n'= \pm r_{uv}\cdot n=\pm M,$$

and similarly for the remaining coefficient of the second fundamental form, i.e. $N$.

The case of dilations is treated analogously. Please check this answer again for the explicit computations.

Answer 2

For the second part of the question, to understand the effect on the second fundamental form of applying a dilation, lets consider a hypersurface in $\mathbb{R}^n$ with the metric induced by the usual inner product $\langle \cdot , \cdot \rangle$.

Then we have a parameterisation $F$ of a surface in $\mathbb{R}^n$, and the dilation $\tilde{F} = \psi F$, where $\psi$ is a constant (independent of the spacial coordinates). We then notice that

\begin{equation} \frac{\partial \tilde{F}}{\partial x_i} = \psi\frac{\partial F}{\partial x_i}, \quad \mbox{and}\quad \frac{\partial^2 \tilde{F}}{\partial x_j\partial x_i} = \psi\frac{\partial^2 F}{\partial x_j\partial x_i}, \end{equation}

from which follows \begin{equation} \tilde{g}_{ij} = \left\langle \frac{\partial \tilde{F}}{\partial x_i} , \frac{\partial \tilde{F}}{\partial x_j} \right\rangle = \left\langle \psi \frac{\partial F}{\partial x_i} ,\psi \frac{\partial F}{\partial x_j} \right\rangle = \psi^2 g_{ij}. \end{equation}

Since the unit normal vector $\nu$ of the hypersurface $F$ is also a normal unit vector for the hypersurface $\tilde{F}$, we compute the second fundamental form as \begin{equation} \tilde{h}_{ij} = \left\langle \frac{\partial^2 \tilde{F}}{\partial x_j\partial x_i} , \nu\right\rangle = \psi \left\langle \frac{\partial^2 F}{\partial x_j\partial x_i} , \nu\right\rangle = \psi h_{ij}. \end{equation}