# Curvature With Parallel Surfaces

• Aug 11th 2010, 11:04 PM
lllll
Curvature With Parallel Surfaces
I have a two part question, one of which I have no clue of how to approach, the other has been giving me a difficult time for the last three days. The questions are as follow:

Le M be an arbitrairy orientable surface, with $U$ the unit normal vector. Define the following map: $F: M \mapsto \mathhbb{R}^3$ by $F(p)=p+\varepsilon U(p)$

i) Show that the canonical isomorphisms of $\mathhbb{R}^3$ make $U$ a unit normal on $\overline{M}=F(M)$ for which $\overline{S}(\overline{v})=S(v)$, where $v$ are vectors, and $S(v)$ indicates the shape operator.

I have no idea what's being asked here.

ii)Derive the following formulas of the Gaussian and Mean Curvature

$\overline{K}(F) = \frac{K}{J}$ where $J = (1-\varepsilon k_1)(1-\varepsilon k_2)$ and $k_i$ denote the principal vectors

Before I begin I managed to deduce from a previous problem that $(\overline{v})\times(\overline{w}) = J(p)v\times w$.

Now I would imagine that since $K=det(k_1,k_2)$ that I would simply have $F(k_1)\times F(k_2)$ wich would simply give $J(p)k_1 \times k_2$ which isn't the desired result.

So looking at both RHS and LHS I think I would get:

$F(k_1)\times F(k_2) = (k_1+\varepsilon k_1) \times (k_2+\varepsilon k_2)= k_1k_2(1-\varepsilon)^2$

then from the RHS I get :

$\frac{k_1k_2}{(1-\varepsilon k_1)(1-\varepsilon k_2)}$ which has lead nowhere.

I'm not looking for the answer, but any hints would be greatly appreciated.
• Aug 12th 2010, 04:08 AM
HallsofIvy
Quote:

Originally Posted by lllll
I have a two part question, one of which I have no clue of how to approach, the other has been giving me a difficult time for the last three days. The questions are as follow:

Le M be an arbitrairy orientable surface, with $U$ the unit normal vector. Define the following map: $F: M \mapsto \mathhbb{R}^3$ by $F(p)=p+\varepsilon U(p)$

i) Show that the canonical isomorphisms of $\mathhbb{R}^3$ make $U$ a unit normal on $\overline{M}=F(M)$ for which $\overline{S}(\overline{v})=S(v)$, where $v$ are vectors, and $S(v)$ indicates the shape operator.

I have no idea what's being asked here.

So F "moves" M a short distance parallel to itself. What are "the canonical isomorphisms of $\mathhbb{R}^3$" and what is "the shape operator"?

Quote:

ii)Derive the following formulas of the Gaussian and Mean Curvature

$\overline{K}(F) = \frac{K}{J}$ where $J = (1-\varepsilon k_1)(1-\varepsilon k_2)$ and $k_i$ denote the principal vectors

Before I begin I managed to deduce from a previous problem that $(\overline{v})\times(\overline{w}) = J(p)v\times w$.

Now I would imagine that since $K=det(k_1,k_2)$ that I would simply have $F(k_1)\times F(k_2)$ wich would simply give $J(p)k_1 \times k_2$ which isn't the desired result.

So looking at both RHS and LHS I think I would get:

$F(k_1)\times F(k_2) = (k_1+\varepsilon k_1) \times (k_2+\varepsilon k_2)= k_1k_2(1-\varepsilon)^2$

then from the RHS I get :

$\frac{k_1k_2}{(1-\varepsilon k_1)(1-\varepsilon k_2)}$ which has lead nowhere.

I'm not looking for the answer, but any hints would be greatly appreciated.
• Aug 12th 2010, 05:13 AM
xxp9
The first question is to prove that M and $\overline{M}$ have the same normal vectors, that is, $\overline{U} \circ F=U$. Note that the differential of F,
$d\mathbf{F}(v) = v + \varepsilon \cdot d\mathbf{U}(v)$, where v is any vector tangent to M at p.
Also note that U is nothing but the Gauss map
$\mathbf{U} : M \mapsto S^2$
Thus $-d\mathbf{U}$ is the Weingarten map $\mathbf{S}$.

Identifying the tangent plane of M at $\mathbf{p}$ and the tangent plane of $S^2$ at $\mathbf{U}(\mathbf{p})$( since they're parallel), we can see that
$d\mathbf{U}(v)=-\mathbf{S}(v)$ is tangent to M.
Thus $d\mathbf{F}(v)$ is orthogonal to U(p), we proved that the tangent plane of $\overline{M}$ is orthogonal to U(p).
We're done for i).
I'm going to prove ii) in another post.
• Aug 12th 2010, 06:02 AM
xxp9
To see ii), note that for the two principle directions $v_1$ and $v_2$,
$\mathbf{S}(v_1) = k_1 \cdot v_1$ and $\mathbf{S}(v_2) = k_2 \cdot v_2$, where $k_1$ and $k_2$ are the principle curvatures.
So $d\mathbf{F}(v_1) = v_1 - \varepsilon \mathbf{S}(v_1) = (1-\varepsilon k_1)\mathbf{v_1}$
$d\mathbf{F}(v_2) = (1-\varepsilon k_2)\mathbf{v_2}$
Since $v_1$ and $v_2$ are orthogonal we can easily deduce that the pull back of the volume element
$\mathbf{F}^{*}(dV_{\overline{M}}) = (1-\varepsilon k_1)(1-\varepsilon k_2)dV_{M}$, where $dV_{\overline{M}}$ and $dV_M$ are the volume elements of $\overline{M}$ and M, respectively.
Note that the Gaussian curvature $K=\frac{\mathbf{U}^*(d\sigma)}{dV_M}$, where $d\sigma$ is the volume element of $S^2$ and the similar formula for $\overline{M}$, and $\overline{U} \circ F = U$ we're done.