# Differential geometry - proving an expression for the principal curvature

I have tried to solve the following problem for some time but cannot get it right.

Let $X: U \rightarrow \mathbb{R}^{3}$ be a regular parametrized surface in $\mathbb{R}^{3}$ with Gauss map $N: M\rightarrow S^{2}$ and principal curvatures $\kappa_{1} = \frac{1}{r_{1}}$ and $\kappa_{2} = \frac{1}{r_{2}}$, respectively. Let $r \in \mathbb{R}$ be such that $X^{r}(u,v): U \rightarrow \mathbb{R}^{3}$ with

$X^{r}(u,v) = X(u,v) + rN(u,v)$

is a regular parametrized surface in $\mathbb{R}^{3}.$ Prove that the principal curvatures of $X^{r}$ satisfy

$\kappa_{1} = \frac{1}{r_{1} - r }$ and $\kappa_{2} = \frac{1}{r_{2} - r }$

My approach:

Let $\gamma(t)$ be the curve parametrized by arclength in $X(U)$ such that $\dot{\gamma}(t_{0}) = t_{1}$ where $t_{1}$ is an eigenvector for the shape operator with eigenvalue $\kappa_{1}$.

Now, $X^{r}(U)$ has the same Gauss map as $X(U)$, this can be seen by taking the scalar products $\langle X^{r}_{u}, N \rangle$ and $\langle X^{r}_{v}, N \rangle$ and using that $\frac{d}{dv} \langle N, N \rangle = 2\langle N_{v}, N\rangle = 0 = 2\langle N_{u}, N \rangle$. Since both scalar products are zero, the normal N is orthogonal to both $X^{r}_{u}$ and $X^{r}_{v}$.

Hence, the normal map is the same and the shape operator has the same eigenvectors.

Now define $\gamma_{r}(t) = \gamma(t) + rN(\gamma(t))$, which is not parametrized by arclength. Let $t = \phi(s)$ be a reparametrization such that $\tilde{\gamma}_{r}(s)$ is unit-speed, and $\phi(s_{0}) = t_{0}$.

Then we can calculate $\kappa_{1}^{r}$ by taking the scalar product $\langle -dN \dot{\tilde{\gamma}_{r}}(s),\dot{\tilde{\gamma}_{r}}(s)\rangle$, which gives me when evaluated at $s_{0}$,

$\langle -dN (\dot{\gamma}_{r}(\phi(s_{0}))\frac{d\phi}{ds}), \dot{\gamma}_{r}(\phi(s_{0}))\frac{d\phi}{ds} \rangle$ $=$

$\langle -dN (\dot{\gamma}(\phi(s_{0}))\frac{d\phi}{ds}+r\dot{N}(\gamma(\phi(s)))\dot{\gamma}(\phi(s))\frac{d\phi}{ds}), \dot{\gamma}(\phi(s_{0})\frac{d\phi}{ds}+r\dot{N}(\gamma(\phi(s)))\dot{\gamma}(\phi(s))\frac{d\phi}{ds})\rangle =$

$(\frac{d\phi}{ds})^{2} \langle-dN (t_{1} -r\kappa_{1}t_{1}),t_{1} - r\kappa_{1}t_{1} \rangle =$

$(\frac{d\phi}{ds})^{2}(1-r\kappa_{1})^{2}\kappa_{1} = \kappa_{1}$

Since we want $\gamma_{r}(s)$ at unit-speed, I get $|\dot{\tilde{\gamma}}_{r}(s)| = |\dot{\gamma_{r}}(\phi(s))\frac{d\phi}{ds}| = |\dot{\gamma}(\phi(s))\frac{d\phi}{ds}(1-r\kappa_{1})|$, so I conclude that $\frac{d\phi}{ds} = \frac{1}{1-r\kappa_{1}}$.

I don't really see where I go wrong.

Edit: Shall it be $\langle -d\tilde{N} \dot{\tilde{\gamma}_{r}}(s),\dot{\tilde{\gamma}_{r}}(s)\rangle$, instead because of the reparametrization, such that $\langle -d\tilde{N}(\dot{\tilde{\gamma}}_{r}), \dot{\tilde{\gamma_{r}}} \rangle = \langle -dN(\dot{\gamma}_{r} \frac{d\phi}{ds})\frac{d\phi}{ds}, \dot{\gamma}_{r}\frac{d\phi}{ds} \rangle$?

Any help and clarification would be appreciated. A solution would also help me to get thins right.

/ Erik

Here's a less confounding way to approach the problem. Assume that our original parametrization $X(u,v)$ has the property that the $u$-curves and $v$-curves are both lines of curvature. (This can always be arranged away from umbilic points.) Then we have $N_u = -\kappa_1 X_u$ and $N_v = -\kappa_2 X_v$. Now, $X^r_u = X_u + rN_u = (1-r\kappa_1)X_u$ and $X^r_v = X_v + rN_v = (1-r\kappa_2)X_v$. But then we have $$N_u = -\frac{\kappa_1}{1-r\kappa_1}X^r_u \qquad\text{and}\qquad N_v = -\frac{\kappa_2}{1-r\kappa_2}X^r_v\,.$$ This tells us that the principal curvatures of the parallel surface are now $$\kappa^r_1 = \frac{\kappa_1}{1-r\kappa_1} = \frac1{r_1-r} \qquad\text{and}\qquad \kappa^r_2 = \frac{\kappa_2}{1-r\kappa_2} = \frac1{r_2-r}\,.$$

ADDED: Here's the way to make your shape operator approach work. Let $\upsilon = ds/dt = 1-r\kappa$ denote the speed of $\gamma_r$. As you observed, we have $N(\gamma_r(t)) = N(\gamma(t))$ for all $t$. Differentiating this, we have $$dN_{\gamma_r(t)}\dot\gamma_r(t) = dN_{\gamma(t)}\dot\gamma(t)\,.$$ Noting that $\dot\gamma_r = \upsilon\dot\gamma$ (so $\dot\gamma$ is the unit tangent vector to $\gamma_r$) and dotting with $\dot\gamma$, we have $$\kappa^r = \langle - dN_{\gamma_r}\dot\gamma,\dot\gamma\rangle = \langle -\frac1{\upsilon}dN_{\gamma_r}\dot\gamma_r,\dot\gamma\rangle = \langle -\frac1{\upsilon}dN_{\gamma}\dot\gamma,\dot\gamma\rangle = \frac1{\upsilon}\kappa\,.$$ That is, $\kappa^r = \dfrac{\kappa}{1-r\kappa}$, as desired.

• Thank you, when we solved this in class, the solution was to do a reparametrization so I tried to remake it. I suspect that when you do a reparametrization you are also changing the shape operator with a scalar, can that be the case? Nov 4 '13 at 16:13
• @Erik: See my final edit for the proof you wanted. Nov 4 '13 at 21:46
• Are these Bertrand parallel surfaces with r as constant separation along common normals? Sep 5 '14 at 9:28
• @Narasimham: I've never heard parallel surfaces referred to as such. Yes, these are parallel surfaces. They are somewhat like the Bertrand mates for curves, so perhaps that's where you got your terminology. Sep 5 '14 at 16:25
• These things are called "offset surfaces" in some fields. Feb 15 '18 at 9:52

Still trying to figure this out. If $r,r_1, r_2$ are scalars, is it required to prove:

$r_1-r = r_1 ; r_2-r = r_2 ; r = 0 ?$

As it is not so, perhaps it may needs to prove for a parallel tangent plane at distance d along normal:

$r_1-d = r_1' ; r_2-d = r_2' ;$