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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

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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.

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  • $\begingroup$ 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? $\endgroup$ – El_Loco Nov 4 '13 at 16:13
  • $\begingroup$ @Erik: See my final edit for the proof you wanted. $\endgroup$ – Ted Shifrin Nov 4 '13 at 21:46
  • $\begingroup$ Are these Bertrand parallel surfaces with r as constant separation along common normals? $\endgroup$ – Narasimham Sep 5 '14 at 9:28
  • $\begingroup$ @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. $\endgroup$ – Ted Shifrin Sep 5 '14 at 16:25
  • $\begingroup$ These things are called "offset surfaces" in some fields. $\endgroup$ – bubba Feb 15 '18 at 9:52
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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' ; $

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