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Consider the tube of radius $a > 0$ around a unit-speed curve $\gamma$ in $\mathbb{R}^3$ $$\sigma (s, \theta) = \gamma (s) + a(\cos \theta \ n(s) + \sin \theta \ b(s))$$

Show that the parameter curves on the tube obtained by fixing the value of $s$ are circular geodesics on $\sigma$.

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Could you give me some hints how we could show that?

Do we maybe use the fact that any normal section of a surface is a geodesic?

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You can see geometrically that the normal at the surface at the point $\sigma(s,\theta)$ is the vector $N_\sigma(s,\theta) = n(s)\cos \theta + b(s)\sin \theta$. If $\alpha(\theta) = \sigma(s_0,\theta)$, then you can check that $\alpha$ is parametrized by arc-length, so it suffices to check that $\alpha''(\theta)$ is parallel to $N_\sigma(s_0,\theta)$ and you're done.

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    $\begingroup$ Parallel to the normal is closely related to perpendicular to the surface. $\endgroup$ – robjohn Jan 10 '16 at 19:08
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    $\begingroup$ Yes, sure. But noticing that the normal is given by that expression, we don't need to compute cross products. :) $\endgroup$ – Ivo Terek Jan 10 '16 at 19:10
  • $\begingroup$ How did you find the formula of the normal at the surface, $N_{\sigma}$ ? $\endgroup$ – Mary Star Jan 10 '16 at 19:14
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    $\begingroup$ Just think geometrically. The surface is constructed as following: at each point $\gamma(s)$, we consider the normal plane to the curve at that point. The surface consists of all the unit circles in the normal planes. We know that the normal vector to a unit circle in the plane is given by its position vector. The position vector of the unit circle in the normal plane to $\gamma$ at $\gamma(s)$ is given by $n(s)\cos \theta + b(s)\sin \theta$. $\endgroup$ – Ivo Terek Jan 10 '16 at 19:19
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    $\begingroup$ You conclude that they are parallel not because $\widetilde{\sigma}''\cdot N_\sigma = -a$, but because $\widetilde{\sigma}''(\theta) = -a N_\sigma(s_0, \theta)$. $\endgroup$ – Ivo Terek Jan 11 '16 at 0:13
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HINT:

Any curve of intersection produced by normal section of this argument cannot or should not be used for arriving at the result. One special case is a sphere cut along any great circle.

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  • $\begingroup$ Is a sphere cut along any great circle a special case where we can use that? $\endgroup$ – Mary Star Jan 10 '16 at 19:17
  • $\begingroup$ When tube and sphere have a common normal, yes. $\endgroup$ – Narasimham Jan 10 '16 at 19:32
  • $\begingroup$ Ok... Thanks!! :-) $\endgroup$ – Mary Star Jan 11 '16 at 9:57
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Hint: In general, if $\sigma'\times\left(\sigma'\times\sigma''\right)$ is perpendicular to the surface, then $\sigma$ is a geodesic. However, for a constant speed curve, we only need that $\sigma''$ is perpendicular to the surface since $$ a\times(a\times b)=\frac{b\cdot a}{a\cdot a}\,a-b $$ and $\sigma'\cdot\sigma''=0$ for a constant speed curve.


From Comments: Two surface tangents are $$ \partial_s\sigma=t+a(\cos(\theta)\,n'+\sin(\theta)\,t\times n')\tag{1} $$ and $$ \tilde\sigma'=\partial_\theta\sigma=a(-\sin(\theta)n+\cos(\theta)t\times n)\tag{2} $$ Since $$ \tilde\sigma''=\partial_\theta^2\sigma=-a(\cos(\theta)n+\sin(\theta)t\times n)\tag{3} $$ is perpendicular to both, it is perpendicular to the surface. Therefore, $\tilde\sigma$ is a geodesic for each $s$.

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    $\begingroup$ So we have $$\tilde{\sigma}(\theta)=\sigma (s_0, \theta)$$ Do we have to show that $\tilde{\sigma}'\times (\tilde{\sigma}'\times\tilde{\sigma}'')$ is perpendicular to the surface? $\endgroup$ – Mary Star Jan 10 '16 at 18:56
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    $\begingroup$ Since this is a constant speed curve, we only need to show that $\tilde\sigma''$ is perpendicular to the surface. $\endgroup$ – robjohn Jan 10 '16 at 18:58
  • $\begingroup$ We have that $$\tilde{\sigma}'(\theta)=\frac{d}{d\theta}\sigma (s_0, \theta) = a(-\sin \theta \ n(s) + \cos \theta \ b(s)) \Rightarrow \|\tilde{\sigma}'\|^2=\|a(-\sin \theta \ n(s) + \cos \theta \ b(s))\|^2=a^2=\text{ constant } $$ right ? To show that $\tilde{\sigma}''$ is perpendicular to the surface do we have to show that $\tilde{\sigma}''\cdot \sigma=0$ ? $\endgroup$ – Mary Star Jan 10 '16 at 19:10
  • $\begingroup$ Or isn't $\tilde{\sigma}''\cdot \sigma=0$ the condition that should be satisfied? $\endgroup$ – Mary Star Jan 10 '16 at 23:33
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    $\begingroup$ Two surface tangents are $(t+a(\cos(\theta)\,n'+\sin(\theta)\,t\times n'))$ and $a(-\sin(\theta)n+\cos(\theta)t\times n)$. Since $\sigma''=-a(\cos(\theta)\,n+\sin(\theta)\,t\times n)$ is perpendicular to both, it is perpendicular to the surface. $\endgroup$ – robjohn Jan 11 '16 at 0:34

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