I am trying to solve the following problem:

Show that a unit-speed curve $\gamma$ with nowhere vanishing curvature is a geodesic on the ruled surface $\sigma(u,v)=\gamma(u)+v\delta(u)$, where $\gamma$ is a smooth function of $u$, if and only if $\delta$ is perpendicular to the principal normal of $\gamma$ at $\gamma(u)$ for all values of $u$.

Edit (rather large): My professor wrote the question down wrong. I fixed it on here. Sadly, even with it right, I can't get either direction.

Any help would be appreciated. Thanks!


A unit-speed curve $\gamma(u)$ (i.e. parametrized by arc-length) is a geodesic in a surface $S$ iff $\gamma''(u)$ is perpendicular to $S$.

The normal to $\sigma(u,v)=\gamma(u)+v\delta(u)$ is parallel to $$ \frac{\partial\sigma}{\partial u}\times\frac{\partial\sigma}{\partial v} =(\gamma'+v\delta')\times\delta\tag{1} $$ On $\gamma$, $v=0$. Thus, $\gamma$ is a geodesic iff $\gamma''\times(\gamma'\times\delta)=0$. Using Lagrange's formula and the fact that $\gamma''\cdot\gamma'=0$, we get $$ \gamma''\cdot\delta\gamma'-\gamma''\cdot\gamma'\delta=0 \Leftrightarrow\gamma''\cdot\delta=0\tag{2} $$ Thus, $\gamma$ is a geodesic iff $\gamma''\cdot\delta=0$, where $\gamma''$ is the parallel to the principal normal since $\gamma$ is unit-speed.

  • $\begingroup$ How did you get the line "Thus, $\gamma$ is a geodesic iff $\gamma'' \times (\gamma'\times \delta)=0$"? Are you just setting $v=0$, and if so, why? $\endgroup$ – Samuel Reid Dec 5 '12 at 5:25
  • $\begingroup$ As I mentioned, on $\gamma$, the coordinate $v=0$. Thus, the normal to the surface at a point on $\gamma$ is $\gamma'\times\delta$. Thus, $\gamma''$ is perpendicular to the surface when $\gamma''\times(\gamma'\times\delta)=0$. $\endgroup$ – robjohn Dec 5 '12 at 5:32
  • $\begingroup$ Oh, that makes perfect sense. I never thought of that. Thank you! $\endgroup$ – Samuel Reid Dec 5 '12 at 5:48

are you enrolled in UofCalgary PMAT 423? I have the same question as you.

(=>) this is what I was thinking as well. Just remember that we're supposed to show that γ is perpendicular to the principal normal of γ, not to γ". Use γ" = kn (not N which is the standard unit normal of the surface).

(<=) here we have to use the idea that the principal normal of γ is perpendicular to γ at every point on γ. Since kn = t' = γ", where k is the curvature of γ (not the surface), this implies γ" is perpendicular to γ at every point which then implies γ" is perpendicular to γ'. A property of the cross product is that if A x B = C then C is perpendicular to A and to B so (N x γ') is perpendicular to γ'. Since Y' is perpendicular to γ" and γ" is parallel to N (because γ lies on the surface) then (N x γ') must be perpendicular to γ" as well. From this we get γ" • (N x γ') = 0 = kg. Since the geodesic curvature equals 0, then the surface must be geodesic.

I think this is the correct way to do this question but it both directions just seems too simple. If you have any ideas I'd love to here it.

  • $\begingroup$ Yeah I am! You're right about the principal normal thing, thanks. I'm not sure about your claim $\gamma"$ is parallel to $N$ (because $\gamma$ lies on the surface). Why would that necessarily be true? All we know is that $\gamma'' \perp \gamma'$. I guess if we wanted to show $\gamma'' \parallel N$, we could show $\gamma''\perp \sigma_u, \sigma_v$, but for that we need $\sigma_u\cdot\gamma''=0=v\gamma''\cdot\delta'$, and we know nothing about $\delta$. Likewise with $\sigma_v$. So that seems problematic to me... $\endgroup$ – Samuel Reid Dec 2 '12 at 22:31
  • $\begingroup$ Also, if you're in my class anyway, did you get 7.3.9? $\endgroup$ – Samuel Reid Dec 2 '12 at 22:37
  • $\begingroup$ IThat was the one point I wasn't entirely sure about. My reasoning behind it is that if γ lies on the boundary of the surface then γ and S would have parallel normals and γ lies on the ruled surface because it is a ruling. I included that part because I think it's possible that γ" and (N x γ') could be parallel without that but I could be wrong $\endgroup$ – Kevin Dec 2 '12 at 22:56
  • 1
    $\begingroup$ As for 7.3.9, we have γ(t), σ(u,v) and γ lies on the surface. Using u=v=t, set γ(t) = σ(t,t). Kn = γ" • N where N is the standard unit normal of σ. From there it's just a calculation question. When I hit it with a math stick I got -2/sqrt(5). $\endgroup$ – Kevin Dec 2 '12 at 23:09
  • 1
    $\begingroup$ let us continue this discussion in chat $\endgroup$ – Samuel Reid Dec 2 '12 at 23:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.