Let $\gamma_1,\gamma_2:I\to\mathbb{R}^3$ be a pair of regular space curves with the same domain, $I$. Define $\sigma:I\times(0,1)\to\mathbb{R}^3$ as $\sigma(t,u) = u\cdot\gamma_1(t)+(1-u)\cdot\gamma_2(t)$. Assume that $\gamma_1,\gamma_2$ are chosen such that $\sigma$ is regular. Prove that $K(t,1/2)=0$ for all $t\in I$.

In the above, $K$ denotes the Gaussian curvature. To solve the above problem, \begin{align} \sigma_t &= u\gamma_1'+(1-u)\gamma_2'\\ \sigma_u&= \gamma_1-\gamma_2\\ \sigma_t\times\sigma_u &= (u\gamma_1'+(1-u)\gamma_2')\times (\gamma_1-\gamma_2)\\ &= \frac{1}{2}(\gamma_1'+\gamma_2')\times (\gamma_1-\gamma_2)\\ &\text{in this case ($u=\frac {1}{2}$).}\\ \sigma_{uu} &= 0\\ \sigma_{tu} &= \gamma_1'-\gamma_2' \end{align}

So that we need to show $<\gamma_1'-\gamma_2',(\gamma_1'+\gamma_2')\times (\gamma_1-\gamma_2)> = 0$ (This is because, $K = \frac{eg-f^2}{EG-F^2}$ where $e,g,f$ from second fundamental form and $E,F,G$ form the first fundamental form. Since $\sigma_{uu} = 0, g = 0$ and now, to show $K = 0$ is equivalent to show $f = 0$.)

Here is where I'm stuck. Why is $<\gamma_1'-\gamma_2',(\gamma_1'+\gamma_2')\times (\gamma_1-\gamma_2)> =0$?

I used the notation in 'Differential geometry in curves and surfaces' by Tapp.

Definition of e,f,g

definition of E,F,G

  • $\begingroup$ I cleaned up your question a bit. You might want to do the same with the last few lines, to make it easier for us to help you. In particular, it's not clear why you want to show that final equality; also, the second to last sentence lacks an equation --- there's only an expression. $\endgroup$ – John Hughes Nov 15 '20 at 14:50
  • $\begingroup$ @JohnHughes Thank you for your input. I edit the post $\endgroup$ – love_sodam Nov 15 '20 at 15:39
  • $\begingroup$ Yes, use \langle \rangle for inner product. $\endgroup$ – Ted Shifrin Nov 15 '20 at 23:35

I do not believe the result. Think about joining two parallel congruent circles by lines to form a hyperboloid of one sheet. The midpoints of these line segments actually trace out the waist of the hyperboloid, which gives the points of maximum $|K|$. For the record, this surface has negative curvature everywhere.

This surface is particularly fascinating because it is in fact doubly-ruled (through each point there are two lines contained in the surface). Projectively, it's the only such surface other than a plane.

  • $\begingroup$ You mean the problem is wrong? I can't understand your comment $\endgroup$ – love_sodam Nov 16 '20 at 4:23
  • $\begingroup$ You can't understand “I do not believe the result” and a counterexample? $\endgroup$ – Ted Shifrin Nov 16 '20 at 5:07
  • $\begingroup$ What does the example you show suggest? You mean the problem is wrong? $\endgroup$ – love_sodam Nov 16 '20 at 5:13
  • $\begingroup$ Yes, the problem is wrong. Most any ruled surface gives you a counterexample. $\endgroup$ – Ted Shifrin Nov 16 '20 at 5:20
  • $\begingroup$ That's a great example! $\endgroup$ – Math Lover Nov 24 '20 at 13:51

Let's let $a(t) = \gamma_1(t) - \gamma_2(t), b(t) = \gamma_1(t) + \gamma_2(t)$.

Evidently $\gamma_1 = \frac12 (a(t) + b(t)), \gamma_2 = \frac12 (b(t) - a(t))$, so there no loss of generality in this renaming.

Now you're asking why, for any curves $a, b$, we have $$ \langle a', b' \times a \rangle = 0. $$

Let's look at $a(t) = (t, t^2, 0), b(t) = (0, 0, t)$. Then the expression above is \begin{align} \langle a', b' \times a \rangle &= \langle (1, 2t, 0), (0,0,1) \times (t, t^2, 0) \rangle \\ &= \langle (1, 2t, 0), (-t^2,t,0) \rangle \\ &= -t^2 + 2t^2 \end{align} which, for $t = 1,$ for instance, is not zero as desired.

So the thing you're trying to prove is false.

Personally, I suspect that the overall claim might be true, but my instincts aren't great. If they are true, then it means that your computations of the various tangents and normals are wrong somewhere. I don't see where --- but I didn't look too closely --- but I'd recheck carefully if I were you.

I also have the suspicion (using the triple-product identities) that if the $\gamma$s are both unit speed curves, then the proof might be pretty easy. But I leave that to you.

Big picture remark

When you're down in the weeds in a diff'l geometry computation, try an example.

  • $\begingroup$ Thank you for your comment. Well, the computations of tangent and normals are all shown above. So I think my computation isn't wrong. And also I assumed unit speed but it doesn't help that much in this problem $\endgroup$ – love_sodam Nov 16 '20 at 4:22

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