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Given $\alpha(t)$ and $\beta(t)$ two differentialble vector functions with common domain I, we define the secant surface between the two resulting curves by: \begin{equation} \boldsymbol X(t,u)=(1-u) \boldsymbol\alpha(t)+ u \boldsymbol\beta(t) \end{equation} for $(t,u) \in I \times \mathbb{R}$. Assume that the corresponding surface is regular. How can I prove that the Gaussian curvature is zero fore any point with $u=\frac{1}{2}$ ?

Here details of my computation: \begin{equation} \boldsymbol X_{t,u}(t,u)=\boldsymbol \beta^{'}(t)-\boldsymbol \alpha^{'}(t) \end{equation} While the normal vector N is by definition: \begin{equation} \boldsymbol N_{t,u}(t,u)=\frac{\boldsymbol X_t(t,u) \times \boldsymbol X_u(t,u)}{||\boldsymbol X_t(t,u) \times\boldsymbol X_u(t,u)||} \end{equation} Now: \begin{equation} \boldsymbol X_t(t,u) \times \boldsymbol X_u(t,u)=(1-u) \boldsymbol \alpha^{'}(t) \times \boldsymbol \beta(t)-(1-u) \boldsymbol \alpha^{'}(t) \times \boldsymbol \alpha(t)+u \boldsymbol \beta^{'}(t) \times \boldsymbol \beta(t)-u \boldsymbol \beta{'}(t) \times \boldsymbol \alpha(t) \end{equation} The mixed product after few computation reduces to: \begin{equation} ( \boldsymbol X_t(t,u) \times \boldsymbol X_u(t,u)) \cdot \boldsymbol X_{t,}(t,u)= \boldsymbol \alpha^{'}(t) \cdot (\boldsymbol \beta^{'}(t) \times (\alpha(t)- \beta(t)) ) \end{equation} If the mixed product is zero, the three vectors are linearly dependent, i.e. two of the three vectors are parallel.

The Gaussian curvature, when secant surface is as specified above, is zero if $\boldsymbol N \cdot \boldsymbol X_{t,u}=0$, where $\boldsymbol N$ is the normal vector. However $\boldsymbol N \cdot \boldsymbol X_{t,u}$ is independent of $u$.

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  • $\begingroup$ The result you're trying to prove seems very wrong. You can easily make a hyperboloid of one sheet (with $K<0$) by taking $\alpha$ and $\beta$ to be parallel circles, parametrized with a phase shift. $\endgroup$ – Ted Shifrin Mar 29 '16 at 9:32
  • $\begingroup$ Hi Ted, I agree with you. I think that the Gaussaian curvature is not zero for all points with $u=\frac{1}{2}$. However this is a problem from the boot Differential Geometry of Curves and Surfaces by Thomas F. Banchoff, Stephen T. Lovett. $\endgroup$ – Upax Mar 29 '16 at 9:38
  • $\begingroup$ I stand by my statement. However, your argument is flawed, I think. $\mathbf X_{t,u}$ is independent of $u$, but $\mathbf N$ is not. $\endgroup$ – Ted Shifrin Mar 29 '16 at 9:54
  • $\begingroup$ It is the dot product of N and $X_{t,y}$ which is independent of u. $\endgroup$ – Upax Mar 29 '16 at 10:27
  • $\begingroup$ Write down the formula for $\mathbf N$. At any rate, I still think the problem is false as stated. For the hyperboloid I mentioned above, we in fact have maximum Gaussian curvature when $u=1/2$. $\endgroup$ – Ted Shifrin Mar 29 '16 at 16:22
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You're missing the denominator in $\mathbf N$, so even though your scalar triple product is independent of $u$, the actual entry of the second fundamental form is not. As I suggested in the comments, the secant surface is ruled and you should not expect it to have points of zero Gaussian curvature. However, when your scalar triple product is $0$, the surface will be developable, i.e., everywhere flat. This will happen, for example, when $\boldsymbol\alpha$ and $\boldsymbol\beta$ are parallel circles, parametrized in synch (so the secant surface is a cylinder), when $\boldsymbol\beta$ is a constant path (so the secant surface is a cone), or when $\boldsymbol\alpha$ and $\boldsymbol\beta$ both lie in the same plane.

P.S. Textbooks do have erroneous exercises (or even theorems) from time to time. I should know. :)

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