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Suppose $K(P) < 0$ where $K(P)$ is the Gauss curvature at $P$, where $K(P) = \det|S_p|$, the determinant of the shape operator at $P$. If $C$ is an asymptotic curve with $\kappa(P) \neq 0$, prove that its torsion satisfies $|\tau(P)|=\sqrt{-K(P)}$.

Hint: If we choose an orthonormal basis $\{U,V\}$ for $T_p(M)$ with $U$ tangent to $C$, what is the matrix for $S_p$?

The answer to this hint is that the matrix for $S_p$ will be symmetric, and furthermore the matrix representation of the first fundamental form will be a scalar multiple of the identity matrix.

Well, first of all, since $C$ is an asymptotic curve, we have $\kappa N \cdot n=0$ where $N$ is the unit normal vector of the curve and $n$ is the unit normal vector of the surface.

I'm having trouble seeing the connection to torsion here, or how the fact that the matrix for $S_p$ is symmetric is going to be useful.

Insights greatly appreciated!!

Ernie060
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1 Answers1

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Here are three insights. They don't give away the answer immediately, but I hope they are useful.

  1. By definition of asymptotic directions, $S_p U\cdot U = 0$. Note that the matrix of $S_p$ w.r.t. the basis $\{U,V\}$ is $$ \begin{bmatrix} S_p U \cdot U & S_p U\cdot V \\ S_p V\cdot U & S_p V\cdot V \end{bmatrix}. $$ Since $S_p U\cdot U= 0$, one finds $$ K(P) = \det(S_p) = (S_p U\cdot U)(S_p V\cdot V) - (S_p U \cdot U)(S_p V\cdot U) = -(S_p U\cdot V)^2. $$

  2. You already showed that $\kappa N \cdot n = 0$. Use this to show that the surface normal $n$ is $\pm B$, where $B$ is the binormal vector.

  3. Finally, use the fact that $S_p U$ is, by definition, minus the derivative of $n$ along $C$.

Ernie060
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  • $det(S_p)=k_1k_2$, the product of the principal curvatures, the eigenvalues of $S_p$. Does that change since $S_pU \cdot U = 0$?

    I understand step two, I can do that.

    But then as far as using hint #3, I know $S_p(U)$ is the derivative of $n$ along $C$ in the direction of $U$, and since $n$ is in the direction of the binormal of the curve, the derivative will be $\tau N$, correct?

     –  Apr 28 '19 at 17:30
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    I gave an extra hint for the first step in the post. Step 3 is also OK. Strictly speaking, since $n = \pm B$, the derivative will be $\pm \tau N$, but apart from that detail, you are correct. – Ernie060 Apr 28 '19 at 18:48
  • Is $V$ also an asymptotic direction so that $S_p(V) \cdot V=0$? are U and V orthogonal basis vectors of asymptotic directions? –  Apr 28 '19 at 18:53
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    No. The two asymptotic directions on a surface with $K<0$ are in general not orthogonal, so $V$ is not automatically the other asymptotic direction. But you do not need to know what $S_p V \cdot V$ in these calculations. – Ernie060 Apr 28 '19 at 18:56
  • Hmmm... Well $S_P(V) \cdot V$ is then equal to $\kappa$, the projection of the normal curvature in the direction of $V$ to $V$.. I don't see how I am going to end up with having a quantity squared, although I would if $S_P(V) \cdot V=0$ which it won't be all the time. –  Apr 28 '19 at 18:59
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    No, problem, I will add an extra detail in the post. We indeed do not know $S_p V \cdot V$, but because $S_p U\cdot U = 0$, this factor vanishes in the calculation. – Ernie060 Apr 28 '19 at 19:00
  • You are the bomb. Any chance you can help out over here? https://math.stackexchange.com/questions/3205981/show-that-if-kp0-then-the-two-asymptotic-curves-have-torsions-of-opposite –  Apr 28 '19 at 19:06
  • And so, $S_p(U) \cdot V$... $S_p(U)$ will be, by its definition, the derivative of $n$ along $C$, where $n$ is equal to its binormal vector. So it will equal $\tau N$... But then we must take the dot product of this with $V$, which is in the orthogonal to the tangent of $C$ and therefore parallel to $N$... got it!! I think –  Apr 28 '19 at 19:11
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    Yes you got it. One detail: the shape operator $S_p(U)$ is minus the derivative of $n$. – Ernie060 Apr 28 '19 at 19:12