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Consider the torus of revolution generated be rotating the circle $\{(x,y,z) \in \mathbb{R}^{3}: (x − a)^{2} + z^{2} = r^{2}, y = 0$ }, where $a > r > 0$, around the $z$-axis.

The parallels generated by the points $(a + r, 0)$, $(a − r, 0)$, $(a, r)$ are called the maximum parallel, the minimum parallel, and the upper parallel, respectively.

Check which of these parallels is: a. A geodesic. b. An asymptotic curve. c. A line of curvature.

Then compute the geodesic curvature of the upper parallel of the torus.

What is the best way to go about doing this and how do I identify which solutions are generated by those points? Is there any way to "just see it"? Also, can you explicitly show me how to calculate the covariant derivative of these points (all of them, even if they are not geodesics, but so that we do identify the geodesic by such means) and how to compute the geodesic curvature?

Is this parametrization: $X(u, v) = ((a + r \cos{(v)}) \cos{(u)}, (a + r \cos{(v)}) \sin{(u)}, r \sin{(v)})$ for the problem as presented?

In that case, I got that the first fundamental form is given by $E = (a + r \cos{(v)})^{2}$, $F=0$, $G=r^{2}$ and that the second fundamental form is given by $e = -|a + r \cos{(v)}| \cos{(v)}$, $f = 0$, and $g = -r |a + r \cos{v}|$. Therefore, the asymptotic curves are given by $\cos{(v)} \cdot (u')^{2} + r (v')^{2} = 0$; but I am still having issues solving this differential equation. Can you explain in depth how to do so? Meanwhile, since $F=0=f$, $u$-lines and $v$-lines are lines of curvature. Is this option (c.)?

kevin
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1 Answers1

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You should not have absolute values when you compute $X_{uu}\cdot N$ (where $N$ is the surface normal), etc. (Since you haven't given much information, it's hard to know what notation your text is using.) Your formula for $g$ is definitely wrong. Recompute that. You are, however, correct that the $u$- and $v$-curves are all lines of curvature.

In general, the differential equations for asymptotic curves and geodesics are extremely hard (impossible) to solve in closed form and must be solved numerically. However, you should be approaching this problem more geometrically. If your notation for the Frenet frame is $t,n,b$, you should, for each of your parallels, look at the vector $\kappa n$ (the derivative of the unit tangent vector with respect to arclength). Up to sign, geodesic curvature is the tangential component and normal curvature is the normal component, and you can see all of these geometrically. (Also, remember that you can only have asymptotic directions at points where $K\le 0$.)

Ted Shifrin
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