About geodesics on a torus

There are a few types of geodesics on a torus of revolution. There are meridians, which are cirlces going the short way around. There is an inner equator and an outer equator. After those, there are about three types that pass through a point on the outer equator with some starting angle $\theta_0 $ compared with the horizontal.

(A) small $\theta_0,$ in which case Clairaut says that they reach a minimum $r$ away from the $z$ axis, where $\cos \theta = 1$ and $\theta = 0.$ By various symmetry properties, these simply return to the outer equator, and pass through it again at angle $-\theta_0,$ thus making a wave forever

(B) large $\theta_0$ in which they reach the inner equator and pass it at a nonzero angle, in which case they wrap around the torus forever

(C) a critical intermediate value of $\theta_0,$ precisely the one that says that the angle $\theta$ at the inner equator would be $0,$ by Clairaut. In this case, the geodesic never actually reaches the inner equator, it wraps around and around it, getting closer and closer.

do Carmo's first book does not picture this on the torus. On pages 262-263, he talks about this type (C) for a hyperboloid of revolution, with Figure 4-22.

This question was posted a long time ago, but I will write my solution because I spent a lot of time stuck on this question.

$$\sigma(u,v) = ((r\cos(u)+a)\cos(v),(r \cos(u) + a )\sin(v), r \sin(u) ) $$

With $0<r<a$.

If $\gamma(t)= \sigma(u(t),v(t))$ is a geodesic parametrized by arc length, then $(u(t),v(t))$ must satisfy the differential equation (geodesic diffential equation)

$$v'' + 2\frac{(r\cos(u)+a)(r\cos(u)+a)' }{(r\cos(u)+a)^2} u' v' = 0 $$ $$u'' - \frac{(r\cos(u)+a) (r\cos(u)+a)'}{((r\cos(u)+a)')^2 + (\cos(u))^2} (v')^2 + \frac{r^2 \cos(u)\sin(u)}{((r\cos(u)+a)')^2 + (\cos(u))^2} (u')^2 = 0. $$

From the first equation, we conclude that \begin{align*} \frac{d}{dt}(r\cos(u(t)) +a)^2v'(t)) &= (r\cos(u)+a)^2 v''(t) + 2(r\cos(u(t) +a)(r\cos(u)+a)' u' v'\\ &=(r\cos(u) + a)^2\left[ v'' + 2\frac{(r\cos(u)+a)(r\cos(u)+a)'}{(r\cos(u)+a)^2}u' v'\right] \\ &=0, \end{align*} which imply that $\exists$ $c \in \mathbb{R}$, such that

$$(r\cos(u(t) +a)^2 v'(t) = c.$$

Follows from these observations

\begin{align*}\cos(\theta(t)) &:= \frac{\langle \sigma_v(u(t),v(t)) , \gamma'(t)\rangle}{|\sigma_v(u(t),v(t))||\gamma'(t)|} \\ &= \frac{\langle \sigma_v , u'\sigma_u + v' \sigma_v \rangle}{\Vert\sigma_v\Vert}\\ &= v' \Vert \sigma_v \Vert \\ &= v'(t)\cdot \Vert(-(r\cos(u(t))+a)\sin(v),(r \cos(u(t)) + a) \cos(v(t)) , 0) \Vert\\ &= (r \cos(u(t)) +a)v'(t) = \frac{c}{r\cos(u(t)) + a } \end{align*}

$$\Rightarrow (r \cos(u(t)) +a) \cos(\theta(t)) \equiv c. \qquad (1) $$

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a)$\gamma$ is tangent to the parallel $u=\pi/2$

Suppose that $\exists$ $v_0,t_0 \in \mathbb{R}$ satisfying \begin{align*}\gamma(t_0) &= \sigma\left(\left(\frac{\pi}{2}, v_0\right)\right) \\ \gamma'(t_0) &= \pm \frac{\sigma_v \left(\frac{\pi}{2},v_0\right)}{\left|\sigma_v\left(\frac{\pi}{2},v_0\right)\right|},\mbox{ remember that }\gamma \mbox{ is parametrized by arc length}. \end{align*}

From (1)

$$c = (r \cos(u(t)) +a) \cos(\theta(t)) = (r \cos(u(t_0)) +a) \cos(\theta(t_0)) = a \quad (2) . $$

Supose by reductio ad absurdum that there is $t_1 \in \mathbb{R}$, satisfying $\gamma(t_1) = \sigma(u_1,v_1)$ with $u_1 \in [-\pi,\pi]\setminus [-\pi/2,\pi/2]$.

Using (2) $$(r \cos(u(t_1)) +a) \cos(\theta(t_1)) = a, \hspace{0.1cm} \mbox{note that $0\leq \cos(\theta(t))\leq 1$} $$ $$\Rightarrow \cos(\theta(t_1)) = \frac{a}{(r \cos(u(t_1)) +a)}, $$

which implies $\cos(\theta(t_1)) > 1$. Absurd!!

These arguments prove the first part of your question .

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b) $\gamma$ intersect the parallel $u=0$ with angle $0<\theta<\pi/2$ .

$\exists$ $t_0$ $\in$ $\mathbb{R}$, such that

$$\gamma(t_0) = \sigma(0,v_0),\quad \cos(\theta(t_0))<\frac{a-r}{a+r} $$

From (1)

$$c = (r \cos(u(t)) +a) \cos(\theta(t)) = (r \cos(0) +a) \cos(\theta(t_0)) < a-r\qquad (3) $$

Suppose by reductio ad absurdum that $\gamma$ never reaches the parallel $u = \pi$, this is equivalent to say that $-\pi < u(t) < \pi$, $\forall$ $t$.

Note that $v'(t) \neq 0$, for all $t$. If $\exists$ $t_1$, such $v'(t_1) = 0$ $\Rightarrow$ $\cos(\theta(t_1)) = 0$ $\Rightarrow$ $\gamma'(t_1)$ is l.d. with a meridian, by the unicity of the geodesics (given a point and a direction there is a unique geodesic passing through this point) $\gamma$ must be the meridian, it implies that $\cos(\theta(t_0)) = 0$ $\Rightarrow$ $\theta(t_0) = \theta = \pi/2$. Absurd! So $v'(t) \neq 0$, $\forall$ $t$.

Suppose without loss of generality that $v'(t) >0$ $\forall t$.

On the order hand $u'(t) \neq 0$. Suppose that $\exists$ $t_1$ such that $u'(t_1) = 0$, it implies that $\cos(\theta(t_1)) = 1$ (remember that we suppose that $-\pi < u(t)< \pi)$. But

$$c = (r\cos(u(t)) + a )\cos(\theta(t)) = (r \cos(u(t_1))+a) 1 > a-r $$

contradicts (3). So $u'(t) \neq 0$ $\forall$ $t$.

Suppose without loss of generality that $u'(t) >0$ $\forall t$.

Note that by the hypothesis $\gamma$ is parametrized by arc length

$$ |u'| |\sigma_u| + |v'| |\sigma_v| = 1 $$

Using that

$$|\sigma_u| = r, \quad |\sigma_v| > a-r>0 $$

this implies that

$$0< u'(t) \leq 1/r , \qquad 0< v'(t) < 1/(r-a) \quad (4)$$

Now we can conclude that the domain of $\gamma$ contains $[t_0, \infty)$ . If this not occur exists a maximal $\omega^{+}$ $\in$ $\mathbb{R}$ such that $\gamma: [0, \omega^+) \rightarrow \mathbb{R}^3$. Remember that $(u(t),v(t))$ resolves the geodesic differential equation, by ODE theorems we know that $(u(t),v(t))$ $\to$ $\partial \mathbb{R}^2$ when $t \rightarrow \omega^{+}$ (This means that for every compact $K$ $\subset$ $\mathbb{R}^2$, $\exists$ $t_K$ such that $(u(t),v(t)) \not\in K$, forall $t>t_K$).

so $$|(u(t),v(t))| \rightarrow \infty, \quad \mbox{when}\quad t \rightarrow \omega^{+}. $$

But from Mean Value Theorem and (4).

$$|u(t) - u(t_0)| < \frac{\omega^{+} - t_0}{r} \quad |v(t) - v(t_0)| < \frac{\omega^{+}- t_0}{a-r}, \quad \forall t \in [t_0, \omega^{+})$$

Absurd! Then $\gamma$ is well defined forall $t$ $\in$ $[t_0, \infty).$

Note that $u'(t) >0$, and (by the reductio ad absurdum hypothesis) $u(t) < \pi$, $\forall$ $t$ $\in$ $[t_0, \infty]$ as $u (t)$ is limited and increasing, there is $\tilde{u} \leq \pi$ satisfying $$\lim_{t \rightarrow \infty} u(t) = \tilde{u} $$

Now we will prove that $\liminf\limits_{t \rightarrow \infty} u'(t) \rightarrow 0 $ when $t \rightarrow \infty$. In case otherwise, there will be $ c> 0 $ such that $ u '(t)> c $ for all $ t> t_c $ which implies that $$ | u (t) - u (t_c) | > c | t - t_c | $$ $$\Rightarrow u(t) \rightarrow \infty, \qquad \mbox{when} \quad t \rightarrow \infty $$ Absurd! So $\exists$ sequence $(t_k)_{k \in \mathbb{N}}$, such that $t_k \rightarrow \infty$ and $u'(t_k) \rightarrow 0$ when $k\rightarrow \infty$. But this imply that

$$\cos(\theta (t_k)) \rightarrow 1, \quad \mbox{when}\quad k \rightarrow \infty $$ and $$a-r > c = \lim_{k \rightarrow \infty} (r\cos(u(t_k)) +a) \cos(\theta (t_k)) = (r \cos(\tilde{u}) +a ) 1 \geq a-r $$ (Final) Absurd! So we conclude that $u(t)$ must pass to $\pi$ (or $-\pi$, if you suppose that $u'(t) <0$).

Therefore, we proved the theorem.