Finding Geodesics on the Unit Cylinder

Question

I am currently working through a problem in Andrew Pressley's Elementary Differential Geometry.

Question 9.2.1

Let $p$, and $q$ be two distinct points on the unit cylinder. Show that there are either two or infinitely many geodesics whose end points are $p$ and $q$.

The first part of this question is quite easy. Suppose $p$ and $q$ lie on the same circular arc around the unit cylinder. Then there are exactly two geodesics joining $p$ and $q$.

But I can't seem to figure out how there are infinitely many geodesics if $p$ and $q$ don't lie on the same circular arc. The book says that there are infinitely many helices joining the points. That is easy enough to understand, but to be a geodesic don't the helices need to be locally length minimizing? How can this be?

Any help would be greatly appreciated!

Answer 1

$f(\phi,h)=(\cos \phi,\sin\phi,h)$ defines a local isometry from $\mathbb{R}^2$ onto the embedded unit cylinder in $\mathbb{R}^3$ and hence sends geodesics in $\mathbb{R}^2$(=straight lines) to geodesics in the cylinder. Assume $p=f(\phi_p,h_p)$ and $q=f(\phi_q,h_q)$. For each $k\in\mathbb{Z}$ the straight line connecting $(\phi_p,h_p)$ with $(\phi_q + 2\pi k,h_q)$ gets mapped to a geodesic $\gamma_k$ joining $p$ and $q$. If $h_p\neq h_q$, then all the $\gamma_k$ are disjoint.

Answer 2

I found another solution with the geodesic equations.

$$ (G) = \left\{\begin{array}{c} \frac{\mathrm{d}}{\mathrm{d}t} (Eu'+Fv') = \frac12(E_uu'^2+2F_uu'v' + G_uv'^2) \\ \frac{\mathrm{d}}{\mathrm{d}t} (Fu'+Gv') = \frac12(E_vu'^2+2F_vu'v' + G_vv'^2) \end{array}\right. $$

So first we need to find the coefficient of the first fundamental form is easy enough to compute with the parametrization given by Jan Bohr in the accepted answer above: $\sigma(u,v) = (\cos u,\sin u, v)$, we have $E=G=1$, and $F=0$, so both of the geodesic equations vanish. We can deduce that $u''$, and $v''$ are both identically zero: $$ \sigma_{uu}\cdot\sigma_u = (-\cos u, -\sin u,0)\cdot(-\sin u,\cos u,0) = 0 $$ And likewise for $\sigma_{vv}\cdot\sigma_v$.

Then we have $u=a+bt$ and $v=c+dt$. If $b=0$ then this system gives a circular arc around the cylinder, otherwise we obtain a circular helix, and without restrictions on $b$ there are infinitely many solutions when $b\neq 0$, and only two when $b=0$ depending on the orientation of $v$.

Answer 3

The helices $$\gamma_a(t)=(\cos(t),\sin(t),t/a) $$ are all geodesics, since $\gamma_a''(t)=(-\cos(t),-\sin(t),0)$ is perpendicular to the cylinder.

Since you mention that "the book says that there are infinitely many helices joining the points. That is easy enough to understand, but to be a geodesic don't the helices need to be locally length minimizing?", the above should be enough for you.

If someone coming here does not understand why there are infinitely many helices joining the points:

Given $p,q$ not on the same circular arc, we have that $p_3 \neq q_3$. Since rotations along the plane and translations along the $z$-axis are isometries, we can assume that $p=(1,0,0)$ (this is just for simplification, not necessary at all), and thus $q_3 \neq 0$.

Pick $T$ such that $q=(\cos(T),\sin(T),q_3)$. The geodesics $\gamma_{(T+2k\pi)/q_3}$ ($k$ running through the integers) all connect $p$ and $q$. Indeed, $$\gamma_{(T+2k\pi)/q_3}(0)=(1,0,0)=p $$ and $$\gamma_{(T+2k\pi)/q_3}(T+2k\pi)=(\cos(T),\sin(T),q_3)=q.$$