I have solved an exercise on gedeosics on a cylinder but I have a question that makes me doubt the correctness.
My strategy to solve this was as follows:
- Determine and solve the geodesic equations.
- Show that the solutions are parametrizations of helices, circles and straight lines.
- Investigate how many solutions one can have when they are required to run through specific points.
Consider the system of standard coordinate frames $\{\partial_i\}_{i=1,2}$ where $\partial_i=\frac{\partial}{\partial x_i}$. In order to find the geodesic equations we need the Christoffel symbols. We have
$$ \nabla_{\partial_i} \partial_j =\sum_{k=1}^2 \Gamma^k_{ij} \partial_k $$
Then we apply the metric and exploit the orthogonality of the frames to get
$$ \langle \nabla_{\partial_i} \partial_j , \partial_k \rangle =\langle \sum_{l=1}^3 \Gamma^l_{ij} \partial_l, \partial_k \rangle =\sum_{l=1}^2 \Gamma^l_{ij} \langle \partial_l, \partial_k \rangle =\Gamma^k_{ij} $$
The Koszul formula yields the Christoffel identity
$$ \langle \nabla_{\partial_i} \partial_j , \partial_k \rangle =\frac{1}{2} (\partial_i \langle \partial_j, \partial_k \rangle +\partial_j \langle \partial_k, \partial_i \rangle -\partial_k \langle \partial_i, \partial_j \rangle) $$
Since the frames are orthogonal we get
$$ \langle \nabla_{\partial_i} \partial_j , \partial_k \rangle=0 $$
But then all of the Christoffel symbols vanish. The geodesic equations take the form
$$ 0=\gamma^{''}_k+\sum_{i,j=1}^3 \gamma'_i \gamma'_j (\Gamma^k_{ij} \circ \gamma) =\gamma^{''}_k, \ k=1,2 \ , \gamma=(\gamma_1\gamma_2). $$
Thus we have the solutions
$$ \gamma_k(t)=a_k+b_k t $$
The cylinder can be parametrized as
$$ C(x_1,x_2)=(\cos(x_1),\sin(x_1),x_2), x_2 \in \mathbb{R}. $$
so the geodesics arise as
$$ C(\gamma_1(t),\gamma_2(t))=(\cos(\gamma_1(t)),\sin(\gamma_1(t)),\gamma_2(t)) $$
This is
- a helix if $b_1,b_2 \neq 0$
- a circle if $b_1 \neq 0$ and $b_2=0$
- a straight line if $b_1=0$ and $b_2 \neq 0$.
Now consider $p,q \in M$.
If the solution is a helix we have infinitely many possible choices for $b_2$ and therefore infinitely many geodesics connecting $p$ and $q$.
If the solution is a circle we have two choices for the sign of $b_1$ and therefore two geodesics connecting $p$ and $q$.
If the solution is a straight line we have infinitely many possible choices for $b_2$ and therefore infinitely many geodesics connecting $p$ and $q$.
However, this solution does not take into account the $a_k$ which raises a question: Since there are no conditions imposed on these parameters shouldn't there be always infinitely many geodesics?
