How to prove that the Möbius band has geodesics? In my class of Differential Geometry, the teacher defined geodesics as follows:
A regular curve on a regular surface, denoted as $\gamma:I\subset\Bbb{R}\to S$, ($S$ is the surface) is a geodesic if, $\forall t\in I$, the vector $\gamma"(t)$ is a normal vector to $S$ at the point $\gamma(t)$.
With this definition, I must prove for an exposition project that the möbius band can have geodesics.
The problem is that saying that a vector is normal to a surface implies orientation, and the möbius band is un-orientable.
So this is my question: How can i define geodesics on an un-orientable surface like the möbius band, and with that, how do i calculate them?
Can't the möbius band have any geodesics at all, because of its un-orientability?
If you can provide me a reference, i would apreciate it.
 A: Normality is not dependent on orientation. 
In $\mathbb R^3$, if I give you a plane $P$ and a vector $V$ based at a point of $P$, I can tell you whether $V$ is normal to $P$ without mentioning any orientation of $P$: $V$ is normal to $P$ if and only if $V \cdot W = 0$ for all vectors $W$ parallel to $P$. All I've used to formulate this definition is the (standard) inner product on the vector space $\mathbb R^3$.
You can now apply this principle at the point $\gamma(t)$, using the tangent plane $P = T_{\gamma(t)} S$ and the vector $V = \gamma''(t)$.
A: Once you have stablished that $\nabla_{\gamma'}\gamma'=\gamma''-(\gamma\cdot\gamma'')N$ then being $\gamma$ a geodesic with your definition we get
$\nabla_{\gamma'}\gamma'=0$.
You can consider that $\gamma=\Phi\circ\alpha$, where $\Phi$ is the local parametrization for the surface and $\alpha$ is a curve in the domain of $\Phi$, 
hence $\gamma'=J\Phi\circ\alpha'$. If $\alpha(t)=v(t)e_1+w(t)e_2$ then
$\alpha'=v'e_1+w'e_2$ and for $\gamma'$ we get
$$\gamma'=v'\partial_1+w'\partial_2,$$ 
where $\partial_1=J\Phi\ e_1$ and $\partial_2=J\Phi\ e_2$ is the tangent space base.
So, by the properties of a covariant derivative we calculate:
\begin{eqnarray*}
\nabla_{\gamma'}\gamma'&=&\nabla_{v'\partial_1+w'\partial_2}(v'\partial_1+w'\partial_2)\\
&=&v'\nabla_{\partial_1}(v'\partial_1+w'\partial_2)+w'\nabla_{\partial_2}(v'\partial_1+w'\partial_2)\\
&=&v'\nabla_{\partial_1}(v'\partial_1)+v'\nabla_{\partial_1}(w'\partial_2)+w'\nabla_{\partial_2}(v'\partial_1)+w'\nabla_{\partial_2}(w'\partial_2)\\
&=&(v''+v'^2\Gamma^1{}_{11}+2v'w'\Gamma^1{}_{12}+w'^2\Gamma^1{}_{22})\partial_1+
(w''+v'^2\Gamma^2{}_{11}+2v'w'\Gamma^2{}_{12}+w'^2\Gamma^2{}_{22})\partial_2.
\end{eqnarray*}
Therefore
for $\nabla_{\gamma'}\gamma'$ to be null we need the coefficients comply
$$v''+v'^2\Gamma^1{}_{11}+2v'w'\Gamma^1{}_{12}+w'^2\Gamma^1{}_{22}=0,$$
and
$$w''+v'^2\Gamma^2{}_{11}+2v'w'\Gamma^2{}_{12}+w'^2\Gamma^2{}_{22}=0.$$
These two ordinary-second-order-non-linear-homogenous equations, under the suitable conditions on the gammas, guarantee that solutions for the two functions $v=v(t)$ and $w=w(t)$ would exists.
For a parametrization of a Möbius band, like:
\begin{eqnarray*}
x&=&(2+v\cos(w/2))\cos w\\
y&=&(2+v\cos(w/2))\sin w\\
z&=&v\sin(w/2)
\end{eqnarray*}
where $-0.7<v<0.7$ and $0\le w<2\pi$ is likely to fulfill the conditions.
Excuse me for being too sketchy but i am available 24/7, here at MSE to any discussions or questioning. : ) 
