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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.

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    $\begingroup$ if you could show $D_{C'}C'=C''$ for any curve $C$ on any surface and after you project that on the tangent space of the surface, you will get $$\nabla_{C'}C'=C''-(N\cdot C'')N.$$ $\endgroup$ – janmarqz Nov 19 '19 at 3:12
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    $\begingroup$ Here $D$ is the std covariant derivative of $\mathbb R^3$. $\endgroup$ – janmarqz Nov 19 '19 at 3:18
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    $\begingroup$ So, an equivalent criterion for a curve to be a geodesic is $\nabla_{C'}C'=\vec 0$ $\endgroup$ – janmarqz Nov 19 '19 at 3:25
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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. : )

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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 $V = \gamma''(t)$.

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  • $\begingroup$ but $\gamma"(t)$ being normal to S doesn't also mean that $\gamma"(t)=k*(N(\gamma(t)))$, where the vector field N is given by the orientation? $\endgroup$ – Armando Rosas Nov 19 '19 at 1:23
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    $\begingroup$ All your care about is that $\gamma''(t)$ lies on the line spanned by $\pm N(\gamma(t))$. The normal line is well-defined, even for a nonorientable surface. $\endgroup$ – Ted Shifrin Nov 19 '19 at 2:16
  • $\begingroup$ . . . nice . . . $\endgroup$ – janmarqz Nov 19 '19 at 18:29

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