# 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.

• 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 $$\nabla_{C'}C'=C''-(N\cdot C'')N.$$ – janmarqz Nov 19 '19 at 3:12
• Here $D$ is the std covariant derivative of $\mathbb R^3$. – janmarqz Nov 19 '19 at 3:18
• So, an equivalent criterion for a curve to be a geodesic is $\nabla_{C'}C'=\vec 0$ – janmarqz Nov 19 '19 at 3:25

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 $$-0.7 and $$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. : )

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

• 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? – Armando Rosas Nov 19 '19 at 1:23
• 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. – Ted Shifrin Nov 19 '19 at 2:16
• . . . nice . . . – janmarqz Nov 19 '19 at 18:29