# Show that a curve is a geodesic on a surface.

Let $$\alpha:[0,1]\rightarrow S^2$$ be the curve $$\alpha(t)=\left(\cos(e^t),\sin(e^t),0 \right)$$.

Show that $$\alpha$$ is a geodesic on $$S^2$$, but in the latitude-longitude parametrization of $$S^2$$, $$\alpha(t)$$ does not satisfy the geodesic equations.

Definition. A geodesic on a surface $$S$$ is a curve on $$S$$ whose geodesic curvature is identically zero.

The geodesic equations: $$\begin{cases} u''+(u')^{2}\Gamma_{11}^{1}+2u'v'\Gamma_{12}^{1}+(v')^2\Gamma_{22}^{1}=0 \\ v''+(u')^{2}\Gamma_{11}^{2}+2u'v'\Gamma_{12}^{2}+(v')^2\Gamma_{22}^{2}=0 \end{cases}$$

Can anyone help me get started?

## 1 Answer

For $$\alpha(t) = (\cos e^t, \sin e^t, 0), \quad t\in[0,1],$$ we have \begin{align} \dot\alpha(t) &= e^t(-\sin e^t,\cos e^t, 0),\\ \ddot\alpha(t) &= e^t(-\sin e^t,\cos e^t, 0) + e^t(-\cos e^t,-\sin e^t, 0) = \dot\alpha(t)- e^t\alpha(t). \end{align} Since the outer unit normal of $$S^2$$ is just $$N(x,y,z)=(x,y,z)$$, it follows that, along $$\alpha,$$ $$N(t) = N(\alpha(t)) = \alpha(t).$$ Hence $$k_g(t) = \langle N(t) \times\dot\alpha(t), \ddot\alpha(t) \rangle = \langle \alpha(t) \times \dot\alpha(t), \dot\alpha(t) \rangle-e^t \langle \alpha(t) \times \dot\alpha(t),\alpha(t)\rangle.$$ I will let you figure out on your own why the last expression is zero.

Now, if we parametrize $$S^2$$ by $$X(u,v) = (\cos u \cos v, \cos u \sin v, \sin u),$$ we see that $$\alpha$$ is given in the local $$(u,v)$$-coordinates by $$(u(t),v(t))=(0, e^t)$$. Furthermore, $$X_u = (-\sin u\cos v, -\sin u \sin v, \cos u), \quad X_v = (-\cos u \sin v, \cos u \cos v, 0),$$ so that the first fundamental form is given in these coordinates by $$E = 1, \quad F= 0, \quad G=\cos^2u.$$ The only thing that is left to do is to find the Christoffel symbols. and plug everything into the geodesic equations. Can you do that on your own?

EDIT: Notice that, along $$\alpha$$, we have $$u=u'=u'' = 0$$, so that the first geodesic equation becomes $$e^{2t} \Gamma_{22}^1 = 0,$$ so it suffices to show that $$\Gamma_{22}^1$$ is not identically zero to reach the wanted conclusion.

• Thank you :) Yeah I can do that – Alim Teacher Nov 12 '18 at 19:59