# why a geodesic is a regular curve?

In most definitions of the geodesic, it is required to be a regular curve,i.e. a smooth curve satisfying that the tangent vector along the curve is not 0 everywhere. I don't know why.

-
if you allow corners, you can automatically create a shorter segment by taking a short cut across the corner. Draw some pictures. – Will Jagy Dec 15 '12 at 2:46

## 1 Answer

Suppose $\gamma:[a,b]\to M$ be a smooth curve on a Riemannian manifold $M$ with Riemannian metric $\langle\cdot,\cdot\rangle$. Then we have $$\tag{1}\frac{d}{dt}\langle\frac{d\gamma}{dt},\frac{d\gamma}{dt}\rangle=\langle\frac{D}{dt}\frac{d\gamma}{dt},\frac{d\gamma}{dt}\rangle+\langle\frac{d\gamma}{dt},\frac{D}{dt}\frac{d\gamma}{dt}\rangle$$ where $\frac{D}{dt}$ is the covariant derivative along the curvature $\gamma$. By definition, $\gamma$ is geodesic if and only if $\frac{D}{dt}\frac{d\gamma}{dt}=0$ for all $t\in[a,b]$, which implies together with $(1)$ that $$\frac{d}{dt}\langle\frac{d\gamma}{dt},\frac{d\gamma}{dt}\rangle=0\mbox{ for all }t\in[a,b],$$ which implies that the function $\langle\frac{d\gamma}{dt},\frac{d\gamma}{dt}\rangle$ is constant, i.e. $$\langle\frac{d\gamma}{dt},\frac{d\gamma}{dt}\rangle=C$$ for some constant $C$.

Thus, if $C\neq 0$, then by definition $\gamma$ is a regular curve. And if $C=0$, we have $\langle\frac{d\gamma}{dt},\frac{d\gamma}{dt}\rangle= 0$, or equivalently, $\frac{d\gamma}{dt}=0$ for all $t\in[a,b]$, which implies $\gamma(t)=p$ for all $t\in[a,b]$ for some point $p\in M$, i.e. $\gamma$ degenerates to a point in $M$. Therefore, if $\gamma$ is a nontrivial geodesic in the sense that it does not degenerates to a point, $\gamma$ must be a regular curve.

-