# Is this a geodesic?

Let $(M,g)$ be a riemannian manifold. Let $p$ in $M$ and $v,v_{0}$ two vectors in $\mathrm{T}_{p}M$. I am looking at the curve

$$\gamma \, : \, t \, \longmapsto \, \mathrm{Exp}_{p}(tv+v_{0})$$

and I am wondering whether this curve is a geodesic. Here, $\mathrm{Exp}_{p}$ denotes the riemannian exponential map : $c \, : \, t \, \mapsto \, \mathrm{Exp}_{p}(tw)$ is the unique geodesic such that $c(0)=p$ and $c'(0)=w$. To prove that $\gamma$ is a geodesic, I should compute $\nabla_{\dot{\gamma}}\dot{\gamma}$ and see whether this quantity is zero or not. However, I'm having trouble computing $\nabla_{\dot{\gamma}}\dot{\gamma}$ and I would appreciate if someone could shed some light for me on this.

It is clear that $\gamma(0) = \mathrm{Exp}_{p}(v_{0})$ and (if I'm not mistaken) $\gamma'(0) = \mathrm{D}_{v_{0}} \big( \mathrm{Exp}_{p} \big) \cdot v \in \mathrm{T}_{\gamma(0)}M$. So I was wondering whether $\gamma$ wouldn't be the geodesic from $\gamma(0)$ with initial speed $\gamma'(0)$.

## 1 Answer

The curve $\gamma(t) = \operatorname{Exp}_{p}(tv + v_{0})$ is not generally a geodesic.

Let $(M, g)$ be the round unit sphere in $\mathbf{R}^{3}$, and let $p = (0, 0, 1)$. In polar coordinates $(r, \theta)$ in the tangent plane $T_{p} M$, the exponential map is given by $$\operatorname{Exp}_{p}(r, \theta) = (\sin r \cos\theta, \sin r \sin\theta, \cos r).$$ Particularly, the circle of radius $\pi$ maps to the south pole $-p = (0, 0, -1)$. A line not through the origin has polar equation $r = c\sec(\theta + \theta_{0})$ for some real numbers $c > 0$ and $\theta_{0}$. It's easy to verify (both analytically and geometrically) that the image of such a line is not generally an arc of a great circle.