Are Jacobi Fields the geodesics of the tangent bundle?

Let $$(M, g)$$ be a Riemannian manifold. Wikipedia states that "Jacobi fields correspond to the geodesics on the tangent bundle". I'm trying to undrestand this statement. Curves $$c : I \to TM$$ correspond to a curve $$\gamma: I \to M$$ and a vector field along $$\gamma$$. I assume that Wikipedia means this correspondence.

I also assume that the metric on $$TM$$ is given by $$\left<(u, v) , (u', v')\right> = \left + \left$$, where $$T_{(p, w)}(TM)$$ is identified with $$T_p M \oplus T_p M$$.

I have tried to prove the above fact using this but I haven't managed to do so. To be precise, I think I get a wrong result for the Christoffel symbols for $$TM$$. This leads me to conclude that the projection of a geodesic $$c$$ in $$TM$$ to $$M$$ must not necessarily be a geodesic, which I think contradicts the statement I want to prove.

Is what I have written correct? And how do you prove this statement?

EDIT: I think my definition of the scalar product on $$TM$$ is wrong since it is not coordinate-independent (since the identification of $$T_{(p, w)}(TM)$$ with $$T_p M \oplus T_p M$$ is not coordinate-independent, I think).

• I have never seen this claim outside the wikipedia page. The metric I'm used to on $TM$ is given as follows: for $\alpha(t) = (p(t), v(t))$ and $\beta(s) = (q(s), w(s))$, we define $\langle \alpha'(0), \beta'(0)\rangle = \langle p'(0), q'(0)\rangle + \langle v'(0), w'(0)\rangle$, where $v'$ refers to the covariant derivative of $v(t)$, which is a vector field along the curve $p(t)$. With respect to this metric, I think I can prove that a curve of the form $\alpha(t) = ( \gamma(t), J(t))$ (for $\gamma$ a geodesic and $J$ a Jacobi field) only has constant length if $sec(span(J, \gamma')) = 0$. – Jason DeVito Feb 21 at 21:04
• ... with $sec$ referring to sectional curvature. So, at least with respect to the metric I'm used to, the claim can only possibly be true on flat manifolds. – Jason DeVito Feb 21 at 21:05

The correct scalar product:

My definition of the scalar product on $$TM$$ is indeed wrong and, as @Jason DeVito points out, the correct way to do this is the following.

For $$\alpha(t) = (p(t), v(t))$$ and $$\beta(t) = (q(t), w(t))$$, define $$\left<\alpha'(0), \beta'(0)\right> = \left + \left< v'(0), w'(0) \right>$$ where $$v'$$ is the covariant derivative of $$v$$.

This is obviously coordinate indepent. One thing to observe is that, using the above notation $$\newcommand{\eps}{\varepsilon} L(\alpha) \geq L(p)$$ where $$L(\cdot)$$ denotes length. From this we conclude that $$d_{TM}\left( (p, v), (q, w) \right) \geq d_M (p, q) \tag{1}\label{d}$$

The claim from Wikipedia:

The claim in Wikipedia was also wrong (and I have removed it). Take for example the sphere $$\mathbb{S}^2$$ parametrized with spherical coordinates $$\phi(\theta, \phi) = (\cos\theta \cos\phi, \cos\theta \cos\phi, \sin\theta)$$ and consider a neighborhood $$U$$ of $$\phi(0, 0)$$ in $$T\mathbb{S}^2$$. The curve $$\gamma: (-\eps, \eps) \to U, \gamma(t) = \left(\phi(t, 0),\, a \cos t \cdot \partial_\phi \right)$$ consists of a geodesic and a jacobi field for every $$a \in \mathbb{R}$$.

By (\ref{d}) one possible curve with minimal length between $$\gamma(-\eps)$$ and $$\gamma(\eps)$$ is $$\delta: (-\eps, \eps) \to U, \delta(t) = \left(\phi(t, 0), a \cos\eps \cdot \partial_\phi \right)$$ (since it's vector part is parallel and hence doesn't contribute to length).

However, if we choose $$U$$ (and $$a$$) small enough then geodesics in $$U$$ must be the unique shortest curve between their endpoints, so $$\gamma$$ cannot be a geodesic.