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 $\left<(u, v) , (u', v')\right> = \left<u, u'\right> + \left<v, v'\right>$, 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).

  • $\begingroup$ 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 $sec(span(J, \gamma')) = 0$. $\endgroup$ – Jason DeVito Feb 21 at 21:04
  • $\begingroup$ ... 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. $\endgroup$ – 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 $\left<\alpha'(0), \beta'(0)\right> = \left<p'(0), q'(0) \right> + \left< v'(0), w'(0) \right>$ 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.


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