4
$\begingroup$

Suppose $(M,g)$ is a complete Riemannian manifold. We can endow the tangent bundle $TM$ with a natural choice of metric, the so called Sasaki metric given by:

$\alpha(t)=(p(t),v(t))$ , $\beta(t)=(q(t),w(t))$. Then define $\langle\alpha^\prime(0),\beta^\prime(0)\rangle=\langle p^\prime(0),q^\prime(0)\rangle+\langle\frac{Dv}{dt}(0),\frac{Dw}{dt}(0)\rangle$. In other words we identify the vertical vectors at $T_{(p,v)}TM$ with $T_pM$ and the Euclidean metric given on that by $g$. Then we pullback $g$ by $D\pi: H \to TM$ to the horizontal vectors and define $H$ and $VE$ to be orthogonal to each other.

In this metric one can prove if $\gamma(t)$ is a geodesic and $v(t)$ a parallel vector field along $\gamma$, then $(\gamma(t),v(t))$ is a geodesic in $TM$. Now suppose $(p,v)$ and $(q,w)$ are two arbitrary points in $TM$ and $\gamma$ is a geodesic with $\gamma(0)=p, \gamma(1)=q$. Consider $v(t),w(t)$ which are the parallel transports of $v$ and $w$ along $\gamma$.

Is it true that the curve $\delta(t)=(\gamma(t),(1-t)v(t)+tw(t))$ is a geodesic between $(p,v)$ and $(q,w)$? Notice that if $w$ is the parallel transport of $v$ at $t=1$ then $\delta(t)$ is a geodesic.

In other words this is equivalent to say if $\delta(t)=(\gamma(t),v(t))$ is a curve in $TM$ such that $\gamma$ is a geodesic in $M$ and $\frac{D^2v}{dt^2}=0$, then $\delta$ is a geodesic in $TM$.

Do you have any counterexample or proof?

$\endgroup$

1 Answer 1

1
$\begingroup$

Yes, your conjecture is true. To prove it, we should calculate the geodesic equation for the tangent bundle. For a curve $t \mapsto (x(t), y(t)) \in TM$ such that $x(t)$ is not a fixed point, it is a geodesic if and only if $$ \nabla_{\dot x} \dot x = -R (y, \nabla_{\dot x} y) \dot x, \quad \nabla_{\dot x} \nabla_{\dot x} y = 0.$$

This formula is given as Corollary 4.4 in Geodesics on tangent bundles with horizontal Sasaki gradient metric by Abderrahim Zagane, but the condition $\dot x \neq 0$ is missing. Rigorously, you can deduce it from general Levi-Civita connection formulae given in Curvature of the Induced Riemannian Metric on the Tangent Bundle of a Riemannian Manifold by Oldrich Kowalski. In the case when $x(t)$ is a fixed point, the geodeic equation becomes $ y^{\prime\prime}(t) = 0$.

In your conjecture, we have $\nabla_{\dot x} y = 0$ by definition of parellel transport and $\nabla_{\dot x} \dot x = 0$ since $x$ is a geodesic on $M$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .