Completeness of the tangent bundle of riemannian manifold

Question

Let $(M,g)$ be a riemannian manifold and $TM$ its tangent bundle. There are natural riemannian metrics that we can endow the tangent bundle with (for instance the Sasaki metric) and I wonder if for some of them $TM$ is complete (under maybe some suitable additional assumptions about $M$). This means that $TM$ is complete as a metric space, which is equivalent with geodesic completeness (by Hopf-Rinow theorem). Does anyone have an idea or reference?

Answer 1

If $(M,g)$ is complete, then I think that the Sasaki metric $\tilde g$ on $TM$ is complete. This belief is motivated by the following statement, without proof: "In other words, we are sincerely convinced $(TM, g)$ is a complete Riemannian manifold, as long as $M$ is complete" (R. Albuquerque, "Notes on the Sasaki metric", Expositiones Mathematicae, Volume 37, Issue 2, 2019). Here is my attempt at a proof:

Let $\pi:TM\rightarrow M$ denote the projection map of the tangent bundle. Recall that the metric $g$ gives a splitting $T(TM)=V\oplus H$ into the vertical and horizontal distributions. Since fibers of $V$ can be affinely identified with fibers of $TM$, we get a natural metric on $V$. The metric on $H$ is defined by lifting $g$ along the fiberwise-isomorphism $d\pi:H\rightarrow TM$. The Sasaki metric $\tilde g$ then just takes the splitting $T(TM)=V\oplus H$ to be orthogonal. For any $X\in T_{(p,v)}TM$, we will write $X_V$ and $X_H$ for the vertical and horizontal components.

Lemma. Let $\gamma:[0,a]\rightarrow TM$ be a smooth curve with $\gamma(0)=p$. For each $t\in [0,a]$, let $P_t:T_{\gamma(t)}M\rightarrow T_{p}M$ denote parallel transport along $\gamma$. We get a curve $\nu:[0,a)\rightarrow T_{p}M$ given by $\nu(t)=P_t^{-1}\circ\gamma(t)$. Then we have$$\text{length}(\gamma)^2\geq \text{length}(\pi\circ \gamma)^2+\text{length}(\nu)^2.$$

Proof. Reparametrizing $\gamma$, we may assume that $a=1$ and that $|\gamma'(t)|$ is constant. Then \begin{align*} \text{length}(\gamma)^2&=E(\gamma)=\int_0^1|\gamma'(t)|^2dt\\&=\int_0^1|\gamma'(t)_H|^2+|\gamma'(t)_V|^2dt\\&=\int_0^1|(\pi\circ \gamma)'(t)|^2+|\nu'(t)|^2dt\\&=E(\pi\circ\gamma)+E(\nu)\geq \text{length}(\pi\circ \gamma)^2+\text{length}(\nu)^2. \quad\square \end{align*}

This lemma has the following consequences:

1. For any $v,w\in TM$, we have $d(v,w)\geq d\big(\pi(v),\pi(w)\big).$
2. In the notation of the lemma, we have $\text{length}(\nu)\geq \big||\gamma(a)|-|\gamma(0)|\big|,$ as the right-hand side is the distance between the circles containing the endpoints of $\nu$.

Now let $(p_n,v_n)\in TM$ be a Cauchy sequence. Then $(p_n)$ is a Cauchy sequence in $M$ by (1), and $|v_n|$ is bounded by (2). Since $M$ is complete, we may define $p=\lim p_n$. Since $p_n\rightarrow p$ and $|v_n|$ is bounded, there exists some $R>0$ such that the set $$K=\{(q,w)\in TM:d(p,q)\leq R\text{ and }|w|\leq R\}$$ contains every $(p_n,v_n)$ in our sequence. But $K$ is compact, so we can now be guaranteed that our Cauchy sequence converges, proving completeness.