Parallel Jacobi fields in a Hadamard manifold

Question

Let $M$ be a Hadamard manifold and let $c: \mathbb{R}\rightarrow M$ be a geodesic. A Jacobi field $Y$ along $c$ is called parallel if $Y'(t) = 0$ for every $t\in \mathbb{R}$. If we assume that $M$ is cocompact (i.e. there exists a compact subset $K$ of $M$ such that for any $p\in M$ there is an isometry $\varphi$ of $M$ such that $\varphi(p)\in K$) does it then hold that if $Y$ is a Jacobi field along $c$ such that $Y'(t) = 0$ for every $t\geq 0$, then $Y'(t) = 0$ for every $t\in \mathbb{R}$?

Answer 1

The answer is "no".

Suppose $D$ be a disc with one handle, nonpositive curvature, geodesic boundary with flat collar. Note that there is a geodesic $\gamma\colon[0,\infty)\to D$ that starts at $\partial D$ in the direction perpendicular to $\partial D$.

We can assume that $\partial D$ is isometric to $\mathbb{S}^1$. Take two copies of the product $D\times \mathbb{S}^1$ and glue them along the boundary so that the $\mathbb{S}^1$-factor is identified with to the $\partial D$-factor and the other way around. We can assume that two copies of $\gamma$ form an infinite geodesic in the glued space.

It produces a $C^\infty$-smooth example. On the othere hand, (evidently), there is no analytic example.