Let's work on a Riemannian manifold $M$ of nonpositive sectional curvature. 


Fix a unit-speed geodesic $\beta$, and a Jacobi field $\eta$ over it. Assume that $\eta(0)$ is nonzero and orthogonal to $\beta'(0)$, and that $\eta'(0)$ (i.e. $\nabla\_{\beta'} \eta (0)$) equals $0$. 


Under these conditions, it's known that $\eta(t) \neq 0$ for every $t$ (indeed, the minimum of $\|\eta(t)\|$ is attained at $t=0$). However, I'd like to show that $\eta(t)$ stays nonzero in a constant direction''; more precisely:


QUESTION: Let $\zeta(t)$ be the parallel transport of $\eta(0)$ along $\beta$. Is it true that $\langle \eta(t), \zeta(t) \rangle > 0$ for every $t$?


If the answer is no, then it looks like the Jacobi field ``turns around'' the geodesic. Is it possible? In that case, I ask:


QUESTION 2: What if we additionally assume that $M$ is a symmetric space?



About Jacobi fields on nonpositive curvature
I think that the answer to quetion 2 is yes. The basic property of Jacobi fields defined along a geodesic $\gamma$ (parameterized at speed $1$) is that they satisfy the equation
$$ \eta''(t)=\pm R(\gamma',\eta)\gamma'~, $$
where $R$ is the curvature operator, $\eta''$ is with respect to the covariant derivative along $\gamma$, and the sign depends on the convention used. 


Note that $H:x\mapsto \pm R(\gamma', x)\gamma'$ is self-adjoint, and positive semi-definite if the sectional curvature is non-positive. Moreover if $M$ is a symmetric space then its curvature operator is parallel, so $H$ is also parallel. So if you bring everything back at a point by parallel transport, the Jacobi field is solution of an equation of the form $\eta''=H\eta$ with $H$ constant and positive semi-definite. So your positivity condition should hold.


