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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?

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I'm pretty rusty, but I think you can control how much the Jacobi field twists, but it might require a lower bound on the sectional curvature. This would allow you to estimate how long $\eta(t)$ remains nonzero along the geodesic. I don't think I know (alas, I don't even know what I know or don't know anymore) any way to ensure that it remains nonzero along the entire geodesic, and I'm not sure I believe that it's true unless the curvature approaches zero. When I learned about Jacobi fields, I found the papers of Karcher and Jost-Karcher extremely useful. –  Deane Yang Aug 27 '11 at 2:44
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2 Answers

up vote 4 down vote accepted

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.

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Question 1: answer is "NO".

Let $\mathbb H^2$ be Lobachevsky plane. Consider space $\mathbb R\times \mathbb H^2$ with geodesic $\beta$ in $0\times \mathbb H^2$. Note that generic Jacoby field tends to approach direction of $\mathbb H^2$.

Now slice your space by Euclidean planes perependicular to $\beta$ and glue together back so that the direction of $\mathbb H^2$ rotates. This way you can produce a $C^{1,1}$ metric so that your Jacoby field rotates as you want. If needed the metric can be smoothed.

I think that is a good way to make one believe that Jacoby field rotates, but it would be pain to write down.

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That should be a comment to Anton Petrunins post. Where do you know, that the parallel transport won't be rotating either? –  kostja Aug 28 '11 at 16:56
    
@kostja, there is no absolute for vector field along a geodesic, but one can compare two vector fields. For example a Jacoby field and parallel field as in this question. –  Anton Petrunin Aug 29 '11 at 3:30
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