most recent 30 from http://mathoverflow.net 2013-05-21T11:04:17Z http://mathoverflow.net/feeds/question/99221 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99221/relation-with-jacobifields-in-a-small-neighbourhood pascal 2012-06-10T08:01:56Z 2012-06-26T22:50:16Z

<p>hi,</p> <p>I have the following question: Let $(M,g)$ be a complete Riemannian manifold with all sectional curvatures non-positive. Let $p \in M$ and consider the function $d(x)=dist_{g}(x,p)$ in a sufficiently small neigbourhood of $p$ (one can assume that this neigbourhood is actually given by normal coordinates, hence by the exponential map $exp_{p}$). Let $\gamma : [0, \epsilon] \rightarrow M$ be a geodesic in this neighbourood starting at $p$. Let furthermore $J$ be a Jacobifield along with $J(0) = 0$ and $\frac{D}{dt}|_{t=0}J = v$ and $J(\epsilon) = w \not= 0$ and orthogonal to $\gamma$. Can one then make the following approximation: $\frac{d(\gamma(t))}{2} \cdot \frac{d}{dt}|J|^{2} \geq |J|^{2}$ ? If yes, why is this so? Hope for answers and tanks in advance.</p> <p>greetings pascal</p>

http://mathoverflow.net/questions/99221/relation-with-jacobifields-in-a-small-neighbourhood/99231#99231 Misha 2012-06-10T11:26:00Z 2012-06-26T22:50:16Z

<p>Your original question is false for the round sphere, since for the unit vector $v$ orthogonal to $\gamma'(0)$ you get $|J(t)|^2=\sin^2(t)$, while $\tan(t)>t$ for $t\in (0, \pi/2)$. </p> <p>It follows immediately from the Taylor expansion for $|J(t)|$ that your inequality holds if curvature of $M$ is negative. The inequality is probably also true for manifolds of nonpositive curvature, but would require a bit more work. </p> <p>Edit: Here is the proof of the inequality in the case of nonpositive sectional curvature. I will assume that $J'(0)$ is a unit vector orthogonal to $\gamma'(0)$ (since the proof in the case of the tangential Jacobi field $t\gamma'(t)$ is clear: you get the equality). Let $v(t)=|J(t)|^2$ and let $\tilde{v}=|\tilde{J}(t)|^2$, where $\tilde{J}$ is a Euclidean Jacobi field (so that $\tilde{J}(0)=0$, and $\tilde{J}'(0)$ is also a unit vector orthogonal to $\tilde{\gamma}'(0)$, where $\tilde{\gamma}$ is a Euclidean geodesic). Then you get the comparison inequality: $$ v' \tilde{v} \ge v \tilde{v}' $$<br> for all $t$, see do Carmo's book "Riemannian Geometry", proof of Rauch comparison theorem, pages 216-217. You get: $\tilde{v}= t^2, \tilde{v}'=2t$ and the above comparison inequality becomes the inequality $$ \frac{t}{2} v'\ge v $$ for all $t$, which is exactly the inequality that you are asking for (since for small $t$, $d(\gamma(t))=t$). </p>