## relation with jacobifields in a small neighbourhood

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hi,

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.

greetings pascal

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## 1 Answer

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 $t\in (0, \pi/2)$.

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.

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: $$v' \tilde{v} \ge v \tilde{v}'$$
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 $t$, $d(\gamma(t))=t$).

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I see. so one will need some asumptions in the curvatures. but what if the curvatures are all non-positive. I will edit this. – pascal Jun 10 at 12:37
I dont see how the curvatures comes in here ? – pascal Jun 10 at 12:40
Curvature comes via Jacobi equation.  – Misha Jun 10 at 13:39
yes but in this case can it be applied to the statement above ? – pascal Jun 10 at 13:55
@Pascal, look at the Taylor expansion for $|J(t)|^2$ at $t=0$. Curvature will appear as the degree 4 term.  – Misha Jun 10 at 14:24
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