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I'm introducing myself to Jacobi fields and conjugate points along geodesics by studying Chapter 5 of Do Carmo's "Differential Geometry of Curves and Surfaces".

I was trying this exercise (exercise 5, section 5.5 of the book mentioned above) without much success after several days thinking about it:

Let $S$ be a complete surface with $K \geq \delta >0$, where $K$ denotes the Gaussian curvature of $S$ and $\delta$ is a constant. Show that any geodesic $\gamma : [0, \infty ) \longrightarrow S$ has a point conjugate to $\gamma (0)$ in the interval $\left(0, \frac{\pi}{\sqrt \delta} \right]$.

I tried to show that the exponential map $\exp_{p} : T_{p}S \longrightarrow S$ (where $p = \gamma(0)$) has a critical point at $s \gamma'(0)$ for some $s$ in the interval $\left(0, \frac{\pi}{\sqrt \delta} \right]$ since this property would imply that the point $q = \exp_{p}(s \gamma'(0))$ is conjugate to $p = \gamma(0)$ along $\gamma(t) = \exp_{p}(t \gamma'(0))$, but I didn't find an easy way to prove that since I don't see a clear reason why the exponential map should have a critical point.

I was also thinking about using Bonnet's Theorem which states that if $S$ is a complete surface with $K \geq \delta >0$, then $S$ is compact and the diameter of $S$ is lower or equal than $\frac{\pi}{\sqrt \delta}$ because the bound for the diameter of $S$ is precisely the upper bound of the interval in which the conjugate point to $\gamma(0)$ should be. So maybe I could somehow prove that if $\gamma(0)$ doesn't have a conjugate point on the stated interval then there exist two points in $S$ that are at a distance greater than $\frac{\pi}{\sqrt \delta}$ (which would contradict the definition of diameter of $S$) or maybe I could show that the fact that any two points in $S$ lie at a distance lower or equal than $\frac{\pi}{\sqrt \delta}$ implies the existence of said conjugate point? However, that's a very wild guess without any solid argument behind it.

I'm kind of lost about the way in which I should attempt to solve this exercise. Am I on the right track towards getting a solution or should I use some other results about conjugate points that I'm not considering?

Any hints or ideas would be greatly appreciated. Thanks in advance.

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The key idea in this problem is to use an ODE comparison theorem, since you're given a curvature inequality. But first, we need to set up an appropriate ODE. Note that you're given a geodesic $\gamma$, and you want to show that there is a conjugate point in $\left(0, \frac{\pi}{\sqrt{\delta}} \right]$, which means you need to show that some non-trivial Jacobi field $J$ such that $J(0) = 0$ must vanish again before $\frac{\pi}{\sqrt{\delta}}$. Without loss of generality, assume that the Jacobi field $J$ is normal, i.e. perpendicular to $\gamma'$. Also assume that the geodesic $\gamma$ is parameterized by arc length. In this scenario, the Jacobi field $J(t)$ can be written in a nice manner (see Do Carmo exercise 1, ch. 5.5). $$J(t) = \mu(t) e(t)$$ Here, $e(t)$ is the parallel transport of $J'(0)$ along $\gamma$, and $\mu(t)$ is just a smooth real valued function. The point of writing the Jacobi field in this manner is to make the Jacobi equation particularly nice. It becomes the following. $$\mu''(t) + K(t)\mu(t) = 0$$ Here, $K(t)$ is the sectional curvature of the plane spanned by $J(t)$ and $\gamma'(t)$.

We know that $K(t)$ is bounded below by the constant function $\delta$. That means we can compare the above ODE to the following ODE. $$ \mu''(t) + \delta\mu(t) = 0 $$ This ODE can be solved explicitly, and the solution to this vanishes at $0$ and $\frac{\pi}{\sqrt{\delta}}$. A standard ODE comparison theorem then tells us that the roots of the solution to the previous ODE lie between $0$ and $\frac{\pi}{\sqrt{\delta}}$ (see Chapter 8 of Coddington's book on ODEs). This solves the problem.

EDIT: The ODE I had in the previous version was incorrect and didn't work. I replaced it with the correct ODE, and the arguments for the previously incorrect ODE work with the new correct ODE as well.

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  • $\begingroup$ Almost everything's clear but how did we go from the Jacobi equation to the equality $\frac{d^2}{dt^2}|J(t)|^2 + \langle R(\gamma'(t), J(t)) \gamma'(t), J(t) \rangle = 0$? Sorry if I'm doing something wrong in my calculations but what I got is $\frac{d}{dt}|J(t)|^2 = 2 \langle J, J' \rangle$, so $\frac{d^2}{dt^2}|J(t)|^2 = 2 \langle J', J' \rangle + 2 \langle J'', J \rangle$ (by compatibility with the metric) and according to what I understood, we would need $\frac{d^2}{dt^2}|J(t)|^2$ to be equal to $\langle J'', J \rangle$ but I don't see why this is the case? $\endgroup$ – user594756 May 15 at 0:04
  • $\begingroup$ Also, are we assuming that $\gamma$ is parameterized by arc-length to get the equality $\langle R(\gamma'(t), J(t)), \gamma'(t), J(t) \rangle = K(t) | \gamma'(t) |^2 |J(t)|^2 = K(t) |J(t)|^2$ or why does the equality $\langle R(\gamma'(t), J(t)), \gamma'(t), J(t) \rangle = K(t) |J(t)|^2$ hold? $\endgroup$ – user594756 May 15 at 0:09
  • $\begingroup$ @user594756 You are correct in noticing the missing $2\langle J', J' \rangle$ term. That was a mistake on my part, and I have replaced it with the correct ODE. Also yes, I'm assuming that the geodesic is unit speed, i.e. parametrized by arc length. $\endgroup$ – sayantankhan May 15 at 3:00
  • $\begingroup$ OK, I think I got it now. Thank you very much for your very detailed answer. $\endgroup$ – user594756 May 15 at 4:58

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