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I posted this question on MSE first but it seems I'm not getting an answer. I'm reading the paper on the generalized sphere theorem by Grove and Shiohama, and there is an observation they made that I'm struggling to prove.

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We have two unit geodesics $\tau,\sigma:[0,L]\to M$, where $M$ has sectional curvatures $sec\geq\delta>0$, and let $c_t:[0,1]\to M$ be the geodesic joining $\tau(t)$ to $\sigma(t)$, for $t\in[0,L]$. Here $\tau$ and $\sigma$ are close, in the sense that each $c_t$ has length smaller than the convexity radius of $M$. Then they claim that the Jacobi fields $J_t$ induced by $\{c_t\}$ has norm bounded by $2$, but I can't rigorously prove it.

Intuitively, this sounds correct, since the boundary values of the Jacobi fields are the unit tangent vectors of $\tau$ and $\sigma$, so nearby geodesics should have points at the same proportional distance stretched by at most $2$. By this I mean that $d(c_{t'}(s),c_t(s))$ should be controlled, since for $t'$ close to $t$, the length of $c_{t'}$ is bounded by the length of $c_t+2(t'-t)$, by the 1st variation formula, but I don't have a quantitative estimate. Since there are no conjugate points in the convexity radius, $J_t$ is uniquely determined by the endpoint values. Then the initial velocities are also known (they are initial velocities of the bridge geodesics).

I also tried to use Rauch or Berger comparison, but for this we need to split the Jacobi field into $2$ parts, each with a zero end point value, and then we lose information in the triangle inequality and it doesn't seem to work.

Any help is appreciated, thanks

Edit: I will work out the answer by user @anything

In a convex neighborhood $W$ of $p\in M$, the distance function $r(q):=d(p,q)$ is convex along geodesics, and smooth on $W\backslash\{p\}$. Hence its Hessian is positive semi-definite. Plugging a Jacobi field $J$ along a radial geodesic $c$, we have $0\leq Hess_r(Y,Y)=(\nabla_Y\nabla r,Y)=(\nabla_{\nabla r}Y,Y)+([\nabla r,Y],Y)=(Y',Y)+0$ The term containing the Lie bracket is $0$ because $Y$ is the variation field and $\nabla r=\partial/\partial t$ along $c$.\ It follows that $(|Y|^2)'=2(Y',Y)\geq0$, so $|Y|$ is monotone.

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The idea from your last paragraph looks fine. But probably the comparison are made against the convexity of the ball, not against the curvature inequality $sec > \delta$ (usually convexity is ensured by the upper bound of the curvature).

The convexity of the ball means that the distance function $t\mapsto d(\tau(t_0), \sigma(t))$ or $t\mapsto d(\tau(t),\sigma(t_0))$ is convex function. Hence the corresponding Hessian is positive and given by the Jacobi field $Y$ along the geodesic $c_{t_0}$ as $(\nabla_{s}Y,Y)$. This Jacobi field vanishes at one endpoint and tangent to $\dot\tau$ or $\dot\sigma$ at the other end point. Thus the Jacobi field $J_{t_0}$ in question is given by the sum of such Jacobi fields, say $J_{t_0}=Y_1+Y_2$. Since $(\nabla_{s}Y,Y)\geq 0$ by convexity, $\vert Y_1\vert$, $\vert Y_2\vert$ is monotone along $c_{t_0}$, thus both are bounded by $1$. Hence norm of $J_{t_0}$ is bounded by $2$.

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  • $\begingroup$ I see, thank you so much for your answer, it really helped a lot. I did think using convexity, but only with the distance function and not the Hessian. $\endgroup$ Commented Dec 6 at 22:34

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