I've been trying to solve the following exercise:
Let $M$ be a Hadamard manifold (simply connected, complete and with sectional curvature $K \leq 0$). Show that:
i) Let $p \in M, v, w \in T_pM$ linearly independent, $\gamma_v$ the geodesic with initial condition $v$, and $E_w$ the parallel vector field along $\gamma_v$ with $E_w(0) = w$. If $\|\mathrm{d}(\exp_p)_v(w)\| = \|w\|$, then $K(\gamma_v'(t), E_w(t)) = 0$ for all $0 \leq t \leq 1$.
ii) Every metric ball $B_r(p)$ in $M$, $r \geq 0$, is strictly convex, i.e every geodesic segment connecting two points of $B_r(p)$ is contained in $B_r(p)$.
But I haven't had any good ideas so far. I know the exponential map is a global diffeomorphism in this case and I think maybe some solution could come from using Jacobi fields/variations but I couldn't think of anything concrete along those lines. I'd appreciate any help. Thanks in advance!
EDIT: Since Hadamard manifolds don't have any conjugate points, I see now how i) is a straightforward consequence of Rauch's comparison theorem. I'm still stuck on ii) though and I realize now I should've split it into two posts, so I'm going to ask it in another post.
Best Answer
Not an answer but definitely too long for a comment. I'll edit this whenever something new comes up. Any hint, help or comment is appreciated.
Question i) We consider $(M,g)$ a Riemannian manifold. Define $\gamma(t) = \exp_p(tv)$, (suppose $\|v\|=1$ for convinience) and $Y(t) = \mathrm{d}\exp_p(tv)\cdot tw$. It is well known that $Y$ is a Jacobi field along $\gamma$, that is: $$ Y'' =-R_{\gamma'}Y \left(:= -R(\gamma',Y)\gamma'\right), $$ and that it is the unique one satisfying $Y(0) = 0, Y'(0) = w$. You may have another convention for $R$, but what is important is that $g\left(R_{\gamma'}X,X\right) = \sec(\gamma',X)\|X\|^2$.
Finally, we suppose that $M$ is Hadamard, (and it follows that $Y(t) \neq 0$ if $t \neq 0$) and that $\|Y(1)\| = \|w\|$.
Step 1. Let $f(t) = \|Y(t)\|$ for $t\in [0,1]$. Then $f$ is a convex function. Indeed, $f$ is smooth whenever $Y \neq 0$, thus is smooth on $]0,1]$. Moreover, on $]0,1]$: \begin{align} f' &= \frac{g(Y',Y)}{\|Y\|} \\ f'' &= \frac{\left(g(Y'',Y) + \|Y'\|^2 \right)\|Y\| - g(Y',Y) \frac{g(Y',Y)}{\|Y\|}}{\|Y\|^2} \\ &= \frac{-\sec(\gamma',Y) \|Y\|^4 + \|Y'\|^2\|Y\|^2 - g(Y',Y)^2}{\|Y\|^3}\\ &= -\sec(\gamma',Y)\|Y\| + \frac{\|Y'\|^2\|Y\|^2 - g(Y',Y)^2}{\|Y\|^3}. \end{align} From the Cauchy-Schwarz inequality, the last term is $\geqslant 0$, and from the Hadamard hypothesis, $\sec(\gamma',Y) \leqslant 0$, hence: $$f'' \geqslant 0,$$ and $f$ is convex.
Step 2. We have an upper bound on $f$. It is clear that $f(0) = 0$ and $f(1) = \|w\|$. By convexity: $$ \forall t \in [0,1], f(t) \leqslant (1-t)f(0) + tf(1) = t\|w\|, $$ hence, $f(t) \leqslant t \|w\|$.
Step 3. This latter inequality is an equality. To see this, define: $$ h(t) = \begin{cases} \frac{f(t)}{t\|w\|} & \text{if } t>0 \\ 1 & \text{if } t =0 \end{cases}. $$ Then $h$ is continuous (because $Y(t) \sim_0 tY'(0)$), smooth on $]0,1]$, and: $$ h'(t) = \frac{tf'(t) - f(t)}{t^2\|w\|}. $$ Write $tf'(t) - f(t) = tf'(t) - f(t) - (0\times f'(0) - f(0))= \int_0^t \left(sf'(s) - f(s)\right)'\mathrm{d}s = \int_0^t sf''(s)\mathrm{d}s$. It follows, from the convexity of $f$, that $h' \geqslant 0$. Hence, $h$ is non-decreasing and: $$ \forall t \in [0,1], h(t) \geqslant h(0) = 1 $$ that is: $$ \forall t \in [0,1], f(t) \geqslant t\|w\|. $$ Finally, $f(t) = t\|w\|$ for all $t \in [0,1]$.
Step 4. $\forall t \in ]0,1]$, $\sec(\gamma'(t),Y(t)) = 0$. Indeed, $f$ is linear, hence $f''=0$. it follows that: $$ 0 = f'' = -\sec(\gamma',Y)\|Y\| + \frac{\|Y'\|^2\|Y\|^2 - g(Y',Y)^2}{\|Y\|^3}\geqslant 0 $$ and as all terms are $\geqslant 0$ on the RHS, we can conclude that $\sec(\gamma',Y) = 0$.
Note that we are in presence of the equality case in the Cauchy-Schwarz inequality, and we can therefore claim that $Y(t)$ and $Y'(t)$ are linearly dependant.
From now, I cannot conclude that $Y$ is colinear to $E_w(t)$, the parallel transport of $w$ along $\gamma$ ; it seems natural in this case, but I guess I'm missing the trick.
Update 1.
We have already shown that $Y'$ and $Y$ are colinear. Hence, there exists $\alpha$, a function defined on $(0,1]$, with $Y' = \alpha Y$. Differentiating this equality gives: $$ Y'' = (\alpha' + \alpha^2)Y, $$ and taking the scalar product with $Y$ gives, recalling that $\sec(\gamma',Y) = 0$: $$ \alpha' + \alpha^2 = 0 $$ It follows that $\alpha(t) = \frac{1}{t - c_0}$ for a constant $c_0$. The initial data give $c_0=0$, i.e $\alpha(t) = \frac{1}{t}$.
Summary of what we have shown for now.