Monotonicity of Riemannian Logarithm

Question

Let $(X,g)$ be a complete, connected, Riemannian manifold of non-positive curvature. Let $(z_n)_{n=1}^{\infty}$ be a sequence in $X$ converging to some $z \in X$ such that $$ d_g(z,z_{n+1})\leq d_g(z,z_n), $$ for every $n\geq 1$; where $d_g$ is the intrinsic metric induced by the riemannian metric $g$. By the Cartan-Hadamard theorem, we know that in this case each $\exp^{-1}_x$ is globally well-defined on $X$.
Is it then true, that the sequence of functions $$ \begin{aligned} F_n:X^2 &\rightarrow [0,\infty)\\ (x,y) & \mapsto \|\exp_{z_n}^{-1}(y)-\exp_{z_n}^{-1}(x)\|, \end{aligned} $$ converges to $(x,y)\mapsto \|\exp_{z}^{-1}(y)-\exp_{z}^{-1}(x)\|$ uniformly and monotonically (ie $F_{n+1}(x,y)\leq F_{n}(x,y)$ for each $x,y \in X$?)

Answer 1

This is not true. Assume $X$ is the hyperbolic plane (with constant curvature $-1$). Fix any $n$, let $w=d(z_n, z)$. Extend the geodesic from $z_n$ to $z$ by distance $b$ and reach a point $x$; and at $z$ draw a geodesic of length $b$ that is perpendicular to the first geodesic, reach a point $y$. Here $b$ is a large number depending on $w$. Write $a=d(z_n, y)$. So $$ |\exp_z^{-1}(y)-\exp_z^{-1}(x)|=\sqrt 2 b. $$ Now by the hyperbolic law of cosine, $$ \cosh a=\cosh b \cdot \cosh w. $$ Now $a, b$ are both big, $\cosh b=\frac{e^{b}}{2}(1+e^{-2b})$, so $\log \cosh b=b-\log 2+O(e^{-2b})$. $w$ is small, so $\log\cosh w=\log(1+\frac{w^2}{2}+...) =\frac{w^2}{2}+O(w^4)$. So $$ a-\log 2+O(e^{-2b})=b-\log 2+O(e^{-2b})+\frac {w^2}{2}+O(w^4), $$ i.e. $a=b+\frac{w^2}{2}+O(e^{-2b})+O(w^4)$. We can take $b$ so big that we have $e^{-2b}\ll w^4$, so $$ a=b+\frac{w^2}{2}+O(w^4). $$ Apply hyperbolic law of cosine to the same thin triangle $z_nzy$ we get $$ \cos\theta=\frac{\cosh a\cosh w-\cosh b}{\sinh a\sinh w} =\frac{\cosh b\cosh w\cosh w-\cosh b}{\sinh a\sinh w} =\frac{\cosh b\sinh^2 w}{\sinh a\sinh w}=\frac{\cosh b\sinh w}{\sinh a}. $$ From $a=b+\frac{w^2}{2}+O(w^4)$ and $b$ is sufficiently big, this implies $$ \cos\theta=w+O(w^2). $$ This is the key estimate, saying that $\theta$ differs from the right angle $\pi/2$ by $O(w)$, much bigger than the Euclidean case $O(w/a)$.

So, finally, $$ \begin{aligned} |\exp_{z_n}^{-1}(y)-\exp_{z_n}^{-1}(x)|=&\sqrt{(b+w)^2+a^2-2(b+w)a\cos\theta}\\ =&\sqrt{b^2+w^2+2bw+b^2+b\cdot O(w^2)-2(b+w)(b+O(w^2))(w+O(w^2))}\\ =& \sqrt{2b^2+w^2+2bw+b\cdot O(w^2)-2b^2w+b^2O(w^2)}\\ =& \sqrt{2b^2+2bw-2b^2w+b^2O(w^2)}\\ =& \sqrt 2 b\sqrt{1+\frac wb-w+O(w^2)}\\ =& \sqrt 2 b\Big(1-\frac w2 +\frac {w}{2b} +O(w^2)\Big); \end{aligned} $$ this differs from the limit $\sqrt 2 b$ by at least $\frac{\sqrt 2 b w}{3}>1$, if $b$ is sufficiently big.

So the converge is not uniform. And there is no reason to think monotone (try travel backwards in a direction from $z$ to $z_n$...)