Geodesic axis and displacement function So let us assume our manifold is complete, simply connected, and has nonpositive sectional curvature. If we assume that the displacement function $f(x)=d(x,\phi(x))$, for an isometry $\phi:M\rightarrow M$, is bounded from below by a positive value, show that there exists a geodesic axis. That is there exists a geodesic $\gamma$ such that $\phi\circ \gamma(t)=\gamma(t+t_0)$ assuming that $f(x)=\inf(f)$ for some $x\in M$.
So here is what I have so far. It seems that I should select the geodesic that goes from $x$ to $\phi(x)$ where $f(x)=\inf(f)$. I had two ideas for showing that this is an axis. Namely we could show that $\phi\circ\gamma(-t_0)=\gamma(0)$, where $\gamma(0)=x$ and $\gamma(t_0)=\phi(x)$, then we could use that geodesics are unique in this type of manifold, or we could show that $f(\gamma)$ is constant which could also establish our result, but I'm not sure how to go about with either. Any suggestions? Thanks.
 A: Your first idea (using the uniqueness of geodesics) is good. The key formula is the derivative of the distance function: we are given a minimum $x_0$ of $f$, so we'd like to extract some information from the critical point condition $Df_{x_0} = 0$. The hypotheses on the manifold guarantee the distance function will be smooth away from the diagonal, so this is definitely true.
If we let $\gamma$ be an arclength-parametrized minimizing geodesic joining $x$ to $y$ and $L$ be its length, then you should be able to derive from the first variation formula that
$$ (Dd_{(x,y)})(u,v) =\langle\dot \gamma(L),v\rangle - \langle \dot \gamma(0), u\rangle.$$
Applying this with $x = x_0, y = \phi(x_0), v = D\phi(u)$ we get
$$ Df_{x_0}(u) = (Dd_{(x_0,\phi(x_0)))})(u,D\phi(u))=\langle\dot \gamma(L), D\phi(u)\rangle - \langle \dot \gamma(0), u\rangle=0$$
for all vectors $u \in T_{x_0} M$. Since $\phi$ is an isometry, we can rewrite this as $$\langle\dot \gamma(L) -  D\phi(\dot \gamma(0)), D\phi(u)\rangle=0$$
for all $u$, and thus the fact that $D\phi_{x_0}$ is bijective tells us that $\dot \gamma(L) = D\phi(\dot \gamma(0))$. Since $\phi$ is an isometry we know $\phi\circ \gamma$ is geodesic, so the uniqueness of geodesics extends this equality of velocities at a point to equalities of the entire geodesics; i.e. $\phi(\gamma(t)) = \gamma(t + L)$.
