# Is the vector field associated with an element of the boundary at infinity on a Hadamard manifold smooth?

A Hadamard manifold $$M$$ (complete, simply connected, non-positive sectional curvature) has a so-called boundary at infinity $$\partial M$$ whose elements are equivalence classes of unit-speed geodesic rays that are asymptotic i.e. two unit-speed geodesic rays $$\gamma_1,\gamma_2:[0,\infty)\rightarrow 0$$ are members of the same element of $$\partial M$$ iff $$d(\gamma_1(t),\gamma_2(t))$$ is bounded on the domain. For each $$p\in M$$ and $$u\in \partial M$$, there exists a unique element of $$u$$ associated with $$p$$ i.e. a unique unit geodesic ray $$\gamma_p$$ satisfying $$\gamma_p(0)=p$$ and $$\gamma_p\in u$$. Define a vector field $$M\rightarrow TM$$ by $$p\rightarrow (p,\gamma_p'(0))$$; that is, $$p$$ is mapped to the initial velocity of the element of $$u$$ associated with $$p$$. Is this vector field smooth?

I feel like this should be a known result but cannot find it in the literature. Any references would be appreciated.

• It is known to be $C^1$-smooth. It fails to be $C^2$ in general. Sep 12 at 1:25
• @MoisheKohan Thank you. Are there any more details you could provide? I am specifically interested in conditions under which it is once or twice continuously differentiable, but even just a reference would be a big help. Sep 12 at 1:44

This vector field is the negative of the gradient vector field of the/a Busemann function $$b_u$$ associated with $$u$$. (Below, I am assuming that the Riemannian metric is $$C^\infty$$.) The function is known to be $$C^2$$ for Hadamard manifolds but not $$C^3$$ (even for general Hadamard surfaces). See:
From this, one sees that the vector field is $$C^1$$ but not, in general, $$C^2$$-smooth.