Let $M$ be a simply connected Hadamard manifold. That is, $M$ is a complete Riemannian manifold with sectional curvature bounded above by 0. Let $f(x,y)=\frac{1}{2}d^2(x,y)$, where $x,y \in M$, and $d(.,.)$ is geodesic distance. I'm curious if $f:M^2 \to \mathbb{R}$ is a Morse-Bott function, and how to show it.


I know that $\nabla f(x,y)=(-\log\_x(y),-\log\_y(x))=0$ if and only if $x=y$. If I understand correctly, this is because $\exp\_x:T\_xM \to M$ and $\exp\_y:T\_yM \to M$ are both diffeomorphisms. Following from this, the critical points of $f$ are precisely $C^\* =\{(x,x):x \in M\}$, hence a closed submanifold. It follows then to show that the kernel of $\nabla^2 f(x,x)$ is precisely the tangent bundle of $C^\*$. This is something I find difficult as I do not really know how to compute the hessian of $f(x,y)=\frac{1}{2}d^2(x,y)$. Could I possibly get some help on this?



Is squared geodesic distance a Morse-Bott function on simply-connected Hadamard manifolds?
Choose $V,W \in T\_p M$ and note that
$$
D^2f((V,W),(V,W)) = \frac{d^2}{dt^2}\Big|\_{t=0} \frac 12 d(\exp\_p(tV),\exp\_p(tW))^2 .
$$
By the computation in [Section 1.3 here](https://www2.math.upenn.edu/%7Ewziller/math660/TopogonovTheorem-Myer.pdf)
$$
\frac 12 d(\exp\_p(tV),\exp\_p(tW))^2 = \frac{t^2}{2} |V-W|^2 + O(t^3). 
$$
Thus we find
$$
D^2f((V,W),(V,W)) = |V-W|^2.
$$


Now, since both sides are quadratic bilinear forms the parallelogram identity implies
$$
D^2f((V\_1,W\_1),(V\_2,W\_2)) = \langle V\_1-W\_1,V\_2-W\_2\rangle.
$$


Now, recall that $(V\_1,V\_2)$ is in the kernel of the Hessian if and only if $D^2 f((V\_1,V\_2),(W\_1,W\_2))$ vanishes for ALL $(V\_1,V\_2)$ (this is the definition of the kernel of a symmetric bilinear form). In particular, we see that $(V\_1,W\_1)$ is in the kernel if and only if $V\_1-W\_1=0$ (by nondegeneracy of the form).


Thus, we see that $\ker D^2 f\_p =\{(V,V) : V\in T\_pM\} = T\_{(p,p)}\Delta$.


