Is squared geodesic distance a Morse-Bott function on simply-connected Hadamard manifolds?

Question

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?

Answer 1

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 $$ \frac 12 d(\exp_p(tV),\exp_p(tW))^2 = \frac{t^2}{2} |V-W|^2 + O(t^3). $$ 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 $\ker D^2 f_p =\{(V,V) : V\in T_pM\} = T_{(p,p)}\Delta$.