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

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?

• The distance squared function isn't smooth on a flat cylinder Jun 5 at 0:27
• @OtisChodosh A cylinder is not a Hadamard space. Jun 5 at 16:45
• Do you mean to add "simply connected" ? Otherwise a cylinder is complete flat Riemannian manifold. Jun 5 at 18:27
• @OtisChodosh You’re right. In that case, yes I should add simply connected. Jun 5 at 18:44
• Hadamard manifolds are simply-connected by definition. Jun 5 at 23:49

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 $$\ker D^2 f_p =\{(V,V) : V\in T_pM\} = T_{(p,p)}\Delta$$.
• Thank you, I appreciate your answer. It's good to see a derivation of this stuff. Can you explain how this proves $\ker \nabla^2 f(p,p) =T \Delta$? If I assume $\nabla^2 f(p,p)((V_1,V_2),(W_1,W_2))=0$, how does that imply $V_1=V_2$ and $W_1=W_2$? I agree that $\nabla^2 f(p,p)((V,0),(0,W))=0$ implies $V=W$. Jun 6 at 20:55
• No worries, and thank you for that resource. From what I can gather, the only property of Hadamard manifolds utilized was that fact $f(x,y)=0$ if and only if $x=y$. What you wrote above seems to hold for all Riemannian manifolds, no? Jun 7 at 16:19
• Yes you are correct: what's written proves that the diagonal $\Delta \subset M\times M$ (which is the minimum of $f$ for any $M$) always has the Morse--Bott property. Of course, for a general $M$ it's unlikely that $f$ will be smooth at points far from $\Delta$. Jun 7 at 20:11