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?

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

1 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 $\ker D^2 f_p =\{(V,V) : V\in T_pM\} = T_{(p,p)}\Delta$.

  • $\begingroup$ 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 $V=W$. $\endgroup$ Jun 6 at 20:55
  • $\begingroup$ @SpencerKraisler . Very sorry, I was writing too quickly this morning and made a huge mess of things (I put the wrong vectors in the domain of the Hessian). I think it's now fixed and now with the correct computation it should be easy to see what you want. (But please ask if anything is unclear!). $\endgroup$ Jun 6 at 23:02
  • $\begingroup$ 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? $\endgroup$ Jun 7 at 16:19
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    $\begingroup$ 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 $\Delta$. $\endgroup$ Jun 7 at 20:11

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