I am reading Petersen's Riemannian geometry.He uses distance function to establish three equations.

1.$L_{\partial_{r}} g=2 \operatorname{Hess} r$

2.$\left(\nabla_{\partial_{r}} \operatorname{Hess} r\right)(X, Y)+\operatorname{Hess}^{2} r(X, Y)=-R\left(X, \partial_{r}, \partial_{r}, Y\right)$

3.$\left(L_{\partial_{r}} \operatorname{Hess} r\right)(X, Y)-\operatorname{Hess}^{2} r(X, Y)=-R\left(X, \partial_{r}, \partial_{r}, Y\right)$

Then he says

if we assume that the curvature is bounded, then equation (2) tells us that, if the Hessian blows up, then it must be decreasing as r increases, hence it can only go to $-\infty$:

I'm not sure what 'blow up'here means here, and I cannot understand why this implies that Hessian is decreasing.

He also says

A conjugate point occurs when the Hessian of r becomes undefined as we solve the differential equation for it.

I cannot imagine when and why will the Hessian of r becomes undefined.

Other places in this section are difficult to understand too since there's only literal explanation without examples or calculations.

Any help will be thanked.

  • 1
    $\begingroup$ @JeanMarie Sorry and I have corrected it. $\endgroup$
    – Tree23
    Jan 23 at 15:00

1 Answer 1


Fix a unit tangent vector $v$ that is parallel along the geodeic. Let $H$ be the Hessian of $r$. Then equation 2 implies that \begin{align*} \partial_r(H(v,v)) &= (\nabla_rH)(v,v)\\ & = -R(v,\partial_r,\partial_r,v) - H^2(v,v)\\ &= -R(v,\partial_r,\partial_r,v) - g^{ij}H(v,\partial_i)H(v,\partial_j) \end{align*} Sayiing $H$ blows up along $R$ means that the last term approaches infinity. In particular, it becomes arbitrarily negative. That curvature is bounded implies that the first term is bounded. Therefore, as $r \rightarrow \infty$, the right side becomes negative. Therefore, $H(v,v)$ is decreasing for any parallel unit vector $v$.

As for an example where the Hessian blows up and therefore there is a conjugate point, I suggest looking at the sphere as described in 4.2.1 and see if you can figure out why, if $r$ is the distance from the north pole, then the Hessian of $r$ blows up at the south pole.

  • $\begingroup$ I tried this example, if I take metric on $S^{2}$ as $g=dt^{2}+\sin^{2}(t)d\theta^{2}$,then I take the distance on sphere to North Pole as distance function $r=t$,then the gradient of r is $∂r=∂t.$Then I calculate the hessian and get$Hessr=\sin(r)\cos(r)d\theta^{2}$(I may make mistake in this step).Then I get some trouble, I cannot see why this Hess blow up at $r=\pi$,where did I make mistake? $\endgroup$
    – Tree23
    Jan 27 at 2:16
  • $\begingroup$ If I write the metric as $g=dt^{2}+\sin^{2}(t)d\theta^{2}=dt^{2}+g_{t}$ then $Hessr=\frac{\cos(r)}{\sin(r)}g_{t}$.I take a parallel unit vector field $v$ so $g_{t}(v,v)=1$ so $Hessr(v,v)=\frac{\cos(r)}{\sin(r)}$.This time it seems that $Hessr$ blow up.Is this right? $\endgroup$
    – Tree23
    Jan 27 at 2:43

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