# How to understand the explanation of conjugate point in Petersen's Riemannian geometry

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.

• @JeanMarie Sorry and I have corrected it. Jan 23 at 15:00

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.
• 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? Jan 27 at 2:16
• 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? Jan 27 at 2:43