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.
Best Answer
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.