3
$\begingroup$

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.

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

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.