Jacobi fields in singular metric on quotient space

Question

Consider the square $\Omega = (0,\pi) \times (0,\pi/2) \ni (r,\theta)$ endowed with the Riemannian metric \begin{equation} f^2 \big(\mathrm{d} r^2 + \sin^2(r) \, \mathrm{d} \theta^2 \big), \end{equation} where \begin{equation} f = (\sin r)^4 (\sin 2 \theta)^2. \end{equation} This arises when quotienting the unit sphere $S^6 \subset \mathbf{R}^7 = \mathbf{R}^6 \times \mathbf{R}$ by the group action of $SO(3) \times SO(3)$, with $f$ being to the volume of the fibre lying above $(r,\theta)$.

Let $\gamma$ be a geodesic segment with $\gamma(0) \in \partial \Omega$. Can $\gamma$ admit a bounded, normal Jacobi field $J$ with $J(s) \not \to 0$ as $s \to 0$?

I am confused about this point because of the following two observations.

Hsiang [Hsi83] explains that the geodesic equation is uniquely solvable for every point $(r,\pi/2) \in \partial \Omega$ with $0 < r < \pi$, and the geodesic $\gamma_r$ is orthogonal to the boundary at $\gamma_r(0) = (r,\pi/2)$. Fix some $r_0$, and let $\gamma := \gamma_{r_0}$. Varying $r$ yields a one-parameter family of geodesics, which induces the normal Jacobi field $J = \frac{\partial \gamma_r} {\partial r} $ on $\gamma$ with $J(0) = 1, J'(0) = 0$.

However, the Jacobi equation along $\gamma$ is $J'' + K_g J = 0$. Because $K \to +\infty$ near the boundary, the only bounded solutions should have $J(s) \to 0$ as $s \to 0$.

Wu-Yi Hsiang. Minimal cones and the spherical Bernstein problem. Annals of Mathematics, Vol. 118 (1983), pp. 61-73.

Answer 1

I don't think your analyses are correct.

Consider the following toy problem which has the same boundary behavior as yours: let domain be $(r,\theta)\in \mathbb{R}\times (0,\infty)$ and set the metric to be $\theta^4 (dr^2 + d\theta^2)$.

The geodesic equation becomes (after computing explicitly the Christoffel symbols)

$$ \ddot{r} + \frac{4}{\theta} \dot{r}\dot{\theta} = 0 $$

and

$$ \ddot{\theta} + \frac{2}{\theta} (\dot{\theta}\dot{\theta} - \dot{r}\dot{r}) = 0 $$

The first equation shows (as expected, since $\partial_r$ is Killing) that $ \theta^4 \dot{r} $ is a constant of motion. The second equation shows that any critical point of $\theta$ is a local minimum, and hence if $\liminf \theta = 0$ we must have $\theta$ is monotonic decreasing. The second equation can be rearranged to read

$$ \frac{d}{ds} (\theta^2 \dot{\theta}) = \frac{2C^2}{\theta} > 0 $$

where $C$ is the constant of motion for $\dot{r}$. this shows that $|\dot{\theta}| < \frac{C'}{\theta^2}$. Hence

$$ \Big|\frac{\dot{r}}{\dot{\theta}}\Big| \geq \frac{C''}{\theta^2} $$

which shows that the only bounded geodesic has $C = \dot{r} = 0$ and $\theta^2\dot{\theta}= C'''$, which implies (after reparametrization)

$$ r = r_0 , \theta(s) = \sqrt[3]{s} $$

Up to this part, we see that our model problem has the same behavior as Hsiang's: the only geodesics that terminate on the boundary are those that are normal to the boundary.

The Jacobi field $J$ therefore has the form $J(s) = \partial_r$ everywhere (again, not surprising, since Killing fields are always Jacobi). We can normalize the geodesics so that the velocity vector is $\theta^{-2}\partial_\theta$. So the first derivative of $J$ is

$$ J' = \theta^{-2}\nabla_{\partial_\theta} \partial_r = \theta^{-2} \Gamma_{r\theta}^r \partial_r = 2\theta^{-3} \partial_r $$

and

$$ J'' = \theta^{-2}\nabla_{\partial_\theta} J' = -2\theta^{-6} \partial_r $$

So we see two things:

1. $J'(0) \neq 0$ (contrary to your claim)
2. Both $J'$ and $J''$ blow up (both in coordinate expression and in norm) as you approach the boundary.

The first claim should also be expected: even in the standard, non-singular geometry, if you launch a family of normal geodesics from a hypersurface, then the Jacobi field generated should satisfy $J'(0) = S_{\gamma(0)}J(0)$, where $S$ is the shape operator / Wirtinger map / second fundamental form of the hypersurface. Unless you have a geodesic boundary, in general $J'$ is not zero.

So no, other than the incorrect claim on the initial velocity of the Jacobi field, there does not appear to be any contradiction.