1
$\begingroup$

Let two geodesic segments in an Alexandrov space with curvature bounded from below start at the same point and the angle between them equals $\pi$. It is possible that these segments are not the two complementary pieces of a larger geodesic?

Remark. In a smooth riemannian manifold this is clearly impossible.

$\endgroup$

1 Answer 1

3
$\begingroup$

Yes, it is possible.

Consider the graph $\Gamma$ of $$z=\phi\left(\sqrt{x^2+y^2}\right),$$ where $\phi\colon\mathbb R\to \mathbb R$ is a convex even function with $\phi(0)=0$. With its intrinsic metric, $\Gamma$ forms an Alexandrov space.

Note that the curve $\gamma$ on the graph described by $y=0$ is formed by two geodesics starting at the origin.

For an appropriate choice of $\phi$ there is always a shortcut which goes around the origin, from say from $(\varepsilon,0,\phi(\varepsilon))$ to $(-\varepsilon,0,\phi(\varepsilon))$. That is, arbitrary small segment of $\gamma$ containing the origin does not minimize the length.

Roughly, $\phi$ has to have infinite second derivative at $0$ in a strong sense.

$\endgroup$
5
  • $\begingroup$ If one takes $\phi(t)=|t|$, will it work? In other words, one takes the graph of $z=\sqrt{x^2+y^2}$. $\endgroup$
    – asv
    Jul 24, 2015 at 11:48
  • $\begingroup$ No, but it works for $\phi(t) = \beta |t|$ when $\beta$ is large enough, I believe any $\beta > \sqrt{3}$ works. (Make cones out of paper. Lines drawn before folding are geodesics.) The border line case is a ``cone over a circle of circumference $\pi$''. The metric of a cone over a circle of circumf $2 \pi \lambda$ is $ds^2 = dr^2 + \lambda^2 r^2 d \theta ^2$ in polar coord based at the cone point. The border-line case is the cone made from a standard rectangular piece of paper (fold a half-plane), so $\lambda =1/2$, and I believe corresponds to $\beta = \sqrt 3$. $\endgroup$ Jul 24, 2015 at 14:34
  • 1
    $\begingroup$ For $\phi(t)=|t|$ the angle is $<\pi$, something like $\phi(t)=|t|^{1.00001}$ should work. $\endgroup$ Jul 24, 2015 at 14:51
  • $\begingroup$ @RichardMontgomery, it works for any $\beta>0$, but the angle in this case is $<\pi$... $\endgroup$ Jul 24, 2015 at 14:54
  • $\begingroup$ @AntonPetrunin. Thanks for the correction! Yes. Of course! I didn't take my own advice and draw lines on paper of a plane minus a sector before folding the paper to make a cone. Any small sliver of a sector deleted shows you that as you say any $\beta > 0$ does the trick $\endgroup$ Jul 25, 2015 at 20:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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