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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.

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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.

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  • $\begingroup$ 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
    Commented 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 $\beta = \sqrt 3$. $\endgroup$
    – Richard Montgomery
    Commented Jul 24, 2015 at 14:34
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    $\begingroup$ For $\phi(t)=|t|$ the angle is $<\pi$, something like $\phi(t)=|t|^{1.00001}$ should work. $\endgroup$
    – Anton Petrunin
    Commented Jul 24, 2015 at 14:51
  • $\begingroup$ @RichardMontgomery, it works for any $\beta>0$, but the angle in this case is $<\pi$... $\endgroup$
    – Anton Petrunin
    Commented 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 $\endgroup$
    – Richard Montgomery
    Commented Jul 25, 2015 at 20:22

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