10
$\begingroup$

Given a small enough convex triangle $(abc)$ in a (smooth or Alexandrov) surface $(X,d)$ of curvature greater than $-1$, let $(\overline{abc})$ be its comparison triangle in $\mathbb{H}^2$. Then there exist a $1$-Lipschitz map $\varphi$ from the interior of $(abc)$ to the interior of $(\overline{abc})$.

Using Kirszbraun theorem (for Alexandrov spaces, as stated by Lang and Schroeder or Alexander, Kapovitch and Petrunin) this is simple. Just identify the sides of $(abc)$ and $(\overline{abc})$ in the usual way, this map is $1$-Lipschitz because of the curvature bound and we can apply Kirszbraun theorem to extend it to the interior of the triangle.

I was wondering if one could explicitly build $\varphi$ in this specific case.

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ I would be surprised if it is an overkill, but I like surprises. $\endgroup$
    – Anton Petrunin
    Commented Feb 29, 2012 at 23:58
  • $\begingroup$ I let it go for the moment, but if I found a construction I'll let you know. The reason I thought it may be possible is that I think, correct me if I'm wrong, it is possible to 1-Lipschitz fill a triangle in a $CAT(0)$ space by its comparison triangle in the plane using a explicit construction. I think choosing a vertex $a$ of the triangle and parametrize $[bc]$ by a constant speed curve $\gamma(t)$ for $t\in[0,1]$, then let $\sigma_t(s)$ be the constant speed geodesic from $a$ to $\gamma$, and associate $(t,s)$ to $\sigma_t(s)$ both in the $CAT(0)$ and the model space does the job. $\endgroup$
    – Thomas Richard
    Commented Mar 1, 2012 at 8:39
  • 1
    $\begingroup$ @Thomas: No, it does not work. The map which you are suggesting is called line-of-sight map; look at the Example on page 71 in our book. math.psu.edu/petrunin/papers/alexandrov-geometry/the-book.pdf $\endgroup$
    – Anton Petrunin
    Commented Mar 1, 2012 at 18:45

0

Reset to default

You must log in to answer this question.