9
$\begingroup$

$X$ be an Alexandrov's space with lower curvature bound and $C$ be a totally convex subset, i.e. for any $x,y \in C$ and any geodesic $\gamma$ (that is a locally shortest path) connecting $x$ and $y$ we have $\gamma \subseteq C$. For $p \in C$ the tangent cone $K_pC \subset K_p X$ is thus well defined. My question is:

Is $K_pC$ totally convex as well?

It is not hard to see that $K_pC$ is convex in the sense that any unique shortest connection between points in $K_pC$ also lies within $K_pC$, solving this problem for example in the riemannian case. (In fact let $v,w \in K_pC$ together with a unique shortest geodesic $\gamma$ connecting the two points. Using the scaling invariance of the problem together with $(K_pC,0) = \lim_{\lambda \to \infty} (\lambda C,p)$ one may approximate $\gamma$ by geodesics contained in $C$. But i think in general it might not be possible to approximate arbitrary geodesics like this).

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ The totally convex subset $C$ usually appears as a sublevel set of a convex function (I do not know other sources of totally convex subsets). In this case the $K_pC$ is also a sublevel set of a convex function. $\endgroup$
    – Anton Petrunin
    Nov 22, 2012 at 18:05
  • $\begingroup$ @ Anton Petrunin: Thanks a lot. Indeed i encountered this problem for sublevel sets of a convex function, say $f$. If a sublevel $C$ corresponds to a nonminimal value $a$ i see that $K_pC$ is a sublevel of the differential $df_p$. But this is wrong if $a$ is minimal. Any hint what function to consider here? ps. Anyhow the general question might be of interest $\endgroup$
    – Wolfgang Spindeler
    Nov 23, 2012 at 12:19

1 Answer 1

Reset to default
5
$\begingroup$

(Too long for a comment)

The question is interesting and it might be hard.

From the comments: The totally convex subset $C$ usually appears as a sublevel set of a locally Lipschitz convex function (I do not know other sources of totally convex subsets). If $C$ is a sublevel set of a convex function for a not mimimal value $a$ then so is $K_pC$, in particular $K_pC$ is totally convex.

Related stuff. Instead of tangent cone you might consider the same question for a (noncollapsing) Gromov--Hausdorff convergence $A_n\to A_\infty$. (In particular you may think that $A=A_n=A_\infty$ for all $n$ and $C_n$ is a sequence of totally convex sets.) Here some relevant statements which might be useful.

  • Any minimizing geodesic in $A_\infty$ can be approximated by minimizing geodesics in $A_n$. (Any minimizing geodesic can be approximated by unique minimizing geodesic, which is approximated by minimizing geodesic in $A_n$.)
  • If $A_n$ are Riemannian then any geodesic in $A_\infty$ can be approximated by geodesic in $A_n$. (You approximate a minimizing piece and then extend the approximation.) The general case would follow if the geodesic in Alexandrov space without boundary have infinite extension with probability 1 (this is not known now).
  • You might consider version of definition of totally convex set with quasigeodesics instead of geodesics. In this case the answer is NO; take $A_n=A_\infty$ to be a 2-dimensional cone and the sets $C_n$ which which lie on distance $\ge 1$ from the tip, but for its limit $C_\infty$ there is a quasigeodesic which pass through the tip.
Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ One example of interest if one considers general noncollapsing GH-limits: Let A_n = (M,g) be a fixed riemannian manifold such that there exists precisely one closed geodesic, say $c$. Then any sequence of points not contained in $c$ is a sequence of totally convex sets. If a limit point lies within $c$ the limit is not totally convex anymore. There should be an example like this among complete metrics on the 2-dimensinal plane. $\endgroup$
    – Wolfgang Spindeler
    Nov 24, 2012 at 17:48
  • $\begingroup$ @wspin, A point is totally convex if it is not on the tip of a geodesic loop (not a closed geodesic). So even you have just one closed geodesic $c$ the points off the $c$ do not have to form a totally convex. $\endgroup$
    – Anton Petrunin
    Nov 24, 2012 at 21:01
  • $\begingroup$ @ Anton Petrunin, Of course you are right, i was careless there. Let´s say there exists precisely one geodesic loop, which is in fact a simple closed geodesic ;) $\endgroup$
    – Wolfgang Spindeler
    Nov 25, 2012 at 0:25

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.