$X$ 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).

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    $\begingroup$ 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. $\endgroup$ 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 $\endgroup$ Nov 23, 2012 at 12:19

1 Answer 1


(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.
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    $\begingroup$ 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. $\endgroup$ 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. $\endgroup$ 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 ;) $\endgroup$ Nov 25, 2012 at 0:25

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