1

Let $(M,g)$ be a compact complete Riemannian manifold. Consider the geodesic quadrupel $ABCD$ where $l:=d(A,B)=d(B,C)=d(C,D)=d(A,D) \geq \frac{1}{2}$ and $d(B,D),d(A,C) \geq 1$. Let $x \in CD$ and $y \in AB$. Is it true that then $d(x,y)\geq \frac{1}{4}$ ? I have tried a lot of things but non of them worked, mostly I tried to do it by the triangle inequaity. I even do not know if its true.

Greetings Lena

  • Is it true or does one need further assumptions? –  Jun 30 '13 at 07:19
  • It seems to me that it might be a good idea to try and prove or disprove a local version, for $A$, $B$, $C$, $D$ in a small neighborhood; it might not be true in the large. – Robert Lewis Jun 30 '13 at 07:28
  • what do you mean by local version? –  Jun 30 '13 at 07:29
  • in a neighbourhood of what ? what would be the advantage if one considers the problem locally? –  Jun 30 '13 at 07:30
  • If necessary letting the distance constraints become smaller (though perhaps proportinately so), and keeping the points sufficiently close together, one can look at the effect of curvature etc. without worrying too much about the manifold "bending back upon itself". – Robert Lewis Jun 30 '13 at 07:31
  • but I do not make any assumtions on curvature. I actually ment, if it is possible, somehow by triangle inequality, or comparison triangles etc. , to show this inequality. Do you think that by comparison triangles is a good approach ? –  Jun 30 '13 at 07:50
  • The problem is that in a curved space, "triangles" can be warped; you might not be able to compare them easily. You need, I think, to start be looking at short geodesics so problems with multiple length minimizing geodesics are avoided; have you checked out the notion of conjugate points? – Robert Lewis Jun 30 '13 at 08:08
  • how can one apply the conjugate points here? –  Jun 30 '13 at 08:11
  • 2
    Note that the inequality does not hold in the plane. But in general, if triangle inequality does not work for you then nothing will --- any compact length-metric space can be approximated by a Riemannian manifold. – Anton Petrunin Jun 30 '13 at 12:22

2 Answers2

2

No (unless you assume something like a lower curvature bound).

Consider two very long segments (say, of length 100) in $\mathbb R^3$ that meet orthogonally at their midpoints. Let $M$ be a smoothened boundary of an $\varepsilon$-neighborhood of the union of these segments. It is a Riemannian manifold invariant under a 90 degrees rotation. Let $A$ be a point of $M$ near one of the segments' endpoints and $B$, $C$, $D$ be its consecutive images under this rotation. Due to rotation invariance, one has $d(A,B)=d(B,C)=d(C,D)=d(D,A)$. And all distances between $A,B,C,D$ are greater than 50.

On the other hand, geodesic segments $AC$ and $BD$ go through the central part of the construction, namely an $O(\varepsilon)$ neighborhood of the center. Take $x$ and $y$ in that region and observe that $d(x,y)=O(\varepsilon)$.

  • But I do not take $x$ and $y$ in the region of the center. As I meantioned, I take $x \in CD$ and $y \in AB$, where by $CD$ and $AB$ I mean the geodesic segment with endpoints $C$ and $D$, and $A$ and $B$ respectively. –  Jun 30 '13 at 11:33
  • In this construction, the segments AB and CD pass really close to the "center". –  Jun 30 '13 at 11:49
  • @Lena: take $x$ in the intersection of $CD$ with the central region (this is possible because the intersection is nonempty) and $y$ in the intersection of $AB$ with the central region. Is this clearer? –  Jun 30 '13 at 13:44
  • yes, thanks ;). –  Jun 30 '13 at 13:46
0

Surface   $z = y^2 - x^2$,   with proper scaling (say, one scaling per   $\epsilon > 0$),   will provide counter-examples, with   $\frac 14$   replaced by arbitrary   $\epsilon > 0$.   (Perhaps something like this was suggested by Anton Petrunin in his comment above).

By the way, all four mid-points of "edges" of   $ABCD$   will form a set of diameter   $\epsilon$   or less.

Włodzimierz Holsztyński
  • 391