6
$\begingroup$

Let $X=\mathbb T^d=\mathbb R^d/\mathbb Z^d$ and let $E$ be a proper open subset of $X$. We say $E$ is geodesically convex if for any $x,y\in E$ the shortest geodesic connecting $x$ and $y$ lies in $E$.

Question. How large can the Haar/Lebesgue measure of $E$ can be?

For example, is $d=2$, then it seems that this cannot exceed $1/2$. Say, $[0,1)\times [0,s)$ is geodesically convex if and only if $s\leq1/2$. (If $s>1/2$, then $[x,x+\delta]$ is not the shortest geodesic for any $\delta\in(1/2,s)$ and any $x\in(0,1)$.)

Is it true for any $d\ge2$ that the measure of such an $E$ cannot exceed $1/2$?

$\endgroup$
7
  • $\begingroup$ No, it doesn't - that's my point exactly. $\endgroup$ Jan 16, 2012 at 21:13
  • $\begingroup$ Why isn't it? in your example the shortest one is of length $1/4$, `going round' on level $y=1/4$ but staying inside $E$. Am I wrong? $\endgroup$ Jan 16, 2012 at 21:28
  • $\begingroup$ Your question is obvious for the circle. Partition the torus into geodesic circles and apply fubini. It seems to work... $\endgroup$ Jan 16, 2012 at 21:31
  • $\begingroup$ Thought about it but couldn't figure out how to do it rigorously. It is not true that each section has measure $\le1/2$, that's for sure. It sort of depends on whether we're trying to `stretch' it horizontally or vertically... $\endgroup$ Jan 16, 2012 at 21:37
  • $\begingroup$ Nikita, I misunderstood, you did type it in correctly and I missed that you were wrapping around in one direction, although not the other. I imagine your figure of 1/2 is correct. I did a long thing with a youngster named Dror Atariah in Germany, there was a little bit on MO and a bunch of email. We never did decide the thing. My main impression is that you can do what you like as long as the boundary of the set $E$ is totally geodesic and other conditions (such as symmetry) are favorable. Proving you cannot do better than 1/2 is another matter. $\endgroup$
    – Will Jagy
    Jan 16, 2012 at 21:39

3 Answers 3

4
$\begingroup$

(This is a new answer; my original answer was completely wrong.)

Assume $\mathop{\rm vol}E>\tfrac12$. Then it contains two opposite points say $x$ and $x'=x+(\tfrac12,\tfrac12,\dots,\tfrac12)$. WLOG we can assume that $x=0$. Taking minimizing geodesics form $(\tfrac12,\tfrac12,\dots,\tfrac12)$ to $y\approx 0$, we get that all main diagonal of unit cube $$\square^n=(0,1)\times(0,1)\times\dots\times(0,1)$$ lie in $E$. Then apply the following lemma:

Trivial Lemma. Let $\square^n$ be open unit cube in $\mathbb R^n$ and $E\subset \square^n$ be a locally convex open set which contains all main diagonals of $\square^n$ then $E=\square^n$.

To prove the lemma, note that local convexity + conectedness in $\mathbb R^n$ implies convexity.

$\endgroup$
0
2
$\begingroup$

Let me write down the steps.

  1. Consider the case d=2, the generalization is straightforward.
  2. There is an open ball B that doesn't intersect E.
  3. Consider two families of geodesic circles F_1 and F_2. F_1 has slope 1/p and F_2 has slope 1-1/p. Number p is chosen in such a way that each circle from F_1 nd F_2 intersects B.
  4. Claim: $Leb(E\cap F_1(x))\le 1/2 length(F_1(x))$ OR $Leb(E\cap F_2(x))\le 1/2 length(F_1(x))$ for each x.
  5. Proof: $E\cap F_1(x)$ is a proper union of open intervals that are separated by gaps of length at least $\sqrt{p^2+1}/2p$. So it remains to show that there are at least $p$ gaps. If there are less than $p$ gaps then there is an interval from $E\cap F_1(x)$ of length greater than $\sqrt{p^2+1}/2p$. It follows that one can find a simple closed curve C_1 in E which is C^0 close to a horizontal generator. Assume that in the same way we also can find C_2 in E which is C^0 close to the vertical generator. Then one can easily see that E is the torus and the claim follows.
  6. Apply Fubini.
$\endgroup$
2
  • $\begingroup$ Sorry, don't understand it, starting with 3. Geodesic circles are either vertical or horizontal, aren't they? So, what do you mean by `slope'? With a slope geodesics will not be even closed. $\endgroup$ Jan 17, 2012 at 2:29
  • $\begingroup$ Nikita, take any line through the origin with rational slope, so $y = \frac{m}{n} \; x.$ This line also passes through the lattice point $(n,m).$ This point is identified with the origin in the torus you define, so it is a closed curve. For example, the line $y = x$ is a closed geodesic in the torus. It is instructive to draw the image of, say, $y = \frac{5}{3} \; x$ in the original 1 by 1 square under identification. $\endgroup$
    – Will Jagy
    Jan 17, 2012 at 3:36
1
$\begingroup$

A bit long for a comment. You can have a geodesically convex set that is an arbitrarily large proportion of the area of a surface. What I have in mind resembles a bulb_thermometer or turkey_baster but is at least $C^\infty.$ It is rotationally symmetric. One end is long, cigar shaped, half of something that approximates a prolate spheroid. It differs from an actual prolate spheroid in that it is necessary for the Gauss curvature to be 0 along the "equator," the closed geodesic where the half cigar joins the bulb. Therefore the curvature must approach 0 near the equator. Immediately upon entering the bulb section, the curvature is slightly negative, which is the reason geodesics leaving the equator cannot quickly return.

$\endgroup$

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.