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 $1/2$?

  • $\begingroup$ 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


(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.


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.
  • $\begingroup$ 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

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.


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