8
$\begingroup$

If $[a, b]$ is a compact interval of $\mathbb{R}$ and $\gamma: [a, b] \to \mathbb{C}$ is continuous, denote the connected, compact set $\gamma([a, b])$ by $[\gamma]$. If $h$ is a complex number of unit modulus, define the curve $h\gamma \colon [a, b] \to \mathbb{C}$ by $(h\gamma)(t) = h(\gamma(t))$. (If $h = -1$, write $h\gamma$ as $-\gamma$.)

If $\gamma(b) = -\gamma(a)$, must we have $[\gamma] \cap [h\gamma] \ne \emptyset$?

This trivially true for $h = \pm1$, and it has been proved (although not easily) for $h = \pm i$. We can assume without loss of generality that $h = e^{ic}$, where $0 < c < \pi$.

The question for the case $h = \pm i$ was first asked by Herman Tulleken, with the proviso that an answer was allowed to make almost arbitrary special assumptions about $\gamma$, subject only to applicability to a particular problem related to polyominoes.

Accordingly, @YiFan asked the question again, still for the case $h = \pm i$, but this time only specifying that $\gamma$ is injective (and of course continuous), but should otherwise be arbitrary. @Jens observed, in a comment on this question, that the result seemed to be true for a general rotation, not just $90^\circ$.

I had made the same comment on the first question, not knowing that a second question had been asked - but I have also made many badly mistaken comments on both questions!

Although my judgement on this problem is evidently suspect, I am fairly confident of the validity of Hagen von Eitzen's answer to the second question, and even of my own answer to it. Both answers (independently) use the same construction, which is to concatenate $\gamma$ with $-\gamma$, forming a closed curve; it is then necessary to argue that if $z$ belongs to $[\gamma] \cup [-\gamma]$ and $[i\gamma] \cup [-i\gamma]$, then one of $\pm z, \pm iz$ belongs to $[\gamma] \cap [i\gamma]$. My initial vague impression that both arguments could handle the case of a general rotation seems, sadly, to be just another mistake.

Neither of these answers to the second question used the assumption of injectivity, so the case $h = \pm i$ (equivalently, w.l.o.g., $c = \frac{\pi}{2}$) seems to be settled.

The evidence for the general case is admittedly weak. I don't find it intuitively obvious (even though I keep kidding myself that I glimpse "reasons" why it should be true). The mere fact that it holds for $h = \pm i$ is not compelling. Beyond this, I only have the conviction gained from doing simple experiments with GeoGebra: creating $\gamma$ as a polyline object, moving the vertices around, and observing the intersections with one or more rotated copies.

Nevertheless, for what it's worth, I feel quite strongly that the general result is true. I also feel, slightly less strongly, that it probably has a simple proof using only elementary results about continuity. I don't think that anything as deep as the Jordan Curve Theorem is needed. Even my proof for the case $h = \pm i$ (which is short, if you allow the use of a lemma that seems to be of more general use) is probably more complex than the "Book" proof of the general result.

However, if no elementary proof is forthcoming, I will accept an answer using advanced methods. I won't be competent to judge such an answer myself, but I will happily take advice (in a chatroom, if there isn't space in the comments).

$\endgroup$
  • $\begingroup$ Looks like this one problem sparked $4$ questions! $\endgroup$ – YiFan Feb 16 at 9:20
  • $\begingroup$ The answer by Hagen von Eitzen to the other question relies on the fact that intersection of the closed paths implies intersection of the original curve and copy, which only follows if the rotation is 90 degrees. We could preserve this if we change the construction from $\gamma$ to $\tilde{\gamma}$ to use a reflection instead of a 180 rotation. I have not checked if the rest of the proof still works in this case. $\endgroup$ – Herman Tulleken Feb 18 at 21:01
  • $\begingroup$ I mean a reflection about the line trough $\gamma(a)$ and $\gamma(b)$. (But like I said, I have not tried testing it properly.) $\endgroup$ – Herman Tulleken Feb 18 at 21:40
  • $\begingroup$ 0.o yes I commented too quickly. (Furthermore, it also looks like even if that part did work we would not have the Jordan curve positioned such that we can prove paths cannot escape from 0 so the whole thing falls flat there too.) $\endgroup$ – Herman Tulleken Feb 18 at 22:09
4
$\begingroup$

Inspired by your counterexample, here's a counterexample with the rotation angle less than $90^\circ$:

enter image description here

Seems my intuition was wrong.

$\endgroup$
  • $\begingroup$ My flabber is quite gasted. $\endgroup$ – Calum Gilhooley Feb 23 at 13:45
  • $\begingroup$ Nice :) I managed to construct counterexamples using this idea for 50 and 35 degrees... but have not been able to do 45 degrees... and I am wondering whether that may be impossible. (Quite likely I just did not try long enough.) $\endgroup$ – Herman Tulleken Feb 23 at 22:03
  • $\begingroup$ (Oh I just saw the edits to the other answer now.) $\endgroup$ – Herman Tulleken Feb 23 at 22:04
6
$\begingroup$

No - but I'll stubbornly carry on trying to prove it for $|c| \leqslant \frac{\pi}{2}$.

It was from work on a promising-looking idea for a proof that this embarrassing counterexample emerged -

Counterexample with c = \frac{3\pi}{4}.


Following on from Jens's coup de grâce, I haven't yet been able to find a counterexample for a rotation by $\frac{\pi}{3}$, $\frac{\pi}{4}$, or $\frac{\pi}{5}$.

In logarithmic polar coordinates (with $\theta$ increasing to the right, and $\log r$ increasing either up or down, it doesn't matter which), mimicking Jens's example for an angle between $\frac{\pi}{3}$ and $\frac{\pi}{2}$, it was easy to construct a counterexample for $\frac{9\pi}{40}$, lying between $\frac{\pi}{5}$ and $\frac{\pi}{4}$:

Counterexample with c = \frac{5\pi}{40}.

But this is what happened when I tried to do something similar for $\frac{\pi}{5}$:

Failing to construct a counterexample with c = \frac{\pi}{5}.

Here is a similarly unsuccessful attempt to construct a counterexample for $c = \frac{\pi}{3}$:

Failing to construct a counterexample with c = \frac{\pi}{3}.

Can anyone find a counterexample with $c = \frac{\pi}{n}$, for any positive integer $n$?

(As this is a Community Wiki post, anyone should feel free to contribute pertinent observations.)


Here is a counterexample for 35 degrees, which is close to 36.)enter image description here

$\endgroup$
  • $\begingroup$ Good luck! your conjecture looks plausible. $\endgroup$ – Lubin Feb 23 at 2:44
  • $\begingroup$ Please do update this page if you make progress, I would be very interested to find out! $\endgroup$ – YiFan Feb 23 at 4:16
  • 1
    $\begingroup$ @YiFan I only have experimental evidence from playing with GeoGebra. I've been dead wrong about this problem on the basis of such evidence before. My experiments never seem to be done thoroughly enough, and I'll have to try to tighten up a bit before leaping to conclusions. That's why I'm holding back from asking yet another question on the same subject! It's not at all unlikely that someone will find a counterexample for some $\frac{\pi}{n}$. $\endgroup$ – Calum Gilhooley Feb 23 at 22:09
  • 1
    $\begingroup$ I managed to construct counter-examples for 35 and 50 degrees quite quickly... but 45 degrees eluded me... so I agree that there is something going on more than just the size. $\endgroup$ – Herman Tulleken Feb 23 at 22:09
  • 1
    $\begingroup$ @Jens Your wish is my command! :) Future counterexamples may be posted as answers here: Must a curve $\eta \colon [a, b] \to \mathbb{R}^2$ intersect the curves $\eta + \frac{\eta(b) - \eta(a)}{n}$ ($n \geqslant 1$)?. $\endgroup$ – Calum Gilhooley Feb 25 at 3:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.