7
$\begingroup$

Let $\gamma$ be a continuous curve in the complex plane without self-intersections and let $\lambda$ be a complex non-real number less than 1 in modulus. Put $\gamma'=\lambda\gamma$.

Question. Is it true that the sumset $\gamma+\gamma'$ has a non-empty interior?

$\endgroup$
  • 1
    $\begingroup$ It's instructive to look at some specific examples. Setting $\gamma $ to $y = 0$ and $y = 1$ are instructive. More generally, if you parameterize both $\gamma$ and $\gamma'$ with two different parameters, say, $s$ and $t$, then you get a map from $\mathbb{R}^2$ to the Minkowski sum, and you can compute the rank of this map. $\endgroup$ – Deane Yang Jan 20 '15 at 16:02
  • 1
    $\begingroup$ Part of the issue here is that $\gamma$ may be nowhere differentiable... $\endgroup$ – Nikita Sidorov Jan 20 '15 at 16:23
  • 2
    $\begingroup$ Why the conditions "without self-intersections" and "less than 1 in modulus"? Is there any counterexample without them? $\endgroup$ – Lutz Mattner Jan 20 '15 at 18:14
  • 1
    $\begingroup$ Lutz: no particular reason, it's just the way it is in the context. In fact, I'm sure this is true for any "reasonable" pair $\gamma, \gamma'$ but it's possible "my" case is easier - I just don't know. $\endgroup$ – Nikita Sidorov Jan 20 '15 at 20:56
  • 3
    $\begingroup$ Surely it's true that if $\alpha,\beta:[0,1]\to\mathbb{R}^2$ are continuous curves such that $\alpha{+}\beta$ has no interior, then the vectors $\alpha(s_2)-\alpha(s_1)$ and $\beta(t_2)-\beta(t_1)$ must be parallel for all $0\le s_1,s_2,t_1,t_2\le 1$. This is certainly the case if $\alpha$ and $\beta$ are $C^1$; just compute the integral $$ \int_{t_1}^{t_2}\int_{s_1}^{s_2} \det\bigl(\alpha'(s),\beta'(t)\bigr) \,\mathrm{d}s\,\mathrm{d}t =\det\bigl(\alpha(s_2)-\alpha(s_1),\beta(t_2)-\beta(t_1)\bigr) $$ and note that if the integral is nonzero then $\alpha{+}\beta$ has nonempty interior. $\endgroup$ – Robert Bryant Jan 22 '15 at 6:53
3
$\begingroup$

Well, I cannot call it a complete answer, but it seems an idea that can work.

First: take any loop in the (s,t) coordinates, and consider its image on the complex plane under $\gamma(s)+\gamma'(t)$. Any point having nonzero index w.r.t. this loop is then contained in $\gamma+\gamma'$.

Second: take the loop to be the boundary of a rectangle, $[s_1,s_2]\times [t_1,t_2]$, and assume that along the second coordinate it is very thin, $t_1$ is very close to $t_2$. Then, the boundary consists of two translated images $\gamma([s_1,s_2])+\gamma'(t_1)$ and $\gamma([s_1,s_2])+\gamma'(t_2)$, and two very small curves $\gamma(s_j)+\gamma'([t_1,t_2])$.

Third: I would say that the two translated images should more or less coincide, up to what happens at the endpoints. As if they do not, there is a part of one "somewhere in the middle", where it is far away from the other one, and then, crossing the curve, we change the index -- so on one of two sides there will be an open set of points with nonzero index.

Last: making $t_2$ go towards $t_1$, in the limit of the almost-translation-invariance above we get that $\gamma([s_1,s_2])$ is a straight line. And this allows to conclude easily, as $\lambda$ is not real.

Well, as I've said, it's a sketch, but it has a good chance of working.

Upd.: I'm now sure that this argument works. For "somewhere in the middle", I mean the following.

Lemma. Let $\delta$ be the diameter of $\gamma'([t_1,t_2])$, and assume that there is no point with nonzero index w.r.t. the loop described in the "Second." step. Then, the sets $\gamma([s_1,s_2])+\gamma'(t_1)$ and $\gamma([s_1,s_2])+\gamma'(t_2)$ coincide outside $\delta$-neighborhoods of $\gamma'(t_{j})$.

Sketch of the proof. Assume the contrary and let $z$ be a point of a curve $\gamma([s_1,s_2])+\gamma'(t_1)$ that lies outside the above neighborhoods and that does not belong to the $\gamma([s_1,s_2])+\gamma'(t_2)$. Due to the continuity, there is $\epsilon$-neighborhood of $z$ that the latter curve does not intersect.

Now, in this neighborhood one can find two points "on different sides" with respect to the first curve (yes, this phrase should also be more formally stated with an appropriate reference to the Jordan curve theorem). Such two points have thus different index w.r.t. all the closed loop (image under $\gamma(s)+\gamma'(t)$ of the boundary of the rectangle $[s_1,s_2]\times [t_1,t_2]$). Hence, at least one of them in an internal point of the sum $\gamma+\gamma'$. $\square$

Passing to the limit as $t_2\to t_1$ says that the curve $\gamma([s_1,s_2])$ should admit (if there is no internal point) arbitrarily small translational symmetries outside arbitrarily small neighborhoods of its endpoints. Hence, it is a straight segment.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I don't understand the part about "somewhere in the middle", sorry. $\endgroup$ – Nikita Sidorov Jan 22 '15 at 17:08
  • $\begingroup$ I've updated the text. In fact, this argument does not use that the curves $\gamma$ and $\gamma'$ are related by a multiplication by $\lambda$ -- as Robert Bryant has pointed out in his comment above, the only "bad" situation is two parallel segments. $\endgroup$ – Victor Kleptsyn Jan 23 '15 at 1:18

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.