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during some geometrical speculations I came across the next problem:

Let $\mathcal{P}$ be the class of all continuous spatial closed curves $r:\mathbb{S}^{1}\rightarrow\mathbb{R}^{3}$, in coordinates $r(t)=(r_{1} (t),r_{2}(t),r_{3}(t))$, $t\in\mathbb{S}^{1}$, with the following property:

$\prod\limits_{i=1}^{3}(r_{i}(-t)-r_{i}(t)-\alpha)=0$ for any $t\in \mathbb{S}^{1}$,

where $\alpha\neq0$ is some fixed constant. So, the problem is whether the class $\mathcal{P}$ of such curves is path-connected in the usual (sup-norm) topology or not. Any suggestions and references are welcome.


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I really wonder how you came up with this problem... I get the thing where you want points at opposite parameters $t$ in the circle to be distant of $\alpha$, but how did you come up with this by "geometrical speculations"? It's actually quite interesting to think of. –  Patrick Da Silva Jun 1 '11 at 6:31
Of course, we may take $\alpha=1$. To show that this class is rich enough, here is a way to construct such curves: Let $\mathbb{S}^{1}=\cup_{i=1} ^{3}A_{i}$, where $A_{i}$ are closed sets such that $A_{i}\cap(-A_{i} )=\varnothing$. Now take $r_{i}:\mathbb{S}^{1}\rightarrow\mathbb{I}^{1}$ so that $r_{i}|_{A_{i}}=0$, $r_{i}|_{-A_{i}}=1$. Then $r=(r_{1},r_{2},r_{3})$ is such a curve (with $\alpha=1$). It lies on the edges of the unit cube. –  t22 Jun 1 '11 at 6:31
I'm a little tired here XD Can you give at least an example of one such curve? I'm trying to make it continuous at $t = 0$ and I can't, because your product gives me $-\alpha^3 \neq 0$ when $t = 0 = -t$ so that $r_i(-t) = r_i(t)$. Problem here. Maybe I just got it wrong though. –  Patrick Da Silva Jun 1 '11 at 6:40
@Patrick, here $t\in\mathbb{S}^{1}$, it cannot be $0$. –  t22 Jun 1 '11 at 6:44
The "geometric speculations" are as a result of attempts to define a sort of "distinguishing map" for a given homeomorphism (too vague for the moment...), and I prefer to ask here a concrete question in some particular case concerning the antipodal map on the circle. –  t22 Jun 1 '11 at 6:57

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