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Consider a continous map from $S^2$ to $C$.

Is it true that there exists 3 points equially spaced on a great circle, $x_1,x_2,x_3$, such that if $w$ is the third root of unity, $f(x_1)+wf(x_2)+w^2f(x_3)=0$?

More generally I'm asking this if we take nth unity roots.

Maybe I should add slight motivation: In my topology course we were shown a proof of borsuk ulam that goes through defining $g(x)=(f(x)-f(-x))/(|f(x)-f(-x)|)$, then by looking at it on the great circle, we can lift it to a function to $R$ satisfying (here we view $g$ as a function from $[0,1]$ instead of from the circle, and taking $x+1/2$ mod 1 in the next expression) $g(x+1/2)=n+1/2+g(x)$ for some natural $n$, but since it is continous, this is the same $n$ for all $x$. In particular $g(1)=2n+1+g(0)$ and thus this is a nontrivial path on the circle, but it homotopic to the constant one via returning to the sphere and wrapping the circle around to a constant function.

Then notice the argument after the lift works the same when we know $g(x+1/3)=n+1/3+g(x)$. However there is no direct analong we can do to reach the part of lifting, because that would involve choosing for each point a great circle it is contained in a way that partions the sphere into great circles which is clearly impossible.

If this turns out to be false, is there a space we can do this trick on to get this cute result?

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    $\begingroup$ Does "big circle" means "great circle"? $\endgroup$ – Igor Rivin Oct 17 '17 at 15:14
  • $\begingroup$ @IgorRivin Yes, I'll update. $\endgroup$ – Andy Oct 17 '17 at 15:16
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    $\begingroup$ I think that your first question can be answered (in the affirmative) by resorting to the first theorem in Shizuo Kakutani's "A proof that there exists a circumscribing cube around any bounded closed convex set in $\mathbb{R}^{3}$" (Annals of Mathematics, Vol. 43, #4, Oct. 1942). You are welcome to take a look at a related question I posed several years ago here: mathoverflow.net/questions/26318/points-on-a-sphere $\endgroup$ – José Hdz. Stgo. Oct 17 '17 at 15:54
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    $\begingroup$ @JoséHdz.Stgo. I'm not sure I understand, the paper considers functions to $R$ and proves that there are 3 points which are like the basis for $R^3$ that have the same value, and the question you quote says that we can take arbitary equilateral triangles (on the sphere?). How does this prove my question? $\endgroup$ – Andy Oct 17 '17 at 16:34
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$\require{AMScd}$ First, suppose that $G$ acts freely on connected spaces $X$ and $Y$, and that $p\colon X\to Y$ is $G$-equivariant. We then have a diagram of fibrations \begin{CD} X @>>> X/G @>>> BG \\ @VpVV @VVV @VV1V \\ Y @>>> Y/G @>>> BG \end{CD} This gives a diagram of fundamental groups \begin{CD} \pi_1(X) @>>> \pi_1(X/G) @>>> G \\ @Vp_*VV @VVV @VV1V \\ \pi_1(Y) @>>> \pi_1(Y/G) @>>> G \end{CD} and it is not hard to see that the rows are short exact sequences.

Now specialise to the case where $Y=\mathbb{C}^\times$ and $G=C_3$ acting by rotation, and $\pi_1(X)$ is finite. Using the top row we see that $\pi_1(X/G)$ is also finite, so the only homomorphism to $\mathbb{Z}$ is trivial. On the other hand, bottom row is then $\mathbb{Z}\xrightarrow{3}\mathbb{Z}\to\mathbb{Z}/3$, and this quickly leads to a contradiction.

Now suppose that $f\colon S^2\to\mathbb{C}$ is a counterexample to the OP's question. Fix three points $x_1$, $x_2$ and $x_3$ equally spaced on a great circle. Let $R$ be the rotation that sends $x_1$, $x_2$ and $x_3$ to $x_2$, $x_3$ and $x_1$. Define $p\colon SO(3)\to \mathbb{C}^\times$ by $$ p(T) = f(Tx_1) + \omega f(Tx_2) + \omega^2 f(Tx_3), $$ and note that this satisfies $g(TR)=\omega^{-1}g(T)$. In other words, if we let $C_3$ act on $SO(3)$ using $R$ and on $\mathbb{C}^\times$ by $\omega^{-1}$, we see that $p$ is equivariant. This is impossible by our initial lemma, because $\pi_1(SO(3))=\mathbb{Z}/2$.

All the above still works if we replace $3$ by another integer $n>1$.

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    $\begingroup$ Hi, I am very sad to say I don't know enough about topology to understand your answer, but it motivates very much to learn about it. (I accepted your answer assuming that upvotes from other people verify your solution). $\endgroup$ – Andy Oct 18 '17 at 20:19
  • $\begingroup$ Are you proving that there is no a free action of $\mathbb{Z}/3\mathbb{Z}$. If yes how this solve the OP question? Moreover I think there is an easy argument for non existence of such free action, as I learn in the book by Allen Hatcher: n=2 is the only $n$ such that $\mathbb{Z}/n\mathbb{Z}$ can act freely on a even domensional sphere. The proof is simply based on the degree of maps. The "degree" is a group isomirphism from $G$ to$ {1,-1}$ (Counting $g\in G$ as a homeomorphism of even sphere. On the other hand a fixed point free homeomorphism has degree -1. $\endgroup$ – Ali Taghavi Oct 19 '17 at 6:47
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    $\begingroup$ @AliTaghavi There are no even spheres involved here at all. There is a free action of $C_3$ on $SO(3)$ (which is almost the same as $S^3$) and a free action of $C_3$ on $\mathbb{C}^\times$ (which is homotopy equivalent to $S^1$). There is no issue about the existence of these actions, the point is to prove that there is no $C_3$-equivariant map $SO(3)\to \mathbb{C}^\times$. $\endgroup$ – Neil Strickland Oct 19 '17 at 7:06
  • $\begingroup$ @NeilStrickland So my apology for not reading your answer, carefully.I read it again.I am on my phon. $\endgroup$ – Ali Taghavi Oct 19 '17 at 10:00

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