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?



Is this generalization of Borsuk Ulam true? Roots of unity
$\require{AMScd}$
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 $n>1$.


