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Is the Borsuk-Ulam theorem valid for a torus? In other words, for any map $f: S^1 \times S^1 \rightarrow \mathbb{R^2}$ there is a point $(x,y) \in S^1 \times S^1$ which $f(x,y)=f(-x,-y)$

I'm very stuck on this task. Can someone give a hint? Or there can be a detailed solution, if suddenly this task is easy enough.

Thank you in advance for help!

Kaj Hansen
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mathmaniac
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1 Answers1

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No. With the usual torus embedded in $\mathbb{R}^3$, lying on the $OXY$ plane, one has a natural projection onto that plane, $p:S^1×S^1\to \mathbb{R}^2$, which is continuous.

Two points on the torus have the same image if they are one above the other, in the same vertical line. In particular, they are in the same meridian of the torus, i.e. they have the same first coordinate. So, if $p(a,b) = p(c,d)$, $a = c$. This implies that the Borsuk-Ulam theorem fails on the torus because if $x=-x$, and then $x=0\notin S^1$.

Javi
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    Could you, please, be more explicit - what "natural projection" are you talking about? UPD: Ah, I think I got it: you mean $S^1\times S^1$ embedded as a usual 3D torus into $\mathbb R^3$, and the projection goes onto the plane parallel to the big circle of the torus. It would help to mention this explicitly, I think. – lisyarus Jun 08 '18 at 11:21
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    Thank you very much! Very simple and intuitive solution I like it. – mathmaniac Jun 08 '18 at 11:32
  • Yes, with the usual torus embedded in $\mathbb{R}^3$, lying on the $OXY$ plane, $p$ is just the projection onto that plane. I'll add it to the answer. – Javi Jun 08 '18 at 11:34
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    Way easier than the answer I was cooking up (and then abandoned). +1 – Randall Jun 08 '18 at 11:42
  • What do you mean by $x\not\in S^1$? Is the circle a line segment with ends attached, or a subset of $\mathbb R^2$? – mr_e_man Jun 08 '18 at 18:09
  • @mr_e_man $S^1$ as a subset of $\mathbb{R}^2$ or $\mathbb{C}$. In any definition of $S^1$ where $-x$ means the antipodal of $x$, $x\neq -x$ for all $x\in S^1$. I just took advantage of the notation $-x$ to use it as the opposite point. – Javi Jun 08 '18 at 18:12
  • I was thinking $S^1 = \mathbb R^1/(x+1\sim x)$, in which $-x = 1-x$. Then $\frac12 = -\frac12$, and $0=1$. The other definition that you use is valid, though it wasn't mentioned in the OP. – mr_e_man Jun 08 '18 at 18:14
  • @mr_e_man that's right, I'll edit it to make it more general – Javi Jun 08 '18 at 18:20
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    My mistake; I wasn't familiar with the Borsuk-Ulam Theorem, which does specify that $-x$ is the antipodal point, not an opposite in $\mathbb R^1/(x+1\sim x)$.... I think the answer was better before the edit. – mr_e_man Jun 08 '18 at 19:29
  • @Javi: Can we also prove by finding 3 closed sets whose union is the torus and none of which contains a pair of antipodal points? Since any subspace in $\mathbb R^3$ satisfying Borsuk-Ulam theorem must also satisfy this property? – William Leynoid Oct 22 '22 at 08:33
  • @WilliamLeynoid What is the Borsuk-Ulam theorem for an arbitrary subspace of $\mathbb{R}^3$? – Javi Oct 23 '22 at 13:07
  • @Javi: I mean the property that " for any continuous map from the space $X$ to $\mathbb R^2$, there is a pair of antipodal points taking the same value". For any space satisfing this property, we can apply it to the distance map for the three closed sets cover $X$. – William Leynoid Oct 23 '22 at 16:15
  • @WilliamLeynoid but the point is that the torus does not satisfy that property. And I don't really understand what you get out of covering the spaces with three closed subset with no antipodal points. Are you implying that because on each subset there are not antipodal points the function cannot reach the same value on antipodal points? – Javi Oct 25 '22 at 11:14
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    @Javi: Yes, the torus dosn't satisfies that property, and I mean we can prove it by contradiction. Assume the torus satisfies the property that "for any continuous map from the space $T$ to $\mathbb R^2$, there is a pair of antipodal points taking the same value", we show that if $A_1,A_2,A_3$ are closed sets cover $T$, one of them have a pair of antipodal points. Let $d_j(x)=d(x,A_j)$, then $(d_1,d_2)$ takes the same value at some $x,-x$, hence $d(x,A_1)=d(-x,A_1), d(x,A_2)=d(-x,A_2)$. If one of them is $0$, we are done. If both are positive then $x$ and $-x$ must contains in $A_3$. – William Leynoid Oct 25 '22 at 15:46
  • Since such closed covering $A_1,A_2,A_3$ clearly exist for $T$, it cannot satisfies that property! – William Leynoid Oct 25 '22 at 15:52
  • @WilliamLeynoid I see, that's interesting. Maybe not so appealing for some people as it is non-constructive and probably more complex than a simple counterexample. But it's cool. – Javi Oct 25 '22 at 20:30