Generalization of Borsuk-Ulam - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T16:47:10Z http://mathoverflow.net/feeds/question/61141 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61141/generalization-of-borsuk-ulam Generalization of Borsuk-Ulam Jens Reinhold 2011-04-09T14:05:06Z 2011-04-11T16:54:56Z <p>Let $n$ be a positive intger. Is the following true? For continuous maps $f: \mathbb S^n \rightarrow \mathbb S^n$ and $g: \mathbb S^n \rightarrow \mathbb R^n$, there exists a point $x \in \mathbb S^n$ such that $g(x) = g(f(x))$.</p> http://mathoverflow.net/questions/61141/generalization-of-borsuk-ulam/61171#61171 Answer by Sergei Ivanov for Generalization of Borsuk-Ulam Sergei Ivanov 2011-04-09T22:37:42Z 2011-04-10T22:05:21Z <p>This is not true for $n=2$.</p> <p>Represent $\mathbb S^2$ as the cylinder $\mathbb S^1\times[-1,1]$ with two discs $D_+$ and $D_-$ attached to the boundary components <code>$\mathbb S^1\times\{1\}$</code> and <code>$\mathbb S^1\times\{-1\}$</code>, resp. Denote by $\mathbb S^2_+$ and $\mathbb S^2_-$ the "positive" and "negative" hemispheres: $\mathbb S^2_+=(\mathbb S^1\times[0,1])\cup D_+$ and $\mathbb S^2_-$ is the opposite.</p> <p>First we define $g:\mathbb S^2\to\mathbb R^2$. Consider a smooth $\infty$-shaped loop $\gamma:\mathbb S^1\to\mathbb R^2$, namely $$\gamma(t) = (\sin t,\tfrac12\sin2t)$$ (here $\mathbb S^1=\mathbb R/2\pi\mathbb Z$). Note that the velocity of $\gamma$ is separated away from the vertical vector $e_2=(0,1)$, namely $\angle(\dot\gamma(t),e_2)\ge\pi/4$ for all $t\in\mathbb S^1$. Choose $\varepsilon>0$ so small that $\angle (\gamma(t+\varepsilon)-\gamma(t),e_2)> \pi/5$ for all $t$.</p> <p>For $x=(t,s)$ from the cylinder, define $$\begin{cases} g(x) =\gamma(t)+1000s\cdot\overrightarrow{(1,10)} , &amp;\qquad s\ge 0 \cr g(x) =\gamma(t)+1000|s|\cdot\overrightarrow{(-1,10)} , &amp;\qquad s\le 0 . \end{cases}$$ (its image of the cylinder consists of two almost vertical strips above the $\infty$-figure). Observe that the image of each boundary component <code>$\mathbb S^1\times\{\pm 1\}$</code> is separated away from the image of the other half of the cylinder (by distance at least 100). Extend $g$ to $D_+$ and $D_-$ so as to fill these boundary components within their neighborhoods of radius 2. Then $g(D_+)\cap g(\mathbb S^2_-)=\emptyset$ and $g(D_-)\cap g(\mathbb S^2_+)=\emptyset$.</p> <p>Since $\gamma(t+\varepsilon)-\gamma(t)$ never forms a small angle with $e_2$, the construction guarantees that $g(t,s)\ne\gamma(t+\varepsilon)$ for all $t\in\mathbb S^1$, $s\in[-1,1]$.</p> <p>Now we define $f:\mathbb S^n\to\mathbb S^n$. For $x=(t,s)$ from the cylinder, define $f(x)=(t+\varepsilon,0)$, so the cylinder is projected to its equator and slightly rotated. Extend $f$ to $D_+$ and $D_-$ so that $f(D_+)\subset \mathbb S^2_-$ and $f(D^-)\subset \mathbb S^2_+$.</p> <p>For these $f$ and $g$, we have $g(x)\ne g(f(x))$ for all $x\in\mathbb S^2$. Indeed, if $x=(t,s)\in\mathbb S^1\times[-1,1]$, then $f(g(x))=\gamma(t+\varepsilon)\ne g(x)$ as noted above. For $x\in D^+$ this follows from the fact that $g(D^+)\cap g(f(D^+))=\emptyset$, and similarly for $D^-$.</p> http://mathoverflow.net/questions/61141/generalization-of-borsuk-ulam/61172#61172 Answer by Faisal for Generalization of Borsuk-Ulam Faisal 2011-04-09T23:07:00Z 2011-04-10T23:08:31Z <p>This is false for all $n \geq 2$, but true for $n=1$.</p> <p>A map $g \colon X \to Y$ of topological space is said to be <strong>free</strong> if there is a map $f \colon X \to X$ such that $g(x) \neq g(f(x))$ for all $x\in X$. This definition appears to be due to Hopf, and was given in</p> <blockquote> <p>H. Hopf, <em>Freie Überdeckungen und freie Abbildungen</em>, Fund. Math. <strong>28</strong> (1937), 33--57.</p> </blockquote> <p>In that same paper, Hopf apparently proved that:</p> <ol> <li>There is a free map $g \colon \mathbb S^n \to \mathbb R^n$ for every $n \geq 3$.</li> <li>There is no free map from $\mathbb S^n$ into $\mathbb R$ if $n \geq 1$.</li> </ol> <p>A generalization was obtained by Pannwitz in</p> <blockquote> <p>E. Pannwitz, <em>Eine freie Abbildung der n-dimensionalen Sphäre in die Ebene.</em> Math. Nachr. <strong>7</strong> (1952), 183–-185.</p> </blockquote> <p>According to the MR, she proved that, for every $n \geq 0$, there exists a free map $g \colon \mathbb S^n \to \mathbb R ^2$.</p> <p>Unfortunately, I'm having trouble accessing the papers, so I can't provide any more details.</p> <hr> <p><strong>Update.</strong> Still no luck getting Pannwitz's paper. However, I have managed to find the following related paper</p> <blockquote> <p>C. Biasi, D. de Mattos, E. dos Santos, <em>Applications of the non-standard version of the Borsuk--Ulam theorem</em>, JP J. Geom. Topol. <strong>9</strong> (2009), no. 3, 273--284.</p> </blockquote> <p>(available online <a href="http://www.icmc.usp.br/~sma/cadernos/toc8.2/287.pdf" rel="nofollow">here</a>), which mentions the above papers of Hopf and Pannwitz in its Introduction. More interesting however is Theorem 2.1, which I believe can be applied to Sergei's construction to yield Pannwitz's result for $n\geq 2$. This would show that the question in the OP has a negative answer if $n\geq 2$.</p> <p>It remains to show that the answer is "yes" if $n=1$. (As Harry notes, the answer is also "yes" if $n=0$.) The following is Hopf's argument. Let $g \colon \mathbb S^1 \to \mathbb R$ and $f \colon \mathbb S^1 \to \mathbb S^1$ be given. Since $\mathbb S^1$ is compact, $g$ attains its maximum, resp. minimum, at some $x$, resp. $y$, in $\mathbb S^1$. In particular, $g(x) \geq g(f(x))$ and $g(y) \leq g(f(y))$. The intermediate value theorem then asserts that there is a $z \in \mathbb S^1$ for which $g(z) = g(f(z))$.</p> http://mathoverflow.net/questions/61141/generalization-of-borsuk-ulam/61311#61311 Answer by Michael Renardy for Generalization of Borsuk-Ulam Michael Renardy 2011-04-11T16:54:56Z 2011-04-11T16:54:56Z <p>On the other hand, it is known to be true if f is an involution, i.e. f(f(x))=x. This was proved in a paper by C.T. Yang in Annals of Mathematics in 1954.</p>