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I am trying to understand the proof of the Borsuk-Ulam theorem for $S^2$ given in Hatcher's "Algebraic Topology" (Th. 1.10), as another person does here, but we are stuck at different places, so I hope this question is not considered a duplicate.

Borsuk-Ulam: If $f:S^2\rightarrow\mathbb R^2$ continuous, then there exists $x\in S^2$, such that $f(x)=f(-x)$.

Arguing by contradiction, assume there is no such $x$, then we can define

$g:S^2\rightarrow S^1$,$\quad$$g(x)=\frac{f(x)-f(-x)}{\|f(x)-f(-x)\|}$. Plugging in $-x$, we see that $g(-x)=-g(x)$.

Define a loop circling the equator $\eta:[0,1]\rightarrow S^2,\quad\eta(s)=(\cos(2\pi s), \sin(2\pi s),0)$ and let $h:=g\circ \eta:[0,1]\to S^1$ be the composed loop. Direct check gives:

$h(s+\frac12)=g(\eta(s+\frac12))=g((\cos(2\pi s+\pi),\sin(2\pi s+\pi),0))=g(-\eta(s))=-h(s)$

for all $s\in[0,\frac12]$. Let $\tilde h:[0,1]\to \mathbb R$ be a lifting of $h$.

Then there goes the phrase: 'the equation $h(s+\frac12)=-h(s)$ implies that $\tilde h(s+\frac12)=\tilde h(s)+\frac q2$ for $q$ some odd integer...'

Did anyone try to understand why this cryptic statement may be true?

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    $\begingroup$ This is just a computation. Try to do it first for the equal points on the circle: Lift to the real line and compute the difference. You see that it is an integer. Now, try the same for antipodal points on the circle, the difference between their lifts is a half-integer. $\endgroup$ – Moishe Kohan Sep 19 '14 at 1:05
  • $\begingroup$ @studiosus: "the difference between the lifts of antipodal points is a half-integer". This is not necessarily true, as covers are not considered local isometries in the topological setting. $\endgroup$ – mathreader Sep 19 '14 at 1:26
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OK: The universal cover of the circle that Hatcher uses is $$ p: t\mapsto e^{2\pi i t}, p: {\mathbb R}\to S^1\subset {\mathbb C}, $$ where we think of the circle $S^1$ as the unit circle in the complex plane. Now, it is just a direct calculation to show that for any two points $z, -z\in S^1$, any two elements $t_1\in p^{-1}(z), t_2\in p^{-1}(-z)$, differ by a "half-integer", i.e. a number of the form $n + \frac{1}{2}$, where $n$ is an integer.

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This is a more detailed explanation of studiosus answer.

Lemma 1: Let $p : \mathbb R \to S^1$ be the standard covering map defined by $p(s) = e^{2\pi i s}$ and let $z \in \mathbb S^1$. If $t \in p^{-1}(z)$ and $s \in p^{-1}(-z)$, then $t - s = q/2$ where $q$ is an odd integer.

Proof: Let $p : \mathbb R \to S^1$ be the standard covering map defined by $p(s) = e^{2\pi i s}$ and let $z \in \mathbb S^1$. Let $t \in p^{-1}(z)$ and $s \in p^{-1}(-z)$. Define $x = t - s$ and observe that $x = t - s$ if and only if $2 \pi i x = 2 \pi i (t - s)$ if and only if $$ e^{2 \pi i x} = e^{2 \pi i (t - s)} = \frac{e^{2 \pi i s}}{e^{2 \pi i t}} = \frac{p(t)}{p(s)} = \frac{z}{-z} = -1.$$

Taking the natural log of both sides we get $$\begin{align} 2 \pi i x = \ln (-1) (\star) \end{align}$$ Recall by Euler's Identity, that $e^{i \pi} = e^{\pi i (2n + 1)} = -1$ for $n \in \mathbb Z$, so $$\begin{align} \ln(-1) = \pi i (2n + 1) (\star \star) \end{align}$$ for some $n \in \mathbb Z$. Putting both Equations $(\star)$ and $(\star \star)$, we get $\pi i (2n + 1) = 2 \pi i x$ for some $n \in \mathbb Z$. Hence $x = \frac{q}{2}$ for some $n \in \mathbb Z$ where $q = 2n + 1$ is odd.


Let $p : \mathbb R \to S^1$ be the standard covering map defined by $p(t) = e^{2\pi i t}$.

Let $\tilde h$ be the lifting of $h$ from $S^1$ to $\mathbb R$ via $p$, so we get the following $p \circ \tilde h = h$.

Note that we have $\tilde h(s + 1/2) \in p^{-1}\big(h(s + 1/2) \big)$ and $\tilde h(s) \in p^{-1}\big(h(s) \big)$.

But since $h(s + 1/2) = -h(s)$, they are antipodal points, which means Lemma 1 tells us $\tilde h(s + 1/2) = \tilde h(s) + q/2$ where $q$ is an odd integer.

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    $\begingroup$ I would avoid using $\ln(z)$ as it is not a single-valued function. Instead, one can use the identity $e^{\phi i}=\cos\phi+i\sin\phi$. $\endgroup$ – mathreader Dec 18 '14 at 8:21
  • $\begingroup$ Have you figured out why $q$ odd implies $h$ not nullhomotopic? $\endgroup$ – mathreader Dec 18 '14 at 8:22
  • $\begingroup$ @mathreader, yes! Thanks! $\endgroup$ – Robert Cardona Dec 18 '14 at 15:08

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