MAS61015 Algebraic Topology 23 Transfers, coefficients and homology of projective spaces

24. Borsuk-Ulam and related results

Definition 24.1.

A continuous map $f\colon S^{n}\to S^{m}$ is odd (or antipodal) if $f(-x)=-f(x)$ for all $x\in S^{n}$ .

Example 24.2.

If $n\leq m$ then it is easy to produce examples of odd continuous maps $f\colon S^{n}\to S^{m}$ . Most obviously, we can just define

f(x_{0},\dotsc,x_{n})=(x_{0},\dotsc,x_{n},0,\dotsc,0).

Remark 24.3.

If $f\colon S^{n}\to S^{m}$ is odd, then we have a well-defined map $\overline{f}\colon{\mathbb{R}}P^{n}\to{\mathbb{R}}P^{m}$ given by $\overline{f}([x])=[f(x)]$ .

Proposition 24.4.

Suppose that $f\colon S^{n}\to S^{n}$ is continuous and odd. Then the induced map $f_{*}\colon H_{n}(S^{n};{\mathbb{Z}}/2)\to H_{n}(S^{n};{\mathbb{Z}}/2)$ is the identity.

Proof.

As $f$ is odd we have a well-defined map $\overline{f}\colon{\mathbb{R}}P^{n}\to{\mathbb{R}}P^{n}$ given by $\overline{f}([x])=[f(x)]$ . This satisfies $p\circ f=\overline{f}\circ p\colon S^{n}\to{\mathbb{R}}P^{n}$ , so the right-hand square below commutes:

We claim that the left hand square commutes as well. To see this, define $\chi\colon S^{n}\to S^{n}$ by $\chi(x)=-x$ , so $f\circ\chi=\chi\circ f$ and $p\circ\chi=p$ . Consider a continuous map $u\colon\Delta_{k}\to{\mathbb{R}}P^{n}$ . Choose a lift $v\colon\Delta_{k}\to S^{n}$ . The other lift is then $\chi\circ v$ , so $\tau(u)=v+(\chi\circ v)$ , so

f_{\#}(\tau(u))=(f\circ v)+(f\circ\chi\circ v)=(f\circ v)+(\chi\circ f\circ v).

Here $f\circ v$ and $\chi\circ f\circ v$ are the two lifts of the map $p\circ f\circ v=\overline{f}\circ p\circ v=\overline{f}\circ u$ , so we see that $(f\circ v)+(\chi\circ f\circ v)=\tau(\overline{f}_{\#}(u))$ . This shows that $f_{\#}\circ\tau=\tau\circ\overline{f}_{\#}$ , so the left hand square commutes as claimed.

As the diagram commutes, we see that the maps $f_{*}$ and $\overline{f}_{*}$ are compatible with the maps in the exact sequence obtained in Lemma 23.14. In particular, we have commutative squares

for $1\leq i\leq n$ . It is clear that $\overline{f}_{*}$ gives the identity on $H_{0}({\mathbb{R}}P^{n};{\mathbb{Z}}/2)$ , and we have seen that all of the maps $\Delta$ are isomorphisms, so it follows inductively that $\overline{f}_{*}$ gives the identity on $H_{i}({\mathbb{R}}P^{n};{\mathbb{Z}}/2)$ for $1\leq i\leq n$ . We also have a commutative square

We have seen that $\tau_{*}$ is an isomorphism and $\overline{f}_{*}$ is the identity so $f_{*}$ is also the identity, as claimed. ∎

Theorem 24.5 (Borsuk-Ulam).

If $n>m$ , then there are no odd continuous maps from $S^{n}$ to $S^{m}$ .

Proof.

Suppose that $f\colon S^{n}\to S^{m}$ is odd and continuous. Define $i\colon S^{m}\to S^{n}$ by

i(x_{0},\dotsc,x_{m})=(x_{0},\dotsc,x_{m},0,\dotsc,0).

It is clear that $i$ is odd, so the composite $f\circ i\colon S^{m}\to S^{m}$ is odd, so the induced map $(f\circ i)_{*}\colon H_{m}(S^{m};{\mathbb{Z}}/2)\to H_{m}(S^{m};{\mathbb{Z}}/2)$ must be the identity by Proposition 24.4. In particular, $(f\circ i)_{*}$ is nonzero.

On the other hand, we can define $h\colon[0,1]\times S^{m}\to S^{n}$ by

h(t,x)=\cos(\pi t/2)i(x)+\sin(\pi t/2)e_{m+1}.

(Using the fact that $i(x)$ and $e_{m+1}$ are orthogonal, we see that this does indeed lie in $S^{n}$ .) This gives a homotopy between $i$ and a constant map, which implies that $f\circ i$ is also homotopic to a constant map, so $(f\circ i)_{*}=0$ . This contradiction shows that no such map $f$ can exist. ∎

Corollary 24.6.

Let $g\colon S^{n}\to{\mathbb{R}}^{m}$ be a continuous map, with $0<m\leq n$ . Then there is a point $x\in S^{n}$ with $g(x)=g(-x)$ .

Proof.

Suppose (for a contradiction) that no such point exists, so $g(x)-g(-x)$ is always nonzero. We can then define $f\colon S^{n}\to S^{n-1}$ by $f(x)=(g(x)-g(-x))/\|g(x)-g(-x)\|$ . It is easy to check that this is continuous and antipodal, which contradicts Theorem 24.5, as required. ∎

Example 24.7.

There are two opposite points on the Earth’s surface that have the same temperature and also the same atmospheric pressure, as we see by considering the map $f\colon S^{2}\to{\mathbb{R}}^{2}$ given by

f(a)=(\text{ temperature at }a,\text{ pressure at }a).

Theorem 24.8 (Sandwich Slicing Theorem).

Let $A_{1}$ , $A_{2}$ and $A_{3}$ be three reasonable subsets of ${\mathbb{R}}^{3}$ . Then there is a plane $P\subset{\mathbb{R}}^{3}$ such that for each set $A_{i}$ , half of the volume lies on one side of $P$ , and half of the volume lies on the other side.

Example 24.9.

We could have a sandwich, with the top slice of bread filling the set $A_{1}$ , and cheese filling the set $A_{2}$ , and the bottom slice of bread filling $A_{3}$ . The theorem then says that we can make a single straight cut with a knife to share all three components equally.

Example 24.10.

Suppose that the sets $A_{i}$ are solid balls, with centres $a_{i}$ in general position. Then there is a unique possible choice for $P$ , namely the plane passing through $a_{1}$ , $a_{2}$ and $a_{3}$ .

We will not be very rigorous about what “reasonable” means, but we will make some comments here and in the body of the proof. To start with, each set $A_{i}$ should be bounded (which implies that the volume is finite) and the volume should also not be zero.

Proof of Theorem 24.8.

For any unit vector $u=(u_{0},u_{1},u_{2},u_{3})\in S^{3}$ and $v=(v_{1},v_{2},v_{3})\in{\mathbb{R}}^{3}$ we put $m(u,v)=u_{0}+u_{1}v_{1}+u_{2}v_{2}+u_{3}v_{3}\in{\mathbb{R}}$ , so $m(-u,v)=-m(u,v)$ . We then put

	$\displaystyle P(u)$	$\displaystyle=\{v\in{\mathbb{R}}^{3}\;\|\;m(u,v)=0\}$
	$\displaystyle H(u)$	$\displaystyle=\{v\in{\mathbb{R}}^{3}\;\|\;m(u,v)>0\},$

so $P(u)$ is a plane and $H(u)$ is the half-space on one side of that plane. Note that the plane $P(-u)$ is the same as $P(u)$ , and $H(-u)$ is the half-space on the opposite side to $H(u)$ .

We now define $g\colon S^{3}\to{\mathbb{R}}^{3}$ by

g(u)_{i}=\text{vol}(A_{i}\cap H(u)).

It is not too hard to check that this is continuous. By Corollary 24.6, there exists $u\in S^{3}$ with $g(-u)=g(u)$ , which means that $\text{vol}(A_{i}\cap H(u))=\text{vol}(A_{i}\cap H(-u))$ for $i=1,2,3$ . In other words, the plane $P(u)$ bisects each of the sets $A_{i}$ . ∎