# Lecture 6: Borsuk-Ulam and some combinatorial applications

Now we’ll move away from spectral methods, and into a few lectures on topological methods.  Today we’ll look at the Borsuk-Ulam theorem, and see a stunning application to combinatorics, given by Lovász in the late 70’s.  A great reference for this material is Matousek’s book, from which I borrow heavily.  I’ll also discuss why the Lovász-Kneser theorem arises in theoretical CS.

### The Borsuk-Ulam Theorem

We begin with a statement of the theorem.  We will think of the n-dimensional sphere as the subset of $\mathbb R^{n+1}$ given by

$S^n = \left\{ (x_1, \ldots, x_{n+1}) : x_1^2 + \cdots + x_{n+1}^2 = 1\right\}.$

Borsuk-Ulam Theorem: For every continuous mapping $f : S^n \to \mathbb R^n$, there exists a point $x \in S^n$ with $f(x)=f(-x)$.

Pairs of points $x,-x \in S^n$ are called antipodal.

There are a couple of common illustrative examples for the case $n=2$.  The theorem says that if you take the air out of a basketball, crumple it (no tearing), and flatten it out, then there are two points directly on top of each other which were antipodal before.  Another common example states that at every point in time, there must be two points on the earth which both have exactly the same temperature and barometric pressure (assuming, of course, that these two parameters vary continuously over the surface of the eath).

The n=1 case is completely elementary.  For the rest of the lecture, let’s use $N = (0,0,\ldots,1)$ and $S=(0,0,\ldots,-1)$ to denote the north and south poles (the dimension will be obvious from context).  To prove the n=1 case, simply trace out the path in $\mathbb R$ starting at $f(N)$ and going clockwise around $S^1$.  Simultaneously, trace out the path starting at $f(S)$ and going counter-clockwise at the same speed.  It is easy to see that eventually these two paths have to collide:  At the point of collision, $f(x)=f(-x)$.

We will give the sketch of a proof for $n \geq 2.$  Let $g(x)=f(x)-f(-x)$, and note that our goal is to prove that $g(x)=0$ for some $x \in S^n$.  Note that $g$ is antipodal in the sense that $g(x)=-g(-x)$ for all $x \in S^n$.  Now, if $g(x) \neq 0$ for every $x$, then by compactness there exists an $\epsilon > 0$ such that $\|g(x)\| > \epsilon$ for all $x \in S^n$.  Because of this, we may approximate $g$ arbitrarily well by a smooth map, and prove that the approximation has a 0.  So we will assume that $g$ itself is smooth.

Now, define $h : S^n \to \mathbb R^n$ by $h(x_1, \ldots, x_{n+1}) = (x_1, \ldots, x_n)$, i.e. the north/south projection map.  Let $X = S^n \times [0,1]$ be a hollow cylinder, and let $F : X \to \mathbb R^n$ be given by $F(x,t)=t g(x) + (1-t)h(x)$ so that $F$ linearly interpolates between $g$ and $h$.

Also, let’s define an antipodality on $X$ by $\nu(x,t) = (-x,t)$.  Note that $F$ is antipodal with respect to $\nu$, i.e. $F(\nu(x,t))=-F(x,t)$, because both $g$ and $h$ are antipodal.  For the sake of contradiction, assume that $g(x) \neq 0$ for all $x \in S^n$.

Now let’s consider the structure of the zero set $Z = F^{-1}(0)$.  Certainly $(N,0), (S,0) \in Z$ since $h(N)=h(S)=0$, and these are h’s only zeros.  Here comes the sketchy part:  Since $g$ is smooth, $F$ is also smooth, and thus locally $F$ can be approximated by an affine mapping $F_{\mathrm{loc}} : \mathbb R^{n+1} \to \mathbb R^n$.  It follows that if $F_{\mathrm{loc}}^{-1}(0)$ is not empty, then it should be a subspace of dimension at least one.  By an arbitrarily small perturbation of the initial $g$, ensuring that $F$ is sufficiently generic, we can ensure that $F_{\mathrm{loc}}^{-1}(0)$ is either empty or a subspace of dimension one.  Thus locally, $F^{-1}(0)$ should look like a two-sided curve, except at the boundaries $t=0$ and $t=1$, where $F^{-1}(0)$ (if non-empty) would look like a one-sided curve.  But, for instance, $F^{-1}(0)$ cannot contain any Y-shaped branches.

It follows that $Z$ is a union of closed cycles and paths whose endpoints must lie at the boundaries $t=0$ and $t=1$.  ($Z$ is represented by red lines in the cylinder above.)  But since there are only two zeros on the $t=0$ sphere, and none on the $t=1$ sphere, $Z$ must contain a path $\gamma$ from $(N,0)$ to $(S,0).$  Since $F$ is antipodal with respect to $\nu$, $\gamma$ must also satisfy this symmetry, making it impossible for the segment initiating at N to ever meet up with the segment initiating at S.  Thus we arrive at a contradiction, implying that $g$ must take the value 0.

Notice that the only important property we used about $h$ (other than its smoothness) is that is has a number of zeros which is twice an odd number.  If $h$ had, e.g. four zeros, then we could have two $\nu$-symmetric paths emanating from and returning to the bottom.  But if $h$ has six zeros, then we would again reach a contradiction.

### The Kneser conjecture

While the Borsuk-Ulam theorem seems initially far removed from any combinatorial applications, the following result of Lyusternik and Shnirel’man starts to hint at the combinatorial significance.

LS Lemma: Let $F_1, F_2, \ldots, F_{n+1}$ be a cover of $S^n$ by closed sets.  Then some $F_i$ contains an antipodal pair.

Proof: Consider the continuous map $f : S^n \to \mathbb R^n$ given by $f(x)=(\mathsf{dist}(x,F_1), \ldots, \mathsf{dist}(x, F_n))$, where $\mathsf{dist}$ denotes e.g. the geodesic distance on $S^n$.  By the Borsuk-Ulam theorem, there exists a $y \in S^n$ with $f(y)=f(-y)$.  Now if $f(y)_i = 0$ for some $i$, then $y, -y \in F_i$.  Otherwise, if this holds for no $i = 1, \ldots, n$, then $y,-y \in F_{n+1}$.

In fact, it is not difficult to show that the LS Lemma is equivalent to Borsuk-Ulam.  For our application to the Kneser conjecture, we will need to consider both open and closed sets.

Corollary 1: Let $F_1, \ldots, F_{n+1}$ be a cover of $S^n$ by open sets.  Then some $F_i$ contains an antipodal pair.

Proof: This follows from the LS Lemma by choosing a family of closed sets $U_i \subseteq F_i$ such that $\{U_i\}$ forms a cover of $S^n$.  The existence of such sets follows from applying compactness to the following open cover of $S^n$:  For every $x \in S^n$, choose an open set $N_x$ whose closure lies completely within some $F_i$.  By compactness, there is a finite subcover.  By piecing together elements of its closure, we can form the sets $\{U_i\}$.

Finally, we address the case of both open and closed sets.

Corollary 2: Let $F_1, \ldots, F_{n+1}$ be a cover of $S^n$ with each $F_i$ open or closed.  Then some $F_i$ contains an antipodal pair.

Proof: Suppose that $F_1, \ldots, F_k$ are closed and $F_{k+1}, \ldots, F_{n+1}$ are open.  Consider the open cover $(F_1)_{\varepsilon}, \ldots, (F_k)_{\varepsilon}, F_{k+1}, \ldots, F_{n+1}$, where $(F_i)_{\varepsilon}$ denotes the $\varepsilon$-neighborhood of $F_i$.  Taking $\varepsilon \to 0$ and applying Corollary 1 to each cover results in a sequence $\{x_j\} \subseteq S^n$.  If $x_j,-x_j \in F_i$ for some $i \geq k+1$, we are done.  Thus we may pass to a subsequence for which $x_j,-x_j \in (F_i)_{\varepsilon}$ for some fixed $i \leq k$.  Choose a convergent subsequence $\{x'_j\}$, and observe that since $F_i$ is closed, we have $x = \lim x'_j \in F_i$.  We conclude that there exists an $x \in S^n$ with $x,-x \in F_i$.

The Kneser graphs.

Now, we introduce the Kneser graphs $KG_{n,k}$.  The vertex set is ${[n] \choose k}$, i.e. the vertices are k-element subsets of $\lbrack n\rbrack = \{1,2, \ldots, n\}$.  There is an edge between two sets $S,S' \in {[n] \choose k}$ if and only if $S$ and $S'$ are disjoint.  In 1955, Kneser posed the following conjecture.

Conjecture: For every $k > 0$ and $n \geq 2k-1$, $\chi(KG_{n,k})=n-2k+2$, where $\chi$ denotes the chromatic number.

First, we note that the upper bound is easy:  Construct a coloring $c : {[n] \choose k} \to \{1,2,\ldots,n-2k+2\}$ by defining

$\displaystyle c(S) = \min \left\{ \min(S), n-2k+2 \right\}.$

Clearly if $c(S)=c(S') < n-2k+2$, then $S$ and $S'$ have the same minimum element, and thus they are not disjoint.  On the other hand, if $c(S)=c(S')=n-2k+2$, then $S,S' \subseteq \{n-2k+2, n-2k+3, \ldots, n\}$, and the latter set contains only $2k-1$ elements.  Thus again $S \cap S' \neq \emptyset$.

The conjecture stood open until Lovász resolved it, using the Borsuk-Ulam theorem.  Recently, Greene (an undergraduate at the time) gave a particularly simple proof which we reproduce here.

Proof: Consider $KG_{n,k}$ and let $d = n-2k+1$.  Let $X \subseteq S^d$ be a set of $n$ points in general position, so that no $(d-1)$-dimensional hyperplane through the origin contains more than $d$ points of $X$.  Clearly we may assume that $KG_{n,k}$ actually has vertex set ${X \choose k}$.  For every $x \in S^d$, let

$\displaystyle H(x) = \{ y \in S^d : \langle x,y \rangle > 0 \}$

denote the open hemisphere centered at $x$.

Now, suppose we have a proper coloring of $KG_{n,k}$, and define sets $A_1, A_2, \ldots, A_d \subseteq S^d$ as follows:  $A_i$ is the set of all points $x \in S^d$ such that $H(x)$ contains a k-set $S \subseteq X$ colored $i$.  Finally, $A_{d+1} = S^d \setminus (A_1 \cup \cdots \cup A_d)$.

By Corollary 2 above, there must exist an $i$ and a $x \in S^d$ such that $x,-x \in A_i$.  If $i \leq d$, then since $H(x)$ and $H(-x)$ are disjoint, there must exist two disjoint k-sets colored $i$, which means we did not start with a proper coloring.

If, instead $x,-x \in A_{d+1}$, it implies that both $|H(x) \cap X|, |H(-x) \cap X| \leq k-1$.  But then the $(d-1)$-dimensional hyperplane $S^d \setminus (H(x) \cup H(-x))$ contains $n-2k+2 = d+1$ points, contradicting the fact that $X$ was chosen in general position.

### Appearances in TCS

One of the most useful properties of the Kneser graphs is that there is a large gap between their chromatic number $\chi(KG_{n,k})$ and their fractional chromatic number $\chi_f(KG_{n,k})$.  The fractional chromatic number of graph $G$ is the minimum value $\displaystyle \frac{a}{b}$ such that $G$ can be covered by $a$ independent sets, where each vertex occurs in at least $b$ of them.

It is easy to see that $\chi_f(KG_{n,k}) \leq n/k$ by choosing the n independent sets

$\displaystyle V_i = \left\{ S \in {[n] \choose k} : i \in S \right\}.$

The Kneser graphs are a particularly natural family to have a large gap between $\chi$ and $\chi_f$ because we can define $\chi_f(G)$ for an arbtirary G as the minimum ratio $n/k$ such that G admits a homomorphism into $KG_{n,k}$.  One can also see that $\chi_f(KG_{n,k}) = n/k$ using the Erdős–Ko–Rado theorem.

See e.g. this FOCS’08 paper of Alon, et. al. for a recent application which uses this gap between fractional and actual chromatic numbers.  The Kneser graphs also arise very naturally in PCP constructions, as one can see in this paper of Dinur, Regev, and Smyth about the computational hardness of coloring hypergraphs.

## One thought on “Lecture 6: Borsuk-Ulam and some combinatorial applications”

1. Thanks for the great posts!

In the defn of S^n near the top, I think you mean it’s a subset of R^{n+1}.