Lecture 6: Borsuk-Ulam and some combinatorial applications


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


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.


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.

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