Borsuk-Ulam Theorem on the Two-Sphere

The Antipodal Equality Problem

For a map from a circle to itself, the antipodal condition \(g(-u)=-g(u)\) forces an odd degree. The two-sphere has a natural equator, and its northern hemisphere is a disk bounded by that equator. These facts let us turn a question about values of a map to the plane into a contradiction about the degree of a circle map.

We use the standard subspace topologies on \(S^2=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2=1\}\), \(S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\), and \(\mathbb{R}^2\). The antipode of \(p\in S^2\) is \(-p\). The equator is \(E=\{(x,y,0)\in S^2\}\), and the closed northern hemisphere is \(H=\{(x,y,z)\in S^2:z\geq 0\}\).

Theorem (Borsuk-Ulam for \(S^2\)): For every continuous map \(F:S^2\to\mathbb{R}^2\), there is a point \(p\in S^2\) such that \(F(p)=F(-p)\).

The proof proceeds by contradiction. If \(F(p)\ne F(-p)\) at every point, the difference \(F(p)-F(-p)\) is never zero. Dividing by its length then produces a continuous map from the sphere to the circle. On the equator this map is odd, so its degree is odd. But the map on the equator also extends over the northern hemisphere, so its degree must be zero. We give the details of both implications.

The Normalized Antipodal Difference

Definition: Given a continuous \(F:S^2\to\mathbb{R}^2\), its antipodal difference is the map \(D:S^2\to\mathbb{R}^2\) defined by \(D(p)=F(p)-F(-p)\). Wherever \(D(p)\ne 0\), its normalized difference is \(G(p)=D(p)/\|D(p)\|\), which takes values in \(S^1\).

Lemma: If \(F(p)\ne F(-p)\) for every \(p\in S^2\), then \(G:S^2\to S^1\) is continuous and satisfies \(G(-p)=-G(p)\) for every \(p\in S^2\).

Proof. Under the stated assumption, \(D(p)\ne 0\) for every \(p\). The map \(p\mapsto -p\) is continuous, so \(D\) is continuous by continuity of \(F\), subtraction, and composition. The norm \(p\mapsto\|D(p)\|\) is continuous and is positive everywhere. Dividing by this positive continuous function shows that \(G\) is continuous; moreover, \(\|G(p)\|=1\), so \(G(p)\in S^1\). Finally, \(D(-p)=F(-p)-F(p)=-D(p)\), and \(\|D(-p)\|=\|D(p)\|\). Therefore \(G(-p)=-G(p)\). \(\square\)

The assumption matters: without it, \(D\) may vanish, and normalization is then undefined. The theorem’s proof assumes the absence of such zeros only to construct \(G\); the contradiction will show that this assumption cannot hold.

Worked Example: Projection onto the First Two Coordinates

Let \(F:S^2\to\mathbb{R}^2\) be \(F(x,y,z)=(x,y)\). We find all points with equal values at the antipodes by computing the difference.

For \(p=(x,y,z)\), we have \(F(-p)=(-x,-y)\), so \(D(p)=(x,y)-(-x,-y)=(2x,2y)\). Thus \(F(p)=F(-p)\) exactly when \(2x=0\) and \(2y=0\), or equivalently \(x=y=0\). Since \(p\in S^2\), the equation \(x^2+y^2+z^2=1\) then gives \(z^2=1\), so the two solutions are \((0,0,1)\) and \((0,0,-1)\), an antipodal pair.

On the equator, write \(e(t)=(\cos(2\pi t),\sin(2\pi t),0)\), where \(0\leq t\leq 1\). The difference there is \(D(e(t))=(2\cos(2\pi t),2\sin(2\pi t))\). Its norm is \(2\), since \(4\cos^2(2\pi t)+4\sin^2(2\pi t)=4\). Hence the normalized difference on the equator is \(G(e(t))=(\cos(2\pi t),\sin(2\pi t))\), the identity circle map, of degree \(1\).

Check: the normalized difference does not extend over the whole sphere in this example: at both poles \(D=0\). That is consistent with the proof, which only constructs a global \(G\) under the contradictory assumption that no antipodal values agree.

The Degree and Extension Facts

We use two results about circle maps established in the earlier material on degree and winding number. We state them precisely to make clear which properties of the equatorial map are needed.

Standard result (odd circle maps have odd degree): If \(h:S^1\to S^1\) is continuous and \(h(-u)=-h(u)\) for every \(u\in S^1\), then its degree is an odd integer, and in particular is not zero. This is the odd circle map degree theorem.

Standard result (extension over the disk forces zero degree): If a continuous map \(h:S^1\to S^1\) is the restriction to the boundary of a continuous map \(K:D^2\to S^1\), then \(\deg(h)=0\). This is the extension-over-the-disk criterion for degree.

These are the specific earlier results used below; their proofs belong to the earlier study of circle degree. The first requires an odd map from the circle to itself. The second requires a genuine continuous extension to the entire disk, not merely a map defined on the boundary.

Worked Example: The Boundary Map and Its Disk Extension

Consider \(h:S^1\to S^1\) defined by \(h(u)=u\). It is odd because \(h(-u)=-u=-h(u)\). In the parameterization \(u(t)=(\cos(2\pi t),\sin(2\pi t))\), its lift has angle \(2\pi t\), so the angle changes by \(2\pi\) as \(t\) goes from \(0\) to \(1\); therefore \(\deg(h)=1\).

There cannot be a continuous extension \(K:D^2\to S^1\) with \(K|_{S^1}=h\). If such a \(K\) existed, the extension criterion would give \(\deg(h)=0\), contradicting the computed value \(1\). This illustrates the incompatibility that will arise from the equatorial restriction of \(G\): oddness forces nonzero degree, whereas extension across a hemisphere forces degree zero.

Proof of the Borsuk-Ulam Theorem

Proof. Let \(F:S^2\to\mathbb{R}^2\) be continuous. Suppose, for contradiction, that \(F(p)\ne F(-p)\) for every \(p\in S^2\). By the lemma, the normalized difference \(G:S^2\to S^1\) is continuous and odd.

Parameterize the equator by \(e:S^1\to E\), \(e(u_1,u_2)=(u_1,u_2,0)\), and define \(h:S^1\to S^1\) by \(h(u)=G(e(u))\). This is continuous as a composition. Because \(e(-u)=-e(u)\) and \(G\) is odd, \(h(-u)=G(e(-u))=G(-e(u))=-G(e(u))=-h(u)\). The odd circle map degree theorem therefore implies that \(\deg(h)\) is odd, and hence \(\deg(h)\ne 0\).

Now identify the closed unit disk \(D^2=\{(u_1,u_2)\in\mathbb{R}^2:u_1^2+u_2^2\leq 1\}\) with \(H\) using \(\phi(u_1,u_2)=(u_1,u_2,\sqrt{1-u_1^2-u_2^2})\). This is a homeomorphism: it is continuous, its inverse is the projection \((x,y,z)\mapsto(x,y)\), and these maps are inverse to one another. On the boundary \(S^1\), \(\phi(u)=e(u)\). Thus \(K=G\circ\phi:D^2\to S^1\) is continuous and, for every \(u\in S^1\), \(K(u)=G(e(u))=h(u)\). It is an extension of \(h\) over the disk. The extension-over-the-disk criterion gives \(\deg(h)=0\), contradicting \(\deg(h)\ne 0\).

The supposition that \(F(p)\ne F(-p)\) for all \(p\) is therefore false. Consequently there is some \(p\in S^2\) for which \(F(p)=F(-p)\). \(\square\)

Worked Example: An Affine Map with a Shift

Let \(F:S^2\to\mathbb{R}^2\) be \(F(x,y,z)=(3x+z,-2y+4)\). The constant \(4\) cancels in the antipodal difference. Directly, \(F(-x,-y,-z)=(-3x-z,2y+4)\), so \(D(x,y,z)=(6x+2z,-4y)\).

Equality of the two values requires \(6x+2z=0\) and \(-4y=0\), giving \(z=-3x\) and \(y=0\). Substitution into the sphere equation gives \(x^2+9x^2=1\), so \(10x^2=1\) and \(x=\pm1/\sqrt{10}\). The corresponding points are \(p_+=(1/\sqrt{10},0,-3/\sqrt{10})\) and \(p_-=(-1/\sqrt{10},0,3/\sqrt{10})=-p_+\).

Check: \(F(p_+)=(3/\sqrt{10}-3/\sqrt{10},4)=(0,4)\), and \(F(p_-)=(-3/\sqrt{10}+3/\sqrt{10},4)=(0,4)\). Both points lie on \(S^2\), since \(1/10+9/10=1\), and their images agree as required.

Why the Argument Needs Both Ingredients

The proof does not claim that every map \(F:S^2\to\mathbb{R}^2\) is odd, or that its restriction to the equator is a circle map. Instead, it builds a new map from \(F\), and only under the assumption that \(F\) never agrees at antipodes. The difference reverses sign at antipodal points; normalization then places it in \(S^1\) without changing that sign relation.

The equator and the hemisphere play different roles. Oddness is used on the equator, where the odd circle map theorem controls degree. Extension is obtained from the hemisphere, which is a disk. The contradiction is not that an odd circle map cannot exist: the identity map is an odd circle map. It is that an odd circle map of nonzero degree cannot at the same time extend over the disk.

Common pitfall. A tempting but invalid argument is to normalize \(D\) on the equator and immediately claim that this normalized map extends over the hemisphere. Normalization requires \(D(p)\ne0\) at every point of the hemisphere, not just on its boundary. For the projection \(F(x,y,z)=(x,y)\), the equatorial difference is nonzero and normalizes to the identity circle map, but \(D(0,0,1)=(0,0)\), so no normalized difference is defined at the north pole. The correct proof first assumes that no antipodal pair anywhere on \(S^2\) has equal values. Only then is \(D\) nonzero on the whole hemisphere, and only then does \(G\) provide the required disk extension.

Check Your Understanding

Use the normalized difference and the role of the hemisphere to organize your answers.

For \(F(x,y,z)=(x,y)\), compute the antipodal difference and determine all points where the values agree.
Why does \(D(p)\ne0\) for every \(p\in S^2\) imply that \(G(p)=D(p)/\|D(p)\|\) is continuous and odd?
Suppose the equatorial restriction of \(G\) is an odd circle map. What does the odd circle map degree theorem imply about its degree?
Explain why the northern hemisphere gives an extension over a disk and identify the resulting degree contradiction.
For \(F(x,y,z)=(x^2-y^2,2xy)\), verify directly that \(F(p)=F(-p)\) for every \(p\in S^2\).