From Separated Antipodes to a Circle-Valued Map
Suppose a continuous map assigns a point of the plane to every point of the sphere, and assigns different values to each pair of antipodal points. The difference between the two values is then a nonzero vector in the plane. Dividing that vector by its length gives a point on the unit circle. This simple normalization has two important features: it varies continuously with the point on the sphere, and reversing the point to its antipode reverses the resulting circle point.
We use the Euclidean subspace topologies on \(S^2=\{x\in\mathbb{R}^3:\|x\|=1\}\), \(\mathbb{R}^2\), and \(S^1=\{u\in\mathbb{R}^2:\|u\|=1\}\). The antipode of \(x\in S^2\) is \(-x\), which is again in \(S^2\). A map \(F:S^2\to\mathbb{R}^2\) separates antipodal pairs if \(F(x)\ne F(-x)\) for every \(x\in S^2\). A map \(g:S^2\to S^1\) is odd if \(g(-x)=-g(x)\) for every \(x\in S^2\).
The numerator records the direction from \(F(-x)\) to \(F(x)\), while the denominator removes its length. The result is a unit vector. The order of subtraction matters: changing it everywhere replaces \(g\) by \(-g\), which is still odd, but changing the order at only one of a pair of antipodal points would break the defining relation.
Why the Formula Works
Proof. Define \(D:S^2\to\mathbb{R}^2\) by \(D(x)=F(x)-F(-x)\). The antipodal map \(a:S^2\to S^2\), \(a(x)=-x\), is continuous: for \(x,y\in S^2\), \(\|a(x)-a(y)\|=\|x-y\|\). Therefore \(F\circ a\) is continuous. Subtracting its coordinate functions from those of \(F\) shows that \(D\) is continuous. By the separation hypothesis, \(D(x)\ne(0,0)\) at every \(x\).
For a nonzero vector \(v=(v_1,v_2)\in\mathbb{R}^2\), its norm \(\|v\|=\sqrt{v_1^2+v_2^2}\) is positive, and the normalization map \(N(v)=v/\|v\|\) is continuous on \(\mathbb{R}^2\setminus\{(0,0)\}\). Indeed, each coordinate of \(N(v)\) is a quotient of continuous functions with nonzero denominator. Also, $$ \|N(v)\|^2=\left(\frac{v_1}{\|v\|}\right)^2+\left(\frac{v_2}{\|v\|}\right)^2 =\frac{v_1^2+v_2^2}{\|v\|^2}=1. $$ Thus \(N(v)\in S^1\). Since \(g=N\circ D\), the map \(g\) is continuous and takes values in \(S^1\).
Finally, for every \(x\in S^2\), $$ D(-x)=F(-x)-F(-(-x))=F(-x)-F(x)=-D(x). $$ The Euclidean norm satisfies \(\|-D(x)\|=\|D(x)\|\), so $$ g(-x)=\frac{-D(x)}{\|-D(x)\|}=-\frac{D(x)}{\|D(x)\|}=-g(x). $$ This proves that \(g\) is odd as well as continuous. \(\square\)
The two hypotheses have separate jobs. Continuity of \(F\) gives continuity of the difference, while separation guarantees that normalization is defined everywhere. If separation fails at even one point \(x\), then \(F(x)-F(-x)=0\), and the displayed formula has zero denominator there. Continuity alone does not repair that failure.
Worked Examples
Worked Example: Checking the Formula on One Antipodal Pair
Let \(p=(0,0,1)\) and consider the two-point subspace \(X_0=\{p,-p\}\) of \(S^2\), with its subspace topology. Define \(F:X_0\to\mathbb{R}^2\) by \(F(p)=(3,4)\) and \(F(-p)=(0,0)\). This map separates the only antipodal pair. In fact, \(X_0\) is discrete: the open ball of radius \(1\) around \(p\) in \(S^2\) contains \(p\) but not \(-p\), whose distance from \(p\) is \(2\); the same argument works around \(-p\). Every map from this discrete space is continuous.
At \(p\), the difference is \((3,4)-(0,0)=(3,4)\), whose norm is \(\sqrt{3^2+4^2}=\sqrt{9+16}=5\). Hence $$ g(p)=\frac{(3,4)}{5}=\left(\frac35,\frac45\right). $$ At the antipode, the subtraction reverses: $$ F(-p)-F(p)=(0,0)-(3,4)=(-3,-4). $$ Its norm is \(\sqrt{(-3)^2+(-4)^2}=\sqrt{9+16}=5\), so \(g(-p)=(-3/5,-4/5)\).
Check: \((3/5)^2+(4/5)^2=9/25+16/25=1\), so \(g(p)\in S^1\). Also \(g(-p)=(-3/5,-4/5)=-g(p)\). This example verifies the formula on a two-point domain; it does not assert that these prescribed values extend to a separating map on all of \(S^2\).
Worked Example: The Difference Map on the Equator
Write the equator as \(E=\{(x,y,0)\in S^2:x^2+y^2=1\}\), and let \(F:S^2\to\mathbb{R}^2\) be the coordinate projection \(F(x,y,z)=(x,y)\). This \(F\) is continuous. It need not separate every antipodal pair on all of \(S^2\), but it does separate antipodal pairs on \(E\). For \((x,y,0)\in E\), $$ F(x,y,0)=(x,y),\qquad F(-x,-y,0)=(-x,-y), $$ so their difference is \((2x,2y)\). Its norm is $$ \sqrt{(2x)^2+(2y)^2}=2\sqrt{x^2+y^2}=2, $$ where \(2\) is positive because \(x^2+y^2=1\). The normalized difference on the equator is therefore $$ g(x,y,0)=\frac{(2x,2y)}{2}=(x,y). $$
Check: the output has norm \(\sqrt{x^2+y^2}=1\), and replacing \((x,y,0)\) by \((-x,-y,0)\) changes the output to \((-x,-y)=-g(x,y,0)\). Thus this restricted map is the identity in equatorial coordinates and is odd.
Worked Example: Projection Away from the Poles
Let \(D=S^2\setminus\{(0,0,1),(0,0,-1)\}\), and restrict the same projection \(F(x,y,z)=(x,y)\) to \(D\). Every point of \(D\) has \((x,y)\ne(0,0)\): if \(x=y=0\) and \(x^2+y^2+z^2=1\), then \(z=1\) or \(z=-1\), giving one of the removed poles. Thus for every \((x,y,z)\in D\), $$ F(x,y,z)-F(-x,-y,-z)=(x,y)-(-x,-y)=(2x,2y)\ne(0,0). $$ The restricted map separates every antipodal pair in \(D\). Its normalized difference is $$ g(x,y,z)=\frac{(2x,2y)}{\sqrt{4x^2+4y^2}} =\left(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\right). $$
The denominator is positive on \(D\), as just verified. The output has squared norm $$ \left(\frac{x}{\sqrt{x^2+y^2}}\right)^2+ \left(\frac{y}{\sqrt{x^2+y^2}}\right)^2 =\frac{x^2+y^2}{x^2+y^2}=1. $$ Replacing \((x,y,z)\) by its antipode changes each numerator’s sign and leaves the denominator unchanged, so the value changes to \(-g(x,y,z)\).
Check at a point: take \((x,y,z)=(3/5,4/5,0)\), which lies in \(D\) because \((3/5)^2+(4/5)^2=9/25+16/25=1\). The denominator in the displayed formula is \(\sqrt{9/25+16/25}=1\), giving \(g=(3/5,4/5)\). At the antipode the value is \((-3/5,-4/5)\), as required.
What the Construction Records
The formula converts a comparison of two plane values into a direction on a fixed circle. The length of the difference may vary with \(x\), but the normalized map keeps only its direction. This is useful because properties of maps to \(S^1\), including degree, can then be applied to the directional information.
For example, if \(F\) separates antipodes on an antipodally invariant subset of \(S^2\), the same formula defines an odd map on that subset. When the subset is a circle, this is exactly the setting of the earlier result that odd circle maps have odd degree. The construction itself does not assert that an arbitrary continuous \(F:S^2\to\mathbb{R}^2\) separates antipodes; separation is an assumption, and the normalized map is the consequence.
Common pitfall: forgetting to normalize. In the first worked example the raw difference at \(p\) is \((3,4)\). It is incorrect to call this the circle-valued output: its norm is \(\sqrt{3^2+4^2}=5\), not \(1\). The correction is to divide by \(5\), giving \((3/5,4/5)\), whose squared norm is \(9/25+16/25=1\). Another sign check is essential: at \(-p\), the prescribed difference is \(F(-p)-F(p)=(-3,-4)\), not \((3,4)\). The reversed subtraction is what gives the oddness relation.
Check Your Understanding
For each question, attend to the separation condition and to whether the formula is defined at every point in the stated domain.
- For \(F(p)=(5,12)\) and \(F(-p)=(0,0)\), compute \(g(p)\) and \(g(-p)\), and check that both outputs have norm \(1\).
- For the projection \(F(x,y,z)=(x,y)\), at which points of \(S^2\) does the antipodal difference vanish?
- Suppose \(F:S^2\to\mathbb{R}^2\) is continuous and separates antipodes. Identify which part of the proof establishes that \(g(x)\in S^1\), and which part establishes oddness.
- If the order in the difference is changed everywhere to \(F(-x)-F(x)\), what is the resulting map in terms of \(g\), and is it still odd?
- Explain why continuity of \(F\) alone does not guarantee that the normalized difference formula is defined on all of \(S^2\).