Antipodal Symmetry and Degree
A map of the circle can wrap around the circle several times, in either orientation. Its degree records this signed number of turns. If the map also sends each point’s antipode to the antipode of its image, the number of turns cannot be even. The reason is visible on the real line: antipodal points correspond to parameters separated by one half, and the lift’s change over that half-step must be a half-integer.
We regard \(S^1=\{z\in\mathbb{C}:|z|=1\}\) with the subspace topology inherited from \(\mathbb{C}\), and \(\mathbb{R}\) with its standard topology. Write \(p:\mathbb{R}\to S^1\) for the parametrization \(p(t)=e^{2\pi i t}\). It satisfies \(p(t+1)=p(t)\) and \(p(t+1/2)=-p(t)\). Let \(f:S^1\to S^1\) be continuous. A lift of \(f\) is a continuous function \(F:\mathbb{R}\to\mathbb{R}\) such that \(p(F(t))=f(p(t))\) for every \(t\).
Here is why the degree in this definition is well-defined. Since \(p(t+1)=p(t)\), the lift identity gives \(p(F(t+1))=p(F(t))\). Two real numbers have the same image under \(p\) exactly when their difference is an integer. Thus \(F(t+1)-F(t)\) is integer-valued. It is also continuous as a function of \(t\), so it is constant: if it had two different integer values, the intermediate value theorem would force it to take a non-integer value between them. If \(G\) is another lift, then \(p(G(t))=p(F(t))\), so \(G(t)-F(t)\) is also continuous and integer-valued, hence constant by the same argument. Consequently \(G(t+1)-G(t)=F(t+1)-F(t)\). This verifies the definition without depending on a particular lift.
The Half-Step in a Lift
Proof. Choose a lift \(F:\mathbb{R}\to\mathbb{R}\). For each \(t\in\mathbb{R}\), the parametrization gives \(p(t+1/2)=-p(t)\). Applying the assumed condition on \(f\), we obtain
Since equal values of \(p\) have parameters differing by an integer, the function \(m(t)=F(t+1/2)-F(t)-1/2\) is integer-valued. It is continuous because \(F\) is continuous. It must therefore be constant: if it took distinct integer values, the intermediate value theorem would give it a non-integer value between them. Write the constant as \(k\in\mathbb{Z}\). Then \(F(t+1/2)-F(t)=k+1/2\) for every \(t\). Applying this same equality at \(t+1/2\) gives \(F(t+1)-F(t+1/2)=k+1/2\). Adding the two changes yields $$ \deg(f)=F(t+1)-F(t)=(k+1/2)+(k+1/2)=2k+1. $$ This is an odd integer, including when it is negative. \(\square\)
The constancy matters: the integer \(k\) cannot vary with the starting point \(t\). The proof does not say that the lift changes by exactly \(1/2\) over every half-step. Rather, it changes by \(1/2\) plus one fixed integer. The integer records any additional full turns that occur over that half-step.
Worked Example: The Cubing Map
Let \(f:S^1\to S^1\) be \(f(z)=z^3\). It is continuous, and for every \(z\in S^1\), \(f(-z)=(-z)^3=-z^3=-f(z)\). Find its degree using a lift and check the half-step relation.
For \(p(t)=e^{2\pi i t}\), we have \(f(p(t))=(e^{2\pi i t})^3=e^{2\pi i(3t)}\). Thus \(F(t)=3t\) is a lift. Its full-step change is \(F(t+1)-F(t)=3(t+1)-3t=3\), so \(\deg(f)=3\).
Over a half-step, \(F(t+1/2)-F(t)=3(t+1/2)-3t=3/2=1+1/2\), giving \(k=1\). The theorem’s formula gives \(2k+1=2(1)+1=3\), agreeing with the lift’s full-step change. As a direct check of the symmetry, \((-z)^3=-z^3\) for every \(z\), as required.
Worked Example: Reflection of the Circle
Let \(f:S^1\to S^1\) be complex conjugation, \(f(z)=\overline{z}\). Conjugation is continuous and preserves the unit circle. Also \(f(-z)=\overline{-z}=-\overline{z}=-f(z)\). Compute its degree, keeping track of the sign.
Since \(\overline{p(t)}=\overline{e^{2\pi i t}}=e^{-2\pi i t}=p(-t)\), a lift is \(F(t)=-t\). Hence \(F(t+1)-F(t)=-(t+1)-(-t)=-1\), and \(\deg(f)=-1\). For the half-step, \(F(t+1/2)-F(t)=-(t+1/2)-(-t)=-1/2=-1+1/2\), so \(k=-1\). The odd-degree formula yields \(2(-1)+1=-1\), matching the computed degree. This example shows that “odd” includes negative odd integers; the degree is signed, not merely a count of turns without orientation.
Worked Example: A Nonlinear Map of Degree Five
For \(z\in S^1\), define $$ f(z)=z^5\exp\left(\frac{2\pi i}{7}\operatorname{Re}(z^2)\right). $$ The formula is continuous, and its value has modulus \(1\), since both factors have modulus \(1\). The real part of \((-z)^2\) equals the real part of \(z^2\), while \((-z)^5=-z^5\). Therefore \(f(-z)=-f(z)\).
For \(z=p(t)\), we have \(z^2=e^{4\pi i t}\), so \(\operatorname{Re}(z^2)=\cos(4\pi t)\). Also \(z^5=e^{2\pi i(5t)}\). Thus a lift is \(F(t)=5t+\frac17\cos(4\pi t)\), because \(p(F(t))=e^{2\pi i(5t)}\exp\left(\frac{2\pi i}{7}\cos(4\pi t)\right)=f(p(t))\). Since \(\cos(4\pi(t+1))=\cos(4\pi t+4\pi)=\cos(4\pi t)\), its full-step change is \(F(t+1)-F(t)=5\), so \(\deg(f)=5\).
The half-step change is \(F(t+1/2)-F(t)=5/2+\frac17\bigl(\cos(4\pi t+2\pi)-\cos(4\pi t)\bigr)=5/2=2+1/2\). Hence \(k=2\), and \(2k+1=2(2)+1=5\), as the theorem predicts. For a pointwise check at \(z=1\), \(f(1)=\exp(2\pi i/7)\). At \(z=-1\), \((-1)^5=-1\) and \(\operatorname{Re}((-1)^2)=1\), so \(f(-1)=-\exp(2\pi i/7)=-f(1)\).
A Common Pitfall: Losing the Integer
A tempting but incorrect shortcut is to claim that antipodal symmetry forces \(F(t+1/2)=F(t)+1/2\) exactly. The equation \(p(F(t+1/2))=p(F(t)+1/2)\) only implies that these two real parameters differ by an integer; the parametrization \(p\) repeats every integer. The cubing example makes the error concrete: its lift has \(F(t+1/2)-F(t)=3/2\), not \(1/2\). Replacing \(3/2\) by \(1/2\) would incorrectly produce degree \(1\), while direct calculation gives degree \(3\). The correction is to write the half-step change as \(k+1/2\), prove that the integer \(k\) is constant, and use two half-steps to obtain degree \(2k+1\).
There is a related logical pitfall: the theorem gives a necessary condition, not a converse. An odd degree by itself does not force antipodal symmetry. To check this, define \(F(t)=t+\frac14\sin(2\pi t)\) and \(f(p(t))=p(F(t))\). Because \(F(t+1)=F(t)+1\), this defines a continuous circle map of degree \(1\). But at \(t=1/4\), \(F(1/4)=1/4+1/4=1/2\), so \(f(i)=p(1/2)=-1\). At \(t=3/4\), \(F(3/4)=3/4-1/4=1/2\), so \(f(-i)=-1\), whereas \(-f(i)=1\). Thus \(f(-i)\ne-f(i)\), despite the odd degree.
Why the Parity Conclusion Is Useful
The argument turns a geometric symmetry into an arithmetic restriction. Instead of tracking the image of every point of the circle, it is enough to compare two values of one lift separated by half a unit. That comparison constrains the total change over a full unit and therefore the degree. In applications, this gives an obstruction: if a proposed continuous circle map has the antipodal property but a separately computed degree is even, the two claims cannot both hold.
The hypotheses should remain visible. Continuity supplies a lift and ensures that the integer-valued half-step difference cannot jump between integers. The condition \(f(-z)=-f(z)\) supplies the half-step relation. Neither a formula for a particular map nor an informal picture of its turns replaces these steps. In each worked example, the lift verifies the degree, while the half-step calculation checks exactly how the symmetry enforces odd parity.
Check Your Understanding
For each question, identify the lift or the integer ambiguity being used, and distinguish a necessary condition from a sufficient one.
- For \(f(z)=z^7\), find a lift, its degree, and the integer \(k\) in the half-step formula.
- For \(f(z)=\overline{z}^{\,3}\), verify the antipodal condition and compute the degree from a lift.
- Why does \(p(a)=p(b)\) imply \(a-b\in\mathbb{Z}\), and where is that fact used in the odd-degree proof?
- In the proof, why must \(F(t+1/2)-F(t)-1/2\) be constant rather than merely integer-valued?
- Does every circle map of odd degree satisfy \(f(-z)=-f(z)\)? Explain using the degree-one map in the pitfall discussion.