Degrees of Power Maps, Reflections, and Compositions

Degrees from Lifts

A continuous map of the circle can wind around the circle several times, and it can wind in either direction. The degree records this signed winding. To calculate it, we use the standard covering map from the real line: a lift converts a circle map into a real-valued function whose change over one turn determines the degree.

Throughout, \(S^1=\{z\in\mathbb{C}:|z|=1\}\) has the subspace topology inherited from \(\mathbb{C}\), and \(\mathbb{R}\) has its standard topology. Write \(p:\mathbb{R}\to S^1\) for \(p(t)=e^{2\pi i t}\). Thus \(p(t+1)=p(t)\), and \(p(s)=p(t)\) exactly when \(s-t\in\mathbb{Z}\).

The lift exists by the standard lifting theorem; the degree formula and its independence from the choice of lift follow directly from the definitions.

Proof. Because \(p(t+1)=p(t)\), we have \(p(F(t+1))=f(p(t+1))=f(p(t))=p(F(t))\). Therefore \(F(t+1)-F(t)\) is an integer for every \(t\). This difference is a continuous function of \(t\). A continuous integer-valued function on \(\mathbb{R}\) must be constant: if it took two different integer values, the intermediate value theorem would force it to take a noninteger value between them. Call its constant value \(d\). This proves existence, and uniqueness follows because the difference itself determines \(d\).

Now let \(F\) and \(G\) be two lifts of \(f\). For every \(t\), \(p(F(t))=p(G(t))\), so \(F(t)-G(t)\in\mathbb{Z}\). The same continuity argument shows \(F-G\) is a constant integer. Hence \(F(t+1)-F(t)=G(t+1)-G(t)\) for every \(t\), so both lifts give the same \(d\). \(\square\)

Power Maps and Reflection

For each integer \(n\), define the power map \(P_n:S^1\to S^1\) by \(P_n(z)=z^n\). For \(n<0\), this means \(z^n=1/z^{-n}\), which is defined because every point of \(S^1\) is nonzero; for \(n=0\), it is the constant map with value \(1\). These maps are continuous. The map \(R:S^1\to S^1\) given by \(R(z)=\overline{z}\) is also continuous and is called complex conjugation, or reflection.

Proof. For \(P_n\), define \(F_n(t)=nt\). Then \(p(F_n(t))=e^{2\pi i nt}=(e^{2\pi i t})^n=P_n(p(t))\), so \(F_n\) is a lift. Also \(F_n(t+1)=n(t+1)=F_n(t)+n\), giving \(\deg(P_n)=n\). This computation includes \(n=0\), when the lift is constant and the degree is zero.

For reflection, define \(H(t)=-t\). Then \(p(H(t))=e^{-2\pi i t}=\overline{e^{2\pi i t}}=R(p(t))\), and \(H(t+1)=H(t)-1\). Thus \(\deg(R)=-1\). \(\square\)

Degrees of Compositions

Proof. Let \(F\) be a lift of \(f\) and \(G\) a lift of \(g\), and write \(a=\deg(f)\) and \(b=\deg(g)\). The relation \(G(t+1)=G(t)+b\) and the lift equation for \(F\) give \(p(F(G(t)))=f(p(G(t)))=f(g(p(t)))\). Thus \(F\circ G\) is a lift of \(f\circ g\).

We also need the translation rule for integer shifts. From \(F(s+1)=F(s)+a\), repeated application gives \(F(s+k)=F(s)+ka\) for each positive integer \(k\). Rearranging the same identity as \(F(s-1)=F(s)-a\) proves the formula for negative integers as well; for \(k=0\) it is immediate. Consequently, for every integer \(k\), \(F(s+k)=F(s)+ka\). Now use \(G(t+1)=G(t)+b\):
\(F(G(t+1))=F(G(t)+b)=F(G(t))+ab\).
The lift \(F\circ G\) therefore translates by \(ab\) in one unit, so its degree is \(ab\). \(\square\)

In particular, the degree of a composition is not generally the sum of the degrees. The product rule comes from the fact that one lift changes by \(b\) when its input advances by one, while the other lift changes by \(a\) for each integer unit of input.

A Useful Rule and a Common Pitfall

Combining the power-map formula with the composition theorem gives, for all integers \(m,n\), \(\deg(P_m\circ P_n)=mn\). Indeed, \(P_m\circ P_n=P_{mn}\), since \((z^n)^m=z^{mn}\) for \(z\in S^1\), and the degree is both \(mn\) by the power-map calculation and \(m\cdot n\) by the composition rule. Reflection changes the sign: \(\deg(R\circ P_n)=-n\) and \(\deg(P_n\circ R)=-n\). For the second equality, \((\overline z)^n=\overline{z^n}\) on \(S^1\), so both compositions have the lift \(t\mapsto -nt\).

Common pitfall. It is incorrect to add degrees when composing maps. For example, treating \(R\circ P_3\) as though its degree were \(\deg(R)+\deg(P_3)=-1+3=2\) gives the wrong answer. Its lift is \(t\mapsto-3t\), whose translation is \(-3\), and the correct composition rule gives \((-1)(3)=-3\). Always use multiplication for composition; addition does not describe this operation.

The lift method is useful because it keeps track of signed turns without requiring a map to have a simple geometric formula. Periodic terms may make a lift vary irregularly within each unit interval, but if \(F(t+1)=F(t)+d\), that variation does not affect the degree. The integer \(d\) alone records the net signed change in one turn of the domain.