Trigonometric Circle Maps: A Degree-Calculation Workshop

Degree from One Full Turn

A circle map can make several turns around the circle even when its formula contains oscillating trigonometric terms. The reliable quantity is not the pointwise size of the angle, but the net change in a continuous choice of angle as the input travels once around the circle. This viewpoint turns many degree calculations into endpoint evaluations.

Write \(S^1=\{z\in\mathbb{C}:|z|=1\}\), with its subspace topology from \(\mathbb{C}\), and use the parametrization \(p(t)=e^{it}\). A continuous real-valued function \(\theta:\mathbb{R}\to\mathbb{R}\) is a lift of a map \(f:S^1\to S^1\) when \(f(e^{it})=e^{i\theta(t)}\) for every \(t\). The degree of a circle map can be computed from any lift: if \(\theta(t+2\pi)=\theta(t)+2\pi d\), then \(\deg(f)=d\). This is the standard degree-from-a-lift characterization established in the earlier material on degree of a circle map from its lift.

Proof. Set \(\theta(t)=kt+h(t)\). Since \(k\) is an integer and \(h(t+2\pi)=h(t)\), we have \(\theta(t+2\pi)=k(t+2\pi)+h(t+2\pi)=\theta(t)+2\pi k\). Consequently \(e^{i\theta(t+2\pi)}=e^{i\theta(t)}\), so the formula gives the same value for every parameter representing the same point of \(S^1\). The function \(t\mapsto e^{i\theta(t)}\) is continuous and \(2\pi\)-periodic, so it induces a continuous map on the circle. The lift changes by \(2\pi k\) in one turn; the degree-from-a-lift formula gives \(\deg(f)=k\). \(\square\)

The periodic term can change the route taken by the angle, but it contributes no net change over a full turn. In particular, \(k\) may be positive, negative, or zero. The condition \(k\in\mathbb{Z}\) ensures the formula is well-defined on \(S^1\); the periodicity and continuity of \(h\) ensure the stated lift works.

Piecewise Formulas and Continuous Lifts

A piecewise formula may display different angle branches on different intervals. Since \(e^{i\alpha}=e^{i\beta}\) whenever \(\alpha-\beta\) is an integer multiple of \(2\pi\), the displayed angles can jump by \(2\pi\) even when the circle-valued map is continuous. To calculate degree, first adjust the branches so that the angles join continuously; then compare the lift at the endpoints of one full turn.

Proof. Because the endpoint exponentials agree, the formula assigns the same value to the two parameters \(0\) and \(2\pi\), which represent the same point of \(S^1\). Thus the continuous function \(t\mapsto e^{i\theta(t)}\) on \([0,2\pi]\) defines a continuous map on the circle. Equality of the endpoint exponentials means \(e^{i(\theta(2\pi)-\theta(0))}=1\); for real angles this holds exactly when \(\theta(2\pi)-\theta(0)=2\pi d\) for some \(d\in\mathbb{Z}\). Extending the lift by \(\theta(t+2\pi)=\theta(t)+2\pi d\) gives a lift on \(\mathbb{R}\). Its change in one full turn is \(2\pi d\), so the degree-from-a-lift formula gives \(\deg(f)=d\), as claimed. \(\square\)

More generally, if two displayed angle formulas meet at a parameter \(a\) and their values differ by \(2\pi j\) for some \(j\in\mathbb{Z}\), subtract \(2\pi j\) from one branch when needed to make the real angles agree at \(a\). This produces a continuous lift at that join. It is the lift’s total endpoint change—not the number of pieces, the sign of a local slope, or the chosen initial angle—that determines the degree.

Degrees of Compositions

Proof. Choose lifts \(\theta,\gamma:\mathbb{R}\to\mathbb{R}\) for \(f,g\), respectively. Write \(m=\deg(f)\) and \(n=\deg(g)\), so \(\theta(t+2\pi)=\theta(t)+2\pi m\) and \(\gamma(u+2\pi)=\gamma(u)+2\pi n\). The function \(t\mapsto\gamma(\theta(t))\) is a lift of \(g\circ f\), because exponentiating it applies first \(f\) and then \(g\). The relation for \(\gamma\) implies \(\gamma(u+2\pi m)=\gamma(u)+2\pi mn\) for every integer \(m\): for positive \(m\), apply the relation \(m\) times; for \(m=0\) it is unchanged; and for negative \(m\), use \(\gamma(u-2\pi)=\gamma(u)-2\pi n\) repeatedly. Therefore \(\gamma(\theta(t+2\pi))=\gamma(\theta(t)+2\pi m)=\gamma(\theta(t))+2\pi mn\). The lift of the composition changes by \(2\pi mn\) in one turn, so its degree is \(mn\). \(\square\)

A Common Pitfall: Counting Oscillations as Turns

A trigonometric term may rise and fall several times during one traversal of the domain. Those local changes do not count as extra full turns unless they produce additional net change in the lift over the full interval.

Common mistake. For \(F(e^{it})=e^{i(3t+\sin t)}\), one might count the oscillation of \(\sin t\) as an extra turn and claim degree \(4\). That calculation treats a change in the angle partway around the domain as if it were its net change after one full turn. The lift values are instead \(\Theta(0)=3\cdot0+\sin0=0\) and \(\Theta(2\pi)=3(2\pi)+\sin(2\pi)=6\pi\). Their difference is \(6\pi\), so \(\deg(F)=6\pi/(2\pi)=3\). Equivalently, \(\sin(t+2\pi)-\sin t=0\), so the oscillating term contributes zero to the full-turn change.

For a piecewise map, the corresponding pitfall is to subtract the endpoints of a discontinuous choice of angle. First make the branch choices agree at each join, changing an angle by multiples of \(2\pi\) without changing its exponential. Then use the resulting continuous lift. For compositions, a useful check is to calculate both the lift’s net change and the product of the individual degrees.