Track the Lift, Not Just the Position
A loop in the circle can travel around it several times and still finish where it began. Its endpoint therefore does not record how many turns it made. To count the turns, we follow a continuous angle as the loop moves. This tutorial uses angles measured in turns: an increase of \(1\) means one counterclockwise revolution, and a decrease of \(1\) means one clockwise revolution.
We take \(S^1=\{z\in\mathbb{C}:|z|=1\}\) with the subspace topology inherited from \(\mathbb{C}\), and \(\mathbb{R}\) with its standard topology. Write \(E:\mathbb{R}\to S^1\) for the exponential covering map \(E(t)=\cos(2\pi t)+i\sin(2\pi t)\). Thus \(E(0)=1\), \(E(1/4)=i\), and \(E(-1/4)=-i\). A lift of a path \(\gamma:I\to S^1\) is a continuous path \(\theta:I\to\mathbb{R}\) such that \(E(\theta(t))=\gamma(t)\) for every \(t\in I=[0,1]\).
The existence and uniqueness of a lift with prescribed starting point are standard results for the exponential covering map, established in the earlier material on covering spaces. The endpoint is an integer because \(E(\theta(1))=\gamma(1)=1\), and \(E(s)=1\) exactly when \(s\in\mathbb{Z}\). The lift remembers the signed motion: counterclockwise motion increases its value, while clockwise motion decreases it.
Adding Signed Changes Along Arcs
Proof. The given \(\theta\) is a lift of \(\gamma\), and it starts at \(0\). By uniqueness of the lift starting at \(0\), it is the lift used to define the winding number. Hence that winding number is \(\theta(1)\). For a partition \(0=t_0<t_1<\cdots<t_k=1\), the change on the \(j\)-th interval is \(\theta(t_j)-\theta(t_{j-1})\). Adding these changes cancels every intermediate value: \(\sum_{j=1}^k(\theta(t_j)-\theta(t_{j-1}))=\theta(1)-\theta(0)=\theta(1)\). This calculation keeps clockwise changes negative and counterclockwise changes positive. \(\square\)
Worked Example: Four Arcs with Opposite Directions
Consider a loop made of four equal-duration arcs, described by the angle \(\theta\) at the successive time points \(0,1/4,1/2,3/4,1\). Let the angle values there be \(0,1/4,3/4,1/2,1\), respectively, and let \(\theta\) vary linearly on each interval. Set \(\gamma(t)=E(\theta(t))\).
The successive signed angle changes are \(1/4-0=1/4\), \(3/4-1/4=1/2\), \(1/2-3/4=-1/4\), and \(1-1/2=1/2\). Thus the first, second and fourth arcs move counterclockwise; the third moves clockwise. Their total change is \(1/4+1/2-1/4+1/2=1\).
The arc endpoints on the circle are \(E(0)=1\), \(E(1/4)=i\), \(E(3/4)=-i\), \(E(1/2)=-1\), and \(E(1)=1\). Consecutive arcs join at the stated endpoints, and the path begins and ends at \(1\). Its angle is already a continuous lift starting at \(0\), so its winding number is \(1\).
Check: the total lift ends at \(\theta(1)=1\), and \(E(1)=\cos(2\pi)+i\sin(2\pi)=1\), as required for a loop based at \(1\). The negative change on the third arc is included rather than replaced by its absolute value.
Reading the Winding Number from a Piecewise Formula
A path may be specified by a formula for its angle instead of a verbal list of arcs. The important checks are that the angle is continuous where the formulas meet, begins at \(0\), and ends at an integer when the resulting path is a loop based at \(1\). The angle need not increase throughout: its final value, not its largest value or total distance traveled, determines the winding number.
Worked Example: A Piecewise Angle That Reverses Direction
Define \(\theta:I\to\mathbb{R}\) by linear interpolation through the values \(\theta(0)=0\), \(\theta(1/4)=1/2\), \(\theta(1/2)=-1/4\), and \(\theta(1)=2\). More explicitly, on each interval use the unique linear function with the specified endpoint values, and set \(\gamma(t)=E(\theta(t))\).
On the first interval the angle changes by \(1/2-0=1/2\). On the second it changes by \(-1/4-1/2=-3/4\), so this portion travels clockwise. On the last interval it changes by \(2-(-1/4)=9/4\). The total change is \(1/2-3/4+9/4=2\).
Because the linear pieces agree at their shared endpoints, \(\theta\) is continuous. Therefore \(\gamma=E\circ\theta\) is a continuous path. At its initial and final times, \(\gamma(0)=E(0)=1\) and \(\gamma(1)=E(2)=\cos(4\pi)+i\sin(4\pi)=1\). The path is a loop based at \(1\), and its lift starting at \(0\) is \(\theta\). Its winding number is \(2\).
Check: the middle angle value is negative, but this causes no problem: \(E(-1/4)=-i\). The endpoint calculation uses the lift value \(2\), not a choice of angle restricted to a single revolution.
Proof. Let \(\theta\) and \(\phi\) be the lifts of \(\alpha\) and \(\beta\) starting at \(0\), and write \(n=\theta(1)\), \(m=\phi(1)\). Define the concatenation by \((\alpha*\beta)(t)=\alpha(2t)\) for \(0\le t\le1/2\) and \((\alpha*\beta)(t)=\beta(2t-1)\) for \(1/2\le t\le1\). Define \(\psi(t)=\theta(2t)\) on the first interval and \(\psi(t)=n+\phi(2t-1)\) on the second. At \(t=1/2\), the first formula gives \(n\), and the second gives \(n+\phi(0)=n\), so \(\psi\) is continuous. On the first interval \(E(\psi(t))=\alpha(2t)\). On the second, \(E(\psi(t))=E(n)E(\phi(2t-1))=\beta(2t-1)\), since \(n\) is an integer and \(E(n)=1\). Thus \(\psi\) is the lift of \(\alpha*\beta\) starting at \(0\). Its endpoint is \(n+m\), proving the formula. \(\square\)
In particular, traversing the same based loop \(r\) times, for a positive integer \(r\), multiplies its winding number by \(r\). This follows by applying the concatenation rule repeatedly: after one traversal the number is \(n\); each additional traversal adds \(n\). Reversing a loop changes the sign of its winding number: if \(\theta\) is its lift ending at \(n\), the reversed loop has lift \(t\mapsto\theta(1-t)-n\), which starts at \(0\) and ends at \(-n\).
Worked Example: Repeated Counterclockwise and Clockwise Traversals
Let \(\rho(t)=E(2t)\), which makes two counterclockwise turns, and let \(\sigma(t)=E(-t)\), which makes one clockwise turn. Consider the loop that traverses \(\rho\) three times and then \(\sigma\) twice. Each component is based at \(1\), since \(E(0)=E(2)=E(-1)=1\).
The three traversals of \(\rho\) contribute \(2+2+2=6\). The two traversals of \(\sigma\) contribute \(-1+(-1)=-2\). The concatenation rule gives the total winding number \(6-2=4\).
To check directly, divide the time interval into five equal parts. A continuous lift starts at \(0\), rises by \(2\) on each of the first three parts, and falls by \(1\) on each of the last two parts. Its successive endpoint values are \(0,2,4,6,5,4\). Each is an integer, so the corresponding circle point at each join is \(E(k)=1\) for that integer \(k\); the lift ends at \(4\).
Check: \(E(4)=\cos(8\pi)+i\sin(8\pi)=1\), so the computed lift endpoint agrees with the loop condition. Counting the clockwise traversals as positive would instead give \(6+2=8\), which disagrees with the explicit endpoint value \(4\).
A Common Pitfall: Replacing the Lift by Principal Angles
A principal angle records a point on the circle using a chosen representative, often one in a fixed interval of length one turn. It does not in general vary continuously along a path: when the path crosses the endpoint of that interval, the representative jumps. Summing differences of principal angles can therefore discard a whole revolution.
For a concrete check, take a loop with a continuous angle that moves linearly through the successive values \(0,3/4,5/4,1\). Its three signed changes are \(3/4\), \(5/4-3/4=1/2\), and \(1-5/4=-1/4\), with total \(1\). But principal representatives for the circle points at those times may be written as \(0,-1/4,1/4,0\). Subtracting these representatives gives \(-1/4,1/2,-1/4\), with total \(0\). That is the wrong winding number: the representatives changed at a choice-of-angle boundary, while the continuous lift did not. The correction is to track the continuous angle along the path, preserving its accumulated turns.
The calculation methods here concern specified paths and their lifts. They do not require a claim about what happens when a loop is continuously deformed; that is a separate question about homotopy invariance. For computations, the safe procedure is to establish a continuous lift from the given arc data or angle formula, then read its endpoint.
Check Your Understanding
For each problem, keep track of signed changes in a continuous lift; do not reduce intermediate angles to principal representatives.
- A loop has successive linear angle changes \(1/3,-1/6,5/6\). Find its winding number and check that its endpoint lies over \(1\).
- A continuous piecewise-linear angle starts at \(0\) and ends at \(-3\). What is the winding number, and how many net clockwise turns does this represent?
- If a based loop has winding number \(-2\), what is the winding number after traversing it four times?
- Use the concatenation rule to find the winding number of loops with winding numbers \(3\), \(-5\), and \(1\), traversed in that order.
- Explain why the principal-angle calculation in the common pitfall loses one turn.