From a Loop in the Circle to an Integer
A loop in the circle starts and ends at the same point, but its parametrization may travel around the circle several times, in either direction. The circle itself records where the loop is at each time; a lift to the real line records how far the motion has progressed. The difference between the lift’s endpoint and its starting point will be an integer. That integer is the winding number.
We use \(\mathbb{R}\) with its standard topology and \(S^1=\{z\in\mathbb{C}:|z|=1\}\) with the subspace topology inherited from \(\mathbb{C}\). Consider the map \(p:\mathbb{R}\to S^1\) given by \(p(s)=e^{2\pi i s}=\cos(2\pi s)+i\sin(2\pi s)\). The exponential covering map has already been established as a covering map. We use the path-lifting theorem for covering maps: if \(\gamma:I\to S^1\) is a path and \(s_0\in\mathbb{R}\) satisfies \(p(s_0)=\gamma(0)\), then there is a unique continuous path \(\widetilde{\gamma}:I\to\mathbb{R}\) such that \(p\circ\widetilde{\gamma}=\gamma\) and \(\widetilde{\gamma}(0)=s_0\). Here \(I=[0,1]\).
Why the Displacement Is an Integer
The crucial fact about \(p\) is that two real numbers have the same image exactly when their difference is an integer. Indeed, \(e^{2\pi i s}=e^{2\pi i t}\) if and only if \(2\pi(s-t)\) is an integer multiple of \(2\pi\), which is equivalent to \(s-t\in\mathbb{Z}\). Thus each fiber of \(p\) has the form \(p^{-1}(z)=s+\mathbb{Z}\) for any \(s\) with \(p(s)=z\).
Proof. Since \(\widetilde{\gamma}\) is a lift, \(p(\widetilde{\gamma}(t))=\gamma(t)\) for every \(t\in I\). The loop condition gives \(\gamma(1)=\gamma(0)\), so \(p(\widetilde{\gamma}(1))=p(\widetilde{\gamma}(0))\). By the fiber description above, two real numbers with the same image under \(p\) differ by an integer. Therefore \(\widetilde{\gamma}(1)-\widetilde{\gamma}(0)\in\mathbb{Z}\). \(\square\)
There is one more choice to check. A point \(z_0\in S^1\) has infinitely many lifts: if \(p(s_0)=z_0\), then \(p(s_0+k)=z_0\) for every \(k\in\mathbb{Z}\). Starting a lift at a different point in this fiber must not change the winding number.
Proof. Let \(\widetilde{\gamma}\) be the lift starting at \(s_0\), and let \(s_1\) be another lift of the same basepoint. The fiber description gives \(s_1=s_0+k\) for some \(k\in\mathbb{Z}\). Define \(\eta(t)=\widetilde{\gamma}(t)+k\). Since \(p(u+k)=p(u)\) for all \(u\in\mathbb{R}\), we have \(p(\eta(t))=\gamma(t)\), and \(\eta(0)=s_0+k=s_1\). By uniqueness of the lift starting at \(s_1\), \(\eta\) is that lift. Its endpoint displacement is \(\eta(1)-\eta(0)=(\widetilde{\gamma}(1)+k)-(\widetilde{\gamma}(0)+k)=\widetilde{\gamma}(1)-\widetilde{\gamma}(0)\). So both choices give the same integer. \(\square\)
Worked Example: A Constant Loop
Let \(\gamma(t)=i\) for all \(t\in I\). Since \(p(1/4)=e^{2\pi i/4}=i\), choose the lift \(\widetilde{\gamma}(t)=1/4\) for every \(t\). It is a lift because \(p(\widetilde{\gamma}(t))=i=\gamma(t)\) at every time, and it starts at \(1/4\) as required. Its endpoint displacement is \(1/4-1/4=0\), so \(w(\gamma)=0\).
Check: The lift ends where it starts, at \(1/4\); applying \(p\) to both endpoints gives \(i\) in each case. The zero displacement agrees with the fact that this parametrization does not travel around the circle.
Worked Example: Two Positive Turns
Let \(\gamma(t)=\cos(4\pi t)+i\sin(4\pi t)\). Its initial and terminal values are both \(1\), since at \(t=0\) the angle is \(0\), and at \(t=1\) the angle is \(4\pi\). Thus it is a loop based at \(1\). The path \(\widetilde{\gamma}(t)=2t\) starts at \(0\), and \(p(2t)=e^{2\pi i(2t)}=\cos(4\pi t)+i\sin(4\pi t)=\gamma(t)\). Its endpoint displacement is \(2-0=2\), so \(w(\gamma)=2\).
Check: The lifted endpoints project to \(p(0)=1\) and \(p(2)=e^{4\pi i}=1\), as the loop condition requires. The displacement \(2\) is an integer.
Worked Example: One Negative Turn from a Different Basepoint
Set \(\gamma(t)=e^{2\pi i(1/2-t)}\). At \(t=0\), this is \(e^{\pi i}=-1\); at \(t=1\), it is \(e^{-\pi i}=-1\), so \(\gamma\) is based at \(-1\). The path \(\widetilde{\gamma}(t)=1/2-t\) is a lift because applying \(p\) gives \(e^{2\pi i(1/2-t)}=\gamma(t)\). It starts at \(1/2\) and ends at \(1/2-1=-1/2\). Hence \(w(\gamma)=(-1/2)-(1/2)=-1\).
Check: The starting and ending lift points differ by \(-1\), and \(p(-1/2)=e^{-\pi i}=-1=p(1/2)\). The endpoints therefore project to the same basepoint, while the negative displacement records the direction of travel.
The Lift Can Move Unevenly
The winding number does not depend on how quickly the loop moves at different times. What matters in this definition is the total endpoint displacement of the lift, not whether that lift is a straight line. A loop may slow down, speed up, or change direction during its parametrization; one still computes its winding number by lifting the entire path and subtracting the starting value from the ending value.
Worked Example: A Nonuniformly Parametrized Loop
For \(t\in I\), define \(f(t)=1/6+2t+\sin(2\pi t)/(4\pi)\), and let \(\gamma(t)=e^{2\pi i f(t)}\). Since \(f(0)=1/6\) and \(f(1)=1/6+2+0=13/6\), the terminal parameter differs from the initial parameter by \(2\). Consequently \(\gamma(1)=e^{2\pi i(13/6)}=e^{2\pi i(1/6)}=\gamma(0)\), so this is a loop. The path \(\widetilde{\gamma}(t)=f(t)\) is a lift by its definition, and starts at \(1/6\). Its endpoint displacement is \(13/6-1/6=12/6=2\), so \(w(\gamma)=2\).
Check: At both endpoints the circle value is \(\cos(\pi/3)+i\sin(\pi/3)=1/2+i\sqrt{3}/2\); the endpoint angle \(13\pi/3\) differs from \(\pi/3\) by \(4\pi\). Also \(p(13/6)=e^{2\pi i(13/6)}=e^{2\pi i(1/6)}=p(1/6)\), confirming that the lifted endpoints project to the same point.
Proof. By definition, \(w(\gamma)=\widetilde{\gamma}(1)-\widetilde{\gamma}(0)\). If the lift has equal endpoints, this difference is \(0\), so \(w(\gamma)=0\). Conversely, if \(w(\gamma)=0\), then \(\widetilde{\gamma}(1)-\widetilde{\gamma}(0)=0\), which implies \(\widetilde{\gamma}(1)=\widetilde{\gamma}(0)\). \(\square\)
A Common Endpoint Mistake
Common pitfall: The winding number is the displacement between the lifted endpoints, not the terminal coordinate by itself. For example, take \(\gamma(t)=e^{2\pi i(t+1/4)}\), a loop based at \(i\). The lift starting at \(1/4\) is \(\widetilde{\gamma}(t)=t+1/4\), so its endpoint is \(5/4\). The incorrect calculation \(w(\gamma)=5/4\) uses the endpoint alone and even produces a noninteger. The correct calculation subtracts the start: \(w(\gamma)=5/4-1/4=1\). Indeed, \(p(1/4)=e^{\pi i/2}=i\), and \(p(5/4)=e^{5\pi i/2}=i\), as required for a loop.
A related error is to start a lift at \(0\) when the loop is based at \(i\). That cannot be a lift with the prescribed starting point: \(p(0)=1\ne i\). The starting value must lie in the fiber over the loop’s basepoint. Once it does, the path-lifting theorem gives a unique lift, and the endpoint displacement gives an integer independent of which valid starting value was chosen.
The integer winding number condenses a path’s lifted endpoint information into a simple quantity. In these examples it distinguishes a constant loop from loops with positive or negative displacement, even when the basepoint is not \(1\) or the parametrization is nonuniform. This definition alone does not assert that winding number is unchanged under a homotopy of loops; that requires an additional argument.
Check Your Understanding
For each question, keep track of both the starting and ending values of the lift.
- For \(\gamma(t)=e^{2\pi i(3t)}\), identify a lift starting at \(0\) and calculate the winding number.
- For \(\gamma(t)=e^{2\pi i(1/3-2t)}\), verify that it is a loop and calculate its winding number.
- If a lift of a loop starts at \(s_0\) and ends at \(s_0+4\), what is its winding number? What does the loop condition tell you about the projections of these endpoints?
- Explain why replacing a chosen starting lift \(s_0\) by \(s_0+k\), for \(k\in\mathbb{Z}\), does not change the endpoint displacement.
- For a loop based at \(z_0\), why is a real number \(r\) with \(p(r)\ne z_0\) not a valid starting point for its lift?