This exercise assumes some familiarity with analysis on manifolds.
Theorem. Let \(C\) be a simple closed curve in \(\mathbb{R}^2\) that is a smooth submanifold of \(\mathbb{R}^2\); let \(f: I \to C\) be a simple loop smoothly parameterizing \(C\). If \(\mathbf{0}\) is a point of the bounded component of \(\mathbb{R}^2 - C\), then \(n(f, \mathbf{0}) = \pm 1\).
Let \(U\) be the bounded component of \(\mathbb{R}^2 - C\). Let \(B\) be a closed \(\epsilon\)-ball centered at 0 that lies in \(U\); let \(S = \operatorname{Bd} B\). Let \(M\) equal the closure of \(U - B\).
Show \(M\) is a smooth 2-manifold with boundary \(C \cup S\).
Apply Green’s theorem to show that \(\int_C dz/z = \pm \int_S dz/z\), the sign depending on the orientations of \(S\) and \(C\). [Hint: Set \(P = -y/(x^2 + y^2)\) and \(Q = x/(x^2 + y^2)\).]
Show that the second integral equals \(\pm 2\pi i\).
Solution
Goal: Prove that the winding number of any smooth simple closed curve in \(\mathbb{R}^2\) around an interior point \(\mathbf{0}\) is \(\pm 1\) using differential forms and Green’s Theorem.
1. Part (a): Smooth manifold structure of \(M = \overline{U \setminus B}\).Proof:
1.1. The bounded component \(U\) is an open domain in \(\mathbb{R}^2\) with smooth 1-dimensional boundary \(\partial U = C\).
1.2. The ball \(B = \overline{B}(\mathbf{0}, \varepsilon) \subset U\) is a closed disk with smooth boundary \(\partial B = S\).
1.3. The difference \(M = \overline{U \setminus B} = U \setminus \operatorname{Int}(B)\) is a compact subset of \(\mathbb{R}^2\).
1.4. Near any interior point \(p \in \operatorname{Int}(M) = U \setminus B\), \(M\) is locally diffeomorphic to \(\mathbb{R}^2\).
1.5. Near any boundary point \(p \in C\), \(M\) inherits the smooth boundary chart from \(U\); near \(p \in S\), \(M\) inherits the smooth boundary chart from the exterior of \(B\).
1.6. Thus \(M\) is a smooth compact 2-manifold with boundary \(\partial M = C \sqcup S\).
2. Part (b): Application of Green’s / Stokes’ Theorem.Proof:
2.1. Consider the differential 1-form on \(\mathbb{R}^2 \setminus \{\mathbf{0}\}\):\[\omega = \frac{dz}{z} = \frac{x \, dx + y \, dy}{x^2 + y^2} + i \left(\frac{-y \, dx + x \, dy}{x^2 + y^2}\right) = \alpha + i \beta.\]
2.2. Setting \(P = \frac{-y}{x^2 + y^2}\) and \(Q = \frac{x}{x^2 + y^2}\), we compute the exterior derivative:\[\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{(x^2+y^2) - 2x^2}{(x^2+y^2)^2} - \frac{-(x^2+y^2) - (-y)(2y)}{(x^2+y^2)^2} = \frac{y^2 - x^2 + (x^2 - y^2)}{(x^2+y^2)^2} = 0.\]
2.3. Similarly, \(d\alpha = 0\), so \(d\omega = 0\) on all of \(M \subset \mathbb{R}^2 \setminus \{\mathbf{0}\}\).
2.4. By Stokes’ Theorem, \(\int_{\partial M} \omega = \iint_M d\omega = 0\).
2.5. With standard induced boundary orientations where \(C\) is traversed with \(M\) on its left and \(S\) is traversed with \(M\) on its left (clockwise on \(S\)), \(\partial M = C - S\), which gives:\[\int_C \frac{dz}{z} = \pm \int_S \frac{dz}{z}\]depending on whether \(C\) and \(S\) are parameterized with counterclockwise orientation.
3. Part (c): Integration along \(S\) and winding number conclusion.Proof:
3.1. Parameterize the circle \(S = \partial B(\mathbf{0}, \varepsilon)\) counterclockwise by \(z(t) = \varepsilon e^{it}\) for \(t \in [0, 2\pi]\).
3.2. Then \(dz = i \varepsilon e^{it} \, dt\), so:\[\int_S \frac{dz}{z} = \int_0^{2\pi} \frac{i \varepsilon e^{it}}{\varepsilon e^{it}} \, dt = \int_0^{2\pi} i \, dt = 2\pi i.\]
3.3. By 2, \(\int_C \frac{dz}{z} = \pm 2\pi i\).
3.4. By definition, the winding number is:\[n(f, \mathbf{0}) = \frac{1}{2\pi i} \int_C \frac{dz}{z} = \frac{\pm 2\pi i}{2\pi i} = \pm 1.\]Q.E.D.