Let \(f: S^n \to S^n\) be a map of degree zero. Show that there exist points \(x, y \in S^n\) with \(f(x) = x\) and \(f(y) = -y\). Use this to show that if \(F\) is a continuous vector field defined on the unit ball \(D^n\) in \(\mathbb{R}^n\) such that \(F(x) \neq 0\) for all \(x\), then there exists a point on \(\partial D^n\) where \(F\) points radially outward and another point on \(\partial D^n\) where \(F\) points radially inward.
Solution
1. Existence of a fixed point (\(f(x) = x\)):
1.1. Suppose for contradiction that \(f(x) \neq x\) for all \(x \in S^n\).Then \(f(x)\) is never antipodal to \(-x\), so \((1 - t)f(x) - tx \neq 0\) for all \(t \in [0, 1]\) and \(x \in S^n\).
if \((1-t)f(x) - tx = 0\), taking norms gives \(1-t = t \implies t = 1/2\), so \(f(x) = x\), contradiction.
1.2. Define the straight-line homotopy \(H: S^n \times [0, 1] \to S^n\) by:\[ H(x, t) = \frac{(1 - t)f(x) - tx}{\|(1 - t)f(x) - tx\|}. \]\(H\) is a homotopy from \(f\) to the antipodal map \(a(x) = -x\).
\(H(x, 0) = f(x)\) and \(H(x, 1) = -x\).
1.3. Because homotopic maps have the same degree:\[ \deg(f) = \deg(a) = (-1)^{n+1} \neq 0. \]This contradicts \(\deg(f) = 0\).Thus there exists \(x \in S^n\) such that \(f(x) = x\).
degree of antipodal map on \(S^n\).
2. Existence of an antipodal point (\(f(y) = -y\)):
2.1. Suppose for contradiction that \(f(y) \neq -y\) for all \(y \in S^n\).Then \((1 - t)f(y) + ty \neq 0\) for all \(t \in [0, 1]\) and \(y \in S^n\).
identical norm argument.
2.2. Define the straight-line homotopy \(G: S^n \times [0, 1] \to S^n\) by:\[ G(y, t) = \frac{(1 - t)f(y) + ty}{\|(1 - t)f(y) + ty\|}. \]\(G\) is a homotopy from \(f\) to the identity map \(\operatorname{id}_{S^n}\).
\(G(y, 0) = f(y)\) and \(G(y, 1) = y\).
2.3. Thus \(\deg(f) = \deg(\operatorname{id}_{S^n}) = 1\), contradicting \(\deg(f) = 0\).Therefore there exists \(y \in S^n\) such that \(f(y) = -y\).
degree of identity map on \(S^n\).
3. Application to continuous non-vanishing vector fields on \(D^n\):
3.1. Let \(F: D^n \to \mathbb{R}^n \setminus \{0\}\) be continuous.Define the normalized map \(\widetilde{F}: D^n \to S^{n-1}\) by \(\widetilde{F}(z) = \frac{F(z)}{\|F(z)\|}\), and let \(g = \widetilde{F}|_{\partial D^n}: S^{n-1} \to S^{n-1}\).
\(F(z) \neq 0\) ensures \(\widetilde{F}\) is well-defined and continuous.
3.2. Since \(g\) extends to the contractible disk \(D^n\) via \(\widetilde{F}\), \(g\) is null-homotopic, so \(\deg(g) = 0\).
maps extending to a contractible space are null-homotopic.
3.3. Applying 1 and 2 to \(g: S^{n-1} \to S^{n-1}\):
There exists \(x \in S^{n-1} = \partial D^n\) with \(g(x) = x \implies F(x) = \|F(x)\| x\), so \(F\) points radially outward at \(x\).
There exists \(y \in S^{n-1} = \partial D^n\) with \(g(y) = -y \implies F(y) = -\|F(y)\| y\), so \(F\) points radially inward at \(y\).
1 and 2 applied to \(S^{n-1}\).
4. Conclusion: Degree zero maps have fixed and antipodal points, which implies every non-vanishing vector field on \(D^n\) has points on \(\partial D^n\) pointing radially outward and inward.
1 through 3.
Q.E.D.