Let \({ f: S^n \rightarrow S^n }\) be a map of degree zero. Show that there exists 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 \({ R^n }\) such that \({ F(x) \neq 0 }\) for all x, then there exits a point in \({ \partial D^n }\) where F points radially outward and another point in \({\partial D^n }\) where F points radially inward.

Solution:

Suppose g is the antipodal map.

If \({ f(x) \neq x }\) for all \({x \in S^n }\) then \({ f \sim g }\) . This implies \({deg (f) = deg(g) = (-1)^{n+1} \neq 0 }\) hence giving us a contradiction.

Similarly if \({ f(y) \neq – y \forall y \in S^n }\) then \({ g \circ f (y) \neq y }\) for all \({ y \in S^n }\) . This implies \({ g \circ f }\) is homotopic to the antipodal map. But then \({ deg (g \circ f) = (-1)^{n+1} }\) . This is impossible as \({ deg (g \circ f ) = deg (g) \cdot deg (f) = 0 }\) . Hence we again have a contradiction.

Now define \({ \bar{F} (x) = \frac{F(x)}{||F(x)||} , x\in S^{n-1} }\) . It is clearly a continuous vector field from \({ S^{n-1} \rightarrow S^{n-1} }\) . (It is well defined as F(x) never vanishes. Also restriction of F to \({S^{n-1} }\) is continuous as F is continuous. Finally as F is continuous, composing it with absolute value function is continuous. \({||F|| }\) is continuous and F is continuous implies their ratio is continuous).

Note that \({\bar{F} = S^{n-1} \hookrightarrow D^n \rightarrow S^{n-1} }\) where the first map is inclusion \({i }\) and the second map is \({ \frac{F}{||F||}}\) . Since \({ H_n (D^n ) = 0 }\) (as n-disk is contractible), therefore \({ \left ( i \circ \frac{F}{||F||} \right)_* = 0}\) implying \({ \bar{F}_* }\) is of zero degree.

As \({ \bar{F} }\) is of degree zero therefore there is a point \({ x \in S^{n-1} }\) such that \({ \bar{F}(x) = x }\). Hence \({ \frac{F(x)}{||F(x)||} = x \Rightarrow F(x) = c \cdot x }\). This F points radially outward at this point (c is positive constant).

Similarly there is a point \({ y \in S^{n-1} }\) such that \({ \bar{F}(y) = -y }\). This is the point where F points radially inward.