Direction of a vector field

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.

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