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# 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.

## By Ashani Dasgupta

Pursuing Ph.D. in Geometric Group Theory at University of Wisconsin, Milwaukee