Math4202 Topology II (Lecture 21)

Algebraic Topology

Application of fundamental groups

Recall from last Friday, j:S^1\to \mathbb{R}^2-\{0\} is not null homotopic

Hairy ball theorem

Given a non-vanishing vector field on B^2=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq 1\}, (v:B^2\to \mathbb{R}^2 continuous and v(x,y)\neq 0 for all (x,y)\in B^2) there exists a point of S^1 where the vector field points directly outward, and a point of S^1 where the vector field points directly inward.

Proof

By our assumption, then v:B^2\to \mathbb{R}^2-\{0\} is a continuous vector field on B^2.

v|_{S^1}:S^1\to \mathbb{R}^2-\{0\} is null homotopic.

We prove by contradiction.

Suppose v:B^2\to \mathbb{R}^2-\{0\} and v|_{S^1}:S^1\to \mathbb{R}^2-\{0\} is everywhere outward. (for everywhere inward, consider -v must be everywhere outward)

Because v|_{S^1} extends continuously to B^2, then v|_{S^1}:B^2\to \mathbb{R}^2-\{0\} is null homotopic.

We construct a homotopy for functions between v|_{S^1} and j. (Recall j:S^1\to \mathbb{R}^2-\{0\} is not null homotopic)

Define H:S^1\times I\to \mathbb{R}^2-\{0\} by affine combination


H((x,y),t)=(1-t)v(x,y)+tj(x,y)

we also need to show that H is non zero.

Since v is everywhere outward, v(x,y)\cdot j(x,y) is positive for all (x,y)\in S^1.

H((x,y),t)\cdot j(x,y)=(1-t)v(x,y)\cdot j(x,y)+tj(x,y)\cdot j(x,y)=(1-t)(v(x,y)\cdot j(x,y))+t

which is positive for all t\in I, therefore H is non zero.

So H is a homotopy between v|_{S^1} and j.

Corollary of the hairy ball theorem

\forall v:B^2\to \mathbb{R}^2, if on S^1, v is everywhere outward/inward, there is (x,y)\in B^2 such that v(x,y)=0.

Brouwer's fixed point theorem

If f:B^2\to B^2 is continuous, then there exists a point x\in B^2 such that f(x)=x.

Proof

We proceed by contradiction again.

Suppose f has no fixed point, f(x)-x\neq 0 for all x\in B^2.

Now we consider the map v:B^2\to \mathbb{R}^2 defined by v(x,y)=f(x)-x, this function is continuous since f is continuous.

forall x\in S^1, v(x)\cdot x=f(x)\cdot x-x\cdot x=f(x)\cdot x-1.

Recall the cauchy schwartz theorem, |f(x)\cdot x|\leq \|f(x)\|\cdot\|x\|\leq 1, note that f(x)\neq 0 for all x\in B^2, v(x)\cdot x<0. This means that all v(x) points inward.

This is a contradiction to the hairy ball theorem, so f has a fixed point.

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