Question

Several months ago I was browsing through a question posted here ("Applications of Brouwer’s fixed point theorem"); amongst the comments attached it was mentioned that one could derive both Borsuk's Antipodal Theorem as well as the Fundamental Theorem of Algebra from Brouwer's Fixed-Point Theorem. However, to date I haven't seen (correct) proofs of these assertions. Could somebody kindly indicate where such proofs may be found ?

Thanks in advance,

St.

Answer 1

As far as I'm aware, there isn't a compelling direct argument from Brouwer fixed point to imply the fundamental theorem of algebra. Such an argument isn't impossible -- I can imagine some fairly contrived proofs but I don't know of a very natural one. The references like Guillemin and Pollack don't derive FTOA from Brouwer, they derive both FTOA and Brouwer from degree/intersection theory. In particular they only use mod-2 degree theory for Brouwer but oriented degree theory for FTOA.

I had an argument written down here previously that I thought might work but now I realize it can't work. Oh, but it's fixable.

EDIT I've managed to repair the argument. The downside is it's not as simple.

A polynomial without roots produces a polynomial without fixed points. Specifically, $p(z) \neq 0$ for all $z \in \mathbb C$ means $q(z) = p(z)+z$ has no fixed points in $\mathbb C$. So what? Think of $q(z)$ as a map of the Riemann sphere. Now take the real oriented blow-up of the Riemann sphere at infinity (i.e. replace the point at infinity by its unit normal bundle in the sense of smooth real manifolds). This is a disc. So $q(z)$ becomes a smooth map of the disc, denote it $\hat q$, and identify the blow-up with $D^2$, the unit disc in $\mathbb C$ centred at the origin.

If $q(z) = z^n + a_{n-1}z^{n-1} + \cdots + a_1 z + a_0$ and if we conjugate by $z \longmapsto 1/z$ so

$$\frac{1}{q(1/z)} = \frac{z^n}{1+a_{n-1}z + \cdots + a_0z^n}$$

So when you restrict $\hat q$ to the boundary circle, it becomes $z \longmapsto z^n$.

$z \longmapsto z^n$ has fixed points $z^{n-1}=1$, the $(n-1)$-th roots of unity. So we can not directly appeal to Brouwer, since Brouwer's fixed point theorem might give you a pre-existing fixed point on the boundary.

Consider the vector field $v(z) = z-\hat q(z)$ on $D^2$. It is inward-pointing on the boundary circle with the sole exception of $z^{n-1}=1$, the $(n-1)$ roots of unity. But if we remove a small neighbourhood of $\partial D^2$ from $D^2$, the vector field $v$ restricts to an inward pointing vector field. So you could appeal to Poincare-Hopf and say there has to be a zero in the interior, or you could talk about the flow of the vector field, and Brouwer's fixed-point theorem would then tell you the vector field must have a zero in the interior.

So its not a slick proof, but it can be done.

A suitable identification between $D^2$ and the blow-up of the Riemann sphere at infinity is done by the map $X : D^2 \to \hat{\mathbb C}$ given by $X(z) = \frac{1}{1-|z|^2} z$. So $\hat q$ is the unique continuous extension of $X \circ q \circ X^{-1}$ to $D^2$.

Answer 2

Since Ryan has discussed the relationship between Brouwer and FTOA, I would like to say something about Brouwer and Borsuk-Ulam（I prefer this name）. What I know is there are series of classical theorems on this direction, all closely related:

Thm.1(Brouwer) Continuous map $f:B^{n+1} \to B^{n+1}$ has at least one fixed point.

Thm.2(Hirsch) $\partial B^{n+1}$ is not a deformation retract of $B^{n+1}$.

Thm.3 $\mathrm{id}:S^n \to S^n$ is not homotopic to a constant map.

Thm.4 Continuous map $g:S^n \to S^n$ sending antipodal points to antipodal points is not homotopic to a constant map.

Thm.5 There exists continuous map $h:S^n \to S^m$ sending antipodal points to antipodal points if and only if $n \leq m$.

Thm.6(Borsuk-Ulam)For continuous map $k:S^n \to \mathbb{R}^n$，there exists $x \in S^n$ such that $k(x)=k(-x)$.

Let me explain. Equivalence of 1 and 2 is quite standard, for example, see

Milnor Topology from the differentiable viewpoint

By the way, it also contains a proof of FTOA, using the degree of maps.

To see the relation of 2 and 3, simply glue the boundary to a point.

4 is an obvious generalization of 3. You may prove it by studying $H_*(\mathbb{R}P^n)$. This study also leads to a proof of 5.

Equivalence of 6 and (the "only if" part of) 5 is also not hard, you may take it as a exercise:)

All in all, what I have shown is some knowledge of $H_*(\mathbb{R}P^n)$ is sufficient for both Brouwer and Borsuk-Ulam. Hope this helps.