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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.

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I think the book by Guillemin and Pollack, "Differential Topology", has it, but I am not certain. –  Claudio Gorodski Apr 10 '12 at 18:15
    
Could you provide a link to the thread you refer to? –  Ryan Budney Apr 10 '12 at 18:16
    
Sorry, I seem to have forgotten how to add links ... I simply copied the title of the thread, but the hypertext marking isn't preserved ... :-( Some advice would be appreciated ! Kind regards, St. –  Stephan F. Kroneck Apr 10 '12 at 18:20
    
Ryan: Google books gives some of the book. –  Claudio Gorodski Apr 10 '12 at 18:24
    
Stephan: I am quite sure the three results are there, but not that the two implications you want are. Nevertheless, I think those proofs are very nice, based on elementary tranversality arguments. Maybe it is worth having a look. You could try Google books. –  Claudio Gorodski Apr 10 '12 at 18:27

2 Answers 2

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 $D^2$.

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Ryan Budney: Thank you for your interesting answer ! I wouldn't have thought of that trick; I've been trying all sorts of other maps related to the original polynomial, without success. Still I must agree with you; I would also appreciate a simpler argument. Kind regards ! St. –  Stephan F. Kroneck Apr 10 '12 at 19:03
    
Another version can be found here: jstor.org/stable/2305130 . I like these proofs. –  Deane Yang Apr 10 '12 at 20:12
    
@Deane Yang: Thanks for your comment, but I believe the article you are referring to is the faulty one I mentionend above; I think there is a follow-up erratum in the AMM by Arnol'd and Niven ... Kind regards ! St. –  Stephan F. Kroneck Apr 10 '12 at 20:22
    
Stephen, oops. Sorry about that. –  Deane Yang Apr 10 '12 at 21:32
    
Stephen, is the error the same as the one in Ryan's argument? –  Deane Yang Apr 10 '12 at 21:55

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.

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@Zhang Xiao: Thank you very much for your comments ! Basically, though, I believe most of the above is well-known (and is contained for instance in Granas-Dugundji II, §§ 5,7). Naturally, using the homological machinery one can prove Borsuk etc., but this seems like "overkill" in view of elementary proofs of Brouwer (cp. arxiv.org/pdf/1109.4604.pdf) and of the FTA (cp. arxiv.org/pdf/1110.0165.pdf). In all sources I can find, though, Borsuk seems to be the stronger statement (than Brouwer); I'm still lacking substantiation of the claims in the other thread ... Kind regards ! St. –  Stephan F. Kroneck Apr 10 '12 at 22:07
    
@Stephan: You are correct. Although homology helps to make a unified picture, it contains strictly more information than Brouwer and Borsuk, hence of no use in setting the equivalence between two. –  Zhang Xiao Apr 10 '12 at 23:44

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