Talk:Brouwer fixed point theorem

From Wikipedia, the free encyclopedia

You have new messages (last change).

Courant and Robbins provide an accessible proof.

According to Lyusternik Convex Figures and Polyhedra, the theorem was first proved by a Lettish mathematician named Bol. No references are provided. Anyone know what this is about?--192.35.35.36 00:08, 18 Feb 2005 (UTC)

Contents

[edit] Proofs

Why is wikipedia not the place to reroduce a long proof? --anon

What? El_C 20:10, 22 September 2005 (UTC)
See Wikipedia:WikiProject Mathematics/Proofs and the math style manual. Oleg Alexandrov 22:44, 22 September 2005 (UTC)

[edit] Hex in multiple dimensions?

The article says:

A quite different proof can be given based on the game of Hex. The basic theorem about Hex is that no game can end in a draw. This is equivalent to the Brouwer fixed point theorem for dimension 2. By considering n-dimensional versions of Hex, one can prove that in general that Brouwer's theorem is equivalent to the "no draw" theorem for Hex.

But how does one play an "n-dimensional version of Hex"?

Consider the original 2D hexboard as being made from a lattice, where one connects the lower left corner of a square of a lattice to the upper right corner with an edge. These added diagonals make it so you can have 6 neighbors instead of 4 (on the original lattice). It's easy to see how to cut out an n by n Hex board from this modified lattice. In general, to create an n x n .... x n board, consider a lattice in R^m (where m is the number of dimensions of the board) and then in each m-dimensional cube add a diagonal. Then cut out a board as in 2D.
Each player has an opposite pair of sides as before, but now some sides belong to neither player. The game is played the same way as before. It doesn't appear to be so interesting to play, but people have devised other higher dimensional versions which are probably more fun and interesting mathematically. In any case, in the version I described, there can never be a draw, and this no-draw result is equivalent to the Brouwer fixed point theorem. --Chan-Ho (Talk) 07:03, 15 January 2006 (UTC)

[edit] elemantary proof with stokes' theorem

first, the retraction is given by:

Image:Theorem of brouwer-F.png
Illustration of the retraction
F(x):=x + \left( \sqrt{1-|x|^2 + \left\langle x,\frac{x-f(x)}{|x-f(x)|} \right\rangle^2 } - \left\langle x, \frac{x-f(x)}{|x-f(x)|} \right\rangle \right) \frac{x-f(x)}{|x-f(x)|}

and it can be proved using the strokes' theorem for differential forms.

for \omega^{n-1}:= F^1\, dF^2\wedge\cdots\wedge dF^n, dωn − 1 = 0, so you get: 0 = \int_{D^n} \mathrm d\omega^{n-1} = \int_{S^{n-1}} \omega^{n-1} = \int_{S^{n-1}} x_1 dx^2 \wedge \cdots \wedge dx^n = vol (D^n) \neq 0. first = because the jacobian is 0 by theorem of implicit functions.

See the german page as example. this could be integrated. ~ibotty

Looks a bit messy to me.... Oleg Alexandrov (talk) 16:24, 9 February 2006 (UTC)
It's not that messy! Don't be scared by the symbols, Oleg :-) I think with a little reworking it would make a fine addition. This is a pretty famous proof. Unfortunately, I take a bit of an issue with "elementary" describing this proof. The proof, as given, only proves Brouwer's theorem for sufficiently smooth maps f, since one needs to take the exterior derivative. Luckily, we can homotope f to be smooth while still keeping the map fixed point free (we can pick a straight-line homotopy that moves every point less than epsilon, where epsilon is smaller than the minimal distance between x and f(x)); however, this is not so trivial to show and can be regarded as a technicality that makes the entire proof not as elementary. And of course, we could debate, if we wished, whether this whole business of Stoke's theorem and smoothing maps is really more elementary than some simple homology (or homotopy) theory. Personally, I think the only kind of proof of Brouwer that really qualifies as elementary are the ones involving some form of coloring trick, e.g. Sperner's lemma or Hex. --Chan-Ho (Talk) 01:46, 2 April 2006 (UTC)
Hmmm...actually I see that since I last closely perused the article, somebody has added another proof of Brouwer for smooth maps. Per my comment right above, this is actually a proof for continuous maps also; I'll add that to the article. --Chan-Ho (Talk) 10:42, 2 April 2006 (UTC)
Stokes theorem may well be derived from the Brouwer Fixed Point Theorem, so I would be weary of using it. But I am not certian of this, as I have not read the proof. --Dark Side of the Moon 17:02, 28 August 2006 (UTC)

[edit] Outline of Proof

I believe that there's an error outline of the proof given - in particular the induced transformation from the disk to its boundary need not be a retraction (the points on the boundary need not be fixed by the transformation). At the same time this is only a small error in that it's true that there's no continuous mapping from the disk to its boundary (retraction or not).

I'm not sure what you mean. The map described is the one that sends x to the point on the boundary given by following the directed line going from f(x) to x. So if x is on the boundary, it gets fixed. --Chan-Ho (Talk) 03:36, 1 May 2006 (UTC)

Sorry, you're completely right, I thought the ray was in the other direction.

However there is still a problem, why does F(x) first have to be continuous (the above section shows that it is, but this should be in the outline). The biggest problem I see is that F : D^2 \to S^1 is not necessarily surjective! --Dark Side of the Moon 16:59, 28 August 2006 (UTC)
F is necessarily surjective, as it maps every element of S1 to itself. As for continuity, this follows from the explicit formula for F(x) that someone has given above. --Zundark 17:57, 28 August 2006 (UTC)
I was mistaken; I had misread and thought that the point F(x) was created by moving from x towards f(x) which would cause problems. The other way, the way it is in the article, will infact work. --Dark Side of the Moon 21:10, 28 August 2006 (UTC)

[edit] Non-constructive

The history section seems to imply that the theorem does not have a constructive proof. Is there a way to formalize (and prove) this statement? AxelBoldt 05:37, 9 June 2006 (UTC)

Ok, I found in [1] that there are in fact algorithms to approximate a fixed point. AxelBoldt 03:34, 12 June 2006 (UTC)
Retrieved from "http://en.wikipedia.org../../../b/r/o/Talk%7EBrouwer_fixed_point_theorem_d4a4.html"