Comments on: Four Color Theorem and Lie Algebras

By: antony

antony — Mon, 16 Nov 2009 11:14:36 +0000

ooops,
i think the theoremsays abt
4 colours by using one could be able to
colour a map.
The map should be xoloured in suchc a way
that no same colours touch,intese,coicide each other.

am a student from chry….

By: Cui Shitai

Cui Shitai — Sat, 07 Jun 2008 07:13:48 +0000

I have a proof for the four color theorem with no computer aid, 6 pages, absolutely correct. I submitted it to many journals but rejected, they haven’t found any mistakes only said it’s too simple and it has not the potential to open new perspectives, to attract broad interest and to have lasting impact. I think a correct proof of the 4 color theorem is surely a lasting impact and attract broad interest. Would you suggest me which journal to submit for publication. If so please email to cuishitai12000 at yahoo.com.cn I believe you are an expert on 4 color theorem, maybe a referee for journals.

By: Jacob Freeze

Jacob Freeze — Tue, 19 Feb 2008 00:39:05 +0000

Thanks, Todd.

I read This Week’s Finds, where Baez says “the only way known to prove this [bracketing] Theorem is to deduce it from the 4-color theorem.”

Hmmmm.

The Four-Color Theorem has a particular appeal for people like me, since an amateur (Jean Mayer) made a significant contribution to Haken and Appel’s proof.

Incidentally, Haken and Appel also share a little credit for “intellectual” work on the theorem with their IBM 360, beginning in 1975:

“At this point the computer began to surprise us. At the beginning we would check its arguments by hand so we could always predict the course it would follow in any situation; but now it suddenly started to act like a chess-playing machine. It would work out compound stategies based on all the tricks it had been ‘taught,’ and often these approaches were far more clever than those we would have tried. Thus it began to teach us things about how to proceed that we never expected. In a sense it had surpassed its creators in some aspects of the ‘intellectual’ as well as the mechanical parts of the work.” (in Steen, Mathematics Today, 1978)

By: Todd Trimble

Todd Trimble — Mon, 18 Feb 2008 20:22:07 +0000

If you follow the link he gives, to This Week’s Finds, you’ll find the reference

Map coloring and the vector cross product, by Louis Kauffman, J. Comb. Theory B, 48 (1990) 45.

in case you’re really interested. My guess is that he, John Baez, is not personally invested in the Four Color Theorem in the first place!

By: Jacob Freeze

Jacob Freeze — Mon, 18 Feb 2008 14:10:45 +0000

Why would John Baez spell out the standard basis for R^3 and yet leave it as an exercise for the reader to figure out how the Four Color Theorem falls out of bracketing vector products? (Someone like Ramanujan shows up on Ars Mathematica knowing only that vectors are like little arrows, and when he sees i=(1,0,0), everything else is obvious to him.) But for me and Beavis here in the back of the classroom, maybe some kind soul will fill in a few of the blanks, or at least provide a little arrow to point us in the right direction. We want to learn!

By: Walt

Walt — Tue, 05 Feb 2008 21:07:31 +0000

I looked into this a little bit, and it seems I would have to log into the hosting server to fix it so that the html command for superscripts works. I’ll talk to the guy hosting it when he gets back from vacation (he’s skiing in France right now).

By: John Baez

John Baez — Fri, 01 Feb 2008 20:49:36 +0000

Could you change the “R3″ above to something nicer, say R^3?

Why is the html command for superscripts disabled on this blog.

By: John Baez

John Baez — Fri, 01 Feb 2008 20:47:57 +0000

There are lots of ways to restate the 4-color theorem. For example: it’s equivalent to this fact about the vector cross product:

Theorem: Consider any two bracketings of a product of any finite number of vectors, e.g.:

L = a x (b x ((c x d) x e) and R = ((a x b) x c) x (d x e)

Let i, j, and k be the standard basis for R^3:

i = (1,0,0) j = (0,1,0) k = (0,0,1).

Then we may assign a,b,c,… values taken from {i,j,k} in such a way that L = R and both are nonzero.

However, these restatements haven’t yet made the result easier to prove.

We can still hope.

By: Mike

Mike — Thu, 31 Jan 2008 21:12:54 +0000

Oops, let me revise or pull that comment! I just realized the paper in ArXiv was the first version of the Comb. paper, submitted in 1996. I remeber reading it in Comb. when it came out.
Thanks and my apologies.

By: Mike

Mike — Thu, 31 Jan 2008 21:07:41 +0000

I haven’t had a chance to read thru the paper yet and my mastery of finite type invariants is weak, to put it mildly. As soon as I glanced at it though, I had a memory of reading a similar article connecting algebra and the 4CT.
To my surprise, when I used MathSciNet, I pulled up an article of Bar-Natan’s from 10 yrs ago, which is almost certainly the one I was thinking of, in Combinatorica:

Lie algebras and the Four Color Theorem
Journal Combinatorica
Publisher Springer Berlin / Heidelberg
ISSN 0209-9683 (Print) 1439-6912 (Online)
Issue Volume 17, Number 1 / March, 1997
DOI 10.1007/BF01196130
Pages 43-52

More later once I have the time to actually digest both articles as well as some of Bar-Natan’s older work.