Four Color Theorem and Lie Algebras
January 26th, 2008 by WaltThanks to Greg Muller, I’m looking at this paper by Dror Bar-Natan that reduces the Four Color Theorem to a plausible statement about Lie algebras. Now we just have to hope this new conjecture does not not require hundreds of pages of computer generated proof.
January 27th, 2008 at 8:31 pm
A poem, not a telephone directory.
January 31st, 2008 at 2:07 pm
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.
January 31st, 2008 at 2:12 pm
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.
February 1st, 2008 at 1:47 pm
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.
February 1st, 2008 at 1:49 pm
Could you change the “R3″ above to something nicer, say R^3?
Why is the html command for superscripts disabled on this blog.
February 5th, 2008 at 2:07 pm
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).
February 18th, 2008 at 7:10 am
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!
February 18th, 2008 at 1:22 pm
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!
February 18th, 2008 at 5:39 pm
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)