Comments on: Just Fokkinga Abide

By: Edward Kmett

Edward Kmett — Wed, 16 Mar 2011 20:37:13 +0000

I’ve been blogging here since around 2006, off and on.

By: Darin Justiniano

Darin Justiniano — Wed, 16 Mar 2011 18:31:52 +0000

Wow, marvelous blog layout! How long have you been blogging for? you make blogging look easy. The overall look of your web site is wonderful, let alone the content!

By: pozorvlak

pozorvlak — Fri, 09 May 2008 19:05:33 +0000

Ah, right, that explains it. I've blogged the 2-d version of the proof here: I think you'll like it. I look forward to seeing your code!

By: Edward Kmett

Edward Kmett — Fri, 09 May 2008 15:43:26 +0000

I assure you there was no malice in the filtering, it was done automatically because of the link.

Fixed!

Great catch regarding the Eckmann-Hilton bit. Encoding it into Haskell for bifunctors as we speak. =)

By: pozorvlak

pozorvlak — Fri, 09 May 2008 15:08:41 +0000

OK, that's totally weird. You seem to have blocked the comment in which I explained the connection to the Eckmann-Hilton argument, and let through my two subsequent comments (in which I tried to reproduce the proof in all its 2-dimensional glory, and in which I complained that my

 tags had been stripped, rendering the proof unintelligible).

So, for those who are feeling confused: if * and + abide and have units 1 and 0, then 0=1, *=+, and * (and hence +) is commutative. The proof's fun and reasonably straightforward, particularly if you use the 2-d notation. This is known as the Eckmann-Hilton argument, and it's famous in category theory, having many beautiful consequences.

By: pozorvlak

pozorvlak — Fri, 09 May 2008 13:33:30 +0000

Aaargh! No

tags. So much for that plan.

By: pozorvlak

pozorvlak — Fri, 09 May 2008 13:32:46 +0000

Because I clearly don't have better things to do right now :-)

0 = 0|0 = 1|0 = 1|0 = 1 = 1 -|- --- - - 0|1 0|1 1 1 x|y = 1|y = 1|y = y = y|1 = y|1 = y|x = 1|x = 1|x = x -|- --- - --- -|- -|- --- - x|1 x|1 x 1|x 1|x y|1 y|1 y

By: pozorvlak

pozorvlak — Fri, 09 May 2008 13:25:13 +0000

It's worth mentioning that if two abiding operations have identities, then they're equal to each other and commutative, and the identities coincide. Let + and * be abiding binary operations, with identities 0 and * respectively. Then 0 = 0+0 = (0*1)+(1*0) = (0+1)*(1+0) = 1*1 = 1 And now: x*y = (0+x)*(y+0) = (0*y)+(x*0) = (1*y)+(x*1) = y+x = (y*1)+(1*x) = (y+1)*(1+x) = (y+0)*(0+x) = y*x = (0+y)*(x+0) = (0*x)+(y*0) = (1*x)+(y*1) = x+y This is the Eckmann-Hilton argument beloved of category theorists: among its implications are that a monoid object in the category of monoids is a commutative monoid, and that all higher homotopy groups are abelian. It becomes particularly elegant when you use the x | y =(x*y)+(y*w) z | w notation - try it! The proof can be written in a circle, and it's then called the Eckmann-Hilton clock. The Catsters have done a video about the Eckmann-Hilton argument.