• Welcome to the Speedsolving.com, home of the web's largest puzzle community!
    You are currently viewing our forum as a guest which gives you limited access to join discussions and access our other features.

    Registration is fast, simple and absolutely free so please, join our community of 40,000+ people from around the world today!

    If you are already a member, simply login to hide this message and begin participating in the community!

How to Notate Commutators and Conjugates?

M

MrRedHair

Guest
I understand how commutators work. How a commutator is an algorithm, AB, and its inverse, A'B', written as [A,B], and also how a conjugate is CBC', or [C:B]. My question is more directed to the notation. [A:B] [C:B'] is the same as A B A' C B' C' - how can those moves turn into [C: [C' A, B]]? The last part makes no sense to me. If anybody could try and explain how commutator/conjugate notation works, it would be most appreciated. Thank you.

This commutator/conjugate example came from
HTML:
http://snk.digibase.ca/k4/7.htm
 

qqwref

Member
Joined
Dec 18, 2007
Messages
7,834
Location
a <script> tag near you
WCA
2006GOTT01
YouTube
Visit Channel
[A:B] [C:B'] = (A B A B') (B C B' C') = A B A' C B' C'

[C: [C' A, B]] = C (C' A B A' C B') C' = A B A' C B' C'

What's going on here isn't a weird identity, but just a different way of dividing up the expression. The first one works because you can do the two halves of the expression as commutators and have the extra two moves cancel each other out; the second one works because the commutator [C' A, B] ends up being just a setup move away from the expression you want.

Incidentally, [C: [C' A, B]] = [A' C', C B C'], by the property that [O: [P, Q]] = [[O:p], [O:Q]]. This identity can be pretty useful to know.
 
M

MrRedHair

Guest
I thought [C: [C A, B']] was read as do C, do the commutator of CA,B', which is C A B' C' A' B, then do C' to finish the conjugate.
 
Joined
Dec 11, 2009
Messages
294
[A:B] [C:B'] = (A B A B') (B C B' C') = A B A' C B' C'

[C: [C' A, B]] = C (C' A B A' C B') C' = A B A' C B' C'

What's going on here isn't a weird identity, but just a different way of dividing up the expression. The first one works because you can do the two halves of the expression as commutators and have the extra two moves cancel each other out; the second one works because the commutator [C' A, B] ends up being just a setup move away from the expression you want.

Incidentally, [C: [C' A, B]] = [A' C', C B C'], by the property that [O: [P, Q]] = [[O:p], [O:Q]]. This identity can be pretty useful to know.

cancelling setup move (conjugate) ~ cyclic shift.

Good post, but the intermediate results are incorrect as you have shown them.
[A:B] [C:B'] = (A B A B') (B C B' C') is wrong.
The 2nd A should be a A', and [A:B] [C:B'] is conjugates - not commutators, and so that extra "B') (B" does not belong here.

Incidentally, [C: [C' A, B]] = [A' C', C B C']

This is wrong too, since that A'C' needs to be AC'

@michael, I sent you a courtesy PM a couple of days ago to inform you of these mistakes so that you could correct them yourself. Did you not get it?
 
Top