Talk:Commutator

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[edit] commutator in algebras

How does this correspond to the functional analysis commutator, such as that in the Uncertainty Principle defined by [A,B] = AB-BA

They are certainly connected: the common spirit is to give a "measure" of how much A and B (or whatever) do not commute. The final remark about commutators being defined also in structures different from groups should definitely be expanded. --Goochelaar

The connection is given by going to infinitesimal generators of Lie group elements. This is explained (rather alluded to....) in Lie algebra#Examples point 5 (currently *sic*). In some sense, consider e.g.

g = exp( x ) , h = exp( y )

then

g h = exp(x + y + 1/2[x,y] +...), thus g h g¯¹ h¯¹ = exp( [x,y] +...) (where ... are higher order terms).

It can also be formulated in terms of the derivative of the elements of a Lie group at the origin. — MFH: Talk 21:09, 21 Jun 2005 (UTC)

[edit] commutator

It seems the element "g\-1 h\-1 gh" would be more clearly stated as "h\-1 g\-1 gh" Or am I missing something?

Ken Krechmer

The second expression you write is the identity always, since the middle two g's cancel. Thus it is trivial, and doesn't say anything about whehter g and h commute. -Lethe | Talk 01:05, Jun 15, 2005 (UTC)

[edit] other convention

Indeed the order must of course be g,h,g,h. However, the convention differ, and often [g,h] is rather defined as g h g¯¹ h¯¹ which I personally prefer, e.g. in view of the "other" ("algebra") commutator, [A,B]=A B - B A (so the "plain" part remains and what is "minus reverse order" in the latter becomes "times the inverses" in the former). — MFH: Talk 20:42, 21 Jun 2005 (UTC)


[edit] another property

Looking at the ring theory definition of commutator, the property AB = BA + [A,B] seems obvious in retrospect, but a usage (Griffiths, Quantum Mechanics) of that property had me scratching my head for a good ten minutes. *shrug* In case anyone else is scratching their head about it at any time. Popefelix 00:15, 14 December 2005 (UTC)

[edit] Matrices in bases

What does this statement mean?

In linear algebra, if two matrices commute in one basis they will commute in any basis.

It seems to be trying to mean that, if A and B are the matrices of two linear operators with respect to one basis, and A' and B' are the matrices of the same two operators (in the same order) with respect to another basis, then A and B commute if and only if A' and B' do. However, the statement as it stands is meaningless. JadeNB 21:18, 10 October 2006 (UTC)

[edit] Standard notation

The standard mathematical notation for commutator is \scriptstyle{ad(x)(y)=[x,y]}. As I see no reason to depart from the standard, I changed the relevant paragraph. AlainD 10:02, 25 January 2007 (UTC)

[edit] notation for ad(x)ad(x)(y)

I think it's misleading to use the notation "ad(x)^3(y) = [x,[x,[x,y]]]", since there are a lot of different algebras floating around. In particular, ad(x) is a derivation, which has both the usual algebra structure of composition, and a Lie algebra structure induced from the former. Of course y lives in a Lie algebra, where something like y^3 is not defined. I admit that ad(x) may be unambiguous mathematically, but I feel that in this context it's best to avoid confusion if possible. Tesseran 07:46, 28 June 2007 (UTC)