Talk:Baker-Campbell-Hausdorff formula

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[edit] Alternate form

This article should also provide the first few terms of W for the left-multiplication variant

eA + B = eAeBeW

as I'm too lazy to perform the inversion myself. I suppose, for completeness, the right-multiplication version as well. linas 16:20, 1 March 2006 (UTC)


I think in the formula for X * Y the power of (-1) should be n-1 and not n

[edit] Misprint

I think there is misprint in the formula because ad(Y)(Y)=0

It looks strange, yes, but I looked it up and it is intentional. I slightly rewrote the formula to get something which is clearer in my opinion, and added an explanation. -- Jitse Niesen (talk) 05:06, 22 May 2007 (UTC)

[edit] Explicit formula

While the first few terms of the BCH formula...

X + Y + \frac{1}{2}[X,Y] + \frac{1}{12}[X,[X,Y]] - \frac{1}{12}[Y,[X,Y]] - \frac {1}{48}[Y,[X,[X,Y]]] - \frac{1}{48}[X,[Y,[X,Y]]] + \cdots

...are correct, they're a bit misleading, because [Y,[X,[X,Y]]] = [X,[Y,[X,Y]]]. You can check this by working them both out in the universal enveloping algebra, where [X,Y] = XY − YX (or by typing the whole thing into GAP). See [[1]] for example. I'll change this in the article unless anyone comes up with a good reason not to —Preceding unsigned comment added by 163.1.148.158 (talk) 23:13, 10 December 2007 (UTC)

In case someone wanted to have GAP check, this is quite quick:
gap> F:=FreeLieAlgebra(Rationals,["x","y"]);x:=F.1;y:=F.2;
<Lie algebra over Rationals, with 2 generators>
(1)*x
(1)*y
gap> y*(x*(x*y))=x*(y*(x*y));
true
One could also use free associative algebras and actually compute commutators, but it is easier just to use a Lie algebra. Thanks for the interesting link. For free groups there are nice bases for the lower central factors given by Hall's commutator formulas, and for free metabelian Lie algebras your link gives a basis. Is there a nice basis for free Lie algebras? I sort of thought Hall's method worked for free Lie algebras too. If there is such a basis, it might be nice to expand in it. JackSchmidt (talk) 00:49, 11 December 2007 (UTC)
I believe Hall's basis for the free group lower central factors transfers directly to the free Lie algebra case, but I'm not sure of the details. It's interesting that the online Magma at [[2]] answers "false" to both of the following
L<x,y>:=FreeLieAlgebra(Rationals(),2);
(y,(x,(x,y)))eq (x,(y,(x,y)));
z:=(x,y)+2*x;
(x,(y,z))+(z,(x,y))+(y,(z,x)) eq 0;

...I think there is something wrong with it. 79.67.200.228 (talk) 01:31, 11 December 2007 (UTC)

This bug may be by design:

The free Lie algebra LF(X) is spanned by M(X). However, the elements of this set are not linearly independent. It is a nontrivial problem to describe a basis of the free Lie algebra. One of several possibilities is the well-known Hall basis. Currently Magma does not support calculations involving bases of the free Lie algebra, as they are of little use for our main problem: the construction of a basis and multiplication table for a finitely-presented Lie algebra. ... Thus, mathematically speaking, in Magma rather than work in the free Lie algebra, we actually work in the free nonassociative anticommutative algebra.

There are other oddly named functions which have similar caveats. JackSchmidt (talk) 04:06, 11 December 2007 (UTC)