# Campbell-Baker-Hausdorff formula for $\log(\exp(X+Y)\exp(X-Y))$

Given $X,Y\in \mathfrak g\mathfrak l_{\mathbb R}(n)$, and the CBH formula for $\log(\exp X\exp Y)$ (wiki), is it possible to derive the general term in the series of $\log(\exp(X+Y)\exp(X-Y))$ that involve only an odd number of $X$? The even term should vanish since $f(X):=\log(\exp(X+Y)\exp(X-Y))$ is an odd function.

If this can be worked out in a nice way, I would also like to have a similar result for $\log(\exp(X+Y)\exp(Z)\exp(X-Y))$. In this case, only terms involving an odd number of $X$'s and/or $Z$'s remain.

See 4.29 (on page 54)

• Peter W. Michor: Topics in Differential Geometry. Graduate Studies in Mathematics, Vol. 93 American Mathematical Society, Providence, 2008.(pdf)

for complete formulas involving the the general terms of the BCH formula. You can use this to derive some expressions for the formula you want.

• I came to this question via meta.mathoverflow.net/q/1954/26935 Sep 9 '15 at 6:44
• Dear Prof.Peter, thank you very much for your reply. Indeed, I have considered what you have mentioned in your nice book (which is also free...). What I really want is not the general terms, but 2 terms, as the sum of the odd and even terms. This question outreaches to Lie triple systems and symmetric subspaces of a Lie group considered as an affine symmetric subspace. In my case, the underlying Lie group is $SE(3)\simeq SO(3)\ltimes\mathbb R^3$. I have been able to solve this only for $SO(3)$, and it seems the two terms have ugly inverse trigonometric expression. Sep 12 '15 at 17:12
• I doubt there will ever be a general expression for all Lie triple subsystems of $\mathfrak{se}(3)$. But I still hope...Please understand my situation and my choice to leave this question open. Sep 12 '15 at 17:12

The systematics of the CBH expansion comport with the structure of the Lie algebra involved. There is no "nice" full expansion beyond Dynkin's unwieldy expression, except in special cases. To fix language, you might consult the bare minimum review for theoretical physicists.

To account for the "number of commutators" in the expansion, introduce the gratuitous parameter t to set equal to 1 in the end, if you wish. it is the number of nested commutators minus 1, so $Z(t;X,Y)\equiv \log( ~ \exp (t X) \exp( tY)~)$. Thus $Z(t;X,Y)=-Z(-t;Y,X)$, that is, odd powers of t are X-Y symmetric, and even powers of t are antisymmetric.

Your $W(t;X,Y)=Z(t;X+Y,X-y)= \log(~\exp(t(X+Y))\exp(t(X−Y))~)$ will then have the symmetry $$W(t;X,Y)=-Z(-t;X-Y,X+Y)=-W(-t;X,-Y),$$ so that odd powers of t will be even in Y and hence odd in X; likewise, even powers of t are odd in Y and thus odd in X, as you already observed.

But that is as far as the train goes: for further simplifications you need special features of the algebra, some of which the cited reference illustrates.

• Dear Prof.Cosmas, WOW...that is very sharp observation. And I think you are absolutely right about the low possibility in finding a formula for the general case. Indeed, this formula is involved in the special case of a Cartan decomposition where $X,Z$ lies in a Lie triple system and $Y$ in its corresponding Cartan subalgebra. Apr 1 '17 at 20:31
• Originally, I was trying to figure out if there is this possibility of having a closed form expression for the series, but then I guess I should first derive that for the terms. Thanks a lot for pointing out your lovely notes. I will now follow your lead and see what I can do. But if you already have an answer to this extended question of mine, please kindly share it in your answer. Apr 1 '17 at 20:36
• If you have a low-dimensional Lie algebra, you could play this game... Apr 1 '17 at 20:57
• Yes..., subalgebras of special Euclidean group $SE(3)$ (I'm an engineer). Apr 1 '17 at 20:58