I am trying to prove a result for which I need the nth term of the Baker-Campbell-Hausdorff formula. I came at this particular result (which is not of significance for the question, but mentioning for context) by hypothesizing and using the first few terms of the Baker formula to verify. In order to prove my result rigorously, I think I need the nth term of the Baker formula. Is there an expression for that – I could not find it through online research. ? It would also help if I could find a proof of the BCH formula which is based on recursion i.e. if I could see how the nth term relates recursively to the (n-1)th term.
[Math] nth term in the Baker-Campbell-Hausdorff formula
dg.differential-geometrylie-algebraslie-groupsnoncommutative-algebrareference-request
Related Solutions
To answer your first question, the composition of two time-1 flows won't necessarily be another time-1 flow.
One way to see this is to note that when a time-1 flow $\phi_X$ has a periodic point $P$ (period > 1), then $P$ can't be hyperbolic since it lies on a closed orbit of the flow for $X$. (The eigenvector of $D\phi_X$ tangent to this orbit has corresponding eigenvalue 1.)
Now, take a flow on $S^2$ whose time-1 map rotates the sphere, switching the north and south poles. Take a second flow for which both poles are hyperbolic attracting fixed points. Composing the two time-1 maps gives you a new diffeomorphism with hyperbolic points of period 2.
Giving your series a name, I'll set $$A(P,Q) \overset{\rm def}= \exp\bigl( \log(1 + P) + \log(1 + Q) \bigr) $$ to be your power series, where $P,Q$ are free (noncommuting) variables. I'm not sure what you want to know about this series: it exists, I don't think it has a name, the first few terms are $$ A(P,Q) = 1 + P + Q + \frac12(PQ + QP) + \frac1{3! \cdot 2}\left(-P^2 Q + 2PQP - Q^2 P-PQ^2 + 2QPQ - Q P^2\right) + $$ $$ + \frac1{4!}\left( P^3 Q - P^2QP -PQP^2 + QP^3 - PQ^2P + PQPQ + \{P\leftrightarrow Q\} \right) + \dots.$$ I don't see much of a pattern in the coefficients, and I haven't worked out the next term. If the cubic term didn't have that $\frac12$, or if the quartic terms had more fractions, I would be happier. As it should, when $[P,Q] = 0$, the series truncates to $A(P,Q) = 1 + P + Q + PQ$. It is left-right symmetric and symmetric in $P\leftrightarrow Q$.
Here's one remark that might be useful for your intended application. Let $K = k\langle\langle X,Y\rangle\rangle$ be the ring of power series in two noncommuting variables. Then there is an algebra homomorphism $\Delta: K \to K \hat\otimes K$, where $\hat\otimes$ denotes that you should complete the tensor product w.r.t. the adic topology, given on generators by $\Delta(X) = 1 \otimes X + X\otimes 1$ and $\Delta(Y) = 1 \otimes Y + Y \otimes 1$. Recall that an element $P \in K$ is primitive if $\Delta(P) = 1 \otimes P + P\otimes 1$. Then the primitive elements form a Lie subalgebra of $K$, and consist precisely of the Lie series: the power series that can be expressed without ever referring to multiplication in $K$, only to the Lie bracket $[x,y] = xy - yx$ (and that begin in degree $1$). Recall also that an element $P\in K$ is grouplike if $\Delta(P) = P\otimes P$. Then the grouplike elements if $K$ form a group under multiplication; in particular, they are all units.
If $P,Q$ are primitive, then there's no particular reason for $A(P,Q)$ to be primitive, and actually I think it never will be. However, by construction $\Delta$ is continuous w.r.t. the adic topology, and it is a homomorphism, and so $\Delta(f(P)) = f(\Delta(P))$ for any power series $f$ in one variable. Also, an element $P\in K$ is primitive iff $\exp(P)$ is grouplike. So it follows that if $1+P$ and $1+Q$ are both grouplike, then so is $A(P,Q)$.
I'll close by saying that, to me anyway, you already have an "explicit expression" for the power series, and even a "geometric interpretation", which is that it moves the additive structure of the Lie algebra to the (formal) group (whereas the BCH series moves the group multiplication to the Lie algebra). You shouldn't strongly hope for a simple description of the coefficients in terms of combinatorics, because similar descriptions for BCH, although they do exist, can be rather complicated.
Update
Above I observed that $A(P,Q)$ is not a Lie series in $P,Q$. This can be seen directly: any Lie series truncates to its linear and constant terms when $[P,Q] = 0$, whereas $A(P,Q) = (1+P)(1+Q) = 1 + P + Q + PQ$ in commuting land.
So the next best thing, as I asked in the comments, is whether $$A(P,Q) - (1+P)(1+Q) = -\frac12[P,Q] - \frac1{12} \bigl( [P,[P,Q]] + [Q,[Q,P]] \bigr) + \dots $$ is a Lie series. Alas, this also fails, as can be seen at the quartic part. Indeed, by Jacobi and antisymmetry, $$ [P,[Q,[P,Q]]] = [[P,Q],[P,Q]] + [Q,[P,[P,Q]] = [Q,[P,[P,Q]] = -[Q,[P,[Q,P]] $$ is the unique Lie monomial (up to scalar) of degree $P^2Q^2$. But it is antisymmetric under $P\leftrightarrow Q$, whereas $A(P,Q)$ is symmetric under the same transposition. Thus if $A(P,Q) - (1+P)(1+Q)$ were a Lie series, then it could not have any terms of degree $P^2Q^2$, whereas by direct calculation: $$ A(P,Q) \ni \frac1{4!} \bigl( PQPQ - PQ^2P - QP^2Q + QPQP \bigr) = \frac1{24} [P,Q] ^2 $$
Best Answer
The Dynkin formula is somewhat cumbersome. Maybe a better choice is Goldberg's version http://projecteuclid.org/euclid.dmj/1077466673 In the commutator form Goldberg's result is reformulated in http://www.ams.org/journals/proc/1982-086-01/S0002-9939-1982-0663855-0/ (Cyclic relations and the Goldberg coefficients in the Campbell-Baker-Hausdorff formula, by Robert C. Thompson).
By the way an interesting early history of the Baker-Campbell-Hausdorff-Dynkin formula can be found in http://link.springer.com/article/10.1007%2Fs00407-012-0095-8 (The early proofs of the theorem of Campbell, Baker, Hausdorff, and Dynkin, by R. Achilles and A. Bonfiglioli).