# Inverse of Baker-Campbell-Hausdorff

This should be quite simple to answer. I have a situation in which I must have an explicit expression for the inverse of Baker-Campbell-Hausdorff. More precisely:

I have two power series $P_1(X,Y)$, $P_2(X,Y)$ in noncommuting variables $X,Y$. I want an expression for

$$exp\left(log(1+P_1) + \ log(1 +P_2) \right) .$$

I searched some books and google; but beyond the fact that it exists, I am unable to find an explicit expression. If it already exists in print, it would save me some work of trying to derive it myself.

Thank you very much.

• Deriving it yourself is fun! By the way, what is the significance of $P_1$ and $P_2$ as opposed to $X$ and $Y$? Even if one or both $P_i$s have a non-zero constant term, it can be absorbed into the constant 1 inside the $\log.$ – Victor Protsak Jul 17 '10 at 19:17
• There is no significance, except that when I needed to work it out, it came up as two arbitrary power series, rather than just X and Y. – Anweshi Jul 17 '10 at 21:06
• I don't remember if you are automatically updated when I edit an answer. If not, this note is to mention that I have added some content, answering a question I asked in the comments. – Theo Johnson-Freyd Jul 19 '10 at 5:37
• I am rather surprised that this question got so many upvotes, albeit pleasantly. – Anweshi Jul 19 '10 at 16:46

The following recursion formula for the homogeneous factors of $A(P,Q)$ exists:

Define:

$A(tP,tQ) = 1 + t \frac{B_1}{1!}+t^2 \frac{B_2}{2!}+t^3 \frac{B_3}{3!}+...$

$C_n = P^n+Q^n$

$\Delta C_n = -n C_{n+1}$

Then:

$B_1 = C_1$

$n! B_n = [(\Delta+C_1) B_{n-1}] _{symm}$

Where the symmetrization is of all words of the type $C_{i1} C_{i2} C_{i3} ...$

This formula can be obtained by differentiation with respect to t. I compared the results up to the cubic term to Theo's expression with a full agreement.

The coefficient of the string $\[(C_{i1} C_{i2} C_{i3} ...]_{symm}$ in $B_n$ is equal to the coefficient of $tr(M^{i1}) tr(M^{i2}) tr(M^{i3})...$ in the coefficient of $t^n$ in $det(I-tM)$. These coefficients are the charcters of $\{{{1^n\}}}$ of $S_n$.

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: $$A(P,Q) \ni \frac1{4!} \bigl( PQPQ - PQ^2P - QP^2Q + QPQP \bigr) = \frac1{24} [P,Q] ^2$$

• Thanks. This was helpful. The application was the following. I wanted to define a sort of "geometric mean" for some finite number of noncommutative power series and see how badly it behaves, and how much can be salvaged. That is why I asked for an explicit expression, and I was imagining that something may exist using nested Lie brackets, like in BCH. I was hoping that such an expression was already worked out in the literature. Thanks for your effort. I will wait for some time and if no better answer comes, I will accept yours. – Anweshi Jul 17 '10 at 21:10
• @Anweshi: well, I hope you get something closer to what you're looking for. But you can see explicitly, and also from the abstract nonsense about primitive versus grouplike elements, that there will not be a formula with nested brackets. OTOH, I don't see immediately whether $A(P,Q) - (1+P)(1+Q)$ is a Lie series. It does vanish when $[P,Q] = 0$, which is promising, but not sufficient. – Theo Johnson-Freyd Jul 18 '10 at 1:48
• Since this is in the group (as opposed to the Lie algebra), the right kind of "bracket" is a group commutator. Anweshi: the "geometric mean" that you allude to is probably related to non-commutative exponential function (given by the multiplicative integral). You may find some general theory in books by Maslov and coauthors. – Victor Protsak Jul 18 '10 at 2:24
• @Victor: Yes, this (use the group commutator) occurred to me. But with only multiplication and inverses, I'm dubious that I'd be able to write down an infinite product that converges to the operation in question. OTOH, the series $A(x,y)$, when $x,y$ are the generators of the noncommutative power series ring, is the limit of $\bigl((1+x)^{1/n}(1+y)^{1/n}\bigr)^n$ --- note that $(1+x)^{1/n}$ converges in the power series ring. There is certainly an infinite product whose $n$th partial product is $\bigl((1+x)^{1/n}(1+y)^{1/n}\bigr)^n$, but I don't find that enlightening. – Theo Johnson-Freyd Jul 18 '10 at 3:25
• Well, I was not saying that group commutator $\textit{would}$ work (cf Hall's identity, which is a nontrivial group-like analogue of the Jacobi identity), merely stating an obvious reason why the Lie bracket wouldn't. – Victor Protsak Jul 18 '10 at 9:11