# Baker-Campbell-Hausdorff formula

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## Statement

### Broad statement

The Baker-Campbell-Hausdorff formula is a formula that tried to express:

$\! \log(\exp(X)\exp(Y))$

in terms of $X$ and $Y$. Here, $X$ and $Y$ are elements of a Lie algebra or Lie ring, $\exp$ is the exponential map from the Lie algebra to its Lie group, the multiplication occurs in the Lie group, and $\log$ is the inverse of the exponential map.

The expression for the Baker-Campbell-Hausdorff formula is a $\mathbb{Q}$-linear combination of $X$, $Y$, and expressions constructed from them using the Lie bracket.

### Clear statement

Fix a positive integer $c$. We will describe here how to obtain the class $c$ truncation of the Baker-Campbell-Hausdorff formula.

Let $\mathcal{A}$ be the free associative $\mathbb{Q}$-algebra on two generators $\tilde{X}$ and $\tilde{Y}$. Let $A = \mathcal{A}/\mathcal{A}^{c+1}$ be the quotient of $\mathcal{A}$ by the ideal generated by products of length $c + 1$ or more. We use the letters $X$ and $Y$ for the images of $\tilde{X}$ and $\tilde{Y}$ respectively in $A$.

Note that both $X$ and $Y$ are nilpotent elements of $A$. They give rise to nilpotent elements with divided powers, and we can therefore compute their exponentials. Explicitly:

$\exp(X) = e^X = 1 + X + \frac{X^2}{2!} + \dots + \frac{X^c}{c!}, \exp(Y) = e^Y = 1 + Y + \frac{Y^2}{2!} + \dots + \frac{Y^c}{c!}$

The class $c$ truncated Baker-Campbell-Hausdorff formula is an expression $w(X,Y)$ such that:

$\exp(w(X,Y)) = \exp(X)\exp(Y)$

Equivalently, it is the expression for $\log(\exp(X)\exp(Y))$.

Note that because of the freeness of our setup, the expression here is unique in the sense that any two different expressions for $w(X,Y)$ would be equal to each other identically.

## Explicit expression

The following explicit expression works when the Lie group is a simply connected group. In the nilpotent case, it therefore works if the Lie group is the naturally chosen group corresponding to the Lie algebra, where the exponential and logarithmic maps are global bijections.

### Explicit formula

The iterated Lie bracket in this version of the formula is right-normed, i.e., $[X_1,X_2,X_3]$ stands for $[X_1,[X_2,X_3]]$:

$\log(\exp X\exp Y) = \sum_{n>0}\frac {(-1)^{n-1}}{n} \sum_{ \begin{smallmatrix} {r_i + s_i > 0} \\ {1\le i \le n} \end{smallmatrix}} \frac{(\sum_{i=1}^n (r_i+s_i))^{-1}}{r_1!s_1!\cdots r_n!s_n!} [ X^{r_1} Y^{s_1} X^{r_2} Y^{s_2} \ldots X^{r_n} Y^{s_n} ]$

### First few terms

$\log(\exp X\exp Y) = X + Y + \frac{1}{2}[X,Y] + \frac{1}{12}[X,[X,Y]] - \frac{1}{12}[Y,[X,Y]] - \frac{1}{24}[Y,[X,[X,Y]]]$ $\! - \frac{1}{720}([[[[X,Y],Y],Y],Y] +[[[[Y,X],X],X],X])$ $+\frac{1}{360}([[[[X,Y],Y],Y],X]+[[[[Y,X],X],X],Y])$ $+ \frac{1}{120}([[[[Y,X],Y],X],Y] +[[[[X,Y],X],Y],X]) + \dots$

### Deducing the formula

The above explicit formula, as well as a few small cases, can be computed by hand. See deducing the Baker-Campbell-Hausdorff formula from associative algebra manipulation.

## Particular cases

### Nilpotent groups

In the case that we have a nilpotent group or equivalently a nilpotent Lie ring, the Baker-Campbell-Hausdorff formula terminates in finitely many steps, because all terms that involve more than a given number of Lie bracket iterations vanish. Below are the formulas for small values of nilpotency class.

For class $c$, the formula makes sense over any field or ring where all the primes less than or equal to $c$ are invertible.

Note that in the abelian case, the exponential map is a homomorphism, and hence the Baker-Campbell-Hausdorff formula just gives $X + Y$.

For explicit computations, see deducing the Baker-Campbell-Hausdorff formula from associative algebra manipulation.

Nilpotency class Details Formula Primes required to be invertible
1 abelian group and abelian Lie ring -- the group and Lie ring structure are the same $\! X + Y$ no condition
2 Baker-Campbell-Hausdorff formula for nilpotency class two $\! X + Y + \frac{1}{2}[X,Y]$ 2 only
3 Baker-Campbell-Hausdorff formula for nilpotency class three $\! X + Y + \frac{1}{2}[X,Y] + \frac{1}{12}[X,[X,Y]] - \frac{1}{12}[Y,[X,Y]]$ 2 and 3 only
4 Baker-Campbell-Hausdorff formula for nilpotency class four $\! X + Y + \frac{1}{2}[X,Y] + \frac{1}{12}[X,[X,Y]] - \frac{1}{12}[Y,[X,Y]] - \frac{1}{24}[Y,[X,[X,Y]]]$ 2 and 3 only
5 Baker-Campbell-Hausdorff formula for nilpotency class five $\! X + Y + \frac{1}{2}[X,Y] + \frac{1}{12}[X,[X,Y]] - \frac{1}{12}[Y,[X,Y]] - \frac{1}{24}[Y,[X,[X,Y]]]$ $\! - \frac{1}{720}([X,[X,[X,[X,Y]]]] + [Y,[Y,[Y,[Y,X]]]])$ $+\frac{1}{360}([X,[Y,[Y,[Y,X]]]]+[Y,[X,[X,[X,Y]]]])$ $+ \frac{1}{120}([Y,[X,[Y,[X,Y]]]] +[X,[Y,[X,[Y,X]]]])$ 2, 3, and 5 only

### Homogeneous components

The formula in the simply connected case (whose truncations give the nilpotent case) can be separated out into homogeneous components. These homogeneous components are given below. Note that the formula for nilpotency class $c$ is simply obtained by adding up the homogeneous components of degree up to and including $c$.

Degree $d$ Homogeneous component of degree $d$ Symmetric or skew symmetric? See symmetry and skew symmetry of homogeneous components of Baker-Campbell-Hausdorff formula -- even degree terms are skew symmetric, odd degree terms are symmetric lcm of denominators Primes required to be invertible to make sense of this component
1 $X + Y$ symmetric 1 no condition
2 $\frac{1}{2}[X,Y]$ skew symmetric 2 2 only
3 $\frac{1}{12}[X,[X,Y]] - \frac{1}{12}[Y,[X,Y]]$ symmetric 12 2 and 3 only
4 $- \frac{1}{24}[Y,[X,[X,Y]]]$ skew symmetric (proving this for the explicit expression requires use of the Jacobi identity) 24 2 and 3 only
5 $\! - \frac{1}{720}([X,[X,[X,[X,Y]]]] + [Y,[Y,[Y,[Y,X]]]])$ $+\frac{1}{360}([X,[Y,[Y,[Y,X]]]]+[Y,[X,[X,[X,Y]]]])$ $+ \frac{1}{120}([Y,[X,[Y,[X,Y]]]] +[X,[Y,[X,[Y,X]]]])$ symmetric 720 2, 3, and 5 only