Deducing the Baker-Campbell-Hausdorff formula from associative algebra manipulation

Infinite version
The Baker-Campbell-Hausdorff formula in the simply connected case (and hence also the nilpotent case, at least in large characteristic) can be computed by the following procedure.


 * Start with the power series for the exponential function $$\exp(X)$$ and $$\exp(Y)$$ viewed as elements of the formal power series ring $$\mathbb{Q}\langle X,Y \rangle$$ with $$X,Y$$ non-commuting variables:

$$\! \exp(X) = \sum_{k=0}^\infty \frac{X^k}{k!} = 1 + X + \frac{X^2}{2!} + \frac{X^3}{3!} + \dots$$

$$\! \exp(Y) = \sum_{l=0}^\infty \frac{Y^l}{l!} = 1 + Y + \frac{Y^2}{2!} + \frac{Y^3}{3!} + \dots$$

Now, multiply both of these:

$$\! \exp(X)\exp(Y) = \sum_{k=0}^\infty \sum_{l=0}^\infty \frac{X^kY^l}{k!l!}$$

Subtract 1 and define $$W = \exp(X)\exp(Y) - 1 = (\exp(X) - 1)(\exp(Y) - 1) + (\exp(X) - 1) + (\exp(Y) - 1)$$. We have:

$$\! W = \sum_{k,l \ge 0, 0 < k + l} \frac{X^kY^l}{k!l!} = \sum_{n=1}^\infty \frac{1}{n!} \sum_{k=0}^n \binom{n}{k}X^kY^{n-k}$$

The Baker-Campbell-Hausdorff formula is obtained by applying the power series for $$\log(1 + W)$$ and plugging the above expression for $$W$$ into that. Ultimately, we have to convert terms that are obtained in terms of $$X,Y$$ into commutators and iterated commutators.

Truncated version for nilpotent groups
In the case of a group of nilpotency class at most $$c$$, we can simplify the computations above as follows: we can work within a ring where all products of length $$c + 1$$ or more are zero. Thus, in here, we can assume that:

$$\! \exp(X) = \sum_{k=0}^{c-1} \frac{X^k}{k!} = 1 + X + \frac{X^2}{2!} + \frac{X^3}{3!} + \dots + \frac{X^{c-1}}{(c - 1)!}$$

$$\! \exp(Y) = \sum_{l=0}^{c-1} \frac{Y^l}{l!} = 1 + Y + \frac{Y^2}{2!} + \frac{Y^3}{3!} + \dots + \frac{Y^{c-1}}{(c - 1)!}$$

Now, multiply both of these and remove all terms where the total exponent is $$c + 1$$ or more:

$$\! \exp(X)\exp(Y) = \sum_{k,l \ge 0, k + l \le c} \frac{X^kY^l}{k!l!}$$

Subtract 1 and define $$W = \exp(X)\exp(Y) - 1$$. We have:

$$\! W = \sum_{k,l \ge 0, 0 < k + l \le c} \frac{X^kY^l}{k!l!} = \sum_{n=1}^c \frac{1}{n!} \sum_{k=0}^n \binom{n}{k} X^kY^{n-k}$$

Now apply the power series for $$\log(1 + W)$$. Note that this power series again can be terminated after $$c$$ terms.

Particular cases
We consider the cases for various nilpotent groups for fixed bounds on nilpotency class.

Case of abelian group
The short version is as follows. Using the quick formula for $$W$$, we obtain:

$$W = X + Y$$

Since we are in the abelian case, $$W^2 = 0$$, so $$\log(1 + W) = W$$, so we get that the Baker-Campbell-Hausdorff formula gives:

$$\! X + Y$$

A more detailed explanation is below:

In this case, we work with non-commuting variables $$X,Y$$ such that $$X^2 = XY = YX = Y^2 = 0$$. Thus:

$$\exp(X) = 1 + X$$

$$\exp(Y) = 1 + Y$$

We thus get:

$$\exp(X)\exp(Y) = (1 + X)(1 + Y) = 1 + X + Y + XY = 1 + X + Y$$

(we drop the $$XY$$ term because it equals zero).

Thus:

$$W = X + Y $$

We thus get:

$$\log(1 + W) = W - \frac{W^2}{2} + \frac{W^3}{3} - \dots = W = X + Y$$

Thus, the Baker-Campbell-Haudorff formula gives $$X + Y$$.

Higher order terms become zero because of all terms involving products of length two or more are zero.