Formula for group commutator in terms of Lie bracket for nilpotency class four

Statement
This article describes the formula for group commutator in terms of Lie bracket for groups of nilpotency class four. In particular, it applies to the class four Lazard correspondence.

There are two notions of group commutator, depending on whether we use the left convention or the right convention. Formulas for both cases are presented.

Facts used

 * 1) uses::Baker-Campbell-Hausdorff formula for nilpotency class four
 * 2) uses::Formula for difference of logarithms of group products in terms of Lie bracket

Proof for commutator with left convention
We already have, by Fact (1):

$$\log(\exp(X)\exp(Y)) = X + Y + \frac{1}{2}[X,Y] + t_3(X,Y) + t_4(X,Y)$$

where:

$$t_3(X,Y) = \frac{1}{12}[X,[X,Y]] - \frac{1}{12}[Y,[X,Y]]$$

and:

$$t_4(X,Y) = \frac{-1}{24}[Y,[X,[X,Y]]]$$

Similarly, we have:

$$\log(\exp(-X)\exp(-Y)) = -(X + Y) + \frac{1}{2}[-X,-Y] + t_3(-X,-Y) + t_4(-X,-Y)$$

Using the alternating nature of the Lie bracket, this simplifies to:

$$\log(\exp(-X)\exp(-Y)) = -(X + Y) + \frac{1}{2}[X,Y] - t_3(X,Y) + t_4(X,Y)$$

We will separately compute four pieces of the right side:

The piece $$\log(\exp(X)\exp(Y)) + \log(\exp(-X)\exp(-Y))$$
The piece $$\log(\exp(X)\exp(Y)) + \log(\exp(-X)\exp(-Y))$$ can be simplified either directly or using Fact (2) to get:

$$\log(\exp(X)\exp(Y)) + \log(\exp(-X)\exp(-Y)) = [X,Y] + 2t_4(X,Y)$$

Plugging in the expression for $$t_4$$ gives:

The piece $$\frac{1}{2}[\log(\exp(X)\exp(Y)) ,\log(\exp(-X)\exp(-Y)]$$
We have:

$$\frac{1}{2}[\log(\exp(X)\exp(Y)),\log(\exp(-X)\exp(-Y))] = \frac{1}{2}[X + Y + \frac{1}{2}[X,Y] + t_3(X,Y) + t_4(X,Y), -(X + Y) + \frac{1}{2}[X,Y] - t_3(X,Y) + t_4(X,Y)]$$

If we distribute and expand, the only terms that survive the expansion are those where the total degree is four or lower. Of these, the term of the form $$\frac{1}{2}[X + Y, -(X+ Y)]$$ is zero because of the alternating condition. We thus get the following terms:

$$\frac{1}{2}[X + Y, \frac{1}{2}[X,Y]] + \frac{1}{2}[\frac{1}{2}[X,Y],-(X + Y)] + \frac{1}{2}[X + Y, -t_3(X,Y)] + \frac{1}{2}[t_3(X,Y),-(X + Y)]$$

The last two terms cancel by the alternating condition, and the first two terms are equal, so double up, and we get:

The piece $$t_3(\log(\exp(X)\exp(Y)),\log(\exp(-X)\exp(-Y)))$$
Again by degree considerations, when we expand out this term, we only need to care about the products of degree four or less. Thus, we can ignore $$t_3$$ and $$t_4$$ on the inside completely, and we thus get:

$$t_3(\log(\exp(X)\exp(Y)),\log(\exp(-X)\exp(-Y))) = t_3(X + Y + \frac{1}{2}[X,Y], -(X + Y) + \frac{1}{2}[X,Y])$$

This further simplifies to:

$$t_3(\log(\exp(X)\exp(Y)),\log(\exp(-X)\exp(-Y))) = \frac{1}{12}[(X + Y + \frac{1}{2}[X,Y]) - (-(X + Y) + \frac{1}{2}[X,Y]),[X + Y + \frac{1}{2}[X,Y],-(X + Y) + \frac{1}{2}[X,Y]]]$$

This becomes:

$$t_3(\log(\exp(X)\exp(Y)),\log(\exp(-X)\exp(-Y))) = \frac{1}{6}[X + Y,[X + Y + \frac{1}{2}[X,Y],-(X + Y) + \frac{1}{2}[X,Y]]]$$

The right Lie bracket was computed in the preceding step (off by a factor of 1/2) to be $$[X + Y,[X,Y]]$$, so we get:

The piece $$t_4(\log(\exp(X)\exp(Y)) ,\log(\exp(-X)\exp(-Y))$$
By degree considerations, this piece becomes zero. Note that the only terms that are of degree four are of the form where all inputs are $$X + Y$$, and these are zero by the alternating condition.

Simplification of the total
$$[X,Y] + \frac{1}{2}[X + Y,[X,Y]] + \frac{1}{6}[X + Y,[X + Y,[X,Y]]] - \frac{1}{12}[Y,[X,[X,Y]]]$$

The degree four component of this can be simplified. Let's concentrate on this part. It originally reads:

$$\frac{1}{6}[X + Y,[X + Y,[X,Y]]] - \frac{1}{12}[Y,[X,[X,Y]]]$$

By expanding the first part linearly, we get:

$$\frac{1}{6}[X,[X ,[X,Y]]] + \frac{1}{6}[X,[Y,[X,Y]]] + \frac{1}{6}[Y,[X ,[X,Y]]] + \frac{1}{6}[Y,[Y ,[X,Y]]]- \frac{1}{12}[Y,[X,[X,Y]]]$$

We now use the fact that $$[Y,[X,[X,Y]]] = [X,[Y,[X,Y]]]$$ proof: From the Jacobi identity applied to $$X,Y,[X,Y]$$, we obtain that:

$$[X,[Y,[X,Y]]] + [Y,X,Y],X + X,Y],[X,Y =0$$

Simplifying gives:

$$[Y,X,Y],X = -[X,Y,[X,Y]]$$

Using the alternating condition on the left side, we get:

$$[Y,[X,[X,Y]]] = [X,Y,[X,Y]]$$

Plug this back into the expression and get:

$$\frac{1}{6}[X,[X ,[X,Y]]] + \frac{1}{6}[Y,[X,[X,Y]]] + \frac{1}{6}[Y,[X ,[X,Y]]] + \frac{1}{6}[Y,[Y ,[X,Y]]]- \frac{1}{12}[Y,[X,[X,Y]]]$$

Combine terms and rearrange to get:

$$\frac{1}{6}[X,[X ,[X,Y]]] + \frac{1}{6}[Y,[Y ,[X,Y]]] + \left(\frac{1}{6} + \frac{1}{6} - \frac{1}{12}\right)[Y,[X,[X,Y]]]$$

Simplify, and obtain:

$$\frac{1}{6}[X,[X ,[X,Y]]] + \frac{1}{6}[Y,[Y ,[X,Y]]] + \frac{1}{4}[Y,[X,[X,Y]]]$$

Plugging back into the original expression, we obtain:

$$\! [X,Y] + \frac{1}{2}[X + Y,[X,Y]] + \frac{1}{6}[X,[X,[X,Y]]] + \frac{1}{6}[Y,[Y,[X,Y]]] + \frac{1}{4}[Y,[X,[X,Y]]]$$