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

Statement
This article describes the formula for group commutator in terms of Lie bracket for groups of nilpotency class three. In particular, it applies to the class three 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 three
 * 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):

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

where:

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

and:

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

This simplifies to:

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

Use Fact (1) again on the values $$\exp(X)\exp(Y)$$ and $$\exp(-X)\exp(-Y)$$ to get:

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

Here $$t_3$$ denotes the degree three part of the Baker-Campbell-Hausdorff formula.

We will separately compute three pieces of the right side:

It remains to show that each of the pieces simplifies as indicated above.

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:

This is because the degree one and degree three parts cancel when adding and the degree two part doubles up.

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

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

By the constraint of nilpotency class three, all the brackets involving the $$t_3$$ terms vanish, so the expression becomes:

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

We now distribute and simplify to get:

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

Using the alternating nature of the Lie bracket and simplifying, we get:

Now plug back in the expression:

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

giving the desired answer.

The piece $$t_3(\log(\exp(X)\exp(Y)),\log(\exp(-X)\exp(-Y)))$$
We argue that this piece is zero. Note that it involves Lie products of degree three in terms that of the form $$\pm (X + Y)$$ plus higher degree Lie products. When expanding each product, any new Lie product that involves a nontrivial Lie product as one of its pieces becomes zero. This leaves only Lie products of the form $$[\pm(X + Y),[\pm(X + Y),\pm(X + Y)]]$$, which are zero by the alternating nature of the Lie bracket.