Proof of class three Lazard correspondence from Lie ring to group

This article describes the Lie ring to group direction of the class three Lazard correspondence, a particular case of the Lazard correspondence.

Statement
Suppose $$L$$ is a $$\{2,3\}$$-powered Lie ring of nilpotency class three, with addition denoted $$+$$ and the Lie bracket denoted $$[ \, \ ]$$. We can give $$L$$ the structure of a group of nilpotency class three as follows:

With these operations, $$L$$ acquires the structure of a group of nilpotency class three.

Related facts

 * Proof of class three Lazard correspondence from group to Lie ring
 * Proof of Baer construction of Lie ring from Baer Lie group
 * Proof of Baer construction of Lie group from Baer Lie ring

Associativity
To prove: $$(xy)z = x(yz)$$

Proof: We will first simplify the left side:

$$(xy)z = \left(x + y + \frac{1}{2}[x,y] + \frac{1}{12}[x,[x,y]] - \frac{1}{12}[y,[x,y]] \right)z$$

This simplifies to:

$$x + y + \frac{1}{2}[x,y] + \frac{1}{12}[x,[x,y]] - \frac{1}{12}[y,[x,y]] + z + \frac{1}{2}\left[\left(x + y + \frac{1}{2}[x,y] + \frac{1}{12}[x,[x,y]] - \frac{1}{12}[y,[x,y]] \right),z\right] + $$

$$\frac{1}{12}\left[\left(x + y + \frac{1}{2}[x,y] + \frac{1}{12}[x,[x,y]] - \frac{1}{12}[y,[x,y]] \right),\left[\left(x + y + \frac{1}{2}[x,y] + \frac{1}{12}[x,[x,y]] - \frac{1}{12}[y,[x,y]] \right),z\right]\right] $$

$$- \frac{1}{12}\left[z,\left[\left(x + y + \frac{1}{2}[x,y] + \frac{1}{12}[x,[x,y]] - \frac{1}{12}[y,[x,y]] \right),z\right]\right]$$

Separate out the parts by degree and obtain:


 * The degree one part is $$x + y + z$$
 * The degree two part is $$\frac{1}{2}[x,y] + \frac{1}{2}[x,z] + \frac{1}{2}[y,z]$$
 * The degree three part is:

$$\frac{1}{12}[x,[x,y]] - \frac{1}{12}[y,[x,y]] + \frac{1}{4}x,y],z] + \frac{1}{12}[x + y,[x + y,z - \frac{1}{12}[z,[x + y,z]]$$
 * All higher degree parts become zero based on our assumption.

We now simplify the right side:

$$x(yz) = x \left(y + z + \frac{1}{2}[y,z] + \frac{1}{12}[y,[y,z]] - \frac{1}{12}[z,[y,z]]\right)$$

This simplifies to:

$$x + y + z + \frac{1}{2}[y,z] + \frac{1}{12}[y,[y,z]] - \frac{1}{12}[z,[y,z]] + \frac{1}{2}\leftx,y + z + \frac{1}{2}[y,z] + \frac{1}{12}[y,[y,z - \frac{1}{12}[z,[y,z]]\right] + $$$$\frac{1}{12}\left[x,x,\left[y + z + \frac{1}{2}[y,z] + \frac{1}{12}[y,[y,z]] - \frac{1}{12}[z,[y,z]]\right]\right]$$$$ - \frac{1}{12}\left[y + z + \frac{1}{2}[y,z] + \frac{1}{12}[y,[y,z]] - \frac{1}{12}[z,[y,z]],\left[x,y + z + \frac{1}{2}[y,z] + \frac{1}{12}[y,[y,z]] - \frac{1}{12}[z,[y,z]]\right]\right]$$

Separate out the parts by degree and obtain:


 * The degree one part is $$x + y + z$$
 * The degree two part is $$\frac{1}{2}[x,y] + \frac{1}{2}[x,z] + \frac{1}{2}[y,z]$$
 * The degree three part is:

$$\frac{1}{12}[y,[y,z]] - \frac{1}{12}[z,[y,z]] + \frac{1}{4}[x,[y,z]] + \frac{1}{12}[x,[x,y + z]] - \frac{1}{12}[y + z,[x,y + z]]$$
 * All higher degree parts become zero based on our assumption.

The equality for the degree one parts is clear. Similarly, the equality for the degree two parts is clear. Let's verify the equality of the degree three parts. First, note that the left side expands to the following, where we group together the expressions with repeated factors and group together the expressions without repeated factors. We get:

$$\frac{1}{12}\left([x,[x,y]] - [y,[x,y]] + [y,[y,z]] - [z,[y,z]]\right) + \frac{1}{12}\left(3x,y],z] + [x,[y,z + [y,[x,z]]\right)$$

The right side becomes:

$$\frac{1}{12}\left([x,[x,y]] - [y,[x,y]] + [y,[y,z]] - [z,[y,z]]\right) + \frac{1}{12}\left(3[x,[y,z]] - [y,[x,z]] - [z,[x,y]]\right)$$

Upon subtracting and using the skew symmetry of the Lie bracket, we obtain:

$$\frac{1}{12}\left(-2[x,[y,z]] - 2[y,[z,x]] - 2[z,[x,y]]\right) = \frac{-1}{6}\left([x,[y,z]] + [y,[z,x]] + [z,[x,y]]\right) = 0$$

where the last step uses the Jacobi identity known to be true for Lie rings. This completes the proof of associativity.