# Exterior square of Lazard Lie group and Lazard Lie ring are in Lazard correspondence

## Statement

Suppose is a Lazard Lie group and is its Lazard Lie ring. Then, the exterior square is a Lazard Lie group, the exterior square is a Lazard Lie ring and in fact, they are in Lazard correspondence with each other.

Further, the commutator map homomorphism is in Lazard correspondence with the Lie bracket homomorphism .

## Proof

The first thing we note before proceeding to the proof is that the argument is essentially a *formal* one in the following sense. The Lazard correspondence gives the formulas for group operations in terms of Lie ring operations and vice versa. We have formulas for the operations in the exterior square of both the group and the Lie ring.

Specifically, our formulas allow for the translations:

The vertical arrows involve using the Baker-Campbell-Hausdorff formula for the Lazard correspondence. The horizontal arrows involve using the description of the exterior square in terms of the original Lie ring or group.

The goal is to show that the "diagram commutes" in the sense that both paths from to give the same ultimate formulas.

We show this in two steps:

- This is true for the Malcev correspondence over , i.e., it is true if we ignore the issue of problematic denominators.
- Since the ultimate formulas are independent of whether we are over , it is always true.