Proof of Baer construction of Lie group for Baer Lie ring

Statement

This is a part of the Baer correspondence.

Suppose $L$ is a Baer Lie ring (?), i.e., a uniquely 2-divisible class two Lie ring, with addition denoted $+$ and Lie bracket denoted $[,]$. We give $L$ the structure of a 2-powered class two group as follows:

Group operation that we need to define	Definition in terms of the Lie ring operations	Further comments
Group multiplication	$xy:=x+y+{\frac {1}{2}}[x,y]$	Since center of uniquely 2-divisible Lie ring is uniquely 2-divisible, we obtain that the element ${\frac {1}{2}}[x,y]$ is central.
Identity element for multiplication	Same as the zero element $0$ of the Lie ring.
Multiplicative inverse $x^{-1}$.	Same as the additive inverse $-x$.
Group commutator $[x,y]=xyx^{-1}y^{-1}$	Same as the Lie bracket $[x,y]$.

Related facts

Baer correspondence and its other parts

• Baer correspondence
• Proof of Baer construction of Lie ring for Baer Lie group
• Proof of mutual inverse nature of the Baer constructions between group and Lie ring

Generalizations

• Proof of generalized Baer construction of Lie group for class two 2-Lie ring with a suitable cocycle

Proof

Multiplication is associative

To prove: $\!(xy)z=x(yz)$

Proof: We have:

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

We use that the linearity of the Lie bracket, and also use that the Lie ring has nilpotency class two to simplify $[[x,y],z]$ to zero. The expression simplifies to:

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

Similarly, we can show that:

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

Thus, associativity holds.

Agreement of identity and inverses

Since $[x,0]=[0,x]=0$, we have $x\cdot 0=x+0+{\frac {1}{2}}[x,0]=x$, and similarly, $0\cdot x=x$.

Further, since $[x,-x]=0$, we have $x\cdot (-x)=x+(-x)+{\frac {1}{2}}[x,-x]=0$, and similarly, $(-x)\cdot x=0$.

Commutator agrees with Lie bracket

To prove: The multiplicative commutator $[x,y]=(xy)(yx)^{-1}$ equals the Lie bracket $[x,y]$

Proof: We have:

$(xy)(yx)^{-1}=(xy)\cdot (-yx)$

which becomes:

$\!(xy)(yx)^{-1}=xy+(-yx)+{\frac {1}{2}}[xy,-yx]$

which simplifies to:

$\!(xy)(yx)^{-1}=x+y+{\frac {1}{2}}[x,y]-(y+x+{\frac {1}{2}}[y,x])+{\frac {1}{2}}[x+y+{\frac {1}{2}}[x,y],-(y+x+{\frac {1}{2}}[y,x])]$

Simplifying, and using the fact that the Lie ring has nilpotency class two (so applying a Lie bracket twice gives zero), we are left with the Lie bracket $[x,y]$.

Class two

Since the commutator agrees with the Lie bracket, the class two condition on the Lie ring forces the class two condition on the Lie ring.

2-powered

The squaring map for the group operation agrees with the doubling map of the additive group of the Lie ring. Explicitly:

$x^{2}=x+x+{\frac {1}{2}}[x,x]=2x$

We know that the Lie ring is 2-powered, i.e., the doubling map of the additive group of the Lie ring is bijective. Hence, the squaring map of the group is also bijective, i.e., the group is 2-powered.