Proof of Baer construction of Lie group for Baer Lie ring

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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

Generalizations

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.