Proof of Baer construction of Lie group for Baer Lie ring

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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



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.


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.