Lie algebras from other algebras

1 Lie algebras from associative algebras

Given an associative (unital) algebraMathworldPlanetmathPlanetmathPlanetmath A over a commutative ring R, we define A- as the R-module A together with a new multiplication [,]:A×AA derived from the associative multiplication as follows:


This operationMathworldPlanetmath is commonly called the commutator bracket on A.

Proposition 1.

A- is a Lie algebraMathworldPlanetmath.


We already know A- is a module so we need simply to confirm that the commutator bracket is a bilinear mapping and then demonstrate that it is alternating and satisfies the Jacobi identityMathworldPlanetmath.

Given a,b,cA, and lR then


The similarPlanetmathPlanetmath argumentPlanetmathPlanetmath in the second variable shows that the operation is bilinear.

Next, [x,x]=xx-xx=0 so [,] is alternating. Finally for the Jacobi identity we compute directly.


We notice this produces a functorMathworldPlanetmath from the categoryMathworldPlanetmath of associative algebras to the category of Lie algebras. However, to every commutative algebra A, A- is a trivial Lie algebra, and so this functor is not faithfulPlanetmathPlanetmathPlanetmath. More generally, the center of an arbitrary associative algebra A is lost to the Lie algebra structureMathworldPlanetmath A-.

We do observe some relationships between the algebraic structurePlanetmathPlanetmath of A and that of A-.

Theorem 2.

If IA then IA-.


We observe that a submoduleMathworldPlanetmath of A is a submodule of A- as the two are identitcal as modules. It remains to show [I,A-]I. So given aI and bA, then [a,b]=ab-ba and as ab,baI we conclude [a,b]I. ∎

2 Associative envelopes

Given a Lie algebra 𝔤 it is often desirable to reverse the process described above, that is, to provide an associative algebra A for which 𝔤=A-. In general this is impossible as we will now explain.

Let V be a vector spaceMathworldPlanetmath and A the endomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmath algebra on V. Then we give the name 𝔤𝔩(V) to the Lie algebra A- (noting that A is associative under the composition of functions operation.) Then we can also define a subalgebraMathworldPlanetmathPlanetmath 𝔰𝔩(V) as the set of linear transformations with trace 0.

Now we claim that 𝔰𝔩2() is not equal to B- for any associative (unital) algebra B. For it is easy to see 𝔰𝔩2() has a basis of three elements:


Therefore B would also be 3-dimensional. We also know that 𝔰𝔩2() is a simple Lie algebraMathworldPlanetmath, that is, it has no proper idealsMathworldPlanetmath. Therefore by Theorem 2, B can have no ideals either, so B must be simple. However the finite dimensional simple ringsMathworldPlanetmath over are isomorphic to matrix rings (by the Wedderburn-Artin theorem) Mn() and thus cannot have dimensionPlanetmathPlanetmath 3.

This forces the weaker question as to whether a Lie algebra can be embedded in A- for some associative algebra A. We call such embeddingsPlanetmathPlanetmath associative envelopes of the Lie aglebra. The existence of associative envelopes of arbitrary Lie algebras is answered by a corollary to the Poincare-Birkhoff-Witt theoremMathworldPlanetmath.

Theorem 3.

Every Lie algebra g embeds in the universal enveloping algebra U(g)-, where U(g) is an associative algebra.

Finite dimensional analogues also exist, some of which are simpler to observe. For instance, a Lie aglebra 𝔤 can be represented in 𝔤𝔩(𝔤) by the adjoint representationMathworldPlanetmath. The representation is not faithful unless the center of 𝔤 is trivial. However, for semi-simple Lie algebras, the adjoint representation thus suffices as an associative envelope.

Remark 4.

This result is in contrast to Jordan algebrasMathworldPlanetmathPlanetmath where there are isomorphismMathworldPlanetmathPlanetmathPlanetmath types (for example 3×3 matrices over the octonions) which cannot be embedded in A+ for any associative algebra A. [A+ is the derived algebra of A under the productMathworldPlanetmathPlanetmath a.b=ab+ba.]

2.1 Lie algebra from non-associative algebras

If A is not an associative algebra to begin with then we may still determine the commutator bracket is bilinear and alternating. However, the Jacobi identity is in question. If we define the associatorMathworldPlanetmath bracket as [a,b,c]=(ab)c-a(bc) then we can write the computation for the Jacobi identity as:


We can write this right hand side using permutations on the set {a,b,c} as:


That is, in a non-associative algebra the corresponding Jacobi identity is the possibly non-trivial sum over all permutations of associators. We consider a few non-associative examples.

  • If A is a commutativePlanetmathPlanetmathPlanetmath non-associative algebra (perhaps a Jordan algebra) then


    so the Jacobi identity holds. However, if A is commutative then [a,b]=0 to begin with so the associated Lie algebra product is trivial.

  • If A is an alternative algebraMathworldPlanetmath, so [a,b,c]=-[b,a,c], then again the Jacobi identity holds. So A- is a Lie algebra. The typicall non-associative examples of an alternative algebra are the octonion algebras. These produce a non-trivial Lie algebra.

  • We can also consider beginning with a Lie algebra A and producing A-. To avoid confusing the bracket of A and that of A- we let the multiplication of A be denoted by juxtaposition, ab, a,bA. Recall that in a Lie algebra of characteristicPlanetmathPlanetmath 0 or odd then ab=-ba so that [a,b]=ab-ba=2ab in A-. So we have simply scaled the original product of A by 2. To see the Jacobi identity still holds we note


    So once again the associators cancel.

Title Lie algebras from other algebras
Canonical name LieAlgebrasFromOtherAlgebras
Date of creation 2013-03-22 16:37:33
Last modified on 2013-03-22 16:37:33
Owner Algeboy (12884)
Last modified by Algeboy (12884)
Numerical id 8
Author Algeboy (12884)
Entry type Example
Classification msc 17B99
Related topic Algebras2
Defines associative envelope