Lie algebras from associative algebras

Given an associative (unital) algebra $A$ over a commutative ring $R$ , we define $A^-$ as the $R$ -module $A$ together with a new multiplication $[,]:A\times A\rightarrow A$ derived from the associative multiplication as follows: $$ [a,b]=ab-ba $$ This operation is commonly called the commutator bracket on $A$ .

Proof. 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 identity.

Given $a,b,c\in A$ , and $l\in R$ then $$ [la+b,c]=(la+b)c-c(la+b)=l(ac-ca)+(bc-cb)=l[a,c]+[b,c] $$ The similar argument 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 functor from the category 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 faithful. More generally, the center of an arbitrary associative algebra $A$ is lost to the Lie algebra structure $A^-$ .

We do observe some relationships between the algebraic structure of $A$ and that of $A^-$ .

Theorem 2   If then .

Proof. We observe that a submodule of $A$ is a submodule of $A^-$ as the two are identitcal as modules. It remains to show $[I,A^-]\leq I$ . So given $a\in I$ and $b\in A$ , then $[a,b]=ab-ba$ and as $ab,ba\in I$ we conclude $[a,b]\in I$ .

Associative envelopes

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

Let $V$ be a vector space and $A$ the endomorphism algebra on $V$ . Then we give the name $\mathfrak{gl}(V)$ to the Lie algebra $A^-$ (noting that $A$ is associative under the composition of functions operation.) Then we can also define a subalgebra $\mathfrak{sl}(V)$ as the set of linear transformations with trace 0.

Now we claim that $\mathfrak{sl}_2(\mathbb{C})$ is not equal to $B^-$ for any associative (unital) algebra $B$ . For it is easy to see $\mathfrak{sl}_2(\mathbb{C})$ has a basis of three elements: $$ e=\begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix},\qquad h=\begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix},\qquad f=\begin{bmatrix} 0 & 0 \\ 1 & 0\end{bmatrix} $$ Therefore $B$ would also be 3-dimensional. We also know that $\mathfrak{sl}_2(\mathbb{C})$ is a simple Lie algebra, that is, it has no proper ideals. Therefore by Theorem 2, $B$ can have no ideals either, so $B$ must be simple. However the finite dimensional simple rings over $\mathbb{C}$ are isomorphic to matrix rings (by the Wedderburn-Artin theorem) $M_n(\mathbb{C})$ and thus cannot have dimension 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 embeddings 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 theorem.

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

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 associator bracket as $[a,b,c]=(ab)c-a(bc)$ then we can write the computation for the Jacobi identity as:

• If $A$ is a commutative non-associative algebra (perhaps a Jordan algebra) then $$ [a,b,c]=(ab)c-a(bc)=(ba)c-(bc)a=c(ba)-(cb)a=[c,b,a $$ 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 algebra, 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, $ab$ , $a,b\in A$ . Recall that in a Lie algebra of characteristic 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 $$ [a,b,c]=(ab)c-a(bc)=-(ba)c+(bc)a=c(ba)-(cb)a=[c,b,a] $$ So once again the associators cancel.