Lie algebras from other algebras

1 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 $[,]:A\times A\to A$ derived from the associative multiplication as follows:

$$[a,b]=ab-ba.$$

This operation is commonly called the commutator bracket on $A$.

Proposition 1.

$A^{-}$ is a Lie algebra.

Proof.

We already know $A^{-}$ is a module so we need simply to confirm that the commutatorbracket 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 isbilinear.

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

$$\begin{array}{c}[[a,b],c]+[[b,c],a]+[[c,a],b]=(ab-ba)c-c(ab-ba)+(bc-cb)a-a(bc-cb)+(ca-ac)b-b(ca-ac)\hfill \\ \hfill =abc-bac-cab+cba+bca-cba-abc+acb+cab-acb-bca+bac=0.\end{array}$$

∎

We notice this produces a functor from the category of associative algebrasto 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 lostto the Lie algebra structure $A^{-}$.

We do observe some relationships between the algebraic structure of $A$ andthat of $A^{-}$.

Theorem 2.

If $I\mathrm{⊴}A$ then $I\mathrm{⊲}{A}^{\mathrm{-}}$.

Proof.

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

2 Associative envelopes

Given a Lie algebra $𝔤$ it is often desirable to reverse the processdescribed 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 space and $A$ the endomorphism algebra on $V$. Then wegive the name $𝔤𝔩(V)$ to the Lie algebra $A^{-}$ (noting that $A$ isassociative under the composition of functions operation.) Then we can alsodefine a subalgebra $𝔰𝔩(V)$ as the set of linear transformationswith trace 0.

Now we claim that $𝔰{𝔩}_2(ℂ)$ is not equal to $B^{-}$ for any associative (unital) algebra $B$. For it is easy to see $𝔰{𝔩}_2(ℂ)$ has a basis ofthree elements:

Therefore $B$ would also be 3-dimensional. We also know that $𝔰{𝔩}_2(ℂ)$ is a simple Lie algebra, that is, it has noproper ideals. Therefore by Theorem 2, $B$ can have noideals either, so $B$ must be simple. However the finite dimensionalsimple rings over $ℂ$ are isomorphic to matrix rings (by the Wedderburn-Artin theorem)$M_n(ℂ)$ and thus cannot have dimension 3.

This forces the weaker question as to whether a Lie algebra can be embeddedin $A^{-}$ for some associative algebra $A$. We call such embeddings associative envelopes of the Lie aglebra. The existence of associativeenvelopes of arbitrary Lie algebras is answered by a corollaryto the Poincare-Birkhoff-Witt theorem.

Theorem 3.

Every Lie algebra $\mathrm{g}$ embeds in the universal enveloping algebra $\mathrm{U}\mathit{}{\mathrm{(}\mathrm{g}\mathrm{)}}^{\mathrm{-}}$, where $\mathrm{U}\mathit{}\mathrm{(}\mathrm{g}\mathrm{)}$ is an associative algebra.

Finite dimensional analogues also exist, some of which are simpler toobserve. For instance, a Lie aglebra $𝔤$ can be represented in $𝔤𝔩(𝔤)$ by the adjoint representation. The representation isnot faithful unless the center of $𝔤$ is trivial. However,for semi-simple Lie algebras, the adjoint representation thus sufficesas an associative envelope.

Remark 4.

This result is in contrast to Jordan algebras where there are isomorphism types (for example $\mathrm{3}\mathrm{\times }\mathrm{3}$ matrices over the octonions) which cannot be embeddedin $A^{\mathrm{+}}$ for any associative algebra $A$. [$A^{\mathrm{+}}$ is the derived algebraof $A$ under the product $a\mathrm{.}b\mathrm{=}a\mathit{}b\mathrm{+}b\mathit{}a$.]

2.1 Lie algebra from non-associative algebras

If $A$ is not an associative algebra to begin with then we may still determinethe commutator bracket is bilinear and alternating. However, the Jacobiidentity 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 Jacobiidentity as:

$$\begin{array}{c}[[a,b],c]+[[b,c],a]+[[c,a],b]=(ab-ba)c-c(ab-ba)+(bc-cb)a-a(bc-cb)+(ca-ac)b-b(ca-ac)\hfill \\ \hfill =(ab)c-(ba)c-c(ab)+c(ba)+(bc)a-(cb)a-a(bc)+a(cb)+(ca)b-(ac)b-b(ca)+b(ac)\\ \hfill =((ab)c-a(bc))+((bc)a-b(ca))+((ca)b-c(ab))-((ba)c-b(ac))-((cb)a-c(ba))-((ac)b-a(cb))\\ \hfill =[a,b,c]+[b,c,a]+[c,a,b]-[b,a,c]-[c,b,a]-[a,c,b].\end{array}$$

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

$$\sum _{\sigma \in \mathrm{Alt}(\{a,b,c\})}[[a\sigma ,b\sigma ],c\sigma ]=[a,b],c]+[[b,c],a]+[[c,a],b]=\sum {}_{\sigma \in \mathrm{Sym}(\{a,b,c\})}\mathrm{sign}(\sigma )[a\sigma ,b\sigma ,c\sigma ].$$

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

• 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 commutativethen $[a,b]=0$ to begin with so the associated Lie algebra product istrivial.

• If $A$ is an alternative algebra, so $[a,b,c]=-[b,a,c]$, then againthe 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 themultiplication of $A$ be denoted by juxtaposition, $ab$, $a,b\in A$. Recallthat 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 productof $A$ 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.