# 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 $[,]:A\times A\rightarrow A$ derived from the associative multiplication as follows:

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

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

###### Proposition 1.

$A^{-}$

###### 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

 $[la+b,c]=(la+b)c-c(la+b)=l(ac-ca)+(bc-cb)=l[a,c]+[b,c].$

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

 $\displaystyle[[a,b],c]+[[b,c],a]+[[c,a],b]=(ab-ba)c-c(ab-ba)+(bc-cb)a-a(bc-cb)% +(ca-ac)b-b(ca-ac)\\ \displaystyle=abc-bac-cab+cba+bca-cba-abc+acb+cab-acb-bca+bac=0.$

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

###### Theorem 2.

If $I\unlhd A$ then $I\lhd A^{-}$.

###### 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 $[a,b]\in I$. ∎

## 2 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  .

###### Theorem 3.

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

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.

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

 $\displaystyle[[a,b],c]+[[b,c],a]+[[c,a],b]=(ab-ba)c-c(ab-ba)+(bc-cb)a-a(bc-cb)% +(ca-ac)b-b(ca-ac)\\ \displaystyle=(ab)c-(ba)c-c(ab)+c(ba)+(bc)a-(cb)a-a(bc)+a(cb)+(ca)b-(ac)b-b(ca% )+b(ac)\\ \displaystyle=((ab)c-a(bc))+((bc)a-b(ca))+((ca)b-c(ab))-((ba)c-b(ac))-((cb)a-c% (ba))-((ac)b-a(cb))\\ \displaystyle=[a,b,c]+[b,c,a]+[c,a,b]-[b,a,c]-[c,b,a]-[a,c,b].$

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

 $\sum_{\sigma\in\operatorname{Alt}(\{a,b,c\})}[[a\sigma,b\sigma],c\sigma]=[a,b]% ,c]+[[b,c],a]+[[c,a],b]=\sum_{\sigma\in\operatorname{Sym}(\{a,b,c\})}% \operatorname{sign}(\sigma)[a\sigma,b\sigma,c\sigma].$

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.

Title Lie algebras from other algebras LieAlgebrasFromOtherAlgebras 2013-03-22 16:37:33 2013-03-22 16:37:33 Algeboy (12884) Algeboy (12884) 8 Algeboy (12884) Example msc 17B99 Algebras2 associative envelope