# Lie Bracket on Suitable Associative Algebra

This question is somewhat naive. Please see the following proof before reading the question itself.

Let $L$ be a vector space over a field $\Bbb{F}.$ Assume there is an associative multiplication on $L$ so that $L$ is an associative ring, and denote this multiplication simply as $xy$ for $x,y\in L$. Furthermore, suppose this multiplication to be compatible with scaling, that is $\alpha\in\Bbb{F},$ then $\alpha x = x\alpha\in L$ for all $x\in L$ ($L$ should be a left and right $\Bbb{F}$ vector space). Define $[x,y]=xy-yx$ for all $x,y\in L.$ Then $[-,-]:L\times L \longrightarrow L$ is a Lie bracket.

Proof: It is clear that $[-,-]$ is anti-commutative, and bilinear. We only need to show that it satisfies the Jacobi identity. That is

$$[[x,y],z]+[[y,z],x]+[[z,x],y]=0.$$

One readily sees that this is equivalent to

$$(xy)z-(yx)z-z(xy)+z(yx)+(yz)x-(zy)x-x(yz)+x(zy)+(zx)y-(xz)y-y(zx)+y(xz)=0.$$

The previous equation is certainly true when the multiplication defined earlier is associative. Therefore $L$ is a Lie algebra. $\Box$

So here's the question:

1) Is it true that an algebra defined in this way is a Lie algebra?

An example of such an associative algebra is the $M_{n\times n}(\Bbb{F}).$

2) If so, is there any name which separates these from the classical Lie algebras?

There is rich theory surrounding classical Lie algebras, especially their connection to Lie groups. This motivates the next question.

3) Is there any significance in viewing an associative algebra in this way?

• Well, most of the time this associative multiplication you're assuming will not exist. See en.wikipedia.org/wiki/… – Matthew Leingang May 2 '17 at 20:13
• You did build a functor from the associative algebra to the Lie algebra. There is another functor from the Lie algebra to associative algebra, which send $\mathfrak g$ to $U \mathfrak g$, the universal enveloping algebra. There are adjoint functors. – user171326 May 2 '17 at 20:15
• I understand this is not the case for division algebras. But for an arbitrary vector space? You can always define one, although it may not be interesting. – Chickenmancer May 2 '17 at 20:15

Note that 1) is even true in the more general case that $(A,\cdot)$ is a pre-Lie algebra, where the product $x\cdot y$ need not be associative, but just semi-associative. If $(a,b,c)=a\cdot(b\cdot c)-(a\cdot b)\cdot c$ denotes the associator, then $A$ is called a pre-Lie algebra with associated Lie algebra $L(A)$, if $$x\cdot y-y\cdot x=[x,y],$$

$$(x,y,z)=(y,x,z)$$

Note that $[x,y]$ is a Lie bracket; the proof is similar to the associative case. We have \begin{align} [x,[y,z]]+[y,[z,x]]+[z,[x,y]] &= (y,x,z)+(z,y,x)+(x,z,y)-(x,y,z)-(y,z,x)-(z,x,y)\\ & =0. \end{align}

Of course, if we take $A$ as an associative algebra, we have $(x,y,z)=(y,x,z)=0$.

For 2), be carefull with the term "classical Lie algebras", which denotes the series $A_n,B_n,C_n,D_n$ of simple Lie algebras. By Ado's theorem, any finite-dimensional Lie algebra is isomorphic to a matrix Lie algebra with bracket $[A,B]=AB-BA$. Over fields of prime characetristic this has been shown by Iwasawa.

3) Yes, there is a significance in geometry and physics to consider "Lie-admissible algebras".

1) Yes it's true if the multiplication is associative.

2) I don't know if such algebras have a name. Notice that over $\mathbb C$, every finite-dimensional Lie algebra is a "concrete" Lie algebra, i.e is a Lie subalgebra of $\mathfrak gl(V)$ for $V$ a finite dimensional vector space over $\mathbb C$, so in particular it is obtained from your construction, starting with an associative algebra and forgetting multiplication.

3) There is a link between associative algebras and Lie algebra : you have a functor in both way which gives you an adjunction $\text{Hom}_{\rm Lie}(\mathfrak g, Lie(A)) \cong \text{Hom}_{\rm Ass}(U \mathfrak g, A)$. But this is in no way an equivalence of category as usually $U \mathfrak g$ is much bigger that $\mathfrak g$ (see the PWB theorem).

Probably looking at the universal enveloping algebra and quantum groups can answer most of your questions.