# The Unapologetic Mathematician

## Lie Algebras Revisited

Well it’s been quite a while, but I think I can carve out the time to move forwards again. I was all set to start with Lie algebras today, only to find that I’ve already defined them over a year ago. So let’s pick up with a recap: a Lie algebra is a module — usually a vector space over a field $\mathbb{F}$ — called $L$ and give it a bilinear operation which we write as $[x,y]$. We often require such operations to be associative, but this time we impose the following two conditions: \displaystyle\begin{aligned}{}[x,x]&=0\\ [x,[y,z]]+[y,[z,x]]+[z,[x,y]]&=0\end{aligned}

Now, as long as we’re not working in a field where $1+1=0$ — and usually we’re not — we can use bilinearity to rewrite the first condition: \displaystyle\begin{aligned}0&=[x+y,x+y]\\&=[x,x]+[x,y]+[y,x]+[y,y]\\&=0+[x,y]+[y,x]+0\\&=[x,y]+[y,x]\end{aligned}

so $[y,x]=-[x,y]$. This antisymmetry always holds, but we can only go the other way if the character of $\mathbb{F}$ is not $2$, as stated above.

The second condition is called the “Jacobi identity”, and antisymmetry allows us to rewrite it as: $\displaystyle[x,[y,z]]=[[x,y],z]+[y,[x,z]]$

That is, bilinearity says that we have a linear mapping $x\mapsto[x,\underline{\hphantom{X}}]$ that sends an element $x\in L$ to a linear endomorphism in $\mathrm{End}(L)$. And the Jacobi identity says that this actually lands in the subspace $\mathrm{Der}(L)$ of “derivations” — those which satisfy something like the Leibniz rule for derivatives. To see what I mean, compare to the product rule: $\displaystyle\frac{d}{dt}\left(fg\right)=\frac{df}{dt}g+f\frac{dg}{dt}$

where $f$ takes the place of $y$, $g$ takes the place of $z$, and $\frac{d}{dt}$ takes the place of $x$. And the operations are changed around. But you should see the similarity.

Lie algebras obviously form a category whose morphisms are called Lie algebra homomorphisms. Just as we might expect, such a homomorphism is a linear map $\phi:L\to L'$ that preserves the bracket: $\displaystyle\phi\left([x,y]\right)=\left[\phi(x),\phi(y)\right]$

We can obviously define subalgebras and quotient algebras. Subalgebras are a bit more obvious than quotient algebras, though, being just subspaces that are closed under the bracket. Quotient algebras are more commonly called “homomorphic images” in the literature, and we’ll talk more about them later.

We will take as a general assumption that our Lie algebras are finite-dimensional, though infinite-dimensional ones absolutely exist and are very interesting.

And I’ll finish the recap by reminding you that we can get Lie algebras from associative algebras; any associative algebra $(A,\cdot)$ can be given a bracket defined by $\displaystyle [x,y]=x\cdot y-y\cdot x$

The above link shows that this satisfies the Jacobi identity, or you can take it as an exercise.

Advertisements

August 6, 2012 - Posted by | Algebra, Lie Algebras

## 7 Comments »

1. […] now that we’ve remembered what a Lie algebra is, let’s mention the most important ones: linear Lie algebras. These are ones that arise from […]

Pingback by Linear Lie Algebras « The Unapologetic Mathematician | August 7, 2012 | Reply

2. […] examples of Lie algebras! Today, an important family of linear Lie […]

Pingback by Special Linear Lie Algebras « The Unapologetic Mathematician | August 8, 2012 | Reply

3. Glad to see you back! And this is the exact topic I was hoping you’d cover next! Comment by Joe English | August 10, 2012 | Reply

4. […] first defining (or, rather, recalling the definition of) Lie algebras I mentioned that the bracket makes each element of a Lie algebra act by derivations on itself. We […]

Pingback by Derivations « The Unapologetic Mathematician | August 10, 2012 | Reply

5. […] we said, a homomorphism of Lie algebras is simply a linear mapping between them that preserves the bracket. […]

Pingback by Ideals of Lie Algebras « The Unapologetic Mathematician | August 13, 2012 | Reply

6. Great! I’m going to read all the nonassociative material as a memory-refreshing exercise.

I point one little thing: subalgebras by your definition are not subspaces, but submodules! Comment by Jose Brox | September 1, 2012 | Reply

• Yes, they’re submodules in the more general situation of a Lie algebra over a ring, but we’re just going to be looking at Lie algebras over fields — and usually characteristic zero and almost always algebraically closed, to boot. Comment by John Armstrong | September 2, 2012 | Reply