# The Unapologetic Mathematician

## Lie Algebras

One more little side trip before we proceed with the differential geometry: Lie algebras. These are like “regular” associative algebras in that we take a module (often a vector space) and define a bilinear operation on it. This much is covered at the top of the post on algebras.

The difference is that instead of insisting that the operation be associative, we impose different conditions. Also, instead of writing our operation like a multiplication (and using the word “multiplication”), we will write it as $[A,B]$ and call it the “bracket” of $A$ and $B$. Now, our first condition is that the bracket be antisymmetric:

$\displaystyle[A,B]=-[B,A]$

Secondly, and more importantly, we demand that the bracket should satisfy the “Jacobi identity”:

$\displaystyle[A,[B,C]]=[[A,B],C]+[B,[A,C]]$

What this means is that the operation of “bracketing with $A$” acts like a derivation on the Lie algebra; we can apply $[A,\underline{\hphantom{X}}]$ to the bracket $[B,C]$ by first applying it to $B$ and bracketing the result with $C$, then bracketing $B$ with the result of applying the operation to $C$, and adding the two together.

This condition is often stated in the equivalent form

$\displaystyle[A,[B,C]]+[C,[A,B]]+[B,[C,A]]=0$

It’s a nice exercise to show that (assuming antisymmetry) these two equations are indeed equivalent. This form of the Jacobi identity is neat in the way it shows a rotational symmetry among the three algebra elements, but I feel that it misses the deep algebraic point about why the Jacobi identity is so important: it makes for an algebra that acts on itself by derivations of its own structure.

It turns out that we already know of an example of a Lie algebra: the cross product of vectors in $\mathbb{R}^3$. Indeed, take three vectors $u$, $v$, and $w$ and try multiplying them out in all three orders:

\displaystyle\begin{aligned}u\times&(v\times w)\\w\times&(u\times v)\\v\times&(w\times u)\end{aligned}

and add the results together to see that you always get zero, thus satisfying the Jacobi identity.

May 17, 2011 - Posted by | Algebra, Lie Algebras

## 4 Comments »

1. […] Algebras from Associative Algebras There is a great source for generating many Lie algebras: associative algebras. Specifically, if we have an associative algebra we can build a lie algebra […]

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2. […] John Armstrong: Another Existence Proof (of the convergence of the Picard iteration), Gronwall’s Inequality, Lie Algebras […]

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3. […] a Lie group is a smooth manifold we know that the collection of vector fields form a Lie algebra. But this is a big, messy object because smoothness isn’t a very stringent requirement on a […]

Pingback by The Lie Algebra of a Lie Group « The Unapologetic Mathematician | June 8, 2011 | Reply

4. […] 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 […]

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