The Unapologetic Mathematician

Mathematics for the interested outsider

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:


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


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


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


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