What good is the Commutator product?

Question

In geometric algebra, the commutator product is defined as $A \times B = \frac 1 2 (AB - BA)$.

From linear algebra, I remember that the commutator of matrices is $[A, B] = AB - BA$ and the commutator of linear functions is likewise $[f,g]=fg-gf$. Looking at my notes I see that skew transformations are closed under this operation. But why is that important?

Why do we need this product/ operation? Can I use it to prove some cool theorems or is it useful in performing some algorithms? What is the point of the commutator?

Answer 1

The commutator product is mainly used in the context of multiplication by a bivector. In particular, the commutator product of a bivector with a $k$-vector is always a $k$-vector.

Some useful examples of commutator products:

• Line of intersection between two planes. Two planes are represented by $A$ and $B$ as 2-blades. $i(A \times B)$ is the vector along the line of intersection. This is exactly analogous to how one would use the cross product of the normals in traditional vector algebra.
• Covariant derivatives. Using Gull, Doran, and Lasenby's "gauge theory gravity" as a way of writing Cartan formalism for differential geometry, we write covariant derivatives using a bivector-valued function of a vector, which then acts upon fields using the commutator product. This preserves grade, so while that action isn't "differentiation" in any conventional sense, it's a term that arises from the chain rule and it has the correct grade. (cf. Ricci rotation coefficients for a more conventional treatment.)
• Any "dot product" of a vector and a bivector could be written as a commutator product instead (since the result is a vector). Examples would be like the Lorentz force law, or the whole subject of angular momentum.

Some authors put a lot of emphasis on writing $A \times B = (AB-BA)/2$. To me, this has the most connection to the linear algebra idea of commutator, but it's not very useful in practice. While that formula does tell you that the commutator anticommutes (read that over a couple times), I tend to treat it (and most products in GA in general) as merely shorthands for grade projection. That's useful for the usual case of considering 2-vectors, for the only antisymmetric part of the geometric product there is the part that preserves grade of the other multiplicand.

It may seem very non-general to do that, but seriously, 99% of the time I use the commutator product, it's with a bivector. Then again, I don't do a ton of stuff outside 3+1 spacetime or 3d space, either, so there's little need for me to work with higher dimensional commutator products.

Now, what does this core idea of the commutator product have in common with other notions of commutator? Well, it does capture whether two multivectors $A$ and $B$ do or do not commute under the geometric product, and to what degree. You can always decompose the geometric product of two multivectors into symmetric and antisymmetric pieces. It's useful to note that these symmetric and antisymmetric pieces typically are multivectors of mixed grade, however. For instance, even considering a single bivector $F$, representing the electromagnetic field in Minkowski spacetime, $FF$ is fully symmetric, but it has both scalar and 4-vector terms (two well-known invariants of the EM field under Lorentz transformations).

Answer 2

Muphrid's answer addresses the commutator in geometric algebra, so I will instead address the commutator of matrices.

The point is that this defines a Lie bracket on the space of all matrices, turning this space into a what's called a "Lie Algebra". In particular, this is the Lie algebra of the group of all invertible matrices. The fact that the space of skew-symmetric matrices is also closed under this operation says that this space is also a Lie algebra, in this case the Lie algebra of the group of all orthogonal matrices. Lie algebras (and their global counterparts Lie groups) have many diverse applications.

As to what the commutator of matrices really means, here are two answers, which admittedly require a background in differential geometry. In one approach, the Lie algebra of a Lie group is defined as the tangent space at the identity. In this setting, the commutator is then the "infinitesimal" version of the operation $h \mapsto ghg^{-1}$ known as conjugation. In another approach, the Lie algebra of a Lie group is defined as the space of left-invariant vector fields, and in this setting the commutator is nothing but the commutator of vector fields, which measures the extent to which one vector field "changes" along the flow of another.