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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?

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    $\begingroup$ The common form of the Heisenberg Uncertainty Principle is a statement that the commutator of the momentum and position operators is not $0$. More generally, it gives explicit bounds on information based on the size of commutator relations (and if the commutator is $0$, then there might not be a limit) $\endgroup$ – davidlowryduda Jun 20 '14 at 19:51
  • $\begingroup$ @mixedmath Bob Dylan is asking about a very particular operation in geometric algebra, contrasting it with other operations that use the word "commutator" that he understands already. I think your remarks are more pertinent toward the linear algebra idea of commutator, which he seems to already comprehend. $\endgroup$ – Muphrid Jun 20 '14 at 19:52
  • $\begingroup$ @Muphrid Actually, because they look so similar, I assumed they were similar ideas. And I know those definitions from linear algebra, but we never mentioned what they meant. Just in some exercise, it'd be, "and this is called the commutator". So an answer either about the linear algebra "commutator" or the geometric algebra "commutator product" is welcome. $\endgroup$ – user137731 Jun 20 '14 at 19:54
  • $\begingroup$ @BobDylan What you call the "commutator product" usually called the "exterior product". $\endgroup$ – user122283 Jun 20 '14 at 19:58
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    $\begingroup$ @SanathDevalapurkar That would be true for vectors, but $A$ and $B$ are not explicitly vectors here, and the product here is not the tensor product. $\endgroup$ – Muphrid Jun 20 '14 at 19:59
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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).

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  • $\begingroup$ Isn't the commutator product a generalization of the exterior product. That is, if we define the exterior product on blades, we can can extend it to multivectors by either using the formula $A\wedge B = \sum_{r,s}<A>_r<B>_s$ or by the commutator product, right? So the commutator product of a bivector and a $k$-vector should be a $k+2$-vector, unless it's zero! $\endgroup$ – Rodrigo Jan 11 '16 at 16:07
  • $\begingroup$ @Rodrigo The wedge product of a bivector and a $k$-vector is grade $k+2$. But unlike the commutator, this wedge is symmetric for a bivector. $\endgroup$ – Muphrid Jan 11 '16 at 16:11
  • $\begingroup$ Hmm... I understood what you said. But then you are implying that the two ways of extending the wedge product on blades to multivectors that I described are not equivalent, right? $\endgroup$ – Rodrigo Jan 11 '16 at 16:15
  • $\begingroup$ I guess not implying... you gave an example of where they aren't! $\endgroup$ – Rodrigo Jan 11 '16 at 16:16
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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.

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  • $\begingroup$ Does that mean that the geometric algebra $\Bbb G^n$ is a Lie algebra? It does seem to satisfy: bilinearity, alternation, and the Jacobi identity (from the wiki page you linked to). $\endgroup$ – user137731 Jun 20 '14 at 20:45
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    $\begingroup$ Why do only bivectors form a Lie algebra? Doesn't the commutator product of general multivectors satisfy those 3 properties, and thus can be considered a Lie bracket? $\endgroup$ – user137731 Jun 20 '14 at 21:28
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    $\begingroup$ In the context of geometric algebra, the Lie group of the Lie algebra of bivectors consists of all elements of the form $e^B$ and their products, where $B$ spans all bivectors. The exponential function is defined by the Taylor series as usual and the product in the Lie group is the usual geometric product. $\endgroup$ – Andrey Sokolov Jun 25 '14 at 6:09
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    $\begingroup$ The reason that Santiago (and Doran, Hestenes, and Sommen in geocalc.clas.asu.edu/pdf/LGasSG.pdf) singles out the space of bivectors seems to be that any subspace of $\Bbb G^n$ except for bivectors will not be closed under the commutator product. $\Bbb G^n$ itself should also be closed under the commutator product, but the fact that Santiago and those authors don't mention it makes me wonder. $\endgroup$ – user137731 Jun 27 '14 at 21:16
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    $\begingroup$ Let $a$, $b$, $c$ be general multivectors. $a \times (b \times c) + c \times (a \times b) + b \times (c \times a) = \frac 1 2 [a (\frac 1 2 (bc - cb)) - (\frac 1 2 (bc - cb)) a] + \frac 1 2 [c (\frac 1 2 (ab - ba)) - (\frac 1 2 (ab - ba)) c] + \frac 1 2 [b (\frac 1 2 (ca - ac)) - (\frac 1 2 (ca - ac)) b] = \frac 1 4 (abc - acb -bca +cba +cab -cba -abc +bac +bca - bac - cab +acb) = 0$ $\endgroup$ – user137731 Jun 28 '14 at 3:49

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