# Doran/Lasenby. Commutator and symmetric products?

## Main Question or Discussion Point

Geometric Algebra for Physicists, in equation (4.56) introduces the following notation

$$A * B = \langle AB \rangle$$

as well as (4.57) the commutator product:

$$A \times B = \frac{1}{2}\left(AB - BA\right)$$

I can see the value defining the commutator product since this selects all the odd grade terms of a product (this can be used to express the othogonal component of a vector with respect to another (rejection), or the othogonal component of a plane with respect to an intersecting plane (or any plane in $$\mathbb{R}^3$$)).

But what's the point of introducing a second notation for grade zero selection? If one is going to introduce an operator to complement the commutator product, then something like the following would make more sense:

$$A * B = \frac{1}{2}\left(AB + BA\right)$$

ie: select all the even grade components of the product. This is only equivalent to $$\langle AB \rangle$$ for specific cases like the symmetric vector product (dot product), for intersecting bivectors, for trivectors with bivector intersection, ...

Does anybody else think that this is probably a typo in the text? I haven't been reading this book linearly (skipping back and forth between it and Hestenes NFCM) so I haven't seen where or if either this commutator product or this * product are employed (seeing the usage would probably confirm if this is a typo).

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I can see the value defining the commutator product since this selects all the odd grade terms of a product
I don't think this is true.
Suppose A and B are bivectors. Then the product AB has only even-grade terms but the
commutator is not zero.

$$A * B = \frac{1}{2}\left(AB + BA\right)$$

ie: select all the even grade components of the product.
How does this select the even-grade components? If A and B have the same grade (as is assumed in the definition of the scalar product), then there are no odd-grade components in the product, so a selection is superfluous. In the case that A and B are of different grade, then the above expression still does not select even grades: let A be a vector a; then, the rhs = $$a\wedge B$$, which could be odd.

I don't think that A*B is just another notation for <AB>. A*B can be viewed as a mapping from a pair of k-blades to the real numbers. It can be used to express the norm (squared) of a blade as the scalar product of the blade with its reverse. Or it may be used to compute the angle between subspaces. In summary, the equations in the text-book are not typos in my opinion.

Oops. I see that what I was thinking of when I wrote even and odd isn't well described by those words.

Let me clarify what I meant by example, considering the product of two bivectors:

$$AB = {\langle AB \rangle} + {\langle AB \rangle}_2 + {\langle AB \rangle}_4$$

By definition of dot and wedge product as the lowest and highest grade terms of a product respectively, this is:

$$AB = A \cdot B + {\langle AB \rangle}_2 + A \wedge B$$

One can show that:

$${\langle AB \rangle} = {\langle BA \rangle}$$
$${\langle AB \rangle}_4 = {\langle BA \rangle}_4$$

and also can show that:
$${\langle AB \rangle}_2 = -{\langle BA \rangle}_2$$

So, one can write the various grade terms of a bivector product in terms of symmetric and antisymetric sums like so:

$$A \cdot B = \langle \frac{1}{2}\left(AB + BA\right) \rangle$$

$${\langle AB \rangle}_2 = \frac{1}{2}\left(AB - BA\right)$$

$$A \wedge B = {\langle \frac{1}{2}\left(AB + BA\right) \rangle}_4$$

It's only with a restriction that one of the $${\langle \rangle}_{0,4}$$ terms is zero that one can write the other 0 or 4 grade term strictly in using the symmetric sum of products (a good example of that is for intersecting planes, or any planes in $$\mathbb{R}^3$$).

So, when Geometric Algebra for Physicists, in equation (4.57) introduces the commutator product:

$$A \times B = \frac{1}{2}\left(AB - BA\right)$$

One can see the utility of this in describing the geometry of planes. As with an orthonormal decomposition of vectors into projective and rejective components:

$$a = a \cdot b \frac{1}{b} + a \wedge b \frac{1}{b}$$

For bivectors that intersect ($$A \wedge B = 0$$), one has
$$A = A \cdot B \frac{1}{B} + A \times B \frac{1}{B}$$

Notationally, I think this is simpler looking than mixed use of the grade operator and the dot product:
$$A = A \cdot B \frac{1}{B} + {\langle A B \rangle}_2 \frac{1}{B}$$

(of course one could use the grade operator exclusively for vectors too, but given how common the dot product is, that just obfuscates things).

Now, when Geometric Algebra for Physicists, in equation (4.56) introduces the following notation:

$$A * B = \langle AB \rangle$$

I don't see the point of introducing a second notation for grade zero selection. We already have a special notation for that:

$$A \cdot B = \langle AB \rangle.$$

If one is going to introduce an operator to complement the commutator product, then something like the following would make more sense to me:

$$A * B = \frac{1}{2}\left(AB + BA\right)$$

ie: a symmetric sum of products.

For bivectors this selects all the 0,4 grade terms, whereas the commutator selects only the grade two term. For trivector products the symmetric product above selects
the grade 0,4 terms, and the commutator would select the grade 2,6 terms (this general alternation pattern is what I initially described very badly as even and odd).

So my conclusion is that this is a typo in the text, despite not being in the errata.

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I don't think that A*B is just another notation for <AB>. A*B can be viewed as a mapping from a pair of k-blades to the real numbers. It can be used to express the norm (squared) of a blade as the scalar product of the blade with its reverse.In summary, the equations in the text-book are not typos in my opinion.
You don't need this notation for the angle between subspaces. For example the angle between two intersecting bivectors is

$$\cos\theta = - \frac{A \cdot B}{\lvert A \rvert \lvert B \rvert}$$

or equivalently:

$$\cos\theta = - \frac{\langle A B \rangle}{\lvert A \rvert \lvert B \rvert}$$

either of the existing $$\langle \rangle$$, or dot notations can be used for the purposes you describe.

I don't see the point of introducing a second notation for grade zero selection. We already have a special notation for that:

$$A \cdot B = \langle AB \rangle.$$

So my conclusion is that this is a typo in the text, despite not being in the errata.
about the typographical error is still incorrect. What do Doran
et al. actually say? It is "The product in eqn 4.54, <AB>, is
sometimes given the symbol *, so we write A*B=<AB>". So they do
not introduce the notation A*B---they merely point out
that others sometimes use this notation and that it is
the same as <AB>. So there is no typo.

If I have understood you correctly, you appear to be saying
that the notation A*B is not needed for the scalar product
and that it should be used for something else. This would be
very unwise because you would thereby introduce new confusion
into a difficult enough topic.

If one is going to introduce an operator to complement the commutator product, then something like the following would make more sense to me:

$$A * B = \frac{1}{2}\left(AB + BA\right)$$
Perhaps it would be nice to have a complement operator but that
is your suggestion and is not the intention of Doran et al.
I would then plead for a new notation for a new concept.
If (AB-BA)/2 is the commutator, then (AB+BA)/2 would be the
anti-commutator. This usage would be consistent with other
applications, e.g. in thermodynamic Greens functions, where
the notations $$\left[ A,B\right]_-$$ and
$$\left[ A,B\right]_+$$ are used for the commutator
and anti-commutator respectively.

Well, that's a simple enough explaination. I guess I'm guilty of being blind. I couldn't see why this was introduced and imagined my own reason for it. This also accounts for the fact that this redundant notation didn't appear to be used anywhere else in the book!

Although logically something to complement the commutator product would fit, I'd agree that things are complex enough without introducing one (I'd be inclined to omit most use of the commutator notation too unless there was really good reason because of similarity with the cross product operator).

Once again, thanks for the back and forth on this subject!

ps. Reading your last note made me want to try the calculation for angle between line and intersecting plane in $$\mathbb{R}^N$$ which I hadn't previously tried. I'm continually amazed how simple these sorts calculations become with GA, all the sorts of results we had for 3d vectors in high school we get for vectors and everything else.

It's kind of neat to see the symmetry of these results:

1) Line and line (dir'n vectors u, v):

$$\cos\theta = \hat u \cdot \hat v$$

(standard high school result).

2) line (dir'n u) and plane (dir'n bivector A):

$$\cos\theta = \lvert \hat u \cdot \hat A \rvert$$

here, $$\theta \in [0,\pi/2]$$

3)
two planes, with dir'n bivectors A and B (intersecting or $$\mathbb{R}^3$$).

$$\cos\theta = - \hat A \cdot \hat B$$

($$\theta \in [0,\pi]$$)