In later posts, I want to critique Joy Christian's claim that his counter-example to Bell'sTheorem is “real” (maybe it is, but I suspect not, and will work that out later), but first we have to understand the math he is using. That math is closely related to Quaternions which Doug Sweetser has been busy familiarizing Science 2.0 readers with. So, I will start in the Quaternion “looking glass” and find a way out to the Clifford Algebra CL(3,0) used by Joy Christian in his counter-example to Bell's Theorem.

In my later post where I critique his counter-example, I won't be following step-by-step through his calculations looking for a mistake because we can safely assume he knows the math better than I do. Instead, I want to be able to clearly picture what his math represents. We should be able to picture what the math is doing because the CL(3,0) that he's using is just a way of algebraically encoding the full geometry of ordinary3-D space. From now on I'll refer to the math he's using as Geometric Algebra (of 3-D Euclidean space = G_{3}). Before understanding the “geometry”, let's find the algebraic relationship between the vectors used in Geometric Algebra and Quaternions.

#### Algebra: Finding Vectors in Quaternions

What is the relationship between the basis elements of Quaternion vectors, **i j k**, and the Cartesian basis elements of 3-D Euclidean vectors, **e**_{1}**e**_{2 }**e**_{3}? They aren't exactly the same because the square of a Quaternion basis element is -1 and the square of a Euclidean basis element is +1.

The simplest relationship I can think of to try is **i**=μ**e**_{1}, **j**=μ**e**_{2}, and **k**=μ**e**_{3 }assuming also that μ commutes with **e**_{1}**e**_{2}**e**_{3}. To make any sense of this, we have to expand our concept of Euclidean vectors to include Quaternion multiplication. In particular, we need to realize that

**i j k ***anti-commute implies ***e**_{1}**e**_{2}**e**_{3}* anti-commute*. So, **e**_{1}**e**_{2}=-**e**_{2}**e**_{1} etc.

We have to satisfy the defining equation of the Quaternions:

**i**^{2}=**j**^{2}=**k**^{2}=**ijk**=-1.

Remembering that the Euclidean basis vectors square to 1 and that I've assumed μ commutes with them, we easily translate the Quaternion equation to Euclidean basis vectors:

μ^{2}=μ^{3}**e**_{1}**e**_{2}**e**_{3}=-1

Substituting μ^{2}=-1 into μ^{3}**e**_{1}**e**_{2}**e**_{3}=-1, we get μ**e**_{1}**e**_{2}**e**_{3}=1. Can we solve this last equation for μ? Yes.

Start with

μ**e**_{1}**e**_{2}**e**_{3}=1

and multiply both sides by **e**_{1}**e**_{2}**e**_{3}, to get

μ**e**_{1}**e**_{2}**e**_{3}**e**_{1}**e**_{2}**e**_{3}=**e**_{1}**e**_{2}**e**_{3}.

To simplify the left hand side

μ**e**_{1}**e**_{2}**e**_{3}**e**_{1}**e**_{2}**e**_{3}

anti-commute**e**_{1}forward to meet itself in the middle, squaring to 1. Then do the samefor the other two.

_{-}μ**e**_{2}**e**_{1}**e**_{3}**e**_{1}**e**_{2}**e**_{3}

μ**e**_{2}**e**_{3}**e**_{1}**e**_{1}**e**_{2}**e**_{3}

μ**e**_{2}**e**_{3}**e**_{2}**e**_{3}

-μ**e**_{3}**e**_{2}**e**_{2}**e**_{3}

-μ**e**_{3}**e**_{3}

-μ

What is -μ again? Oh yeah, it's the left hand side of the equation that had **e**_{1}**e**_{2}**e**_{3} on the right side.

So, μ=-**e**_{1}**e**_{2}**e**_{3}. The reader can easily verify it commutes with all the basis vectors (μ**e**_{1}=**e**_{1}μ) and squares to -1.

Obviously, the starting point that μ^{2}=-1 means that μ is also equal to plus or minus the square-root of minus one. Normally the letter *i* is used to denote square-root of minus one, but that looks too much like the Quaternion basis element **i. **So, we'll use *I *instead, not to be confused with the identity element which is just the number 1. We're not doing matrix algebra! Geometric Algebra is alot easier than that.

Now we can write μ=±*I* (because μ^{2}=-1 and *I*=*i*=√-1). If we then define *I*≡**e**_{1}**e**_{2}**e**_{3},then we conclude μ=-*I*. This *I* is sometimes used in Geometric Algebra in exactly the same way the ordinary imaginary number *i* is used in complex numbers, but it's not the only *I *that plays that part. Geometric Algebra has more than one object that can be used as the imaginary part of a complex number; some commute, like this *I* does, but others anti-commute. In terms of it's acting like *i, I *is referred to as a pseudo-scalar.* I *is also a volume (three lengths multiplied together) but it also keeps a sense of direction by the handedness of the vectors. *I *is *right-handed*, and is referred to as a directed unit volume. Finally, it's the product of three vectors, so *I* is also a trivector.

We did all this to come up with **i**=-*I***e**_{1},** j**=-*I***e**_{2}, **k**=-*I***e**_{3}. Is this satisfactory? Not really.

**The looking glass**

Now **i j k **form a left-handed basis because they flip all three of our right-handed directions. To see that flipping all 3 Cartesian directions is the same as flipping just one of them, imagine your hands are guns with index forward and thumb up on both hands. Now bend your middle fingers straight out so they point to each other. Now rotate the left hand tilting the thumb backward until it's pointing down, keepingyour middle fingers pointing at each other. You should have thumbs, index fingers, and middle fingers all opposite each other. In otherwords, flipping all three is the same as flipping one axis and then rotating by 180 about that axis. So, what happens if we reflect the Quaternions through the **e**_{1}**e**_{3} plane, changing the sign on **e**_{2} alone?

Changing to a left-handed system by going from **e**_{1}**e**_{2}**e**_{3 }to **e**_{1}(-**e**_{2})**e**_{3} remembering to flip the **e**_{2} in *I *too

**i**=-*I***e**_{1}→-**e**_{1}(-**e**_{2})**e**_{3}**e**_{1}=(**e**_{1}**e**_{2}**e**_{3})**e**_{1}=*I***e**_{1}

**j**=-*I***e**_{2}→-**e**_{1}(-**e**_{2})**e**_{3}(-**e**_{2})=(**e**_{1}**e**_{2}**e**_{3})(-**e**_{2})=*I*(-**e**_{2})

**k**=-*I***e**_{3}→-**e**_{1}(-**e**_{2})**e**_{3}**e**_{3}=(**e**_{1}**e**_{2}**e**_{3})**e**_{3}=*I***e**_{3}

So, **i**=*I***e**_{1}**j**=*I*(-**e**_{2}) **k**=*I***e**_{3}

and we see **i j k **line up perfectly with a left-handed coordinate system meaning that the Quaternions were definitely a left-handed system. Not only that, but since they didn't transform the way a vector would (because they went from being opposite the vectors to lining up with them) in going from right-handed to left-handed, Quaternions are pseudo-vectors. (They are also *bivectors* because **i**=*I***e**_{1}=(**e**_{1}**e**_{2}**e**_{3})**e**_{1}=**e**_{2}**e**_{3} where I've flipped the sign twice to get the **e**_{1}'s together. The term bivector is obvious because it's two vectors multiplied together.) I chose to keep the minus sign on the y-direction to emphasize this is a change of coordinate system from left-handed to right-handed. In fact, throughout this post, I've relied heavily on using coordinate basis vectors because that's what the structure of Quaternion's forced on me. But in Geometric Algebra, coordinates are not emphasized because most problems can be solved more easily without even choosing basis vectors. How do we indicate a change from left-handed to right-handed without referring to the coordinates?

Remember when first introducing *I*≡**e**_{1}**e**_{2}**e**_{3} I pointed out it was right-handed? Notice, however, that the product of our left-handed basis is **e**_{1}(-**e**_{2})**e**_{3}=-*I* which can obviously be re-written as **e**_{1}**é**_{2}**e**_{3}=-*I* (where**é**_{2}≡-**e**_{2})*.* So the real way the math shows we're dealing with a left-handed coordinate system is that the product of the basis vectors is of the opposite sign as the right-handed volume element. That means we are free to ignore the basis vectors, and then just use -*I* (the left-handed unit volume element) instead of *I* in our equations to indicate we're using a left-handed system. If we had known that from the beginning, we could have gone straight from the right-handed basis **i**=-*I***e**_{1} to the left **i**=-(-*I*)**e**_{1}=*I***e**_{1} without fuss.

But, wait a minute. If *I* is right-handed, shouldn't I define a left-handed pseudo-scalar and use that in my left-handed system? Then the Quaternions would be pointing opposite the basis vectors again (relative to the left-handed pseudo-scalar) and I would be right back where I started. This problem of which pseudo-scalar to use when switching to a left-handed system is analogous to the difference between representing a change from left-handed to right-handed by using your left-hand when applying the cross-product VS staying with your right-hand for the cross-product while looking in a mirror.

It's worth elaborating on handedness because Joy Christian's counter-example is in fact taking the unknown * handedness* of a random unit volume element μ=±

*I*as something that can be measured in a spin-analyzer. So, I want to be sure we are clear on how reversing handedness works. The Wikipedia article on pseudo-vectors has a couple of useful pictures that I was at first confused about. It was only by clearing up my confusion over these pictures that I was able to get the right result in this post when reversing the handedness for the Quaternions.

In both these pictures we are looking edge-on to a mirror placed vertically down the middle so that we can't actually see the mirror. Then we are imagining one side shows the real object and the other side shows the mirror image (as if we would be able to see the reflection even though we can't see the mirror). Notice that the magnetic field **B** flipped when reflected, in obvious conflict with what a real vector would do, and the angular momentum of the van wheels driving away from us stayed in the same direction when reflected, also in obvious conflict with what a real vector would do. But wait a minute (the confused me said to myself), shouldn't I use the left-hand rule when looking at the mirror image? (For those that don't know, the right-hand rule is very easy to apply to anything rotating: just curl your fingers in the direction of rotation and stick your thumb straight out. Your thumb points in the direction of the pseudo-vector.) If you switch to your left hand for the mirror image, the magnetic field and the angular momentum behave just like ordinary vectors.

So, what's wrong with the idea that I should use my left hand when looking at the reflection? I tried thinking about how if I switch to my left hand then the mirror image of me is still using his right hand but I wanted him to use his left hand because he's in the left-handed land of the mirror, but that just left me a little bewildered. For me, a better way to think about this is to go back to what we really do when reversing the handedness of our coordinate system.

If we have graphed something spinning using the standard right-handed coordinates, then to change that graph to a left-handed one, no need to redraw the spinning thing, just erase the arrow pointing right on the y-axis and draw a new arrow pointing left on the left side of the y-axis. That's it! Now in the left-handed coordinates though, we have to use the left-hand rule to determine the direction of angular momentum for the spinning object. Re-labeling the y-axis flips angular momentum direction even though we didn't do anything to the spinning object.

Now if we want to see what happens when switching to a left-handed system without actually referring to a new coordinate system, we can just use a mirror and keep using the right-handed rules. The mirror provides a way to talk about reversing handedness without worrying about coordinates. Using the right-hand rule for both the original object and it's reflection, we get the same result as when we just relabeled the y-axis. The pseudo-vector flips for no physical reason.

Basically, in both ways of looking at reversing handedness, we have to keep oriented by reference to the original handedness. Right-handed coordinates are used by convention just as number-lines increase off to the right by convention. In order to indicate we're not using that convention, we either have to draw the left-handed coordinate system or put a minus sign on the right-hand system. It's mathematically simpler to just use the minus sign so calculations can be done without changing any rules. The mirror is like the minus sign, allowing us to keep using our right hand in picturing what happens. Therefore, we always use the right-handed *I* and just put a minus sign on it to indicate we're using left-handed coordinates.

#### Geometric Algebra

Quaternions aren't used directly in Geometric Algebra since they emphasize coordinates and are left-handed. Now that we've found Alice in the Quaternion looking glass, time to remove the mirror and just talk about Geometric Algebra.

Notice that when you multiply a trivector and a vector, as for example* I***e**_{1}=(**e**_{1}**e**_{2}**e**_{3})**e**_{1}=**e**_{2}**e**_{3}, you don't get a quadvector or something. Instead, you get another bivector (plus a scalar in general). For the same reason, multiplying two bivectors just returns another bivector plus a scalar. That's why the Quaternions could be used as an algebra all on their own – the system is closed with no need for vectors and the pseudo-scalar trivector.

Linear combinations of bivectors and scalars just gives us more of the same, so if we are to get anything new out of introducing *I *(beyond the very important realization that the Quaternions were left-handed pseudo-vectors), we'll need to add it and the vectors separately to the bivectors and scalars. This new thing we get from adding everything together is a Multivector. Using the letter *M* to stand for Multivector, lower case Greek letters (*α,β*...) for scalars, and bold lower case letters for vectors, a Multivector can be written

*M*= *α*+ **a****+ ***I***b****+ ***βI*

The “grade” of *α* is 0, the grade of * a* is 1, the grade of

*I*

*is 2, and the grade of*

**b***βI*is 3. They are linearly independent, but get mixed up in multiplication.

How many basis elements do we need to describe *M*?

1 is the identity and base element of the scalars

**e**_{1},_{ }**e**_{2}, and_{ }**e**_{3} are the (right-handed) orthonormal vectors used as a basis for vectors

**e**_{1}**e**_{2}=*I***e**_{3},_{ }**e**_{2}**e**_{3}=*I***e**_{1}, and_{ }**e**_{3}**e**_{1}=*I***e**_{2} are the orthonormal bivectors (that square to -1, so sometimes used as *i* but they don't commute) used as a basis for bivectors

*I*≡**e**_{1}**e**_{2}**e**_{3} (and *I*=*i*=√-1) is the unit trivector (squares to -1 and does commute) that is right-handed and known as a “directed unit volume” or “pseudo-scalar” and is the only base element needed for trivectors in 3 dimensional space

So, the complete algebra of 3-D Euclidean space is eight dimensional. (If you count up the basis elements, there are eight of them.)

#### Products:Geometric = Inner(Dot) + Outer(*i*Cross)

What exactly are we doing when we multiply two vectors together? Kind of looks like a cross product, but what's the exact relationship? What about the scalar that comes up in a full product of two vectors or two bivectors that aren't orthogonal like the basis vectors? Time for some proper definitions. Might as well just copy and paste definitions and results starting from page 9 of David Hestene's Reformingthe Mathematical Language of Physics (aka Geometric Algebra 1).

The Geometric ProductI take the standard concept of a real vector space for granted and define the

geometric product

abfor vectorsa,b,cby the following rules:(

ab)c=a(bc), associative (1)

a(b+c)=ab+ac, left distributive (2)(

b+c)a=ba+ca, right distributive (3)

a^{2}= |a|^{2}. contraction (4)where|

a| is a positive scalar called the magnitude ofa, and |a| =0 impliesthat

a= 0.From the geometric product

abwe can define two new products, a symmetricinner product

a·b=½(ab+ba) =b·a, (5)and an antisymmetric outer product

a∧b=½(ab−ba) = −b∧a. (6)Therefore,the geometric product has the canonical decomposition

ab=a·b+a∧b. (7)From the contraction rule (4) it is easy to prove that

a·bis scalar-valued, soit can be identified with the standard Euclidean inner product. The formal

legitimacy and geometric import of adding scalars to bivectors as well as to

vectors is discussed in Section VIA.

The geometric significance of the outer product

a∧bshould also be familiarfrom the standard vector cross product

aхb. The quantitya∧bis called abivector, and it can be interpreted geometrically as an oriented plane segment, as

shown in Fig. 2. It differs from

aхbin being intrinsic to the plane containing

aandb, independent of the dimension of any vector space in which the planelies.

#### Skipping ahead to page 15 in Section VIA, where *i *is the same as our *I* =e_{1}e_{2}e_{3}:

...every bivector

Bin G_{3}is the dual of a vectorbas expressed by=

Bib=bi. (27)Thus, the geometric duality operation is simply expressed as multiplication by

the pseudoscalar i . This enables us to write the outer product defined by (6) in

the form

a∧b=iaхb. (28)Thus, the conventional vector cross product

aхbis implicitly defined as thedual of the outer product. Consequently, the fundamental decomposition of the

geometric product (7) can be put in the form

ab=a·b+iaхb. (29)The elements in any geometric algebra are called multivectors. The special

properties of i enable us to write any multivector M in G3 in the expanded form

M= α +a+ib+iβ, (30)where α and β are scalars and

aandbare vectors. The main value of this formis that it reduces multiplication of multivectors in G

_{3}to multiplication of vectorsgiven by (29). Note that the four terms in (30) are linearly independent, so

scalar, vector, bivector and pseudoscalar parts combine separately under multivector

addition, though they are mixed by multiplication. Thus, the Geometric

Algebra G

_{3}is a linear space of dimension 1 + 3 + 3 + 1 = 2^{3}= 8.The expansion (30) has the formal algebraic structure of a “complexscalar”

α +

iβ added to a “complex vector”a+ib, but any physical interpretationattributed to this structure hinges on the geometric meaning of

i. The mostimportant example is the expression of the electromagnetic field F in terms of

an electric vector field

Eand a magnetic vector fieldB:

F=E+iB.

I encourage the interested reader to continue with the following papers from Overview of Geometric Algebra in Physics:

Reforming the Mathematical Language of Physics (aka Geometric Algebra 1).

Spacetime Physics with Geometric Algebra (aka Geometric Algebra 2)

Gauge Theory Gravity with Geometric Calculus

and a lot more – I recommend

Hunting for Snarks in Quantum Mechanics

I've only scratched the surface of Geometric Algebra here (and since I've only recently started learning it, couldn't do more if I wanted to). Since not many people use Quaternions and they just complicate the introduction of Geometric Algebra, it's understandable that the sources I've found don't bother to go into so much detail about the relationship between the two. But developing a concept of bivectors in the form of **a****∧****b** after just a few lines of abstract definitions, and then saying they are Quaternions, leaves any Alice in the looking glass more than a little bewildered_{. }My motivation was just to understand the exact relationship between Quaternions and bivectors. I probably would've stopped once I got as far as μ=-*I*, but then I started reading Handouts, Part 1 from Physical Applications of Geometric Algebra and found they do briefly comment that the Quaternions are left-handed. Working out the details of them being left-handed has been very beneficial for me. For the first time, I understand pseudo-vectors. Also, quite unexpectedly, the study of this relationship between bivectors and Quaternions turned out not to be a side trip to understanding Joy Christian's counter-example, but actually a fairly direct route to many of the concepts needed. I feel comfortable enough with the mathematical concepts now to move ahead with my planned critique of Joy Christian's counter-example to Bell's Theorem in my next post.

Hi Colin,

Great post! I haven’t read through the whole thing yet, but it looks good.

I would like to pre-empt your “critique” of my work however---if I may. The idea that there might be something “non-real” about my counterexample to Bell’s theorem is preposterous. To begin with, there can be nothing non-real about geometric objects such as bivectors and trivectors. They are classical, geometric, objects. Moreover, geometric algebra of the physical space is a real, linear, vector space, albeit of dimension 8. What can be “non-real” about that?

But more importantly, the word “real” has a very specific meaning within the context of Bell’s theorem. Bell adapted the idea of reality from EPR, who proposed a very precise criterion of reality. Essentially, what EPR had in mind was the idea of “objective reality”---i.e., the concern that things may or may not exist independently of us observing them. In quantum mechanics, for example, things do not always have definite values. But in my counterexample to Bell’s theorem everything always has a definite value, independently of anyone observing it. In particular, all measurement results always have definite, predetermined, values. That is to say, everything in my counterexample is as real as EPR would have ever wanted. Therefore, the idea that there might be something non-real about my counterexample to Bell’s theorem is preposterous.

But thanks for your post otherwise.

Cheers!

Joy Christian