I've been reading some papers related to Bell's Theorem which involve Clifford Algebra. I am investigating it for an undergrad project but none of my professors seem to know anything about Clifford Algebras.

In this one paper, I found the wedge product defined by the "well-known identity (Hodge duality)": $$ a\wedge b = I \cdot (a \times b)$$

I've scoured the web for this Hodge duality thing and can only find obscure things about the Hodge star... Which is way out of my league at this point.

For context, $a$ and $b$ are arbitrary directions of Stern-Gerlach apparatuses and the basis vectors are e(x,y,z).

I tried calculating out the wedge of $a$ and $b$ by myself and have gotten this far:

$$a \wedge b = a \cdot b + (a_{1}b_{2}-a_{2}b_{1})e_{x}e_{y}+(a_{1}b_{3}-a_{3}b_{1})e_{x}e_{z}+(a_{2}b_{3}-a_{3}b_{2})e_{y}e_{z}$$

I did out the cross product, which looks similar...

$$a \times b = (a_{1}b_{2}-a_{2}b_{1})e_{z}-(a_{1}b_{3}-a_{3}b_{1})e_{y}+(a_{2}b_{3}-a_{3}b_{2})e_{x}$$

... but isn't exactly what I need.

By inspection, I would think that I would have to have that dot product somehow disappear into the I and one of the vectors in the bivector somehow also disappear into the I in order to get that cross product.

From the paper, I is defined as a trivector:

$$I = e_{x} \wedge e_{y} \wedge e_{z}$$

Am I making a mistake in using $$e_{x} \wedge e_{y} = e_{x}e_{y}$$

to simplify? Or is it something else?

And in case you're wondering, the paper which talked about this "well-known identity (Hodge duality)" did not cite it... Presumably because it is actually well-known. But not to me, an undergrad, or my professors, who don't study this field.

(I did cross post with SE.Math here.)


  1. Main article.

  2. The critic who references the Hodge duality. (There are MANY more critics; I'm just trying to understand more about Bell's theorem and why JC would attempt this.)

  • 1
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Commented Dec 10, 2017 at 20:03

2 Answers 2


First, be aware of my small warning at the end of my answer.

The wedge and cross are the same modulo Hodge dual in 3 dimensions. Yes, this stuff is well know, but let me suggest that you stick to 3 dimensions conceptually for the time being. A directed area in three dimensions, i.e. a section of a plane, is defined by the unit normal to it, and contrariwise. This is a way to visualize the Hodge dual in three dimensions. So think of the wedge as defining the plane spanned by the two vectors that are wedged together; this plane can also be defined by the vector normal to it, which is the cross product; the magnitude of the cross product is the area defined by the parallelogram defined by the two wedged vectors. So they are essentially the same information, and your procedure is essentially correct, but you make a small mistake with ordering of the components. The way to order them is:

$$\hat{e}_x\wedge \hat{e}_y \quad\leftrightarrow\quad \hat{e}_z$$ $$\hat{e}_z\wedge \hat{e}_x \quad\leftrightarrow\quad \hat{e}_y$$ $$\hat{e}_y\wedge \hat{e}_z \quad\leftrightarrow\quad \hat{e}_x$$

where I write $\leftrightarrow$ for the isomorphism that invertibly maps directed area basis members to vector basis members. This isomorphism represented by $\leftrightarrow$ is the Hodge dual in 3 dimensions. Notice that I get each vector on the right as the cross product between the two wedged arguments on the left; alternatively, each formula follows from its forerunner by cyclically permuting the members of the forerunner (where the formula before the first is the last formula).

Notice that this only works in three dimensions. In four dimensions, a plane cannot be defined by a unit normal. Both $\hat{e}_z$ and $\hat{e}_t$ are linearly independent normals to the plane spanned by $\hat{e}_x$ and $\hat{e}_y$. So the cross product is kind of an "accident" in three dimensions, whereby we can visualize the more general notion of Hodge dual by the simpler notion of "normal to a plane".

Once you have grasped this meaning behind the cross product in 3D, it will probably stand you in better stead to read about exterior and Clifford algebras.

Clifford algebras are actually a little more general than the exterior algebra, which concerns itself with the wedge. (Exterior algebras can be thought of as the special case where the quadratic form returns the zero vector along the diagonal). Therefore, if you are trying to understand the latter, then it might help you to read about exterior algebra first and then move on to Clifford algebras. But some people find it easier to plunge into the more general, but more abstract, Clifford situation from the outset; it looks like your texts are taking this approach, which is the one favored by the pedagogical approach to Clifford algebras known as Geometric Algebra. Understand that there are several approaches; try both of them. You might want to read this thread for various opinions about the relative merits of the approaches, but beware of buying into dogma as everyone learns differently.


The wedge product is also called the exterior product.

Your first formula establishes an isomorphism between the exterior product $\wedge$, it only works in 3d. Geometrically, the cross product is the signed area; whereas, the exterior product is the signed parallelopid; thus we can see the exterior product as a generalisation of the cross product to arbitrary dimension.

The exterior product of k vectors gives the k-vectors; this space is also closed under sums. So not only is $u \wedge v$ is a 2-vector for $u$ & $v$ as vectors, but so is $u\wedge v + x\wedge v$, for $x$ and $y$ as vectors. Further, the exterior product of a $k$-vector and $l$-vector is a $k+l$-vector; Hodge duality, via the hodge star, gives an isomorphism between $k$-vectors & $(n-k)$-vectors.

The Clifford product generalises the exterior product via an auxiliary quadratic form; when this quadratic form is just the zero quadratic form then the Clifford product is exactly the exterior product.


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