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.)
References:
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.)