Hodge duality in relating wedge and cross products? I've been reading some quantum mechanics papers 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.Physics here.)
 A: You're mostly on the right track, but it looks like you may be confused about the various products that are involved.  As Jesse Madnick points out, there are already a lot of assumptions built in to what you're looking at.  Clifford algebras are pretty general, but you're dealing with the specific flavor of them called Geometric Algebra (GA), which assumes things like the fact that your coefficients are real numbers (as opposed to complex numbers or objects from more general fields), and the presence of a dot product and an orthonormal basis.  Since those are things you were already assuming about this situation, you can basically ignore them until you want to really get into the dirt with Clifford algebras.  Alternatively, I think GA is very rewarding and surprisingly easy.  There's a great book called "Geometric Algebra for Physicists".
Now, the three important types of product in GA are:
\begin{align}
  a\, b & & &\text{Clifford (or geometric) product} \\
  a \cdot b & & &\text{the usual dot product} \\
  a \wedge b & & &\text{the wedge product}
\end{align}
For vectors at least, the dot product really is just what you're used to.  And I would say that for your situation, you never need to use a dot product with anything other than vectors.  In particular, I would ignore the equation that says
\begin{equation}
  a \wedge b = I \cdot (a \times b),
\end{equation}
and use the equation from the appendix that says
\begin{equation}
  a \wedge b = I(a \times b).
\end{equation}
Note the missing dot!  The first equation happens to be correct using more advanced notational standards in GA, but it's really just a distraction.
As for the wedge product, the important point is that it is antisymmetric: $a \wedge b = - b \wedge a$.  Geometrically, it represents the plane spanned by the vectors $a$ and $b$.  If the vector magnitudes are $\lvert a \rvert$ and $\lvert b \rvert$ and the angle between them is $\theta$, then the magnitude of $a \wedge b$ is just $\lvert a \rvert\, \lvert b \rvert\, \sin\theta$.  And since the cross product has that magnitude, but is a vector perpendicular to this plane, you shouldn't be surprised that they are related.  (Note that the usual cross product is defined specifically for three dimensions; GA works in any number of dimensions, but this cross product relation is only valid in three.)  This $a \wedge b$ object is called a "bivector".  You can also have a "trivector" $a \wedge b \wedge c$, which represents the volume spanned by those three vectors.  In three dimensions, there's just one three-dimensional volume (as opposed to many two-dimensional planes), so every "trivector" is proportional to every other one.  We usually pick one and call it $I$, as you have found.
Finally, we come to the geometric product.  For vectors, the geometric product is related to the other two products as
\begin{equation}
  a\, b = a \cdot b + a \wedge b.
\end{equation}
I've seen at least one treatment where this is actually taken as the definition of the geometric product (which is continued inductively for bivectors, and so on) — so you might just take this equation as gospel and move on.  For example, if you take two elements of your orthonormal basis, you have
\begin{equation}
  e_x\, e_y = e_x \cdot e_y + e_x \wedge e_y.
\end{equation}
But since they're orthogonal, $e_x \cdot e_y = 0$, so you get
\begin{equation}
  e_x\, e_y = e_x \wedge e_y.
\end{equation}
By the same logic, you also have $e_y\, e_x = e_y \wedge e_x$, so all of these are equal:
\begin{equation}
  e_x\, e_y = e_x \wedge e_y = -e_y \wedge e_x = -e_y\, e_x.
\end{equation}
In general, if you have orthogonal vectors, not only is the wedge product antisymmetric, but also the geometric product — since they're equivalent in that case.  Furthermore, we have
\begin{equation}
  I = e_x \wedge e_y \wedge e_z = e_x\, e_y\, e_z.
\end{equation}
And another useful fact is that for any vector $a$, antisymmetry tells us that $a \wedge a = - a \wedge a = 0$, so $a\, a = a \cdot a$.  (In fact, this is essentially the relationship the defines the geometric product in most treatments.)
So let's look at $I(a \times b)$, expanding $a \times b$ using the formula you've got.  The geometric product is linear, which means that we can break this up into each of the terms, and each coefficient just comes right out front.  So we can just look at $I\, e_z$, for example.  We have
\begin{equation}
  I\, e_z = e_x\, e_y\, e_z\, e_z = e_x\, e_y\, (e_z\, e_z) = e_x\, e_y\, (e_z \cdot e_z) = e_x\, e_y = e_x\wedge e_y.
\end{equation}
So the Hodge dual of the vector $e_z$ is just the plane perpendicular to it.  (No right-hand rule is necessary in GA!)  Remember that ordering is important, though, so you get
\begin{align}
  I e_y &= - e_x\, e_z, \\
  I e_x &= e_y\, e_z.
\end{align}
You can use these facts with every term in $I(a \times b)$ and get something like what you got for $a \wedge b$.
...Except that you've got that $a \cdot b$ at the front of your expression for $a \wedge b$.  That's just wrong.  Actually, the right-hand side of your expression for $a \wedge b$ is just $a\, b$.  So that's why I think you've confused the types of products in GA.  Just remove that $a \cdot b$ and you'll have the correct expression.

As a point of interest, you may have heard of quaternions, which are generally the best way to represent rotations.  They're frequently introduced as "a scalar plus a vector".  Well, it turns out the "vector" thing was an accident of history, and they're "really" bivectors — Hamilton just accidentally used the Hodge duals of the bivectors because he didn't know any better.  So the math you've learned above is precisely quaternion math: the geometric product $a\, b$ gives you a quaternion.  In the same way, these are also Pauli spinors, which were accidentally represented by complex matrices — the $e_x\, e_y$, etc., objects behave algebraically just like the usual Pauli matrices, so you could have learned QM using these instead.  And then, it's really not hard to throw $e_t$ (for the time direction) into the mix.  It's all the same kind of manipulation; you just have to remember $e_t\, e_t = -1$.  Then you get spacetime algebra, which lets you do boosts as rotations, and Dirac spinors.  The list of applications for GA just goes on and on, so you might see why some of us feel it's a very powerful and pedagogically useful tool.  It's also caught on in a big way among the computer graphics crowd for its computationally and conceptually efficient representation of geometry.
