## Hodge star vs. pseudoscalar multiplication.

We find a definition of the hodge star for basic k-forms in .

## Definition 1.7: Hodge star.

Let $$\omega$$ be a basic k-form on $$\mathbb{R}^n$$. The hodge star of $$\omega$$, denoted by $${*} \omega$$ is the unique $$n-k$$-form with the property
\begin{equation*}
\omega \wedge {*} \omega = dx_1 \wedge \cdots \wedge dx_n.
\end{equation*}

I find it interesting that this duality definition is completely free of any notion of metric or inner product. That isn’t the case with the hodge star definition from . This is certainly an easier definition to understand.

Let’s calculate all the duals for the basic forms from $$\mathbb{R}^3$$. We let $$I = dx_1 \wedge dx_2 \wedge dx_3$$, and then by inspection find all the duals satisfying
\begin{equation}\label{eqn:formAndCurl:1110}
\begin{aligned}
I &= 1 \wedge {*} 1 \\
I &= dx \wedge {*} dx \\
I &= dy \wedge {*} dy \\
I &= dz \wedge {*} dz \\
I &= (dx dy) \wedge {*} (dx dy) \\
I &= (dy dz) \wedge {*} (dy dz) \\
I &= (dz dx) \wedge {*} (dz dx) \\
I &= dx dy dz \wedge {*} (dx dy dz).
\end{aligned}
\end{equation}
Those are
\begin{equation}\label{eqn:formAndCurl:1130}
\begin{aligned}
{*} 1 &= dx dy dz \\
{*} dx &= dy dz \\
{*} dy &= dz dx \\
{*} dz &= dx dy \\
{*} (dx dy) &= dz \\
{*} (dy dz) &= dx \\
{*} (dz dx) &= dy \\
{*} (dx dy dz) &= 1.
\end{aligned}
\end{equation}

Now let’s compare this to multiplication of the $$\mathbb{R}^3$$ basis vectors with the pseudoscalar $$I = \Be_1 \Be_2 \Be_3$$. We have
\begin{equation}\label{eqn:formAndCurl:1140}
\begin{aligned}
1 I &= I \\
\Be_1 I &= \Be_{1123} = \Be_{23} \\
\Be_2 I &= \Be_{2123} = \Be_{31} \\
\Be_3 I &= \Be_{3123} = \Be_{12} \\
\Be_{23} I &= \Be_{23123} = – \Be_1 \\
\Be_{31} I &= \Be_{31123} = – \Be_2 \\
\Be_{12} I &= \Be_{12123} = – \Be_3 \\
\Be_{123} I &= \Be_{123123} = -1.
\end{aligned}
\end{equation}
With differential forms, the duals of the duals of all our basic forms recovered the original, that is $$** \omega = \omega$$, but that isn’t the case if we use pseudoscalar multiplication to define duality. We see that to model the Hodge dual, we need to multiply by a grade specific pseudoscalar.

## Definition 1.8: Hodge dual of an $$\mathbb{R}^3$$ multivector

Let $$M$$ be a $$\mathbb{R}^3$$ multivector. The Hodge dual $${*} M$$ of that multivector is
\begin{equation*}
{*} M
=
\end{equation*}

In particular, if $$A$$ is a k-blade in $$\mathbb{R}^3$$, a round trip requires multiplication with different signed unit pseudoscalars.

Let’s step back and consider the $$\mathbb{R}^2$$ case as well. This time we let $$i = dx_1 \wedge dx_2$$. We seek all the duals satisfying
\begin{equation}\label{eqn:formAndCurl:1180}
\begin{aligned}
i &= 1 \wedge {*} 1 \\
i &= dx \wedge {*} dx \\
i &= dy \wedge {*} dy \\
i &= (dx dy) \wedge {*} (dx dy).
\end{aligned}
\end{equation}
Those duals are
\begin{equation}\label{eqn:formAndCurl:1200}
\begin{aligned}
{*} 1 &= dx dy \\
{*} dx &= dy \\
{*} dy &= -dx \\
{*} (dx dy) &= 1 \\
\end{aligned}
\end{equation}

Now let’s compare this to multiplication of the $$\mathbb{R}^2$$ basis vectors with the pseudoscalar $$i = \Be_1 \Be_2$$. We have
\begin{equation}\label{eqn:formAndCurl:1220}
\begin{aligned}
1 i &= i \\
\Be_1 i &= \Be_{112} = \Be_{2} \\
\Be_2 i &= \Be_{212} = -\Be_{1} \\
\Be_{12} i &= \Be_{1212} = -1 \\
\end{aligned}
\end{equation}

## Definition 1.9: Hodge dual of $$\mathbb{R}^2$$ multivector

Let $$M$$ be a $$\mathbb{R}^2$$ multivector. The Hodge dual $${*} M$$ of that multivector is
\begin{equation*}
{*} M
=
\end{equation*}

Neither of these grade specific duality operations are as nice as simply multiplying by a unit pseudoscalar, but if we care about correspondence with the Hodge dual (at least according to the definition in the article), then this is what we need.

Having done that, let’s now look at the Hodge dual that produces the divergence operation.

## Lemma 1.13: Divergence relation to the exterior derivative.

Let $$\omega = f dx + g dy + h dz$$ be a one-form in $$\mathbb{R}^3$$. The exterior derivative of the Hodge dual of $$\omega$$ is a divergence three-form
\begin{equation*}
d({*} \omega) = \lr{ \PD{x}{f} + \PD{y}{g} + \PD{z}{h} } dx \wedge dy \wedge dz.
\end{equation*}
The GA equivalent of this, for a vector corresponding to this one-form $$\Bf = f \Be_1 + g \Be_2 + h \Be_3 \in \mathbb{R}^3$$, is
\begin{equation*}
\end{equation*}

### Start proof:

The dual of the one form is
\begin{equation}\label{eqn:formAndCurl:1280}
{*} \omega =
f dy \wedge dz
+ g dz \wedge dx
+ h dx \wedge dy,
\end{equation}
so the exterior derivative is
\begin{equation}\label{eqn:formAndCurl:1300}
\begin{aligned}
d({*} \omega) &=
\lr{
\PD{x}{f} dx +
\PD{y}{f} dy +
\PD{z}{f} dz
}
\wedge dy \wedge dz \\
\lr{
\PD{x}{g} dx +
\PD{y}{g} dy +
\PD{z}{g} dz
}
\wedge
dz \wedge dx \\
\lr{
\PD{x}{g} dx +
\PD{y}{g} dy +
\PD{z}{g} dz
}
\wedge
dx \wedge dy \\
&=
\lr{
\PD{x}{f} +
\PD{y}{g} +
\PD{z}{h}
}
dx \wedge dy \wedge dz.
\end{aligned}
\end{equation}
We expect that the GA equivalent of this is $$\spacegrad \wedge ({*} \Bf) = \lr{ \spacegrad \cdot \Bf} I$$. Let’s check that this is the case. The dual, for a vector, is
\begin{equation}\label{eqn:formAndCurl:1320}
{*} \Bf
= \Bf I,
\end{equation}
so
\begin{equation}\label{eqn:formAndCurl:1340}
\begin{aligned}