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This is a continuation of yesterday’s post on the relationships between the exterior derivative, and the curl operation (grad-wedge) in geometric algebra.

## Hodge star vs. pseudoscalar multiplication.

We find a definition of the hodge star for basic k-forms in [2].

## Definition 1.7: Hodge star.

\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 [3]. 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

\begin{equation*}

{*} M

=

\gpgrade{M}{0,1} I –

\gpgrade{M}{2,3} I.

\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

\begin{equation*}

{*} M

=

\gpgrade{M}{0,1} i –

\gpgrade{M}{2} i.

\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.

\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*}

\spacegrad \wedge ({*} \Bf) = \lr{\spacegrad \cdot \Bf} I.

\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 \\

&\quad+

\lr{

\PD{x}{g} dx +

\PD{y}{g} dy +

\PD{z}{g} dz

}

\wedge

dz \wedge dx \\

&\quad+

\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}

\spacegrad \wedge ({*} \Bf)

&= \gpgradethree{ \spacegrad (\Bf I) } \\

&= \gpgradethree{ (\spacegrad \Bf) I } \\

&= \gpgradethree{ (\spacegrad \cdot \Bf + \spacegrad \wedge \Bf) I } \\

&= \lr{ \spacegrad \cdot \Bf } I.

\end{aligned}

\end{equation}

### End proof.

# References

[1] Vincent Bouchard. Math 215: Calculus iv: 4.4 the exterior derivative and vector calculus, 2023. URL https://sites.ualberta.ca/ vbouchar/MATH215/section_exterior_vector.html. [Online; accessed 11-November-2023].

[2] Vincent Bouchard. Math 215: Calculus iv: 4.8 hodge star, 2023. URL https://sites.ualberta.ca/ vbouchar/MATH215/section_hodge.html. [Online; accessed 13-November-2023].

[3] H. Flanders. *Differential Forms With Applications to the Physical Sciences*. Courier Dover Publications, 1989.