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The author of a book draft I am reading pointed out if a vector transforms as

\begin{equation}\label{eqn:adjoint:760}

\Bv \rightarrow M \Bv,

\end{equation}

then cross products must transform as

\begin{equation}\label{eqn:adjoint:780}

\Ba \cross \Bb \rightarrow \lr{ \textrm{adj}\, M }^\T \lr{ \Ba \cross \Bb }.

\end{equation}

Bivectors clearly must transform in the same fashion. We also noticed that the adjoint is related to the reciprocal frame vectors of the columns of \( M \), but didn’t examine the reciprocal frame formulation of the adjoint in any detail.

Before we do that, let’s consider a slightly simpler case, the transformation of a pseudoscalar. That is

\begin{equation}\label{eqn:adjoint:800}

\begin{aligned}

M(\Ba) \wedge M(\Bb) \wedge M(\Bc)

&\rightarrow

\sum_{ijk}

\lr{ \Bm_i a_i } \wedge

\lr{ \Bm_j a_j } \wedge

\lr{ \Bm_k a_k } \\

&=

\sum_{ijk}

\lr{ \Bm_i \wedge \Bm_j \wedge \Bm_k } a_i b_j c_k \\

&=

\sum_{ijk}

\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 } \epsilon_{ijk} a_i b_j c_k \\

&=

\Abs{M}

\sum_{ijk} \epsilon_{ijk} a_i b_j c_k \\

&=

\Abs{M} \lr{ \Ba \wedge \Bb \wedge \Bc }.

\end{aligned}

\end{equation}

This is a well known geometric algebra result (called an outermorphism transformation.)

It’s somewhat amusing that an outermorphism transformation with two wedged vectors is a bit more complicated to express than the same for three. Let’s see if we can find a coordinate free form for such a transformation.

\begin{equation}\label{eqn:adjoint:820}

\begin{aligned}

M(\Ba) \wedge M(\Bb)

&=

\sum_{ij} \lr{ \Bm_i a_i } \wedge \lr{ \Bm_j b_j } \\

&=

\sum_{ij} \lr{ \Bm_i \wedge \Bm_j } a_i b_j \\

&=

\sum_{i < j} \lr{ \Bm_i \wedge \Bm_j }

\begin{vmatrix}

a_i & a_j \\

b_i & b_j

\end{vmatrix} \\

&=

\sum_{i < j} \lr{ \Bm_i \wedge \Bm_j } \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_j \wedge \Be_i } }.

\end{aligned}

\end{equation}

Recall that the reciprocal frame with respect to the basis \( \setlr{ \Bm_1, \Bm_2, \Bm_3 } \), assuming this is a non-degenerate basis, has elements of the form

\begin{equation}\label{eqn:adjoint:840}

\begin{aligned}

\Bm^1 &= \lr{ \Bm_2 \wedge \Bm_3 } \inv{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 } \\

\Bm^2 &= \lr{ \Bm_3 \wedge \Bm_1 } \inv{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 } \\

\Bm^3 &= \lr{ \Bm_1 \wedge \Bm_2 } \inv{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }.

\end{aligned}

\end{equation}

This can be flipped around as

\begin{equation}\label{eqn:adjoint:860}

\begin{aligned}

\Bm_2 \wedge \Bm_3 &= \Bm^1 \Abs{M} I \\

\Bm_3 \wedge \Bm_1 &= \Bm^2 \Abs{M} I \\

\Bm_1 \wedge \Bm_2 &= \Bm^3 \Abs{M} I \\

\end{aligned}

\end{equation}

\begin{equation}\label{eqn:adjoint:880}

\begin{aligned}

M&(\Ba) \wedge M(\Bb) \\

&=

\lr{ \Bm_1 \wedge \Bm_2 } \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_2 \wedge \Be_1 } }

+

\lr{ \Bm_2 \wedge \Bm_3 } \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_3 \wedge \Be_2 } }

+

\lr{ \Bm_3 \wedge \Bm_1 } \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_1 \wedge \Be_3 } } \\

&=

I \Abs{M} \lr{

\Bm^3 \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_2 \wedge \Be_1 } }

+

\Bm^1 \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_3 \wedge \Be_2 } }

+

\Bm^2 \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_1 \wedge \Be_3 } }

}

\end{aligned}

\end{equation}

Let’s see if we can simplify one of these double index quantities

\begin{equation}\label{eqn:adjoint:900}

\begin{aligned}

I \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_2 \wedge \Be_1 } }

&=

\gpgradethree{ I \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_2 \wedge \Be_1 } } } \\

&=

\gpgradethree{ I \lr{ \Ba \wedge \Bb } \lr{ \Be_2 \wedge \Be_1 } } \\

&=

\gpgradethree{ \lr{ \Ba \wedge \Bb } \Be_{12321} } \\

&=

\gpgradethree{ \lr{ \Ba \wedge \Bb } \Be_{3} } \\

&=

\Ba \wedge \Bb \wedge \Be_3.

\end{aligned}

\end{equation}

We have

\begin{equation}\label{eqn:adjoint:920}

M(\Ba) \wedge M(\Bb) = \Abs{M} \lr{

\lr{ \Ba \wedge \Bb \wedge \Be_1 } \Bm^1

+

\lr{ \Ba \wedge \Bb \wedge \Be_2 } \Bm^2

+

\lr{ \Ba \wedge \Bb \wedge \Be_3 } \Bm^3

}.

\end{equation}

Using summation convention, we can now express the transformation of a bivector \( B \) as just

\begin{equation}\label{eqn:adjoint:940}

B \rightarrow \Abs{M} \lr{ B \wedge \Be_i } \Bm^i.

\end{equation}

If we are interested in the transformation of a pseudovector \( \Bv \) defined implicitly as the dual of a bivector \( B = I \Bv \), where

\begin{equation}\label{eqn:adjoint:960}

B \wedge \Be_i = \gpgradethree{ I \Bv \Be_i } = I \lr{ \Bv \cdot \Be_i }.

\end{equation}

This leaves us with a transformation rule for cross products equivalent to the adjoint relation \ref{eqn:adjoint:780}

\begin{equation}\label{eqn:adjoint:980}

\lr{ \Ba \cross \Bb } \rightarrow \lr{ \Ba \cross \Bb } \cdot \Be_i \Abs{M} \Bm^i.

\end{equation}

As intuited, the determinant weighted reciprocal frame vectors for the columns of the transformation \( M \), are the components of the adjoint. That is

\begin{equation}\label{eqn:adjoint:1000}

\lr{ \textrm{adj}\, M }^\T = \Abs{M}

\begin{bmatrix}

\Bm^1 & \Bm^2 & \Bm^3

\end{bmatrix}.

\end{equation}