In geometric algebra, the outermorphism of a linear function between vector spaces is a natural extension of the map to arbitrary multivectors. It is the unique unital algebra homomorphism of exterior algebras whose restriction to the vector spaces is the original function.
Let
f
R
V
W
f
style\underline{f
\underline{f
\underline{f
\underline{f
\underline{f
x
A
B
stylewedge(V)
V
The outermorphism inherits linearity properties of the original linear map. For example, we see that for scalars
\alpha
\beta
x
y
z
\begin{align}\underline{f
Let
\underline{f
\overline{f
\overline{f
a
b
⋅
This results in the property that
\overline{f
A
B
*
If geometric calculus is available, then the adjoint may be extracted more directly:
\overline{f
The above definition of adjoint is like the definition of the transpose in matrix theory. When the context is clear, the underline below the function is often omitted.
It follows from the definition at the beginning that the outermorphism of a multivector
A
\underline{f
\langle~\rangler
r
A
Since any vector
x
x=1\wedgex
\underline{f
\underline{f
\detf=\underline{f
The underline is not necessary in this context because the determinant of a function is the same as the determinant of its adjoint. The determinant of the composition of functions is the product of the determinants:
\det(f\circg)=\detf\detg
If the determinant of a function is nonzero, then the function has an inverse given by
\underline{f
\overline{f
The concepts of eigenvalues and eigenvectors may be generalized to outermorphisms. Let
λ
B
r
B
λ
\underline{f
It may seem strange to consider only real eigenvalues, since in linear algebra the eigenvalues of a matrix with all real entries can have complex eigenvalues. In geometric algebra, however, the blades of different grades can exhibit a complex structure. Since both vectors and pseudovectors can act as eigenblades, they may each have a set of eigenvalues matching the degrees of freedom of the complex eigenvalues that would be found in ordinary linear algebra.
R
f(x)=RxR-1
\underline{f
We check that this is the correct form of the outermorphism. Since rotations are built from the geometric product, which has the distributive property, they must be linear. To see that rotations are also outermorphisms, we recall that rotations preserve angles between vectors:
x ⋅ y=(RxR-1) ⋅ (RyR-1)
Next, we try inputting a higher grade element and check that it is consistent with the original rotation for vectors:
\begin{align}\underline{f
l{P}B
B
l{P}B(x\wedgey)=l{P}B(x)\wedgel{P}B(y).
\perp | |
l{P} | |
B |
B
\perp | |
l{P} | |
B(1) |
=1-l{P}B(1)=0\ne1.
\langlex\wedgey\rangle1=0
\langlex\rangle1\wedge\langley\rangle1=x\wedgey