# Transformation groups

In the 4D rigid geometric algebra $$\mathcal G_{3,0,1}$$, every Euclidean isometry of 3D space can be represented by a motor $$\mathbf Q$$ of the form

- $$\mathbf Q = Q_{vx} \mathbf e_{41} + Q_{vy} \mathbf e_{42} + Q_{vz} \mathbf e_{43} + Q_{vw} {\large\unicode{x1d7d9}} + Q_{mx} \mathbf e_{23} + Q_{my} \mathbf e_{31} + Q_{mz} \mathbf e_{12} + Q_{mw} \mathbf 1$$

or by a flector $$\mathbf F$$ of the form

- $$\mathbf F = F_{px} \mathbf e_1 + F_{py} \mathbf e_2 + F_{pz} \mathbf e_3 + F_{pw} \mathbf e_4 + F_{gx} \mathbf e_{423} + F_{gy} \mathbf e_{431} + F_{gz} \mathbf e_{412} + F_{gw} \mathbf e_{321}$$ .

Under the geometric antiproduct $$\unicode{x27C7}$$, arbitrary products of these operators form the Euclidean group $$\mathrm E(3)$$ with $${\large\unicode{x1D7D9}}$$ as the identity, and they covariantly transform any object $$\mathbf x$$ in the algebra through the sandwich products $$\mathbf x' = \mathbf Q \mathbin{\unicode{x27C7}} \mathbf x \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{Q}}}$$ and $$\mathbf x' = -\mathbf F \mathbin{\unicode{x27C7}} \mathbf x \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{F}}}$$.

Symmetrically, every complement Euclidean isometry of 3D space can be represented by a complement motor $$\mathbf Q$$ of the form

- $$\mathbf Q = Q_{vx} \mathbf e_{23} + Q_{vy} \mathbf e_{31} + Q_{vz} \mathbf e_{12} - Q_{vw} {\large\unicode{x1d7d9}} + Q_{mx} \mathbf e_{41} + Q_{my} \mathbf e_{42} + Q_{mz} \mathbf e_{43} - Q_{mw} \mathbf 1$$

or by a complement flector $$\mathbf F$$ of the form

- $$\mathbf F = F_{px} \mathbf e_{423} + F_{py} \mathbf e_{431} + F_{pz} \mathbf e_{412} + F_{pw} \mathbf e_{321} - F_{gx} \mathbf e_1 - F_{gy} \mathbf e_2 - F_{gz} \mathbf e_3 - F_{gw} \mathbf e_4$$ .

Under the geometric product $$\unicode{x27D1}$$, arbitrary products of these operators form the complement Euclidean group $$\overline{\mathrm E}(3)$$ with $$\mathbf 1$$ as the identity, and they covariantly transform any object $$\mathbf x$$ in the algebra through the sandwich products $$\mathbf x' = \mathbf Q \mathbin{\unicode{x27D1}} \mathbf x \mathbin{\unicode{x27D1}} \mathbf{\tilde Q}$$ and $$\mathbf x' = -\mathbf F \mathbin{\unicode{x27D1}} \mathbf x \mathbin{\unicode{x27D1}} \mathbf{\tilde F}$$.

The geometric product corresponds to transform composition in the group $$\overline{\mathrm E}(3)$$, and the geometric antiproduct corresponds to transform composition in the group $$\mathrm E(3)$$. Reflections across planes are represented by antivectors (having antigrade one), and they meet at lower-dimensional invariants under the geometric antiproduct. Symmetrically, complement reflections across points are represented by vectors (having grade one), and they join at higher-dimensional invariants under the geometric product. A sandwich product $$\mathbf Q \mathbin{\unicode{x27D1}} \mathbf x \mathbin{\unicode{x27D1}} \mathbf{\tilde Q}$$ transforms the space of $$\mathbf x$$ with an element of $$\overline{\mathrm E}(3)$$, and it transforms the antispace of $$\mathbf x$$ with the complementary element of $$\mathrm E(3)$$. Symmetrically, a sandwich product $$\mathbf Q \mathbin{\unicode{x27C7}} \mathbf x \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{Q}}}$$ transforms the space of $$\mathbf x$$ with an element of $$\mathrm E(3)$$, and it transforms the antispace of $$\mathbf x$$ with the complementary element of $$\overline{\mathrm E}(3)$$.

The groups $$\mathrm E(n)$$ and $$\overline{\mathrm E}(n)$$ are isomorphic, and they each contain the orthogonal group $$\mathrm O(n)$$ as a common subgroup. The complement operation provides a two-way mapping between transforms associated with members of $$\mathrm E(n)$$ and $$\overline{\mathrm E}(n)$$. The groups $$\mathrm E(n)$$ and $$\overline{\mathrm E}(n)$$ have a number of subgroups, and the hierarchical relationships among them are shown in the figure below. In particular, the Euclidean group $$\mathrm E(n)$$ contains the special Euclidean subgroup $$\mathrm{SE}(n)$$ consisting of all combinations of ordinary rotations and translations, which are covered by the elements of $$\mathcal G_{n,0,1}$$ having even antigrade. Correspondingly, the complement Euclidean group $$\overline{\mathrm E}(n)$$ contains the complement special Euclidean subgroup $$\mathrm S\overline{\mathrm E}(n)$$ consisting of all combinations of complement rotations and complement translations, which are covered by the elements of $$\mathcal G_{n,0,1}$$ having even grade. The subgroups $$\mathrm{SE}(n)$$ and $$\mathrm S\overline{\mathrm E}(n)$$ further contain translation subgroups $$\mathrm T(n)$$ and $$\overline{\mathrm T}(n)$$, respectively.

Transforms about invariants containing the origin are the same in both $$\mathrm E(n)$$ and $$\overline{\mathrm E}(n)$$, and they constitute the common subgroup $$\mathrm O(n)$$. Every member of $$\mathrm O(n)$$ has a representation that transforms elements with the geometric product and a complementary representation that transforms elements with the geometric antiproduct. For example, conventional quaternions $$\mathbf q$$ have two representations, one that transforms any object $$\mathbf x$$ through the sandwich product $$\mathbf x' = \mathbf q \mathbin{\unicode{x27D1}} \mathbf x \mathbin{\unicode{x27D1}} \mathbf{\tilde q}$$ and another that transforms any object $$\mathbf x$$ through the sandwich product $$\mathbf x' = \mathbf q \mathbin{\unicode{x27C7}} \mathbf x \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{q}}}$$.

In terms of matrix multiplication, a general element of the group $$\mathrm E(n)$$ transforms a point by multiplying on the left by an $$(n + 1) \times (n + 1)$$ matrix of the form

- $$\begin{bmatrix} \mathbf M_{n \times n} & \boldsymbol \tau_{n \times 1} \\ \mathbf 0_{1 \times n} & 1 \end{bmatrix}$$ ,

where the $$n \times n$$ submatrix $$\mathbf M$$ is orthogonal. A general element of the corresponding group $$\overline{\mathrm E}(n)$$ transforms points with matrices of the form

- $$\begin{bmatrix} \mathbf M_{n \times n} & \mathbf 0_{n \times 1} \\ \boldsymbol \tau_{1 \times n} & 1 \end{bmatrix}$$ .

In the special subgroups $$\mathrm{SE}(n)$$ and $$\mathrm S\overline{\mathrm E}(n)$$, the submatrix $$\mathbf M$$ has a determinant of +1. In the translation subgroups $$\mathrm T(n)$$ and $$\overline{\mathrm T}(n)$$, $$\mathbf M$$ is the identity matrix. Finally, when $$\boldsymbol \tau = \mathbf 0$$, the two matrices above have the same form and belong to $$\mathrm O(n)$$.

The isomorphic mapping between $$\mathrm E(n)$$ and $$\overline{\mathrm E}(n)$$ is given by the inverse transpose operation on the matrix representatives. That is, if $$\mathbf M$$ is an $$(n + 1) \times (n + 1)$$ matrix representing an element of $$\mathrm E(n)$$, then the corresponding element of $$\overline{\mathrm E}(n)$$ is given by $$(\mathbf M^{-1})^{\text T}$$. Of course, this operation is an involution, and the mapping works both ways.

## In the Book

- Transformation groups are discussed in Section 3.9.2.