Transformation groups - Revision history

Eric Lengyel at 07:12, 9 November 2025

2025-11-09T07:12:22Z

← Older revision		Revision as of 07:12, 9 November 2025
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	Symmetrically, every complement Euclidean isometry of 3D space can be represented by a [[complement motor]] $$\mathbf Q$$ of the form		Symmetrically, every complement Euclidean isometry of 3D space can be represented by a [[complement motor]] $$\mathbf Q$$ of the form

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

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

Eric Lengyel at 07:06, 8 August 2024

2024-08-08T07:06:28Z

Eric Lengyel at 23:52, 13 April 2024

2024-04-13T23:52:40Z

← Older revision		Revision as of 23:52, 13 April 2024
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	The isomorphic mapping between E(''n'') and RE(''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 E(''n''), then the corresponding element of RE(''n'') is given by $$(\mathbf M^{-1})^{\text T}$$. Of course, this operation is an involution, and the mapping works both ways.		The isomorphic mapping between E(''n'') and RE(''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 E(''n''), then the corresponding element of RE(''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.

	== See Also ==		== See Also ==

Eric Lengyel at 00:34, 26 August 2023

2023-08-26T00:34:35Z

← Older revision		Revision as of 00:34, 26 August 2023
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	In terms of matrix multiplication, a general element of the group E(''n'') transforms a point by multiplying on the left by an $$(n + 1) \times (n + 1)$$ matrix of the form		In terms of matrix multiplication, a general element of the group E(''n'') transforms a point by multiplying on the left by an $$(n + 1) \times (n + 1)$$ matrix of the form

	:$$\begin{bmatrix} \mathbf m_{n \times n} & \~~mathbf t_~~{n \times 1} \\ \mathbf 0_{1 \times n} & 1 \end{bmatrix}$$ ,		:$$\begin{bmatrix} \mathbf M_{n \times n} & \boldsymbol \tau_{n \times 1} \\ \mathbf 0_{1 \times n} & 1 \end{bmatrix}$$ ,

	where the $$n \times n$$ submatrix $$\mathbf m$$ is orthogonal. A general element of the corresponding group RE(''n'') transforms points with matrices of the form		where the $$n \times n$$ submatrix $$\mathbf M$$ is orthogonal. A general element of the corresponding group RE(''n'') transforms points with matrices of the form

	:$$\begin{bmatrix} \mathbf m_{n \times n} & \mathbf 0_{n \times 1} \\ \~~mathbf t_~~{1 \times n} & 1 \end{bmatrix}$$ .		:$$\begin{bmatrix} \mathbf M_{n \times n} & \mathbf 0_{n \times 1} \\ \boldsymbol \tau_{1 \times n} & 1 \end{bmatrix}$$ .

	In the special subgroups SE(''n'') and RSE(''n''), the submatrix $$\mathbf m$$ has a determinant of +1. In the translation subgroups T(''n'') and RT(''n''), $$\mathbf m$$ is the identity matrix. Finally, when $$\~~mathbf t~~ = \mathbf 0$$, the two matrices above have the same form and belong to O(''n'').		In the special subgroups SE(''n'') and RSE(''n''), the submatrix $$\mathbf M$$ has a determinant of +1. In the translation subgroups T(''n'') and RT(''n''), $$\mathbf M$$ is the identity matrix. Finally, when $$\boldsymbol \tau = \mathbf 0$$, the two matrices above have the same form and belong to O(''n'').

	The isomorphic mapping between E(''n'') and RE(''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 E(''n''), then the corresponding element of RE(''n'') is given by $$(\mathbf M^{-1})^{\text T}$$. Of course, this operation is an involution, and the mapping works both ways.		The isomorphic mapping between E(''n'') and RE(''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 E(''n''), then the corresponding element of RE(''n'') is given by $$(\mathbf M^{-1})^{\text T}$$. Of course, this operation is an involution, and the mapping works both ways.

Eric Lengyel at 00:25, 26 August 2023

2023-08-26T00:25:01Z

Eric Lengyel: Created page with "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_..."

2023-07-15T05:52:29Z

Created page with "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_..."

New page

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 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 dual Euclidean isometry of 3D space can be represented by a [[dual 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 [[dual 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 dual Euclidean group DE(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 DE(3), and the [[geometric antiproduct]] corresponds to transform composition in the group E(3). [[Reflections]] across planes are represented by [[antivectors]] (having [[antigrade]] one), and they meet at lower-dimensional invariants under the geometric antiproduct. Symmetrically, [[dual 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 DE(3), and it transforms the antispace of $$\mathbf x$$ with the complementary element of 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 E(3), and it transforms the antispace of $$\mathbf x$$ with the complementary element of DE(3).

The groups E(''n'') and DE(''n'') are isomorphic, and they each contain the orthogonal group O(''n'') as a common subgroup. The [[complement]] operation provides a two-way mapping between transforms associated with members of E(''n'') and DE(''n''). The groups E(''n'') and DE(''n'') have a number of subgroups, and the hierarchical relationships among them are shown in the figure below. In particular, the Euclidean group E(''n'') contains the special Euclidean subgroup 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 dual Euclidean group DE(''n'') contains the dual special Euclidean subgroup DSE(''n'') consisting of all combinations of [[dual rotations]] and [[dual translations]], which are covered by the elements of $$\mathcal G_{n,0,1}$$ having even grade. The subgroups SE(''n'') and DSE(''n'') further contain translation subgroups T(''n'') and DT(''n''), respectively.

[[Image:Groups.svg|800px]]

Transforms about invariants containing the origin are the same in both E(''n'') and DE(''n''), and they constitute the common subgroup O(''n''). Every member of 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 E(''n'') transforms a point by multiplying on the left by an $$(n + 1) \times (n + 1)$$ matrix of the form

:$$\begin{bmatrix} \mathbf m_{n \times n} & \mathbf t_{n \times 1} \\ \mathbf 0_{1 \times n} & 1 \end{bmatrix}$$ ,

where the $$n \times n$$ submatrix $$\mathbf m$$ is orthogonal. A general element of the corresponding group DE(''n'') transforms points with matrices of the form

:$$\begin{bmatrix} \mathbf m_{n \times n} & \mathbf 0_{n \times 1} \\ \mathbf t_{1 \times n} & 1 \end{bmatrix}$$ .

In the special subgroups SE(''n'') and DSE(''n''), the submatrix $$\mathbf m$$ has a determinant of +1. In the translation subgroups T(''n'') and DT(''n''), $$\mathbf m$$ is the identity matrix. Finally, when $$\mathbf t = \mathbf 0$$, the two matrices above have the same form and belong to O(''n'').

The isomorphic mapping between E(''n'') and DE(''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 E(''n''), then the corresponding element of DE(''n'') is given by $$(\mathbf M^{-1})^{\text T}$$. Of course, this operation is an involution, and the mapping works both ways.

== See Also ==

* [[Motor]]
* [[Flector]]

← Older revision		Revision as of 07:06, 8 August 2024
Line 7:		Line 7:
	:$$\mathbf F = F_{px} \mathbf e_1 + F_{py} \mathbf e_2 + F_{pz} \mathbf e_3 + F_{pw} \mathbf e_4 + F_{gx} \mathbf e_{423} + F_{gy} \mathbf e_{431} + F_{gz} \mathbf e_{412} + F_{gw} \mathbf e_{321}$$ .		:$$\mathbf F = F_{px} \mathbf e_1 + F_{py} \mathbf e_2 + F_{pz} \mathbf e_3 + F_{pw} \mathbf e_4 + F_{gx} \mathbf e_{423} + F_{gy} \mathbf e_{431} + F_{gz} \mathbf e_{412} + F_{gw} \mathbf e_{321}$$ .

	Under the [[geometric antiproduct]] $$\unicode{x27C7}$$, arbitrary products of these operators form the Euclidean group 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 $$\mathbf x' = -\mathbf F \mathbin{\unicode{x27C7}} \mathbf x \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{F}}}$$.		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 ~~reciprocal~~ Euclidean isometry of 3D space can be represented by a [[~~reciprocal~~ motor]] $$\mathbf Q$$ of the form		Symmetrically, every complement Euclidean isometry of 3D space can be represented by a [[complement motor]] $$\mathbf Q$$ of the form

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

	or by a [[~~reciprocal~~ flector]] $$\mathbf F$$ of the form		or by a [[complement flector]] $$\mathbf F$$ of the form

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

	Under the [[geometric product]] $$\unicode{x27D1}$$, arbitrary products of these operators form the ~~reciprocal~~ Euclidean group RE(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 $$\mathbf x' = -\mathbf F \mathbin{\unicode{x27D1}} \mathbf x \mathbin{\unicode{x27D1}} \mathbf{\tilde F}$$.		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 RE(3), and the [[geometric antiproduct]] corresponds to transform composition in the group E(3). [[Reflections]] across planes are represented by [[antivectors]] (having [[antigrade]] one), and they meet at lower-dimensional invariants under the geometric antiproduct. Symmetrically, [[~~reciprocal~~ 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 RE(3), and it transforms the antispace of $$\mathbf x$$ with the complementary element of 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 E(3), and it transforms the antispace of $$\mathbf x$$ with the complementary element of RE(3).		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 E(''n'') and RE(''n'') are isomorphic, and they each contain the orthogonal group O(''n'') as a common subgroup. The [[complement]] operation provides a two-way mapping between transforms associated with members of E(''n'') and RE(''n''). The groups E(''n'') and RE(''n'') have a number of subgroups, and the hierarchical relationships among them are shown in the figure below. In particular, the Euclidean group E(''n'') contains the special Euclidean subgroup 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 ~~reciprocal~~ Euclidean group RE(''n'') contains the ~~reciprocal~~ special Euclidean subgroup ~~RSE~~(''n'') consisting of all combinations of [[~~reciprocal~~ rotations]] and [[~~reciprocal~~ translations]], which are covered by the elements of $$\mathcal G_{n,0,1}$$ having even grade. The subgroups SE(''n'') and ~~RSE~~(''n'') further contain translation subgroups T(''n'') and RT(''n''), respectively.		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.

	[[Image:Groups.svg\|800px]]		[[Image:Groups.svg\|800px]]

	Transforms about invariants containing the origin are the same in both E(''n'') and RE(''n''), and they constitute the common subgroup O(''n''). Every member of 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 $$\mathbf x' = \mathbf q \mathbin{\unicode{x27C7}} \mathbf x \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{q}}}$$.		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 E(''n'') transforms a point by multiplying on the left by an $$(n + 1) \times (n + 1)$$ matrix of the form		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

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

	where the $$n \times n$$ submatrix $$\mathbf M$$ is orthogonal. A general element of the corresponding group RE(''n'') transforms points with matrices of the form		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

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

	In the special subgroups SE(''n'') and ~~RSE~~(''n''), the submatrix $$\mathbf M$$ has a determinant of +1. In the translation subgroups T(''n'') and RT(''n''), $$\mathbf M$$ is the identity matrix. Finally, when $$\boldsymbol \tau = \mathbf 0$$, the two matrices above have the same form and belong to O(''n'').		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 E(''n'') and RE(''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 E(''n''), then the corresponding element of RE(''n'') is given by $$(\mathbf M^{-1})^{\text T}$$. Of course, this operation is an involution, and the mapping works both ways.		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 ==		== In the Book ==

← Older revision		Revision as of 00:25, 26 August 2023
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	Under the [[geometric antiproduct]] $$\unicode{x27C7}$$, arbitrary products of these operators form the Euclidean group 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 $$\mathbf x' = -\mathbf F \mathbin{\unicode{x27C7}} \mathbf x \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{F}}}$$.		Under the [[geometric antiproduct]] $$\unicode{x27C7}$$, arbitrary products of these operators form the Euclidean group 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 ~~dual~~ Euclidean isometry of 3D space can be represented by a [[~~dual~~ motor]] $$\mathbf Q$$ of the form		Symmetrically, every reciprocal Euclidean isometry of 3D space can be represented by a [[reciprocal motor]] $$\mathbf Q$$ of the form

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

	or by a [[~~dual~~ flector]] $$\mathbf F$$ of the form		or by a [[reciprocal flector]] $$\mathbf F$$ of the form

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

	Under the [[geometric product]] $$\unicode{x27D1}$$, arbitrary products of these operators form the ~~dual~~ Euclidean group DE(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 $$\mathbf x' = -\mathbf F \mathbin{\unicode{x27D1}} \mathbf x \mathbin{\unicode{x27D1}} \mathbf{\tilde F}$$.		Under the [[geometric product]] $$\unicode{x27D1}$$, arbitrary products of these operators form the reciprocal Euclidean group RE(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 DE(3), and the [[geometric antiproduct]] corresponds to transform composition in the group E(3). [[Reflections]] across planes are represented by [[antivectors]] (having [[antigrade]] one), and they meet at lower-dimensional invariants under the geometric antiproduct. Symmetrically, [[~~dual~~ 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 DE(3), and it transforms the antispace of $$\mathbf x$$ with the complementary element of 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 E(3), and it transforms the antispace of $$\mathbf x$$ with the complementary element of DE(3).		The [[geometric product]] corresponds to transform composition in the group RE(3), and the [[geometric antiproduct]] corresponds to transform composition in the group E(3). [[Reflections]] across planes are represented by [[antivectors]] (having [[antigrade]] one), and they meet at lower-dimensional invariants under the geometric antiproduct. Symmetrically, [[reciprocal 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 RE(3), and it transforms the antispace of $$\mathbf x$$ with the complementary element of 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 E(3), and it transforms the antispace of $$\mathbf x$$ with the complementary element of RE(3).

	The groups E(''n'') and DE(''n'') are isomorphic, and they each contain the orthogonal group O(''n'') as a common subgroup. The [[complement]] operation provides a two-way mapping between transforms associated with members of E(''n'') and DE(''n''). The groups E(''n'') and DE(''n'') have a number of subgroups, and the hierarchical relationships among them are shown in the figure below. In particular, the Euclidean group E(''n'') contains the special Euclidean subgroup 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 ~~dual~~ Euclidean group DE(''n'') contains the ~~dual~~ special Euclidean subgroup ~~DSE~~(''n'') consisting of all combinations of [[~~dual~~ rotations]] and [[~~dual~~ translations]], which are covered by the elements of $$\mathcal G_{n,0,1}$$ having even grade. The subgroups SE(''n'') and ~~DSE~~(''n'') further contain translation subgroups T(''n'') and DT(''n''), respectively.		The groups E(''n'') and RE(''n'') are isomorphic, and they each contain the orthogonal group O(''n'') as a common subgroup. The [[complement]] operation provides a two-way mapping between transforms associated with members of E(''n'') and RE(''n''). The groups E(''n'') and RE(''n'') have a number of subgroups, and the hierarchical relationships among them are shown in the figure below. In particular, the Euclidean group E(''n'') contains the special Euclidean subgroup 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 reciprocal Euclidean group RE(''n'') contains the reciprocal special Euclidean subgroup RSE(''n'') consisting of all combinations of [[reciprocal rotations]] and [[reciprocal translations]], which are covered by the elements of $$\mathcal G_{n,0,1}$$ having even grade. The subgroups SE(''n'') and RSE(''n'') further contain translation subgroups T(''n'') and RT(''n''), respectively.

	[[Image:Groups.svg\|800px]]		[[Image:Groups.svg\|800px]]

	Transforms about invariants containing the origin are the same in both E(''n'') and DE(''n''), and they constitute the common subgroup O(''n''). Every member of 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 $$\mathbf x' = \mathbf q \mathbin{\unicode{x27C7}} \mathbf x \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{q}}}$$.		Transforms about invariants containing the origin are the same in both E(''n'') and RE(''n''), and they constitute the common subgroup O(''n''). Every member of 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 E(''n'') transforms a point by multiplying on the left by an $$(n + 1) \times (n + 1)$$ matrix of the form		In terms of matrix multiplication, a general element of the group E(''n'') transforms a point by multiplying on the left by an $$(n + 1) \times (n + 1)$$ matrix of the form
Line 31:		Line 31:
	:$$\begin{bmatrix} \mathbf m_{n \times n} & \mathbf t_{n \times 1} \\ \mathbf 0_{1 \times n} & 1 \end{bmatrix}$$ ,		:$$\begin{bmatrix} \mathbf m_{n \times n} & \mathbf t_{n \times 1} \\ \mathbf 0_{1 \times n} & 1 \end{bmatrix}$$ ,

	where the $$n \times n$$ submatrix $$\mathbf m$$ is orthogonal. A general element of the corresponding group DE(''n'') transforms points with matrices of the form		where the $$n \times n$$ submatrix $$\mathbf m$$ is orthogonal. A general element of the corresponding group RE(''n'') transforms points with matrices of the form

	:$$\begin{bmatrix} \mathbf m_{n \times n} & \mathbf 0_{n \times 1} \\ \mathbf t_{1 \times n} & 1 \end{bmatrix}$$ .		:$$\begin{bmatrix} \mathbf m_{n \times n} & \mathbf 0_{n \times 1} \\ \mathbf t_{1 \times n} & 1 \end{bmatrix}$$ .

	In the special subgroups SE(''n'') and ~~DSE~~(''n''), the submatrix $$\mathbf m$$ has a determinant of +1. In the translation subgroups T(''n'') and DT(''n''), $$\mathbf m$$ is the identity matrix. Finally, when $$\mathbf t = \mathbf 0$$, the two matrices above have the same form and belong to O(''n'').		In the special subgroups SE(''n'') and RSE(''n''), the submatrix $$\mathbf m$$ has a determinant of +1. In the translation subgroups T(''n'') and RT(''n''), $$\mathbf m$$ is the identity matrix. Finally, when $$\mathbf t = \mathbf 0$$, the two matrices above have the same form and belong to O(''n'').

	The isomorphic mapping between E(''n'') and DE(''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 E(''n''), then the corresponding element of DE(''n'') is given by $$(\mathbf M^{-1})^{\text T}$$. Of course, this operation is an involution, and the mapping works both ways.		The isomorphic mapping between E(''n'') and RE(''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 E(''n''), then the corresponding element of RE(''n'') is given by $$(\mathbf M^{-1})^{\text T}$$. Of course, this operation is an involution, and the mapping works both ways.

	== See Also ==		== See Also ==