Geometric constraint: Difference between revisions

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An element $$\mathbf x$$ of a geometric algebra possesses the ''geometric property'' if and only if the [[geometric product]] between $$\mathbf u$$ and its own reverse is a scalar, which is given by the [[dot product]], and the [[geometric antiproduct]] between $$\mathbf u$$ and its own antireverse is an antiscalar, which is given by the [[antidot product]]. That is,
An element $$\mathbf x$$ of a geometric algebra possesses the ''geometric constraint'' if and only if the [[geometric product]] between $$\mathbf u$$ and its own reverse is a scalar, which is given by the [[dot product]], and the [[geometric antiproduct]] between $$\mathbf u$$ and its own antireverse is an antiscalar, which is given by the [[antidot product]]. That is,


:$$\mathbf u \mathbin{\unicode{x27D1}} \mathbf{\tilde u} = \mathbf u \mathbin{\unicode{x25CF}} \mathbf u$$
:$$\mathbf u \mathbin{\unicode{x27D1}} \mathbf{\tilde u} = \mathbf u \mathbin{\unicode{x25CF}} \mathbf u$$
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:$$\mathbf u \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{u}}} = \mathbf u \mathbin{\unicode{x25CB}} \mathbf u$$ .
:$$\mathbf u \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{u}}} = \mathbf u \mathbin{\unicode{x25CB}} \mathbf u$$ .


The set of all elements possessing the geometric property is closed under both the [[geometric product]] and [[geometric antiproduct]].
The set of all elements satisfying the geometric constraint is closed under both the [[geometric product]] and [[geometric antiproduct]].


The following table lists the requirements that must be satisfied for the main types in the 4D rigid geometric algebra $$\mathcal G_{3,0,1}$$ to possess the geometric property. Points and planes do not have any requirements—they all possess the geometric property.
The following table lists the requirements that must be satisfied for the main types in the 4D rigid geometric algebra $$\mathcal G_{3,0,1}$$ due to the geometric constraint. Points and planes do not have any requirements—they have no constraints.


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Revision as of 01:00, 9 February 2024

An element $$\mathbf x$$ of a geometric algebra possesses the geometric constraint if and only if the geometric product between $$\mathbf u$$ and its own reverse is a scalar, which is given by the dot product, and the geometric antiproduct between $$\mathbf u$$ and its own antireverse is an antiscalar, which is given by the antidot product. That is,

$$\mathbf u \mathbin{\unicode{x27D1}} \mathbf{\tilde u} = \mathbf u \mathbin{\unicode{x25CF}} \mathbf u$$

and

$$\mathbf u \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{u}}} = \mathbf u \mathbin{\unicode{x25CB}} \mathbf u$$ .

The set of all elements satisfying the geometric constraint is closed under both the geometric product and geometric antiproduct.

The following table lists the requirements that must be satisfied for the main types in the 4D rigid geometric algebra $$\mathcal G_{3,0,1}$$ due to the geometric constraint. Points and planes do not have any requirements—they have no constraints.

Type Definition Requirement
Magnitude $$\mathbf z = x \mathbf 1 + y {\large\unicode{x1d7d9}}$$ $$xy = 0$$
Point $$\mathbf p = p_x \mathbf e_1 + p_y \mathbf e_2 + p_z \mathbf e_3 + p_w \mathbf e_4$$
Line $$\boldsymbol l = l_{vx} \mathbf e_{41} + l_{vy} \mathbf e_{42} + l_{vz} \mathbf e_{43} + l_{mx} \mathbf e_{23} + l_{my} \mathbf e_{31} + l_{mz} \mathbf e_{12}$$ $$l_{vx} l_{mx} + l_{vy} l_{my} + l_{vz} l_{mz} = 0$$
Plane $$\mathbf g = g_x \mathbf e_{423} + g_y \mathbf e_{431} + g_z \mathbf e_{412} + g_w \mathbf e_{321}$$
Motor $$\mathbf Q = Q_{vx} \mathbf e_{41} + Q_{vy} \mathbf e_{42} + Q_{vz} \mathbf e_{43} + Q_{vw} {\large\unicode{x1d7d9}} + Q_{mx} \mathbf e_{23} + Q_{my} \mathbf e_{31} + Q_{mz} \mathbf e_{12} + Q_{mw} \mathbf 1$$ $$Q_{vx} Q_{mx} + Q_{vy} Q_{my} + Q_{vz} Q_{mz} + Q_{vw} Q_{mw} = 0$$
Flector $$\mathbf F = F_{px} \mathbf e_1 + F_{py} \mathbf e_2 + F_{pz} \mathbf e_3 + F_{pw} \mathbf e_4 + F_{gx} \mathbf e_{423} + F_{gy} \mathbf e_{431} + F_{gz} \mathbf e_{412} + F_{gw} \mathbf e_{321}$$ $$F_{px} F_{gx} + F_{py} F_{gy} + F_{pz} F_{gz} + F_{pw} F_{gw} = 0$$

See Also