Geometric algebra is an extension of vector algebra, providing additional algebraic structures on vector spaces, with geometric interpretations. Vector algebra uses all dimensions and signatures, as does geometric algebra, notably 3+1 spacetime as well as 2 dimensions.
Basic concepts and operations Geometric algebra (GA) is an extension or completion of vector algebra (VA). The reader is herein assumed to be familiar with the basic concepts and operations of VA and this article will mainly concern itself with operations in G 3 {\displaystyle {\mathcal {G}}_{3}} the GA of 3D space (nor is this article intended to be mathematically rigorous). In GA, vectors are not normally written boldface as the meaning is usually clear from the context. The fundamental difference is that GA provides a new product of vectors called the "geometric product". Elements of GA are graded multivectors: scalars are grade 0, usual vectors are grade 1, bivectors are grade 2 and the highest grade (3 in the 3D case) is traditionally called the pseudoscalar and designated I {\displaystyle I} . The ungeneralized 3D vector form of the geometric product is:
a b = a ⋅ b + a ∧ b {\displaystyle ab=a\cdot b+a\wedge b}
that is the sum of the usual dot (inner) product and the outer (exterior) product (this last is closely related to the cross product and will be explained below). In VA, entities such as pseudovectors and pseudoscalars need to be bolted on, whereas in GA the equivalent bivector and pseudovector respectively exist naturally as subspaces of the algebra. For example, applying vector calculus in 2 dimensions, such as to compute torque or curl, requires adding an artificial 3rd dimension and extending the vector field to be constant in that dimension, or alternately considering these to be scalars. The torque or curl is then a normal vector field in this 3rd dimension. By contrast, geometric algebra in 2 dimensions defines these as a pseudoscalar field (a bivector), without requiring a 3rd dimension. Similarly, the scalar triple product is ad hoc, and can instead be expressed uniformly using the exterior product and the geometric product.
Translations between formalisms Here are some comparisons between standard R 3 {\displaystyle {\mathbb {R} }^{3}} vector relations and their corresponding exterior product and geometric product equivalents. All the exterior and geometric product equivalents here are good for more than three dimensions, and some also for two. In two dimensions the cross product is undefined even if what it describes (like torque) is perfectly well defined in a plane without introducing an arbitrary normal vector outside of the space. Many of these relationships only require the introduction of the exterior product to generalize, but since that may not be familiar to somebody with only a background in vector algebra and calculus, some examples are given.
Cross and exterior products
u × v {\displaystyle \mathbf {u} \times \mathbf {v} } is perpendicular to the plane containing u {\displaystyle \mathbf {u} } and v {\displaystyle \mathbf {v} } .
u ∧ v {\displaystyle \mathbf {u} \wedge \mathbf {v} } is an oriented representation of the same plane. We have the pseudoscalar I = e 1 e 2 e 3 {\displaystyle I=e_{1}e_{2}e_{3}} (right handed orthonormal frame) and so
e 1 I = I e 1 = e 2 e 3 {\displaystyle e_{1}I=Ie_{1}=e_{2}e_{3}} returns a bivector and
I ( e 2 ∧ e 3 ) = I e 2 e 3 = − e 1 {\displaystyle I(e_{2}\wedge e_{3})=Ie_{2}e_{3}=-e_{1}} returns a vector perpendicular to the e 2 ∧ e 3 {\displaystyle e_{2}\wedge e_{3}} plane. This yields a convenient definition for the cross product of traditional vector algebra:
u × v = − I ( u ∧ v ) {\displaystyle {u}\times {v}=-I({u}\wedge {v})}
(this is antisymmetric). Relevant is the distinction between polar and axial vectors in vector algebra, which is natural in geometric algebra as the distinction between vectors and bivectors (elements of grade two). The I {\displaystyle I} here is a unit pseudoscalar of Euclidean 3-space, which establishes a duality between the vectors and the bivectors, and is named so because of the expected property
I 2 = ( e 1 e 2 e 3 ) 2 = e 1 e 2 e 3 e 1 e 2 e 3 = − e 1 e 2 e 1 e 3 e 2 e 3 = e 1 e 1 e 2 e 3 e 2 e 3 = − e 3 e 2 e 2 e 3 = − 1 {\displaystyle I^{2}=(e_{1}e_{2}e_{3})^{2}=e_{1}e_{2}e_{3}e_{1}e_{2}e_{3}=-e_{1}e_{2}e_{1}e_{3}e_{2}e_{3}=e_{1}e_{1}e_{2}e_{3}e_{2}e_{3}=-e_{3}e_{2}e_{2}e_{3}=-1}
The equivalence of the R 3 {\displaystyle \mathbb {R} ^{3}} cross product and the exterior product expression above can be confirmed by direct multiplication of − I = − e 1 e 2 e 3 {\displaystyle -I=-{e_{1}}{e_{2}}{e_{3}}} with a determinant expansion of the exterior product
u ∧ v = ∑ 1 ≤ i < j ≤ 3 ( u i v j − v i u j ) e i ∧ e j = ∑ 1 ≤ i < j ≤ 3 ( u i v j − v i u j ) e i e j {\displaystyle u\wedge v=\sum _{1\leq i<j\leq 3}(u_{i}v_{j}-v_{i}u_{j}){e_{i}}\wedge {e_{j}}=\sum _{1\leq i<j\leq 3}(u_{i}v_{j}-v_{i}u_{j}){e_{i}}{e_{j}}}
See also Cross product as an exterior product. Essentially, the geometric product of a bivector and the pseudoscalar of Euclidean 3-space provides a method of calculation of the Hodge dual.
Cross and commutator products
The pseudovector/bivector subalgebra of the geometric algebra of Euclidean 3-dimensional space form a 3-dimensional vector space themselves. Let the standard unit pseudovectors/bivectors of the subalgebra be i = e 2 e 3 {\displaystyle \mathbf {i} =\mathbf {e_{2}} \mathbf {e_{3}} } , j = e 1 e 3 {\displaystyle \mathbf {j} =\mathbf {e_{1}} \mathbf {e_{3}} } , and k = e 1 e 2 {\displaystyle \mathbf {k} =\mathbf {e_{1}} \mathbf {e_{2}} } , and the anti-commutative commutator product be defined as A × B = 1 2 ( A B − B A ) {\displaystyle A\times B={\tfrac {1}{2}}(AB-BA)} , where A B {\displaystyle AB} is the geometric product. The commutator product is distributive over addition and linear, as the geometric product is distributive over addition and linear. From the definition of the commutator product, i {\displaystyle \mathbf {i} } , j {\displaystyle \mathbf {j} } and k {\displaystyle \mathbf {k} } satisfy the following equalities:
i × j = 1 2 ( i j − j i ) = 1 2 ( ( e 2 e 3 e 1 e 3 − e 1 e 3 e 2 e 3 ) = 1 2 ( − e 2 e 3 e 3 e 1 + e 1 e 3 e 3 e 2 ) = 1 2 ( − e 2 e 1 + e 1 e 2 ) = 1 2 ( e 1 e 2 + e 1 e 2 ) = e 1 e 2 = k {\displaystyle \mathbf {i} \times \mathbf {j} ={\tfrac {1}{2}}(\mathbf {i} \mathbf {j} -\mathbf {j} \mathbf {i} )={\tfrac {1}{2}}((\mathbf {e_{2}} \mathbf {e_{3}} \mathbf {e_{1}} \mathbf {e_{3}} -\mathbf {e_{1}} \mathbf {e_{3}} \mathbf {e_{2}} \mathbf {e_{3}} )={\tfrac {1}{2}}(-\mathbf {e_{2}} \mathbf {e_{3}} \mathbf {e_{3}} \mathbf {e_{1}} +\mathbf {e_{1}} \mathbf {e_{3}} \mathbf {e_{3}} \mathbf {e_{2}} )={\tfrac {1}{2}}(-\mathbf {e_{2}} \mathbf {e_{1}} +\mathbf {e_{1}} \mathbf {e_{2}} )={\tfrac {1}{2}}(\mathbf {e_{1}} \mathbf {e_{2}} +\mathbf {e_{1}} \mathbf {e_{2}} )=\mathbf {e_{1}} \mathbf {e_{2}} =\mathbf {k} }
j × k = 1 2 ( j k − k j ) = 1 2 ( ( e 1 e 3 e 1 e 2 − e 1 e 2 e 1 e 3 ) = 1 2 ( − e 3 e 1 e 1 e 2 + e 2 e 1 e 1 e 3 ) = 1 2 ( − e 3 e 2 + e 2 e 3 ) = 1 2 ( e 2 e 3 + e 2 e 3 ) = e 2 e 3 = i {\displaystyle \mathbf {j} \times \mathbf {k} ={\tfrac {1}{2}}(\mathbf {j} \mathbf {k} -\mathbf {k} \mathbf {j} )={\tfrac {1}{2}}((\mathbf {e_{1}} \mathbf {e_{3}} \mathbf {e_{1}} \mathbf {e_{2}} -\mathbf {e_{1}} \mathbf {e_{2}} \mathbf {e_{1}} \mathbf {e_{3}} )={\tfrac {1}{2}}(-\mathbf {e_{3}} \mathbf {e_{1}} \mathbf {e_{1}} \mathbf {e_{2}} +\mathbf {e_{2}} \mathbf {e_{1}} \mathbf {e_{1}} \mathbf {e_{3}} )={\tfrac {1}{2}}(-\mathbf {e_{3}} \mathbf {e_{2}} +\mathbf {e_{2}} \mathbf {e_{3}} )={\tfrac {1}{2}}(\mathbf {e_{2}} \mathbf {e_{3}} +\mathbf {e_{2}} \mathbf {e_{3}} )=\mathbf {e_{2}} \mathbf {e_{3}} =\mathbf {i} }
k × i = 1 2 ( k i − i k ) = 1 2 ( e 1 e 2 e 2 e 3 − e 2 e 3 e 1 e 2 ) = 1 2 ( e 1 e 2 e 2 e 3 − e 3 e 2 e 2 e 1 ) = 1 2 ( e 1 e 3 − e 3 e 1 ) = 1 2 ( e 1 e 3 + e 1 e 3 ) = e 1 e 3 = j {\displaystyle \mathbf {k} \times \mathbf {i} ={\tfrac {1}{2}}(\mathbf {k} \mathbf {i} -\mathbf {i} \mathbf {k} )={\tfrac {1}{2}}(\mathbf {e_{1}} \mathbf {e_{2}} \mathbf {e_{2}} \mathbf {e_{3}} -\mathbf {e_{2}} \mathbf {e_{3}} \mathbf {e_{1}} \mathbf {e_{2}} )={\tfrac {1}{2}}(\mathbf {e_{1}} \mathbf {e_{2}} \mathbf {e_{2}} \mathbf {e_{3}} -\mathbf {e_{3}} \mathbf {e_{2}} \mathbf {e_{2}} \mathbf {e_{1}} )={\tfrac {1}{2}}(\mathbf {e_{1}} \mathbf {e_{3}} -\mathbf {e_{3}} \mathbf {e_{1}} )={\tfrac {1}{2}}(\mathbf {e_{1}} \mathbf {e_{3}} +\mathbf {e_{1}} \mathbf {e_{3}} )=\mathbf {e_{1}} \mathbf {e_{3}} =\mathbf {j} }
which imply, by the anti-commutativity of the commutator product, that
j × i = − k {\displaystyle \mathbf {j} \times \mathbf {i} =-\mathbf {k} }
k × j = − i {\displaystyle \mathbf {k} \times \mathbf {j} =-\mathbf {i} }
i × k = − j {\displaystyle \mathbf {i} \times \mathbf {k} =-\mathbf {j} }
The anti-commutativity of the commutator product also implies that
i × i = j × j = k × k = 0 {\displaystyle \mathbf {i} \times \mathbf {i} =\mathbf {j} \times \mathbf {j} =\mathbf {k} \times \mathbf {k} =0}
These equalities and properties are sufficient to determine the commutator product of any two pseudovectors/bivectors A {\displaystyle \mathbf {A} } and B {\displaystyle \mathbf {B} } . As the pseudovectors/bivectors form a vector space, each pseudovector/bivector can be defined as the sum of three orthogonal components parallel to the standard basis pseudovectors/bivectors:
A = ( A 1 i + A 2 j + A 3 k ) {\displaystyle \mathbf {A} =(A_{1}\mathbf {i} +A_{2}\mathbf {j} +A_{3}\mathbf {k} )}
B = ( B 1 i + B 2 j + B 3 k ) {\displaystyle \mathbf {B} =(B_{1}\mathbf {i} +B_{2}\mathbf {j} +B_{3}\mathbf {k} )}
Their commutator product A × B {\displaystyle \mathbf {A} \times \mathbf {B} } can be expanded using its distributive property:
A × B = ( A 1 i + A 2 j + A 3 k ) × ( B 1 i + B 2 j + B 3 k ) = A 1 B 1 i × i + A 1 B 2 i × j + A 1 B 3 i × k + A 2 B 1 j × i + A 2 B 2 j × j + A 2 B 3 j × k + A 3 B 1 k × i + A 3 B 2 k × j + A 3 B 3 k × k = A 1 B 2 k − A 1 B 3 j − A 2 B 1 k + A 2 B 3 i + A 3 B 1 j − A 3 B 2 i = ( A 2 B 3 − A 3 B 2 ) i + ( A 3 B 1 − A 1 B 3 ) j + ( A 1 B 2 − A 2 B 1 ) k {\displaystyle {\begin{aligned}\mathbf {A} \times \mathbf {B} &=(A_{1}\mathbf {i} +A_{2}\mathbf {j} +A_{3}\mathbf {k} )\times (B_{1}\mathbf {i} +B_{2}\mathbf {j} +B_{3}\mathbf {k} )\\&=A_{1}B_{1}\mathbf {i} \times \mathbf {i} +A_{1}B_{2}\mathbf {i} \times \mathbf {j} +A_{1}B_{3}\mathbf {i} \times \mathbf {k} +A_{2}B_{1}\mathbf {j} \times \mathbf {i} +A_{2}B_{2}\mathbf {j} \times \mathbf {j} +A_{2}B_{3}\mathbf {j} \times \mathbf {k} +A_{3}B_{1}\mathbf {k} \times \mathbf {i} +A_{3}B_{2}\mathbf {k} \times \mathbf {j} +A_{3}B_{3}\mathbf {k} \times \mathbf {k} \\&=A_{1}B_{2}\mathbf {k} -A_{1}B_{3}\mathbf {j} -A_{2}B_{1}\mathbf {k} +A_{2}B_{3}\mathbf {i} +A_{3}B_{1}\mathbf {j} -A_{3}B_{2}\mathbf {i} =(A_{2}B_{3}-A_{3}B_{2})\mathbf {i} +(A_{3}B_{1}-A_{1}B_{3})\mathbf {j} +(A_{1}B_{2}-A_{2}B_{1})\mathbf {k} \end{aligned}}}
which is precisely the cross product in vector algebra for pseudovectors.
Norm of a vector Ordinarily,
‖ u ‖ 2 = u ⋅ u {\displaystyle {\Vert \mathbf {u} \Vert }^{2}=\mathbf {u} \cdot \mathbf {u} }
Making use of the geometric product and the fact that the exterior product of a vector with itself is zero:
u u = ‖ u ‖ 2 = u 2 = u ⋅ u + u ∧ u = u ⋅ u {\displaystyle \mathbf {u} \,\mathbf {u} ={\Vert \mathbf {u} \Vert }^{2}={\mathbf {u} }^{2}=\mathbf {u} \cdot \mathbf {u} +\mathbf {u} \wedge \mathbf {u} =\mathbf {u} \cdot \mathbf {u} }
Lagrange identity In three dimensions the product of two vector lengths can be expressed in terms of the dot and cross products
‖ u ‖ 2 ‖ v ‖ 2 = ( u ⋅ v ) 2 + ‖ u × v ‖ 2 {\displaystyle {\Vert \mathbf {u} \Vert }^{2}{\Vert \mathbf {v} \Vert }^{2}=({\mathbf {u} \cdot \mathbf {v} })^{2