Comparison of vector algebra and geometric algebra

Vector algebra and geometric algebra are alternative approaches to providing additional algebraic structures on vector spaces, with geometric interpretations, particularly vector fields in multivariable calculus and applications in mathematical physics.

Vector algebra is simpler, but specific to Euclidean 3-space, while geometric algebra uses multilinear algebra, but works in all dimensions and signatures, notably 3+1 spacetime, as well as 2 dimensions, and is mathematically more elegant; they are mathematically equivalent in 3 dimensions, though the approaches differ. Vector algebra is more widely used in elementary multivariable calculus, while geometric algebra is used in some more advanced treatments, and is proposed for elementary use as well. In advanced mathematics, particularly differential geometry, neither is widely used, with differential forms being far more widely used.

Basic concepts and operations

In vector algebra the basic objects are scalars and vectors, and the operations (beyond the vector space operations of scalar multiplication and vector addition) are the dot product and the cross product &times.

In geometric algebra the basic objects are multivectors (scalars are 0-vectors, vectors are 1-vectors), and the operations are the dot product/inner product/scalar product, the exterior product, and the Clifford product (here called "geometric product"), which is the sum of the inner product and exterior product.

Most basically, vector algebra uses the cross product, while geometric algebra uses the exterior product (and the geometric product, which is obtained from the inner and exterior product). More subtly, geometric algebra in Euclidean 3-space distinguishes 0-vectors, 1-vectors, 2-vectors, and 3-vectors, while elementary vector algebra identifies 1-vectors and 2-vectors (as vectors) and 0-vectors and 3-vectors (as scalars), though more advanced vector algebra distinguishes these as scalars, vectors, pseudovectors, and pseudoscalars.

The cross product does not generalize to higher dimensions (as a product of two vectors, yielding a third vector), and in higher dimensions not all k-vectors can be identified with vectors or scalars. By contrast, the exterior product (and geometric product) is defined uniformly for all dimensions and signatures, and multivectors are closed under these operations.

Embellishments, ad hoc techniques, and tricks

More advanced treatments of vector algebra add embellishments to the initial picture – pseudovectors and pseudoscalars (in geometric algebra terms, 2-vectors and 3-vectors), while applications to other dimensions use ad hoc techniques and "tricks" rather than a general mathematical approach. By contrast, geometric algebra begins with a complete picture, and applies uniformly in all dimensions.

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. The torque or curl is then a normal vector field in this 3rd dimension. By contrast, geometric algebra in 2 dimensions defines torque and curl as pseudoscalar fields (2-vector fields), 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.

List of analogous formulas

Here are some comparisons between standard ${\mathbb {R} }^{3}$ vector relations and their corresponding wedge and geometric product equivalents. All the wedge 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 wedge product to generalize, but since that may not be familiar to somebody with only a traditional background in vector algebra and calculus, some examples are given.

Algebraic and geometric properties of cross and wedge products

Cross and wedge products are both antisymmetric:

$\mathbf {v} \times \mathbf {u} =-(\mathbf {u} \times \mathbf {v} )$

$\mathbf {v} \wedge \mathbf {u} =-(\mathbf {u} \wedge \mathbf {v} )$

They are both linear in the first operand

$(\mathbf {u} +\mathbf {v} )\times \mathbf {w} =\mathbf {u} \times \mathbf {w} +\mathbf {v} \times \mathbf {w}$

$(\mathbf {u} +\mathbf {v} )\wedge \mathbf {w} =\mathbf {u} \wedge \mathbf {w} +\mathbf {v} \wedge \mathbf {w}$

and in the second operand

$\mathbf {u} \times (\mathbf {v} +\mathbf {w} )=\mathbf {u} \times \mathbf {v} +\mathbf {u} \times \mathbf {w}$

$\mathbf {u} \wedge (\mathbf {v} +\mathbf {w} )=\mathbf {u} \wedge \mathbf {v} +\mathbf {u} \wedge \mathbf {w}$

In general, the cross product is not associative, while the wedge product is

$(\mathbf {u} \times \mathbf {v} )\times \mathbf {w} \neq \mathbf {u} \times (\mathbf {v} \times \mathbf {w} )$

$(\mathbf {u} \wedge \mathbf {v} )\wedge \mathbf {w} =\mathbf {u} \wedge (\mathbf {v} \wedge \mathbf {w} )$

Both the cross and wedge products of two identical vectors are zero:

$\mathbf {u} \times \mathbf {u} =0$

$\mathbf {u} \wedge \mathbf {u} =0$

$\mathbf {u} \times \mathbf {v}$ is perpendicular to the plane containing $\mathbf {u}$ and $\mathbf {v}$.
$\mathbf {u} \wedge \mathbf {v}$ is an oriented representation of the same plane.

Norm of a vector

The norm (length) of a vector is defined in terms of the dot product

${\Vert \mathbf {u} \Vert }^{2}=\mathbf {u} \cdot \mathbf {u}$

Using the geometric product this is also true, but this can be also be expressed more compactly as

${\Vert \mathbf {u} \Vert }^{2}={\mathbf {u} }^{2}$

This follows from the definition of the geometric product and the fact that a vector wedge product with itself is zero

$\mathbf {u} \,\mathbf {u} =\mathbf {u} \cdot \mathbf {u} +\mathbf {u} \wedge \mathbf {u} =\mathbf {u} \cdot \mathbf {u}$

Lagrange identity

In three dimensions the product of two vector lengths can be expressed in terms of the dot and cross products

${\Vert \mathbf {u} \Vert }^{2}{\Vert \mathbf {v} \Vert }^{2}=({\mathbf {u} \cdot \mathbf {v} })^{2}+{\Vert \mathbf {u} \times \mathbf {v} \Vert }^{2}$

The corresponding generalization expressed using the geometric product is

${\Vert \mathbf {u} \Vert }^{2}{\Vert \mathbf {v} \Vert }^{2}=({\mathbf {u} \cdot \mathbf {v} })^{2}-(\mathbf {u} \wedge \mathbf {v} )^{2}$

This follows from expanding the geometric product of a pair of vectors with its reverse

$(\mathbf {u} \mathbf {v} )(\mathbf {v} \mathbf {u} )=({\mathbf {u} \cdot \mathbf {v} }+{\mathbf {u} \wedge \mathbf {v} })({\mathbf {u} \cdot \mathbf {v} }-{\mathbf {u} \wedge \mathbf {v} })$

Determinant expansion of cross and wedge products

$\mathbf {u} \times \mathbf {v} =\sum _{i<j}{{\begin{vmatrix}u_{i}&u_{j}\\v_{i}&v_{j}\end{vmatrix}}{\mathbf {e} }_{i}\times {\mathbf {e} }_{j}}$

$\mathbf {u} \wedge \mathbf {v} =\sum _{i<j}{{\begin{vmatrix}u_{i}&u_{j}\\v_{i}&v_{j}\end{vmatrix}}{\mathbf {e} }_{i}\wedge {\mathbf {e} }_{j}}$

Without justification or historical context, traditional linear algebra texts will often define the determinant as the first step of an elaborate sequence of definitions and theorems leading up to the solution of linear systems, Cramer's rule and matrix inversion.

An alternative treatment is to axiomatically introduce the wedge product, and then demonstrate that this can be used directly to solve linear systems. This is shown below, and does not require sophisticated math skills to understand.

It is then possible to define determinants as nothing more than the coefficients of the wedge product in terms of "unit k-vectors" (${\mathbf {e} }_{i}\wedge {\mathbf {e} }_{j}$ terms) expansions as above.

A one by one determinant is the coefficient of $\mathbf {e} _{1}$ for an $\mathbb {R} ^{1}$ 1-vector.

A two-by-two determinant is the coefficient of $\mathbf {e} _{1}\wedge \mathbf {e} _{2}$ for an $\mathbb {R} ^{2}$ bivector

A three-by-three determinant is the coefficient of $\mathbf {e} _{1}\wedge \mathbf {e} _{2}\wedge \mathbf {e} _{3}$ for an $\mathbb {R} ^{3}$ trivector

...

When linear system solution is introduced via the wedge product, Cramer's rule follows as a side effect, and there is no need to lead up to the end results with definitions of minors, matrices, matrix invertibility, adjoints, cofactors, Laplace expansions, theorems on determinant multiplication and row column exchanges, and so forth.

Equation of a plane

For the plane of all points ${\mathbf {r} }$ through the plane passing through three independent points ${\mathbf {r} }_{0}$, ${\mathbf {r} }_{1}$, and ${\mathbf {r} }_{2}$, the normal form of the equation is

$(({\mathbf {r} }_{2}-{\mathbf {r} }_{0})\times ({\mathbf {r} }_{1}-{\mathbf {r} }_{0}))\cdot ({\mathbf {r} }-{\mathbf {r} }_{0})=0.$

The equivalent wedge product equation is

$({\mathbf {r} }_{2}-{\mathbf {r} }_{0})\wedge ({\mathbf {r} }_{1}-{\mathbf {r} }_{0})\wedge ({\mathbf {r} }-{\mathbf {r} }_{0})=0.$

Projective and rejective components of a vector

For three dimensions the projective and rejective components of a vector with respect to an arbitrary non-zero unit vector, can be expressed in terms of the dot and cross product

$\mathbf {v} =(\mathbf {v} \cdot {\hat {\mathbf {u} }}){\hat {\mathbf {u} }}+{\hat {\mathbf {u} }}\times (\mathbf {v} \times {\hat {\mathbf {u} }})$

For the general case the same result can be written in terms of the dot and wedge product and the geometric product of that and the unit vector

$\mathbf {v} =(\mathbf {v} \cdot {\hat {\mathbf {u} }}){\hat {\mathbf {u} }}+(\mathbf {v} \wedge {\hat {\mathbf {u} }}){\hat {\mathbf {u} }}$

It's also worthwhile to point out that this result can also be expressed using right or left vector division as defined by the geometric product

$\mathbf {v} =(\mathbf {v} \cdot \mathbf {u} ){\frac {1}{\mathbf {u} }}+(\mathbf {v} \wedge \mathbf {u} ){\frac {1}{\mathbf {u} }}$

$\mathbf {v} ={\frac {1}{\mathbf {u} }}(\mathbf {u} \cdot \mathbf {v} )+{\frac {1}{\mathbf {u} }}(\mathbf {u} \wedge \mathbf {v} )$

Area of the parallelogram defined by u and v

If A is the area of the parallelogram defined by u and v, then

$A^{2}={\Vert \mathbf {u} \times \mathbf {v} \Vert }^{2}=\sum _{i<j}{\begin{vmatrix}u_{i}&u_{j}\\v_{i}&v_{j}\end{vmatrix}}^{2},$

and

$A^{2}=-(\mathbf {u} \wedge \mathbf {v} )^{2}=\sum _{i<j}{\begin{vmatrix}u_{i}&u_{j}\\v_{i}&v_{j}\end{vmatrix}}^{2}.$

Note that this squared bivector is a geometric multiplication; this computation can alternatively be stated as the Gram determinant of the two vectors.

Angle between two vectors

$({\sin \theta })^{2}={\frac {{\Vert \mathbf {u} \times \mathbf {v} \Vert }^{2}}{{\Vert \mathbf {u} \Vert }^{2}{\Vert \mathbf {v} \Vert }^{2}}}$

$({\sin \theta })^{2}=-{\frac {(\mathbf {u} \wedge \mathbf {v} )^{2}}{{\mathbf {u} }^{2}{\mathbf {v} }^{2}}}$

Volume of the parallelopiped formed by three vectors

In vector algebra, the volume of a parallelopiped is given by the square root of the norm of the scalar triple product:

$V^{2}={\Vert (\mathbf {u} \times \mathbf {v} )\cdot \mathbf {w} \Vert }^{2}={\begin{vmatrix}u_{1}&u_{2}&u_{3}\\v_{1}&v_{2}&v_{3}\\w_{1}&w_{2}&w_{3}\\\end{vmatrix}}^{2}$

$V^{2}=-(\mathbf {u} \wedge \mathbf {v} \wedge \mathbf {w} )^{2}=-\left(\sum _{i<j<k}{\begin{vmatrix}u_{i}&u_{j}&u_{k}\\v_{i}&v_{j}&v_{k}\\w_{i}&w_{j}&w_{k}\\\end{vmatrix}}{\hat {\mathbf {e} }}_{i}\wedge {\hat {\mathbf {e} }}_{j}\wedge {\hat {\mathbf {e} }}_{k}\right)^{2}=\sum _{i<j<k}{\begin{vmatrix}u_{i}&u_{j}&u_{k}\\v_{i}&v_{j}&v_{k}\\w_{i}&w_{j}&w_{k}\\\end{vmatrix}}^{2}$

Derivative of a unit vector

It can be shown that a unit vector derivative can be expressed using the cross product

${\frac {d}{dt}}\left({\frac {\mathbf {r} }{\Vert \mathbf {r} \Vert }}\right)={\frac {1}{{\Vert \mathbf {r} \Vert }^{3}}}\left(\mathbf {r} \times {\frac {d\mathbf {r} }{dt}}\right)\times \mathbf {r} =\left({\hat {\mathbf {r} }}\times {\frac {1}{\Vert \mathbf {r} \Vert }}{\frac {d\mathbf {r} }{dt}}\right)\times {\hat {\mathbf {r} }}$

The equivalent geometric product generalization is

${\frac {d}{dt}}\left({\frac {\mathbf {r} }{\Vert \mathbf {r} \Vert }}\right)={\frac {1}{{\Vert \mathbf {r} \Vert }^{3}}}\mathbf {r} \left(\mathbf {r} \wedge {\frac {d\mathbf {r} }{dt}}\right)={\frac {1}{\mathbf {r} }}\left({\hat {\mathbf {r} }}\wedge {\frac {d\mathbf {r} }{dt}}\right)$

Thus this derivative is the component of ${\frac {1}{\Vert \mathbf {r} \Vert }}{\frac {d\mathbf {r} }{dt}}$ in the direction perpendicular to $\mathbf {r}$. In other words this is ${\frac {1}{\Vert \mathbf {r} \Vert }}{\frac {d\mathbf {r} }{dt}}$ minus the projection of that vector onto $\mathbf {\hat {r}}$.

This intuitively makes sense (but a picture would help) since a unit vector is constrained to circular motion, and any change to a unit vector due to a change in its generating vector has to be in the direction of the rejection of $\mathbf {\hat {r}}$ from ${\frac {d\mathbf {r} }{dt}}$. That rejection has to be scaled by 1/|r| to get the final result.

When the objective isn't comparing to the cross product, it's also notable that this unit vector derivative can be written

${\mathbf {r} }{\frac {d{\hat {\mathbf {r} }}}{dt}}={\hat {\mathbf {r} }}\wedge {\frac {d\mathbf {r} }{dt}}$

See also

• List of vector identities
• Vector calculus identities