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The Wedge product is the multiplication operation in exterior algebra. The wedge product is always antisymmetric, associative, and anti-commutative. The result of the wedge product is known as a bivector; in $ \R^3 $ (that is, three dimensions) it is a 2-form. For two vectors u and v in $ \R^3 $, the wedge product is defined as

$ \mathbf{u} \wedge \mathbf{v} = \mathbf{u} \otimes \mathbf{v} - \mathbf{v} \otimes \mathbf{u} = \begin{bmatrix} 0 & u_1 v_2 - v_1 u_2 & u_1 v_3 - v_1 u_3 \\ u_2 v_1 - v_2 u_1 & 0 & u_2 v_3 - v_2 u_3 \\ u_3 v_1 - v_3 u_1 & u_3 v_2 - v_3 u_2 & 0 \end{bmatrix} $

where ⊗ denotes the outer product. Note that the bivector has only three indepedent elements; as such, it can be associated with another vector in $ \R^3 $. If the associated vector is defined as

$ \mathbf{n}= \begin{bmatrix} (\mathbf{u} \wedge \mathbf{v} )_{23} \\ -(\mathbf{u} \wedge \mathbf{v} )_{13} \\ (\mathbf{u} \wedge \mathbf{v} )_{32} \end{bmatrix} = \mathbf{u} \times \mathbf{v} $

it is the same as the cross product of u and v. In this sense, the cross product is a special case of the exterior product which is in turn a special case of the commutator product (See below).

Cross product and wedge product when written as determinants are calculated in the same way:

$ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix} \,,\quad \mathbf{a} \wedge \mathbf{b} = \begin{vmatrix} \mathbf{e}_{23} & \mathbf{e}_{31} & \mathbf{e}_{12}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}\ , $

so are related by the Hodge dual:

$ {* (\mathbf a \wedge \mathbf b )} = \mathbf {a \times b} \,,\quad {* (\mathbf {a \times b} )} = \mathbf a \wedge \mathbf b . $
Exterior calc cross product

A unit vector and a unit bivector are shown in red

Bivectors are skew-symmetric matrices which are the type of matrices used to calculate the cross product. Bivectors are not rotation matrices but an infinitesimal bivector (plus the identity matrix) can be used to perform an infinitesimal rotation. $ \mathbf{n} $ would therefore be the axis of rotation. See Rotation matrix#Determining the axis.

Properties

$ (\mathbf{a} \wedge \mathbf{b}) \wedge \mathbf{c} = \mathbf{a} \wedge (\mathbf{b} \wedge \mathbf{c}) $
$ (\mathbf{a} + \mathbf{b}) \wedge (\mathbf{c} + \mathbf{d}) = (\mathbf{a} \wedge \mathbf{c}) + (\mathbf{a} \wedge \mathbf{d}) + (\mathbf{b} \wedge \mathbf{c}) + (\mathbf{b} \wedge \mathbf{d}) $
$ \mathbf{u} \wedge \mathbf{v} = -\mathbf{v} \wedge \mathbf{u} $
$ \mathbf{u} \wedge \mathbf{u} = 0 $

Wedge product is distributive

The formula for the wedge product is:
$ \mathbf{u} \wedge \mathbf{v} = (u_1 \mathbf{e_1} + u_2 \mathbf{e_2} + u_3 \mathbf{e_3}) \wedge (v_1 \mathbf{e_1} + v_2 \mathbf{e_2} + v_3 \mathbf{e_3}) $

Using the distributive property this can be written out in full as:

$ \begin{align} \mathbf{u} \wedge \mathbf{v} = & &u_1 \mathbf{e_1} \wedge v_1 \mathbf{e_1}\quad &+ &u_1 \mathbf{e_1} \wedge v_2 \mathbf{e_2}\quad &+ &u_1 \mathbf{e_1} \wedge v_3 \mathbf{e_3} \\ &+ &u_2 \mathbf{e_2} \wedge v_1 \mathbf{e_1}\quad &+ &u_2 \mathbf{e_2} \wedge v_2 \mathbf{e_2}\quad &+ &u_2 \mathbf{e_2} \wedge v_3 \mathbf{e_3} \\ &+ &u_3 \mathbf{e_3} \wedge v_1 \mathbf{e_1}\quad &+ &u_3 \mathbf{e_3} \wedge v_2 \mathbf{e_2}\quad &+ &u_3 \mathbf{e_3} \wedge v_3 \mathbf{e_3} \\ \end{align} $


$ u_1 \mathbf{e_1} \wedge v_2 \mathbf{e_2} = \begin{bmatrix} 0 &u_1 v_2 &0 \\ -u_1 v_2 &0 &0 \\ 0 &0 &0 \end{bmatrix} $

$ u_1 \mathbf{e_1} \wedge v_3 \mathbf{e_3} = \begin{bmatrix} 0 &0 &u_1 v_3 \\ 0 &0 &0 \\ -u_1 v_3 &0 &0 \end{bmatrix} $

$ u_2 \mathbf{e_2} \wedge v_3 \mathbf{e_3} = \begin{bmatrix} 0 &0 &0 \\ 0 &0 &u_2 v_3 \\ 0 &-u_2 v_3 &0 \end{bmatrix} $

Rotating from v to u is the negative of rotating from u to v:

$ u_2 \mathbf{e_2} \wedge v_1 \mathbf{e_1} = -(v_1 \mathbf{e_1} \wedge u_2 \mathbf{e_2}) = \begin{bmatrix} 0 &-v_1 u_2 &0 \\ v_1 u_2 &0 &0 \\ 0 &0 &0 \end{bmatrix} $

$ u_3 \mathbf{e_3} \wedge v_1 \mathbf{e_1} = -(v_1 \mathbf{e_1} \wedge u_3 \mathbf{e_3}) = \begin{bmatrix} 0 &0 &-v_1 u_3 \\ 0 &0 &0 \\ v_1 u_3 &0 &0 \end{bmatrix} $

$ u_3 \mathbf{e_3} \wedge v_2 \mathbf{e_2} = -(v_2 \mathbf{e_2} \wedge u_3 \mathbf{e_3}) = \begin{bmatrix} 0 &0 &0 \\ 0 &0 &-v_2 u_3 \\ 0 &v_2 u_3 &0 \end{bmatrix} $

Each of these is equivalent to two tensor multiplications:

$ u_1 \mathbf{e_1} \wedge v_2 \mathbf{e_2} = u_1 \mathbf{e_1} \otimes v_2 \mathbf{e_2} - v_2 \mathbf{e_2} \otimes u_1 \mathbf{e_1} = \begin{bmatrix} 0 &u_1 v_2 &0 \\ 0 &0 &0 \\ 0 &0 &0 \end{bmatrix} - \begin{bmatrix} 0 &0 &0 \\ v_2 u_1 &0 &0 \\ 0 &0 &0 \end{bmatrix} $

Cross product

The vector cross product also can be expressed as the product of a skew-symmetric matrix and a vector:[1]

$ \mathbf{a} \times \mathbf{b} = \overline{\mathbf{a}} \mathbf{b} $

where $ \overline{\mathbf{a}} $ is the dual of $ \mathbf{a} $:

$ \overline{\mathbf{a}} \stackrel{\rm def}{=} \begin{bmatrix}\,\,0&\!-a_3&\,\,\,a_2\\\,\,\,a_3&0&\!-a_1\\\!-a_2&\,\,a_1&\,\,0\end{bmatrix} $

One actually has

$ \overline{(\mathbf{a \times b})} = \overline{\mathbf{a}}\overline{\mathbf{b}} - \overline{\mathbf{b}}\overline{\mathbf{a}} = [\overline{\mathbf{a}},\overline{\mathbf{b}}] $

i.e., the commutator of skew-symmetric three-by-three matrices can be identified with the cross-product of three-vectors.

$ [\overline{\mathbf{a}},\overline{\mathbf{b}}] = \overline{\mathbf{a}}\overline{\mathbf{b}} - \overline{\mathbf{b}}\overline{\mathbf{a}} = \left[ \begin{matrix} 0 & a_2b_1-b_2a_1 & a_3b_1-b_3a_1 \\ a_1b_2-b_1a_2 & 0 & a_3b_2-b_3a_2 \\ a_1b_3-b_1a_3 & a_2b_3-b_2a_3 & 0 \end{matrix}\right] = \mathbf{b} \wedge \mathbf{a} $

This result can be generalized to higher dimensions using geometric algebra. In particular in any dimension bivectors can be identified with skew-symmetric matrices, so the product between a skew-symmetric matrix and vector is equivalent to the grade-1 part of the product of a bivector and vector.[2] In three dimensions bivectors are dual to vectors so the product is equivalent to the cross product. In higher dimensions the product can still be calculated but bivectors have more degrees of freedom and are not equivalent to vectors.[2]


$ \overline{\mathbf{e_1}} = \mathbf{e_3} \wedge \mathbf{e_2} = \left[\begin{matrix}0&0&0\\0&0&-1\\0&1&0\end{matrix}\right] $
$ \overline{\mathbf{e_2}} = \mathbf{e_1} \wedge \mathbf{e_3} = \left[\begin{matrix}0&0&1\\0&0&0\\-1&0&0\end{matrix}\right] $
$ \overline{\mathbf{e_3}} = \mathbf{e_2} \wedge \mathbf{e_1} = \left[\begin{matrix}0&-1&0\\1&0&0\\0&0&0\end{matrix}\right] $

References

  1. Shuangzhe Liu; Gõtz Trenkler (2008). "Hadamard, Khatri-Rao, Kronecker and other matrix products". Int J Information and systems sciences (Institute for scientific computing and education) 4 (1): 160–177. http://www.math.ualberta.ca/ijiss/SS-Volume-4-2008/No-1-08/SS-08-01-17.pdf. 
  2. 2.0 2.1 Lounesto, Pertti (2001). Clifford algebras and spinors. Cambridge: Cambridge University Press. pp. 193. ISBN 978-0-521-00551-7. 

See also


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