# User:John Randall/ExteriorAlgebra

[Still under construction]

## Introduction

We define ${\displaystyle \Lambda ^{k}}$, the vector space of ${\displaystyle k}$-vectors in a real vector space ${\displaystyle V}$ of dimension ${\displaystyle n}$. These are essentially lists ${\displaystyle [v_{1},\dots ,v_{k}]}$ for ${\displaystyle v_{i}\in V}$ with an equivalence relation. More precisely, we will defined a ${\displaystyle k}$-vector as the formal symbol ${\displaystyle v_{1}\wedge \dots \wedge v_{k}}$, the wedge product of ${\displaystyle v_{1},\dots ,v_{k}}$.

The wedge product generalizes several ideas. Essentially, the determinant is an ${\displaystyle n}$-vector, the wedge product of its rows, and in the case ${\displaystyle n=3}$, ${\displaystyle a\times b}$ is ${\displaystyle a\wedge b}$. The interpretation of the latter as a vector is misleading: this is an accident of dimension.

The reader who is interested in the concrete formulation can skip the next section.

## Formal description

We define ${\displaystyle \Lambda ^{k}}$, the space of ${\displaystyle k}$-vectors on ${\displaystyle \mathbf {R} ^{n}}$, for ${\displaystyle k=0,1,2,\dots }$ as follows. Each ${\displaystyle \Lambda ^{k}}$ is a vector space. The bottom dimension spaces are given by ${\displaystyle \Lambda ^{0}=\mathbf {R} }$ and ${\displaystyle \Lambda ^{1}=V}$. For ${\displaystyle k>0}$, we define ${\displaystyle \Lambda ^{k}}$, as follows: Every element of ${\displaystyle \Lambda ^{k}}$ can be written (non uniquely) as a formal symbol ${\displaystyle v_{1}\wedge \dots \wedge v_{k},}$ where ${\displaystyle \wedge }$ is multilinear, antisymmetric and associative which satisfies the conditions:

1. ${\displaystyle \wedge }$ is linear in each of the ${\displaystyle v_{i}}$. 2. If ${\displaystyle v}$ and ${\displaystyle w}$ are vector in ${\displaystyle V}$, then ${\displaystyle v\wedge w=-w\times v}$, and ${\displaystyle v\wedge v=0}$ (antisymmetry). 3. If ${\displaystyle v_{1},\dots v_{k}}$ are linearly dependent, then ${\displaystyle v_{1}\wedge \dots \wedge v_{k}=0}$.

A basis for ${\displaystyle \Lambda ^{k}}$ can be given as follows. Let ${\displaystyle e_{1},\dots ,e_{n}}$ be an ordered basis for ${\displaystyle V}$ (the ordering corresponds to a choice of handedness). Then each ${\displaystyle v_{j}}$ can be written as a linear combination ${\displaystyle v_{j}=\sum v_{ji}e_{i}}$, and a basis for ${\displaystyle \Lambda ^{k}}$ is given by of ordered ${\displaystyle k}$-fold wedge products of the basis for ${\displaystyle V}$ given by ${\displaystyle e_{i_{1}}\wedge e_{i_{2}}\wedge e_{i_{k}}}$, where ${\displaystyle i_{1}. This gives the dimension of ${\displaystyle \Lambda ^{k}}$ as ${\displaystyle n \choose k}$. For ${\displaystyle k>n}$, any set of ${\displaystyle k}$ vectors must be linearly dependent, so ${\displaystyle \Lambda ^{k}=0}$.

The exterior algebra of ${\displaystyle \mathbf {R} ^{n}}$ is ${\displaystyle \Lambda ^{0}\oplus \Lambda ^{1}\oplus \dots \oplus \Lambda ^{n}}$.

Regardless of the basis, there is a natural homomorphism ${\displaystyle \wedge :\Lambda ^{p}\times \Lambda ^{q}\to \Lambda ^{p+q}}$. If ${\displaystyle \mathbf {R} ^{n}}$ has an inner product, there is also a natural isomorphism ${\displaystyle *:\Lambda ^{k}\to \Lambda ^{n-k}}$ (Hodge duality). This is important when interpreting cross product in ${\displaystyle \mathbf {R} ^{3}}$.

## Concrete interpretation

A ${\displaystyle k}$-vector in ${\displaystyle \mathbf {R} ^{n}}$ can be represented (non uniquely) as a ${\displaystyle k\times n}$ matrix, so

  ]A=:3 5 $? 15#10 6 4 0 6 4 9 8 5 7 4 0 1 2 7 9  represents a 3-vector in ${\displaystyle \mathbf {R} ^{5}}$. There is a product, called wedge product, that given a ${\displaystyle p}$-vector and a ${\displaystyle q}$-vector, produces a ${\displaystyle (p+q)}$-vector. In terms of matrices, this is just given by concatenation.  wedge=:, ]B=:2 5$ ? 10#10
3 2 1 2 9
5 3 9 6 9

A wedge B
6 4 0 6 4
9 8 5 7 4
0 1 2 7 9
3 2 1 2 9
5 3 9 6 9


This representation is not unique. If ${\displaystyle V}$ is a ${\displaystyle k}$-vector represented by a matrix ${\displaystyle M}$, then

(a) ${\displaystyle V}$=0 if the rows of ${\displaystyle M}$ are linearly dependent

(b) swapping two rows of ${\displaystyle M}$ replaces ${\displaystyle V}$ by ${\displaystyle -V}$ (antisymmetry)

(c) multiplying a row of ${\displaystyle M}$ by a constant multiplies ${\displaystyle V}$ by that constant (multilinearity)

(d) adding a multiple of one row of ${\displaystyle M}$ to another does not change ${\displaystyle V}$

A basis for the space of ${\displaystyle k}$-vectors is given by ${\displaystyle k}$-fold wedge products of distinct vectors of size ${\displaystyle n}$ with exactly ${\displaystyle k}$ ones, lexicographically ordered. Consequently the vector space has dimension  k!n . For example,

0 1 0 1
0 1 1 0


is one of the basis vector for 2-vectors in ${\displaystyle \mathbf {R} ^{4}}$.

With respect to this basis, the coordinates of a ${\displaystyle k}$-vector ${\displaystyle V}$ represented by a matrix ${\displaystyle M}$ are obtained by for a basis vector ${\displaystyle e}$ can be calculated by using ${\displaystyle e}$ to select ${\displaystyle k}$ columns of ${\displaystyle M}$ and then taking the determinant of the resulting matrix.

det=:-/ .*

cols=:{&.|:

comb=: 4 : 0
k=. i.>:d=.y-x
z=. (d$<i.0 0),<i.1 0 for. i.x do. z=. k ,.&.> ,&.>/\. >:&.> z end. ; z ) coords=:det@cols"1 _~ comb/@:$

]C=: 2 5 $i.10 0 1 2 3 4 5 6 7 8 9 coords C _5 _10 _15 _20 _5 _10 _15 _5 _10 _5  ### Determinant The space of ${\displaystyle n}$-vectors in ${\displaystyle \mathbf {R} ^{n}}$ has a single coordinate. If ${\displaystyle M}$ represents an ${\displaystyle n}$-vector ${\displaystyle V}$, then its coordinate is ${\displaystyle \det M}$. ### Levi-Civita symbol The Levi-Civita symbol or complete tensor gives the wedge product of ${\displaystyle n}$ basis vectors in ${\displaystyle \mathbf {R} ^{n}}$. These need not be distinct, and order matters. If I is a list of numbers in i.n of length n , then M =:I {=@. i. n is a matrix representing an ${\displaystyle n}$-vector ${\displaystyle V,}$ the wedge product of the corresponding basis elements, and so has one coordinate ${\displaystyle c}$. By the rules for equivalence of k-vectors: If I has repeated indices, then ${\displaystyle c=0}$. Otherwise I is a permutation of i.n , and ${\displaystyle c}$ is the parity of I { i.n. The calculation is given by the complete tensor CT. CT =: C.!.2 @ (#:i.) @$~

I=:0 2 1

]M=:I { =@i.3
1 0 0
0 0 1
0 1 0
coords M
_1
(< I) { CT 3
_1
J=:0 2 2
coords J { =@i.3
0
(< J) { CT 3
0


### Cross product

If ${\displaystyle a}$ and ${\displaystyle b}$ are vectors in ${\displaystyle \mathbf {R} ^{3}}$, ${\displaystyle a\wedge b}$ is a 2-vector, while ${\displaystyle a\times b}$ is a 1-vector. We can reconcile these by considering ${\displaystyle \mathbf {i} ,\mathbf {j} ,\mathbf {k} }$ the standard ordered orthonormal basis for ${\displaystyle \mathbf {R} ^{3}}$, The ordering corresponds to the right hand rule.

We can then identify a 2-vector with a 1-vector by defining an operation ${\displaystyle *}$ given by its action on the basis:

${\displaystyle *(\mathbf {i} \wedge \mathbf {j} )=\mathbf {k} }$, ${\displaystyle *(\mathbf {j} \wedge \mathbf {k} )=\mathbf {i} }$, ${\displaystyle *(\mathbf {k} \wedge \mathbf {i} )=\mathbf {j} }$.

   i=:,: 1 0 0
j=:,: 0 1 0
k=:,: 0 0 1

star=:1 _1 1 * |. @ coords

cross=:star @ wedge

(i cross j),(j cross k),:(k cross i)
0 0 1
1 0 0
0 1 0

(,:1 2 3) cross (,: 4 5 6)
_3 6 _3


Coming soon.