# User:John Randall/ExteriorAlgebra

[Still under construction]

## Introduction

We define $\Lambda^k$, the vector space of $k$-vectors in a real vector space $V$ of dimension $n$. These are essentially lists $[v_1,\dots,v_k]$ for $v_i\in V$ with an equivalence relation. More precisely, we will defined a $k$-vector as the formal symbol $v_1\wedge \dots\wedge v_k$, the wedge product of $v_1,\dots,v_k$.

The wedge product generalizes several ideas. Essentially, the determinant is an $n$-vector, the wedge product of its rows, and in the case $n=3$, $a\times b$ is $a\wedge b$. 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 $\Lambda^k$, the space of $k$-vectors on $\mathbf{R}^n$, for $k=0,1,2,\dots$ as follows. Each $\Lambda^k$ is a vector space. The bottom dimension spaces are given by $\Lambda^0=\mathbf{R}$ and $\Lambda^1=V$. For $k>0$, we define $\Lambda^k$, as follows: Every element of $\Lambda^k$ can be written (non uniquely) as a formal symbol $v_1\wedge\dots\wedge v_k,$ where $\wedge$ is multilinear, antisymmetric and associative which satisfies the conditions:

1. $\wedge$ is linear in each of the $v_i$. 2. If $v$ and $w$ are vector in $V$, then $v\wedge w=-w\times v$, and $v\wedge v=0$ (antisymmetry). 3. If $v_1,\dots v_k$ are linearly dependent, then $v_1\wedge\dots\wedge v_k=0$.

A basis for $\Lambda^k$ can be given as follows. Let $e_1,\dots,e_n$ be an ordered basis for $V$ (the ordering corresponds to a choice of handedness). Then each $v_j$ can be written as a linear combination $v_j=\sum v_{ji}e_i$, and a basis for $\Lambda^k$ is given by of ordered $k$-fold wedge products of the basis for $V$ given by $e_{i_1}\wedge e_{i_2}\wedge e_{i_k}$, where $i_1. This gives the dimension of $\Lambda^k$ as $n\choose k$. For $k>n$, any set of $k$ vectors must be linearly dependent, so $\Lambda^k=0$.

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

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

## Concrete interpretation

A $k$-vector in $\mathbf{R}^n$ can be represented (non uniquely) as a $k\times n$ 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 $\mathbf{R}^5$. There is a product, called wedge product, that given a $p$-vector and a $q$-vector, produces a $(p+q)$-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 $V$ is a $k$-vector represented by a matrix $M$, then

(a) $V$=0 if the rows of $M$ are linearly dependent

(b) swapping two rows of $M$ replaces $V$ by $-V$ (antisymmetry)

(c) multiplying a row of $M$ by a constant multiplies $V$ by that constant (multilinearity)

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

A basis for the space of $k$-vectors is given by $k$-fold wedge products of distinct vectors of size $n$ with exactly $k$ 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 $\mathbf{R}^4$.

With respect to this basis, the coordinates of a $k$-vector $V$ represented by a matrix $M$ are obtained by for a basis vector $e$ can be calculated by using $e$ to select $k$ columns of $M$ 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 $n$-vectors in $\mathbf{R}^n$ has a single coordinate. If $M$ represents an $n$-vector $V$, then its coordinate is $\det M$. ### Levi-Civita symbol The Levi-Civita symbol or complete tensor gives the wedge product of $n$ basis vectors in $\mathbf{R}^n$. 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 $n$-vector $V,$ the wedge product of the corresponding basis elements, and so has one coordinate $c$. By the rules for equivalence of k-vectors: If I has repeated indices, then $c=0$. Otherwise I is a permutation of i.n , and $c$ 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 $a$ and $b$ are vectors in $\mathbf{R}^3$, $a\wedge b$ is a 2-vector, while $a\times b$ is a 1-vector. We can reconcile these by considering $\mathbf{i},\mathbf{j},\mathbf{k}$ the standard ordered orthonormal basis for $\mathbf{R}^3$, The ordering corresponds to the right hand rule.

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

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

   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.