User:John Randall/ExteriorAlgebra

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[Still under construction]


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<i_2<\dots i_k. 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.

  ]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=:-/ .*


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


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

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

Area and volume

Coming soon.