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exterior algebra

(Definition)

Introductory remarks.

We begin with some informal remarks to motivate the formal definitions found in the next section. Throughout,

is a vector space over a field

. Many of the concepts and constructions discussed below apply verbatim to modules over commutative rings, but we will stick to vector spaces to keep things simple.

The exterior product, commonly denoted by the wedge symbol $\wedge$ and also known as the wedge product, is an antisymmetric variant of the tensor product. The former, like the latter is an associative, bilinear operation. Thus, for all $u,v\in V$ and $a,b,c,d\in K$ , we have

$\displaystyle (au+bv) \wedge (cu + dv )$	$\displaystyle = ac (u\wedge u) + ad (u\wedge v) + bc (v\wedge u) + bd (v\wedge v)$	(1)
$\displaystyle (au+bv) \otimes (cu + dv )$	$\displaystyle = ac (u\otimes u) + ad (u\otimes v) + bc (v\otimes u) + bd (v\otimes v).$	(2)

The essential difference between the two operations is that all squares formed using the exterior product vanish, by definition. Thus,

$\displaystyle v\wedge v =0,$

(3)

whereas $v\otimes v\neq 0$ . Hence, the expressions in (1) are equal to $(ad-bc) u\wedge v$ , but there is no way to simplify further the right-hand side of (2).

A polarization argument shows that for $u,v\in V$ we have

0	$\displaystyle = (u+v)\wedge (u+v)$
	$\displaystyle = u\wedge u + u\wedge v + v\wedge u + v\wedge v$
	$\displaystyle = u\wedge v + v\wedge u.$

Therefore, if the characteristic of the underlying field

is not equal to

, that is if $1+1\neq 0$ , then the key postulate (3) is logically equivalent to the antisymmetry condition

$\displaystyle u\wedge v = -v\wedge u,\quad u,v\in V.$

(4)

However, if the characteristic is 2, that is if

is a field where

, then (3) does not, necessarily, follow from (4). Therefore, to keep things as general as possible, we must use (3) to formulate the essential identity satisfied by the exterior product.

So far so good, but we have not yet given a meaning to the symbol $u\wedge v$ . The geometric interpretation of $u\wedge v$ is that of an oriented area element in the plane spanned by and . Without additional structure, there is no way to assign a area measurement to a parallelogram in a vector space. However, parallelograms that lie in the same plane are commensurate. If we adopt the parallelogram spanned by and as the standard area, we can say that the oriented area of another parallelogram, say one that is spanned by and , has an area that is times the area of the first parallelogram. The exterior product allows us to express this algebraically. To wit,

$\displaystyle (au+bv)\wedge(cu+dv) = (ad-bc) (u\wedge v),\quad u,v\in V,\; a,b,c,d\in K.$

The analogous interpretation for vectors is that of an oriented length element on a line. For this reason, the object $u\wedge v$ is referred to as a bivector.

From a more algebraic point of view, a bivector $u\wedge v$ can be considered as a formal antisymmetric product of vectors and , in much the same way that $u\otimes v$ can be regarded as a formal non-commutative product of two vectors. Such descriptions can hardly serve as rigorous definitions, but an explicit construction is not really the way to go here.

Take the case of the tensor product. Formal sums of formal products $u\otimes v$ , where $u,v\in V$ , form a certain vector space, which we denote as $V\otimes V$ . However, rather than saying that $V\otimes V$ is such and such a thing, it is better to state a certain universal property that describes $V\otimes V$ up to vector space isomorphism. The property in question is that every bilinear map $f:V\times V\to W$ determines a unique linear map from $g:V\otimes V\to W$ such that

$\displaystyle f(u,v) = g(u\otimes v),\quad u,v \in V.$

Similarly, formal sums of bivectors constitute a vector space $\Lambda^2(V)$ , called the second exterior power of

. This vector space is defined, up to isomorphism, by the condition that every antisymmetric, bilinear map $f:V\times V\to W$ determines a unique linear map $g:\Lambda^2(V) \to W$ with

$\displaystyle f(u,v) = g(u\wedge v).$

Thus, in the same way that the tensor product replaces bilinear maps with a certain kind of linear map, the exterior product replaces bilinear, antisymmetric maps with linear maps from $\Lambda^2 V$ .

More generally, -multivectors are -fold products $v_1\wedge\cdots\wedge v_k$ , and the $k^{\text{th}}$ exterior power, $\Lambda^k(V)$ , is the vector space of formal sums of -multivectors. The product of a -multivector and an $\ell$ -multivector is a $(k+\ell)$ -multivector. So, the direct sum $\bigoplus_k \Lambda^k(V)$ forms an associative algebra, which is closed with respect to the wedge product. This algebra, commonly denoted by $\Lambda(V)$ , is called the exterior algebra of .

Again, the analogy with the tensor product is useful. The tensor algebra can be characterized as the associative, non-commutative algebra freely generated by . If the characteristic of is not 2, then the wedge product satisfies the supercommutativity relations

$\displaystyle \alpha\wedge \beta = (-1)^{k+\ell} \beta\wedge\alpha,\quad \alpha\in\Lambda^k(V),\; \beta\in\Lambda^\ell(V).$

Thus, $\Lambda(V)$ can be characterized as the supercommutative algebra which is freely generated by

Formal definitions.

Supercommutative algebras.

For the purposes of this discussion, we define a supercommutative algebra to be an associative, unital

-algebra

with an $\mathbb{Z}_2$ -grading, $A=A^+\oplus A^-$ , such that for all odd $a\in A^-$ we have

$\displaystyle a^2 = 0,$

and such that for all even $b\in A^+$ and all $a\in A$ , we have

$\displaystyle ab = ba.$

Using a polarization argument we see that the first condition implies that for all odd $a,b\in A^-$ we have

$\displaystyle ab = -ba.$

If the characteristic of

is different from

, then the converse is true, and we recover the usual definition of supercommutativity, namely that

$\displaystyle ab = \pm ba,\quad a\in A^{\pm},\; b\in A^{\pm},$

with the minus sign employed if both

and

or odd, and with

employed otherwise.

Exterior algebra.

Let

be a supercommutative algebra and $\iota:V\to E^-$ a linear map. We will say that $(E,\iota)$ is a model for the exterior algebra of

, if every linear map $f:V\to A^-$ , where

a supercommutative algebra, lifts to a unique algebra homomorphism $g:E\to A$ , where “lifts” means that $f = g\circ\iota$ . Diagrammatically:

$\includegraphics[width=4cm]{extalg}$

The above condition on

is a universal property; this implies that all models are isomorphic as algebras. Thus, when we speak of $\Lambda(V)$ , the exterior algebra of

, we are referring to the isomorphism class of all such models. It is also common to identify

with its image $\iota(V)$ , and to write

rather than $\iota(v)$ .

Exterior powers.

For the purposes of the present entry, we define an antisymmetric map to be a

-multilinear map $f:V^{\times k}\to W$ such that $f(\dots, v,v,\dots)=0$ for all $v\in V$ . A polarization argument then implies the usual antisymmetry condition, namely that for every permutation $\pi$ of $\{ 1,2,\dots, k\}$ we have

$\displaystyle f(v_{\pi_1},\dots, v_{\pi_k}) = \operatorname{sgn}(\pi) f(v_1,\dots, v_k),\quad v_1,\dots,v_k \in V.$

As usual, if the characteristic of

is different from

, the two assertions are equivalent. However if

, then the first assertion is stronger, and that is why we adopt it as the definition of antisymmetry.

We now define a model of the $k^{\text{th}}$ exterior power of to be a vector space and an antisymmetric map $\wedge: V^{\times k}\to E^k$ such that every antisymmetric map $f:V^{\times k}\to W$ lifts to a unique linear map $g:E^k\to W$ , where “lifts” means that

$\displaystyle f(v_1,\dots,v_k) = g(v_1\wedge\cdots\wedge v_k),\quad v_1,\dots,v_k\in V.$

As above, all models are isomorphic as vector spaces, and we use $\Lambda^k(V)$ to denote the isomorphism class of all such.

The standard model.

A model of the exterior algebra $\Lambda(V)$ , and the exterior powers $\Lambda^k(V)$ can be easily constructed as the antisymmetrized quotients of the tensor algebra

$\displaystyle T(V)=\bigoplus_{k=0}^\infty V^{\otimes k},\qquad V^{\otimes k} = V\otimes\cdots\otimes V$ (k times) $\displaystyle .$

To that end, let

denote the two sided ideal of

generated by elements of the form $v\otimes v,\; v\in V$ . Then

$\displaystyle S(V)=\bigoplus_{k=0}^\infty S^k(V),\qquad \textrm{where} \quad S^k(V)=S(V)\cap V^{\otimes k},$

and let

$\displaystyle E(V)=T(V)/S(V),\qquad E^k(V)=V^{\otimes k}/S^k(V)$

denote the indicated quotients, with $a:T(V)\to E(V)$ and $a_k: V^{\otimes k} \to E^k(V)$ denoting the corresponding antisymmetrization surjections. It is easy to see that

is the trivial vector space, and hence that $E^1(V) \cong V$ . We leave it as an exercise for the reader to show that $E^k(V),\; k\geq 2$ is a model of the $k^{\text{th}}$ exterior power, while

together with the map $V\to E^1(V)$ is a model of the full exterior algebra.

The canonical grading.

An inspection of the above construction reveals that

$\displaystyle E(V) = \bigoplus_{k=0}^\infty E^k(V).$

Indeed, every model of exterior algebra carries a canonical grading. Let

be a particular model of the exterior algebra of

. For $k=1,2,\dots$ , we will call $\alpha\in E$ , a

-primitive element if $\alpha = v_1\wedge\cdots\wedge v_k,$ for some $v_i\in V$ . We now let $E^k\subset E$ denote the vector space spanned by all

-primitive elements, and let

Proposition 1 The subspace is a model for the $k^{\text{th}}$ exterior power of . Furthermore,

$\displaystyle E = \bigoplus_{k=0}^\infty E^k.$

Categorical formulation.

The above definition of exterior product has a very appealing categorical formulation. Let $\mathbb{S}$ denote the category of supercommutative

-algebras, let $\mathbb{V}$ denote category of vector spaces over

, and let $(\cdot)^-:\mathbb{S}\to\mathbb{V}$ denote the forgetful functor $A\mapsto A^-$ . We may now say that the exterior algebra function $\Lambda:\mathbb{V}\to\mathbb{S}$ is the left adjoint of $(\cdot)^-$ . In other words,

$\displaystyle \operatorname{Hom}_{\mathbb{V}}(V,A^-) \cong \operatorname{Hom}_{\mathbb{S}}(\Lambda(V), A),\quad V\in\mathbb{V},\; A\in \mathbb{S},$

with the isomorphism natural in

and

It is useful to compare the above definition to the categorical definition of the tensor algebra. Let $\mathbb{A}$ denote the category of associative, unital -algebras, and let $F:\mathbb{A}\to\mathbb{V}$ be the forgetful functor that gives the underlying vector space structure of a -algebra. We can then define the tensor algebra of a vector space by saying that $T:\mathbb{V}\to\mathbb{A}$ is the left-adjoint of $F:\mathbb{A}\to\mathbb{V}$ . Thus, whereas as the associative algebra freely generated by , the exterior algebra $\Lambda(V)$ is the supercommutative algebra freely generated by . The antisymmetrization quotient map $a_V\colon T(V)\to\Lambda(V)$ is a natural transformation between these two functors.

Finite dimensional models.

Basis models.

is an

-dimensional vector space, there are some down-to-earth constructions of $\Lambda(V)$ that go a long way to illuminate the nature of the exterior product. Suppose then, that

-dimensional, and let $e_1,\ldots,e_n$ be a basis of

. For every ascending sequence

$\displaystyle 0\leq i_1<i_2<\cdots<i_k\leq n$

let us introduce the symbol $e_I=e_{i_1\dots i_k}$ to represent the primitive

-multivector $e_{i_1}\wedge\ldots\wedge e_{i_k}$ . If

is the empty sequence, we let

denote the unit element of the field

Proposition 2 The $\binom{n}{k}$ -dimensional vector space spanned by $e_{i_1\dots i_k}$ is a model of $\Lambda^k(V)$ .

Note that $\Lambda^1(V)$ is just the

-dimensional space spanned by the basis symbols $e_1,\dots,e_n$ . As such, $\Lambda^1(V)$ is naturally isomorphic to

. For disjoint sequences

and

, let us define

$\displaystyle e_I \wedge e_J = \operatorname{sgn}(IJ) e_{[IJ]},$

where

denotes the ascending sequence composed of the union of

and

, and where $\operatorname{sgn}(IJ)=\pm 1$ denotes the parity of the permutation that takes the sorted list

to the unsorted concatenation

. If

and

have one or more elements in common, we define

$\displaystyle e_I\wedge e_J = 0.$

Here are some examples:

	$\displaystyle e_{3} \wedge e_{12} = e_{123},$
	$\displaystyle e_{2}\wedge e_{14} = - e_{124},$
	$\displaystyle e_{14} \wedge e_{23} = e_{1234},$
	$\displaystyle e_{24} \wedge e_{13} = -e_{1234},$
	$\displaystyle e_{24} \wedge e_{14} =0.$

Proposition 3 The dimensional vector spanned by the symbols , together with the above product and the linear isomorphism from to $\Lambda^1(V)$ is a model of the exterior algebra $\Lambda(V)$ .

Evidently, any list of numbers between and with length greater than will contain duplicates. Thus, an immediate consequence of this construction is that $\Lambda^k(V)=0$ for , and hence that

$\displaystyle \Lambda(V) = \bigoplus_{k=0}^n \Lambda^n(V).$

Alternating forms.

is finite-dimensional, we have the natural isomorphism between

and the double-dual $V^{**}$ . We can exploit this natural isomorphism to construct the following model of exterior algebra. Let

denote the vector space of

-multilinear mappings from $(V^*)\times\cdots\times (V^*)$ (

times) to

. Such an mapping is known as an alternating

-form. Using the above duality we can prove that

is a model for the $k^{\text{th}}$ exterior power of

Given alternating forms $u\in A^k(V)$ and $v\in A^\ell(V)$ , let us define $u\wedge v\in A^{k+\ell}(V)$ according to

$\displaystyle (u\wedge v)(a_1,\dots,a_{k+\ell})= \sum_{\pi}\operatorname{sgn}(\pi) u(a_{\pi_1},\dots,a_{\pi_k}) v(a_{\pi_{k+1}},\dots,a_{\pi_{k+\ell}}),$

where $a_1,\dots,a_{k+\ell}\in V^*,$ and where the sum is taken over all permutations $\pi$ of $\{1,2,\dots, k+\ell\}$ such that $\pi_1< \pi_2 < \cdots < \pi_k$ and $\pi_{k+1} < \cdots < \pi_{k+\ell}$ , and where $\operatorname{sgn} \pi=\pm 1$ according to whether $\pi$ is an even or odd permutation. With this definition, we can show that

$\displaystyle A(V) = \bigoplus_{k=0}^n A^k(V)$

together with the above product, and the linear isomorphism $V\to A^1(V)\cong V^{**}$ is a model for the exterior algebra $\Lambda(V)$ .

Historical Notes.

The exterior algebra is also known as the Grassmann algebra after its inventor Hermann Grassmann who created it in order to give algebraic treatment of linear geometry. Grassmann was also one of the first people to talk about the geometry of an

-dimensional space with

an arbitrary natural number. The axiomatics of the exterior product are needed to define differential forms and therefore play an essential role in the theory of integration on manifolds. Exterior algebra is also an essential prerequisite to understanding de Rham's theory of differential cohomology.

"exterior algebra" is owned by rmilson. [ full author list (2) ]

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