exterior algebra

1 Introductory remarks.

We begin with some informal remarks to motivate the formal definitionsfound in the next section. Throughout, $V$ is a vector space over afield $K$. Many of the concepts and constructions discussed belowapply verbatim to modules over commutative rings, but we will stick tovector 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 ofthe 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

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

$(au+bv)\otimes (cu+dv)$$=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 allsquares formed using the exterior product vanish, by definition.Thus,

$$v\wedge v=0,$$(3)

whereas $v\otimes v\ne 0$. Hence, the expressions in(1) are equal to $(ad-bc)u\wedge v$, but there is noway to simplify further the right-hand side of (2).

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

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

$=u\wedge u+u\wedge v+v\wedge u+v\wedge v$

$=u\wedge v+v\wedge u.$

Therefore, if the characteristic (http://planetmath.org/characteristic) of the underlying field $K$ is not equal to $2$, that is if $1+1\ne 0$, then the key postulate (3) islogically equivalent to the antisymmetry condition

$$u\wedge v=-v\wedge u,u,v\in V.$$(4)

However, if the characteristic is 2, that is if $K$ is a field where$1=-1$, then (3) does not, necessarily, follow from(4). Therefore, to keep things as general as possible,we must use (3) to formulate the essential identitysatisfied 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 ofan oriented area element in the plane spanned by $u$ and $v$. Withoutadditional structure, there is no way to assign a area measurement toa parallelogram in a vector space. However, parallelograms that liein the same plane are commensurate. If we adopt the parallelogramspanned by $u$ and $v$ as the standard area, we can say that theoriented area of another parallelogram, say one that is spanned by$au+bv$ and $cu+dv$, has an area that is $ad-bc$ times the area of thefirst parallelogram. The exterior product allows us to express thisalgebraically. To wit,

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

The analogous interpretation for vectors is that ofan 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 beconsidered as a formal antisymmetric product of vectors $u$ and $v$,in much the same way that $u\otimes v$ can be regarded as a formalnon-commutative product of two vectors. Such descriptions can hardlyserve as rigorous definitions, but an explicit construction is notreally 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 wedenote as $V\otimes V$. However, rather than saying that $V\otimes V$is such and such a thing, it is better to state a certainuniversal property that describes $V\otimes V$ up to vector spaceisomorphism. The property in question is that every bilinearmap $f:V\times V\to W$ determines a unique linear map from$g:V\otimes V\to W$ such that

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

Similarly, formal sums ofbivectors constitute a vector space ${\mathrm{\Lambda }}^2(V)$, called the secondexterior power of $V$. This vector space is defined, up toisomorphism, by the condition that every antisymmetric, bilinear map$f:V\times V\to W$ determines a unique linear map $g:{\mathrm{\Lambda }}^2(V)\to W$ with

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

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

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

Again, the analogy with the tensor product is useful. The tensoralgebra $T(V)$ can be characterized as the associative,non-commutative algebra freely generated by $V$. If thecharacteristic of $k$ is not 2, then the wedge product satisfies thesupercommutativity relations

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

Thus, $\mathrm{\Lambda }(V)$can be characterized as the supercommutative algebra which is freelygenerated by $V$.

2 Formal definitions.

Supercommutative algebras.

For the purposes of this discussion, we define a supercommutativealgebra to be an associative, unital $K$-algebra $A$ with an${\mathbb{Z}}_2$-grading, $A=A^{+}\oplus A^{-}$, such that for all odd$a\in A^{-}$ we have

$$a^2=0,$$

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

$$ab=ba.$$

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

$$ab=-ba.$$

If the characteristic of $K$ is different from $2$, then the converseis true, and we recover the usual definition of supercommutativity,namely that

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

with the minus sign employed if both $a$ and $b$ or odd, and with $+$employed otherwise.

Exterior algebra.

Let $E$ be a supercommutative algebra and $\iota :V\to E^{-}$ a linearmap. We will say that $(E,\iota )$ is a model for the exterior algebraof $V$, if every linear map $f:V\to A^{-}$, where $A$ a supercommutativealgebra, to a unique algebra homomorphism $g:E\to A$, where“lifts” means that $f=g\circ \iota$.Diagrammatically:

The above condition on $E$ is a universal property; this implies thatall models are isomorphic as algebras. Thus, when we speak of$\mathrm{\Lambda }(V)$, the exterior algebra of $V$, we are referring to theisomorphism class of all such models. It is also common to identify$V$ with its image $\iota (V)$, and to write $v$ rather than$\iota (v)$.

Exterior powers.

For the purposes of the present entry, we define an antisymmetric mapto be a $k$-multilinear map $f:V^{\times k}\to W$ such that $f(\mathrm{\dots },v,v,\mathrm{\dots })=0$ for all $v\in V$. A polarization argument then implies theusual antisymmetry condition, namely that for every permutation $\pi$of $\{1,2,\mathrm{\dots },k\}$ we have

$$f(v_{{\pi }_1},\mathrm{\dots },v_{{\pi }_k})=\mathrm{sgn}(\pi )f(v_1,\mathrm{\dots },v_k),v_1,\mathrm{\dots },v_k\in V.$$

As usual, if the characteristic of $K$ is different from $2$, the twoassertions are equivalent. However if $1=-1$, then the firstassertion is stronger, and that is why we adopt it as the definitionof antisymmetry.

We now define a model of the $k^{\text{th}}$ exterior power of $V$ to be avector space $E^k$ 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

$$f(v_1,\mathrm{\dots },v_k)=g(v_1\wedge \mathrm{\cdots }\wedge v_k),v_1,\mathrm{\dots },v_k\in V.$$

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

The standard model.

A model of the exterior algebra$\mathrm{\Lambda }(V)$, and the exterior powers ${\mathrm{\Lambda }}^k(V)$ can be easilyconstructed as the antisymmetrized quotients of the tensor algebra

$$T(V)=\underset{k=0}{\overset{\mathrm{\infty }}{\oplus }}{V}^{\otimes k},V^{\otimes k}=V\otimes \mathrm{\cdots }\otimes V\text{ (k times)}.$$

To that end, let $S(V)$denote the two sided ideal of $T(V)$ generated byelements of the form $v\otimes v,v\in V$. Then

$$S(V)=\underset{k=0}{\overset{\mathrm{\infty }}{\oplus }}{S}^k(V),\text{where}\mathit{\hspace{1em}}{S}^k(V)=S(V)\cap V^{\otimes k},$$

and let

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

denote theindicated quotients, with $a:T(V)\to E(V)$ and $a_k:V^{\otimes k}\to E^k(V)$ denoting the corresponding antisymmetrization surjections. Itis easy to see that $S^1(V)$ is the trivial vector space, and hencethat $E^1(V)\cong V$. We leave it as an exercise for the reader toshow that $E^k(V),k\ge 2$ is a model of the $k^{\text{th}}$ exteriorpower, while $E(V)$ together with the map $V\to E^1(V)$ is a model ofthe full exterior algebra.

The canonical grading.

An inspection of the aboveconstruction reveals that

$$E(V)=\underset{k=0}{\overset{\mathrm{\infty }}{\oplus }}{E}^k(V).$$

Indeed, every model ofexterior algebra carries a canonical grading. Let $E$ be a particularmodel of the exterior algebra of $V$. For $k=1,2,\mathrm{\dots }$, we will call$\alpha \in E$, a $k$-primitive element if $\alpha =v_1\wedge \mathrm{\cdots }\wedge v_k,$ for some $v_i\in V$. We now let$E^k\subset E$ denote the vector space spanned by all $k$-primitiveelements, and let $E^0=K$.

Proposition 1.

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

$$E=\underset{k=0}{\overset{\mathrm{\infty }}{\oplus }}{E}^k.$$

Categorical formulation.

The above definition of exterior product has a very appealingcategorical formulation. Let $𝕊$ denote the category ofsupercommutative $K$-algebras, let $𝕍$ denote category of vectorspaces over $K$, and let ${(\cdot )}^{-}:𝕊\to 𝕍$ denote theforgetful functor $A\mapsto A^{-}$. We may now say that theexterior algebra function $\mathrm{\Lambda }:𝕍\to 𝕊$ is the leftadjoint of ${(\cdot )}^{-}$. In other words,

$${\mathrm{Hom}}_{𝕍}(V,A^{-})\cong {\mathrm{Hom}}_{𝕊}(\mathrm{\Lambda }(V),A),V\in 𝕍,A\in 𝕊,$$

with theisomorphism natural in $V$ and $A$.

It is useful to compare the above definition to the categoricaldefinition of the tensor algebra. Let $𝔸$ denote the category ofassociative, unital $K$-algebras, and let $F:𝔸\to 𝕍$ be theforgetful functor that gives the underlying vector space structure ofa $K$-algebra. We can then define the tensor algebra $T(V)$ of avector space $V$ by saying that $T:𝕍\to 𝔸$ is the left-adjointof $F:𝔸\to 𝕍$. Thus, whereas $T(V)$ as the associativealgebra freely generated by $V$, the exterior algebra $\mathrm{\Lambda }(V)$is the supercommutative algebra freely generated by $V$. Theantisymmetrizationquotient map $a_V:T(V)\to \mathrm{\Lambda }(V)$ is a natural transformation betweenthese two functors.

3 Finite dimensional models.

Basis models.

If $V$ is an $n$-dimensional vector space, there are somedown-to-earth constructions of $\mathrm{\Lambda }(V)$ that go a long way toilluminate the nature of the exterior product. Suppose then, that $V$is $n$-dimensional, and let $e_1,\mathrm{\dots },e_n$ be a basis of $V$. Forevery ascending sequence

let usintroduce the symbol $e_I=e_{i_1\mathrm{\dots }{i}_k}$ to represent the primitive$k$-multivector $e_{i_1}\wedge \mathrm{\dots }\wedge e_{i_k}$. If $I$ is theempty sequence, we let $e_I$ denote the unit element of the field$K$.

Proposition 2.

The $\mathrm{\left(}\genfrac{}{}{0pt}{}{n}{k}\mathrm{\right)}$-dimensional vector space spanned by $e_{i_{\mathrm{1}}\mathit{}\mathrm{\dots }\mathit{}{i}_k}$ is a model of ${\mathrm{\Lambda }}^k\mathit{}\mathrm{(}V\mathrm{)}$.

Note that ${\mathrm{\Lambda }}^1(V)$ is just the $n$-dimensional space spanned bythe basis symbols $e_1,\mathrm{\dots },e_n$. As such, ${\mathrm{\Lambda }}^1(V)$ isnaturally isomorphic to $V$. For disjoint sequences $I$ and $J$, letus define

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

where $[IJ]$ denotes theascending sequence composed of the union of $I$ and $J$, and where$\mathrm{sgn}(IJ)=\pm 1$ denotes the parity of the permutation that takes thesorted list $[IJ]$ to the unsorted concatenation $IJ$. If $I$ and $J$have one or more elements in common, we define

$$e_I\wedge e_J=0.$$

Here are some examples:

$e_3\wedge e_{12}=e_{123},$

$e_2\wedge e_{14}=-e_{124},$

$e_{14}\wedge e_{23}=e_{1234},$

$e_{24}\wedge e_{13}=-e_{1234},$

$e_{24}\wedge e_{14}=0.$

Proposition 3.

The ${\mathrm{2}}^n$ dimensional vector spanned by the symbols $e_I$, togetherwith the above product and the linear isomorphism from $V$ to${\mathrm{\Lambda }}^{\mathrm{1}}\mathit{}\mathrm{(}V\mathrm{)}$ is a model of the exterior algebra $\mathrm{\Lambda }\mathit{}\mathrm{(}V\mathrm{)}$.

Evidently, any list of numbers between $1$ and $n$ with length greaterthan $n$ will contain duplicates. Thus, an immediate consequence ofthis construction is that ${\mathrm{\Lambda }}^k(V)=0$ for $k>n$, and hence that

$$\mathrm{\Lambda }(V)=\underset{k=0}{\overset{n}{\oplus }}{\mathrm{\Lambda }}^n(V).$$

Alternating forms.

If $V$ is finite-dimensional, we have the natural isomorphism between$V$ and the double-dual $V^{**}$. We can exploit this naturalisomorphism to construct the following model of exterior algebra. Let$A^k(V)$ denote the vector space of $k$-multilinear mappings from$(V^{*})\times \mathrm{\cdots }\times (V^{*})$ ($k$ times) to $K$. Such an mappingis known as an alternating $k$-form. Using the above duality we canprove that $A^k(V)$ is a model for the $k^{\text{th}}$ exterior power of$V$.

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

$$(u\wedge v)(a_1,\mathrm{\dots },a_{k+\mathrm{\ell }})=\sum _{\pi }\mathrm{sgn}(\pi )u(a_{{\pi }_1},\mathrm{\dots },a_{{\pi }_k})v(a_{{\pi }_{k+1}},\mathrm{\dots },a_{{\pi }_{k+\mathrm{\ell }}}),$$

where $a_1,\mathrm{\dots },a_{k+\mathrm{\ell }}\in V^{*},$ and where the sum is taken overall permutations $\pi$ of $\{1,2,\mathrm{\dots },k+\mathrm{\ell }\}$ such that and , andwhere $\mathrm{sgn}\pi =\pm 1$ according to whether $\pi$ is aneven or odd permutation. With this definition, we can show that

$$A(V)=\underset{k=0}{\overset{n}{\oplus }}{A}^k(V)$$

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

4 Historical Notes.

The exterior algebra is also known as the Grassmannalgebra after its inventor http://www-groups.dcs.st-and.ac.uk/ history/Mathematicians/Grassmann.htmlHermann Grassmannwho created it to give algebraic treatment of lineargeometry. Grassmann was also one of the first people to talk aboutthe geometry of an $n$-dimensional space with $n$ an arbitrary naturalnumber. The axiomatics of the exterior product are needed to definedifferential forms and therefore play an essential role in the theoryof integration on manifolds. Exterior algebra is also an essentialprerequisite to understanding de Rham’s theory of differentialcohomology.

Title	exterior algebra
Canonical name	ExteriorAlgebra
Date of creation	2013-03-22 12:34:14
Last modified on	2013-03-22 12:34:14
Owner	rmilson (146)
Last modified by	rmilson (146)
Numerical id	35
Author	rmilson (146)
Entry type	Definition
Classification	msc 15A75
Synonym	Grassmann algebra
Related topic	AntiSymmetric
Defines	exterior product
Defines	wedge product
Defines	multivector
Defines	exterior power