# exterior algebra

## 1 Introductory remarks.

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,

 $v\wedge v=0,$ (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

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

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

 $u\wedge v=-v\wedge u,\quad u,v\in V.$ (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 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 $u$ and $v$. 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 $u$ and $v$ as the standard area, we can say that the oriented 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 the first parallelogram. The exterior product allows us to express this algebraically. To wit,

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

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 $u$ and $v$, 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

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

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

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

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, $k$-multivectors are $k$-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 $k$-multivectors. The product of a $k$-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 $V$.

Again, the analogy  with the tensor product is useful. The tensor algebra $T(V)$ can be characterized as the associative, non-commutative algebra freely generated by $V$. If the characteristic of $k$ is not 2, then the wedge product satisfies the supercommutativity relations    $\alpha\wedge\beta=(-1)^{k+\ell}\beta\wedge\alpha,\quad\alpha\in\Lambda^{k}(V),% \;\beta\in\Lambda^{\ell}(V).$

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

## 2 Formal definitions.

### Supercommutative algebras.

For the purposes of this discussion, we define a supercommutative algebra 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 first condition implies that for all odd $a,b\in A^{-}$ we have

 $ab=-ba.$

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

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

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 linear map. We will say that $(E,\iota)$ is a model for the exterior algebra of $V$, if every linear map $f:V\to A^{-}$, where $A$ a supercommutative algebra, 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 that all models are isomorphic as algebras. Thus, when we speak of $\Lambda(V)$, the exterior algebra of $V$, we are referring to the isomorphism 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 map to be a $k$-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

 $f(v_{\pi_{1}},\dots,v_{\pi_{k}})=\operatorname{sgn}(\pi)f(v_{1},\dots,v_{k}),% \quad v_{1},\dots,v_{k}\in V.$

As usual, if the characteristic of $K$ is different from $2$, the two assertions are equivalent     . However if $1=-1$, 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 $V$ to be a vector 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

 $f(v_{1},\dots,v_{k})=g(v_{1}\wedge\cdots\wedge v_{k}),\quad v_{1},\dots,v_{k}% \in V.$

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

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

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

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

and let

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

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 $S^{1}(V)$ 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 $E(V)$ 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

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

Indeed, every model of exterior algebra carries a canonical grading. Let $E$ be a particular model of the exterior algebra of $V$. For $k=1,2,\dots$, we will call $\alpha\in E$, a $k$-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 $k$-primitive elements, and let $E^{0}=K$.

###### Proposition 1.

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

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

### Categorical formulation.

The above definition of exterior product has a very appealing categorical formulation. Let $\mathbb{S}$ denote the category of supercommutative $K$-algebras, let $\mathbb{V}$ denote category of vector spaces over $K$, 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,

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

with the isomorphism natural in $V$ and $A$.

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

## 3 Finite dimensional models.

### Basis models.

If $V$ is an $n$-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 $V$ is $n$-dimensional, and let $e_{1},\ldots,e_{n}$ be a basis of $V$. For every ascending sequence  $0\leq i_{1}

let us introduce the symbol $e_{I}=e_{i_{1}\dots i_{k}}$ to represent the primitive $k$-multivector $e_{i_{1}}\wedge\ldots\wedge e_{i_{k}}$. If $I$ is the empty sequence, we let $e_{I}$ denote the unit element of the field $K$.

###### 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 $n$-dimensional space spanned by the basis symbols $e_{1},\dots,e_{n}$. As such, $\Lambda^{1}(V)$ is naturally isomorphic to $V$. For disjoint sequences $I$ and $J$, let us define

 $e_{I}\wedge e_{J}=\operatorname{sgn}(IJ)e_{[IJ]},$

where $[IJ]$ denotes the ascending sequence composed of the union of $I$ and $J$, and where $\operatorname{sgn}(IJ)=\pm 1$ denotes the parity of the permutation that takes the sorted 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:

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

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

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

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

### Alternating forms.

If $V$ is finite-dimensional, we have the natural isomorphism between $V$ and the double-dual $V^{**}$. We can exploit this natural isomorphism to construct the following model of exterior algebra. Let $A^{k}(V)$ denote the vector space of $k$-multilinear mappings from $(V^{*})\times\cdots\times(V^{*})$ ($k$ times) to $K$. Such an mapping is known as an alternating $k$-form. Using the above duality we can prove 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 $v\in A^{\ell}(V)$, let us define $u\wedge v\in A^{k+\ell}(V)$ according to

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

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

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

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

## 4 Historical Notes.

The exterior algebra is also known as the Grassmann algebra after its inventor http://www-groups.dcs.st-and.ac.uk/ history/Mathematicians/Grassmann.htmlHermann Grassmann who created it to give algebraic treatment of linear geometry. Grassmann was also one of the first people to talk about the geometry of an $n$-dimensional space with $n$ 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.

 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