The Exterior Algebra
Introduction
The exterior powers $\Lambda^n M$ constructed in the companion article Exterior Powers are the alternating analogue of the tensor powers, and each is a module in its own right. They assemble into a single object, the exterior algebra $\Lambda(M) = \bigoplus_{n \geq 0} \Lambda^n M$, on which the direct sum gives the addition and the concatenation of elementary wedges gives a product. The product is not commutative; it is graded-commutative, satisfying $uv = (-1)^{|u||v|} v u$ for homogeneous elements $u, v$ of degrees $|u|$, $|v|$. This single sign rule governs the whole structure and is the source of the algebra's use throughout geometry and analysis: the graded-commutative algebra on the cotangent space at a point is the algebra of differential forms, and the graded-commutative algebra on a space of odd generators is the Grassmann algebra of the Berezin calculus.
This article defines the wedge product, proves that it is associative and unital, establishes graded-commutativity, and shows that $\Lambda(M)$ is the free graded-commutative algebra on $M$, equivalently the quotient of the tensor algebra $T(M)$ by the two-sided ideal generated by the elements $x \otimes x$. It then treats functoriality, the behaviour of $\Lambda$ under direct sums and base change, and the explicit structure of $\Lambda$ of a free module of finite rank.
Throughout, $R$ denotes a commutative ring with identity $1 \neq 0$ and $M$, $N$ are $R$-modules; over a commutative ring left and right modules coincide, so no side is specified. The field is written $K$, and $\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$ when only those two are meant. The base is the commutative ring, so all constructions are carried out over $R$; assumptions on the characteristic, and in particular on the invertibility of $2$, are flagged where they enter. No physics is invoked.
Other articles of the category take up the top exterior power and the alternating forms on it; the applications articlesandare and realise $\Lambda$ in two settings.
The Graded Module of Exterior Powers
Definition
Definition. The exterior algebra of an $R$-module $M$ is the graded $R$-module
$$ \Lambda(M) = \bigoplus_{n \geq 0} \Lambda^n M, $$
with $\Lambda^0 M = R$ and $\Lambda^1 M = M$. An element of $\Lambda^n M$ is called an $n$-vector, or a form of degree $n$ when it is regarded as sitting in the dual construction, and a homogeneous element of $\Lambda^n M$ has degree $|u| = n$. The degree-0 part is the scalar part and the degree-1 part is the vector part; elements of degree $2$ are bivectors.
Because $\Lambda(M)$ is a direct sum, addition is componentwise and a general element is a finite sum $u = u_0 + u_1 + \cdots$ with $u_n \in \Lambda^n M$. The decomposition is unique, so each element has well-defined homogeneous components.
The Poincaré Series
Definition. Let $M$ be free of finite rank $m$. The Poincaré series of $\Lambda(M)$ is the formal power series
$$ P_{\Lambda(M)}(t) = \sum_{n \geq 0} \operatorname{rank}_R(\Lambda^n M)\, t^n. $$
Proposition. $P_{\Lambda(M)}(t) = (1 + t)^m$.
Proof. By the basis theorem of Exterior Powers, $\operatorname{rank}_R \Lambda^n M = \binom{m}{n}$ for $0 \leq n \leq m$ and $0$ afterwards, so the series is $\sum_{n=0}^{m} \binom{m}{n} t^n = (1+t)^m$.
In particular $\operatorname{rank}_R \Lambda(M) = 2^m$. This is the graded dimension of the exterior algebra, and it is the numerical form of the statement that $\Lambda(M)$ has one basis monomial for each subset of a basis of $M$.
The Wedge Product
Definition
The exterior powers carry a product that concatenates wedges.
Definition. The wedge product on $\Lambda(M)$ is the bilinear map
$$ \wedge : \Lambda^p M \times \Lambda^q M \to \Lambda^{p+q} M $$
defined on elementary wedges by
$$ (x_1 \wedge \cdots \wedge x_p) \wedge (y_1 \wedge \cdots \wedge y_q) = x_1 \wedge \cdots \wedge x_p \wedge y_1 \wedge \cdots \wedge y_q $$
and extended bilinearly.
Proposition (well-definedness). The displayed prescription is well defined.
Proof. Fix $u = x_1 \wedge \cdots \wedge x_p \in \Lambda^p M$. The map $M^q \to \Lambda^{p+q} M$, $(y_1, \ldots, y_q) \mapsto x_1 \wedge \cdots \wedge x_p \wedge y_1 \wedge \cdots \wedge y_q$, is alternating $q$-multilinear, since alternation and multilinearity are verified in each $y_j$ with the leading $x_i$ held fixed. By the universal property of Exterior Powers it descends to a linear map $\Lambda^q M \to \Lambda^{p+q} M$, $v \mapsto u \wedge v$. The resulting map is linear in $u$ because $\wedge$ on the first factor is multilinear, so it too descends, giving a well-defined bilinear $\wedge$.
Example. For $M = R^3$ with basis $e_1, e_2, e_3$,
$$ e_1 \wedge e_2 = -e_2 \wedge e_1 \in \Lambda^2 M, \qquad (e_1 \wedge e_2) \wedge e_3 = e_1 \wedge e_2 \wedge e_3 \in \Lambda^3 M, $$
and $e_1 \wedge e_1 = 0$.
The Algebra Structure
Theorem. Let $M$ be an $R$-module. With the wedge product, $\Lambda(M)$ is an associative, unital, graded $R$-algebra, and the unit is the element $1 \in \Lambda^0 M = R$.
Proof. Bilinearity is part of the definition, and $\Lambda^p M \wedge \Lambda^q M \subseteq \Lambda^{p+q} M$, so the product is graded. Associativity is checked on elementary wedges: both $(u \wedge v) \wedge w$ and $u \wedge (v \wedge w)$ are the concatenation of the three lists of generators, so they agree, and both sides extend bilinearly. The element $1 \in R = \Lambda^0 M$ satisfies $1 \wedge u = u \wedge 1 = u$ on elementary wedges, hence by linearity.
The algebra is generated by its degree-1 part: every elementary wedge is a product of elements of $M$, and $\Lambda(M)$ is spanned by elementary wedges.
Universality
The exterior algebra is the free solution to the equation $x^2 = 0$.
Theorem (universal property). Let $A$ be an associative unital $R$-algebra and let $f : M \to A$ be an $R$-linear map with $f(x)^2 = 0$ for all $x \in M$. Then there is a unique algebra homomorphism $\Phi : \Lambda(M) \to A$ with $\Phi|_M = f$.
Proof. By the universal property of the tensor algebra, $f$ extends uniquely to an algebra homomorphism $\tilde{f} : T(M) \to A$. For $x \in M$ one has $\tilde{f}(x \otimes x) = f(x)^2 = 0$; since the elements $x \otimes x$ generate the two-sided ideal $I$ of the quotient $\Lambda(M) = T(M)/I$, the map $\tilde{f}$ kills $I$ and descends to the quotient. Uniqueness holds because $M$ generates $\Lambda(M)$ as an algebra.
Corollary. $\Lambda(M) \cong T(M)/I$, where $I$ is the two-sided ideal of the tensor algebra generated by $\{x \otimes x : x \in M\}$. The ideal $I$ also contains every element $x \otimes y + y \otimes x$, and when $2$ is invertible in $R$ the ideal $I$ is generated by these elements alone.
Proof. Since $x \otimes y + y \otimes x = (x+y) \otimes (x+y) - x \otimes x - y \otimes y$, the ideal $J$ generated by the elements $x \otimes y + y \otimes x$ is contained in $I$. Conversely, putting $y = x$ gives $2\,(x \otimes x) \in J$, and when $2$ is a unit of $R$ this gives $x \otimes x \in J$, hence $I \subseteq J$ and $J = I$. The quotient $T(M)/I$ carries the universal property above and is therefore isomorphic to $\Lambda(M)$.
Remark. The two ideals differ when $2$ is not invertible: if $2 = 0$ in $R$ then $x \otimes y + y \otimes x$ and $x \otimes y - y \otimes x$ are the same element, so $J$ is the ideal that defines the symmetric algebra, and $T(M)/J \cong \operatorname{Sym}(M)$. For $M$ free of rank $m$ the degree-two parts then have ranks $\binom{m+1}{2}$ and $\binom{m}{2}$, so the quotient by the anticommutation relations alone is not $\Lambda(M)$. This is the tensor-algebra counterpart of the distinction between skew-symmetric and alternating maps drawn in Exterior Powers.
Remark. The universal property above is not a statement about graded algebras; it characterises $\Lambda(M)$ as the free algebra on $M$ subject to $x^2 = 0$. There is a graded refinement: $\Lambda(M)$ is also the free graded-commutative algebra on $M$ placed in odd degree with the relations $x \wedge x = 0$. The two universal properties describe the same algebra, and no hypothesis on $2$ is needed for the passage between them: if $x$ is odd in a graded-commutative algebra and $x^2 = 0$, then polarising the square gives $xy + yx = -x^2 - y^2 + (x+y)^2 = 0$, so the Koszul sign rule is already forced by the square-zero relation and the algebra map produced by the ungraded property is automatically graded.
Graded-Commutativity
The Koszul Sign Rule
Theorem. For homogeneous elements $u \in \Lambda^p M$ and $v \in \Lambda^q M$,
$$ u \wedge v = (-1)^{p q}\, v \wedge u. $$
Proof. It suffices to prove the claim on elementary wedges $u = x_1 \wedge \cdots \wedge x_p$ and $v = y_1 \wedge \cdots \wedge y_q$, since both sides are bilinear. Moving each of the $q$ elements $y_j$ leftwards past the $p$ elements $x_i$ costs one sign $-1$ per transposition, so the total sign is $(-1)^{pq}$:
$$ y_1 \wedge \cdots \wedge y_q \wedge x_1 \wedge \cdots \wedge x_p = (-1)^{q p}\, x_1 \wedge \cdots \wedge x_p \wedge y_1 \wedge \cdots \wedge y_q. $$
This is the identity.
Definition. An algebra with a direct-sum decomposition and a product satisfying $uv = (-1)^{|u||v|} vu$ on homogeneous elements is called graded-commutative, or supercommutative. The rule $(-1)^{|u||v|}$ is the Koszul sign rule.
Consequences
Corollary. If $u \in \Lambda^n M$ with $n$ odd, then $u \wedge u = 0$.
Proof. Putting $p = q = n$ in the sign rule gives $u \wedge u = (-1)^{n^2} u \wedge u = -u \wedge u$ when $n$ is odd, so $u \wedge u$ is $2$-torsion; this alone does not force it to vanish over an arbitrary $R$, and the vanishing uses the monomial basis. Since $\Lambda^n$ is right exact and $\wedge$ is natural, it suffices to treat a free $M$ with basis $e_1, \ldots, e_m$ and an element $u = \sum_{|I| = n} c_I e_I$. The terms with $I = J$ vanish, because a basis vector then occurs twice. For $I \neq J$ of size $n$ the products $e_I \wedge e_J$ and $e_J \wedge e_I$ vanish together when $I \cap J \neq \varnothing$, and otherwise are both multiples of $e_{I \cup J}$ whose signs differ, since passing the block $J$ of $n$ factors past the block $I$ of $n$ factors costs $n^2$ transpositions and $n^2$ is odd. The two monomials therefore cancel in pairs, and $u \wedge u = 0$.
Remark. In a general graded-commutative algebra the sign rule gives only $2\,u \wedge u = 0$ for an odd element $u$; the stronger conclusion above is a property of the exterior algebra, where the product of a monomial with itself contains a repeated basis vector.
Corollary. The even part
$$ \Lambda^{\mathrm{ev}}(M) = \bigoplus_{n \text{ even}} \Lambda^n M $$
is a commutative subalgebra. The odd part $\Lambda^{\mathrm{odd}}(M) = \bigoplus_{n \text{ odd}} \Lambda^n M$ is a module over the even part, and the product of two odd elements lies in the even part.
Proof. For homogeneous even $u, v$ the sign is $(-1)^{pq}$ with $p, q$ even, hence $+1$, so $u \wedge v = v \wedge u$; linearity gives commutativity on all of $\Lambda^{\mathrm{ev}}$. The remaining statements are immediate from the gradation.
Remark. A square need not vanish for an even element. In $\Lambda(\mathbb{R}^4)$ with basis $e_1, e_2, e_3, e_4$, put $u = e_1 \wedge e_2 + e_3 \wedge e_4 \in \Lambda^2$. Then
$$ u \wedge u = 2\, e_1 \wedge e_2 \wedge e_3 \wedge e_4, $$
because $(e_1 \wedge e_2)^2 = (e_3 \wedge e_4)^2 = 0$ while $(e_1 \wedge e_2) \wedge (e_3 \wedge e_4) = (e_3 \wedge e_4) \wedge (e_1 \wedge e_2) = e_1 \wedge e_2 \wedge e_3 \wedge e_4$. Over a field of characteristic different from $2$ this is nonzero. Thus even elements of the exterior algebra are not nilpotent in general, in contrast with the odd elements.
The Symmetric and Alternating Quotients
The tensor algebra $T(M)$ admits two distinguished quotients, obtained by imposing commutativity or by imposing $x^2 = 0$:
$$ \operatorname{Sym}(M) = T(M)/(x \otimes y - y \otimes x), \qquad \Lambda(M) = T(M)/(x \otimes x). $$
The symmetric algebra is commutative and is the free commutative algebra on $M$; its theory belongs to the category on symmetric algebras. The exterior algebra is graded-commutative and is the free algebra with $x^2 = 0$.
Remark (characteristic $2$). When $2 = 0$ in $R$, every sign $-1$ equals $+1$, so graded-commutativity reduces to ordinary commutativity and $\Lambda(M)$ becomes a commutative algebra. It is not, however, isomorphic to $\operatorname{Sym}(M)$: in $\Lambda(M)$ one still has $x \wedge x = 0$ for every $x \in M$, whereas in $\operatorname{Sym}(M)$ the square of a nonzero $x$ is nonzero. For $M$ free of rank $m$ the two algebras have very different sizes: $\Lambda(M)$ has rank $2^m$, with basis the squarefree wedges, while $\operatorname{Sym}(M) \cong R[x_1, \ldots, x_m]$ has infinite rank as soon as $m \geq 1$, its degree-two part alone having rank $\binom{m+1}{2}$ against the $\binom{m}{2}$ of $\Lambda^2M$. This is the reason the exterior algebra is defined by the relation $x \wedge x = 0$ rather than by anticommutativity: the former is the relation available over every commutative ring.
Functoriality and Constructions
The Functor $\Lambda$
Definition. Let $f : M \to N$ be $R$-linear. Define $\Lambda(f) : \Lambda(M) \to \Lambda(N)$ on homogeneous components by $\Lambda^n f$ and extend by linearity. Thus
$$ \Lambda(f)(x_1 \wedge \cdots \wedge x_n) = f(x_1) \wedge \cdots \wedge f(x_n). $$
Proposition. $\Lambda(f)$ is a homomorphism of graded unital algebras; $\Lambda(\mathrm{id}_M) = \mathrm{id}_{\Lambda(M)}$; and $\Lambda(g \circ f) = \Lambda(g) \circ \Lambda(f)$. Hence $\Lambda$ is a functor from $R$-modules to graded unital $R$-algebras.
Proof. The induced map preserves the product because on elementary wedges it is concatenation followed by $f$-application on each factor, and $f$ is linear; the unit is fixed because $\Lambda^0 f = \mathrm{id}_R$. The functorial identities are checked on elementary wedges.
The Exterior Algebra of a Direct Sum
Theorem. For all $R$-modules $M, N$ there is a natural isomorphism of graded algebras
$$ \Lambda(M \oplus N) \cong \Lambda(M) \hat{\otimes}_R \Lambda(N), $$
where the right-hand side is the graded tensor product, with multiplication
$$ (u \otimes v)(u' \otimes v') = (-1)^{|v|\,|u'|}\, (u \wedge u') \otimes (v \wedge v') $$
for homogeneous $u, u' \in \Lambda(M)$ and $v, v' \in \Lambda(N)$.
Proof. The linear map $M \oplus N \to \Lambda(M) \hat{\otimes} \Lambda(N)$, $(m, n) \mapsto m \otimes 1 + 1 \otimes n$, sends $(m, n)$ to an odd element whose square is zero, because
$$ (m \otimes 1 + 1 \otimes n)^2 = m \wedge m \otimes 1 + (m \otimes n + (-1)^{|m||n|} m \otimes n) + 1 \otimes n \wedge n = 0 $$
in the graded tensor product: the outer terms vanish because $m \wedge m = 0$ and $n \wedge n = 0$, while the two cross terms are $m \otimes n$ and $(-1)^{|m||n|}m \otimes n$, which are opposite for odd $m, n$ and cancel. By the universal property the map extends to an algebra homomorphism $\Lambda(M \oplus N) \to \Lambda(M) \hat{\otimes} \Lambda(N)$, which in degree $n$ is the isomorphism $\Lambda^n(M \oplus N) \cong \bigoplus_{p+q=n} \Lambda^p M \otimes \Lambda^q N$ of Exterior Powers; hence it is an isomorphism of graded algebras.
Base Change
Proposition. Let $R \to S$ be a ring homomorphism. Then there is a natural isomorphism of graded $S$-algebras
$$ \Lambda(M \otimes_R S) \cong \Lambda(M) \otimes_R S. $$
Proof. The degreewise isomorphism of Exterior Powers assembles into a degree-preserving bijection that respects the product, since both products are concatenation of wedges.
The Exterior Algebra of a Free Module
Theorem. Let $M$ be free with basis $(e_1, \ldots, e_m)$. Then $\Lambda(M)$ is free of rank $2^m$, with basis the monomials
$$ e_{i_1} \wedge e_{i_2} \wedge \cdots \wedge e_{i_k}, \qquad 1 \leq i_1 < i_2 < \cdots < i_k \leq m, $$
including the empty product, which is the unit $1 \in \Lambda^0 M$. Equivalently,
$$ \Lambda(R^m) \cong R\langle x_1, \ldots, x_m \rangle / (x_i^2,\ x_i x_j + x_j x_i), $$
the quotient of the free associative algebra by the relations making the generators anticommute and square to zero.
Proof. The basis statement is the degreewise basis theorem of Exterior Powers summed over $n$. For the presentation, let $J$ be the two-sided ideal of the free algebra generated by the elements $x_i^2$ and $x_ix_j + x_jx_i$. Every generator of $J$ lies in the ideal $I = (x \otimes x)$ of the corollary above, since $x_i^2$ and $(x_i+x_j)^2 - x_i^2 - x_j^2$ do; hence $J \subseteq I$. Conversely, for $x = \sum_i a_i x_i$ one has
$$
x \otimes x = \sum_i a_i^2\, x_i^2 + \sum_{i so $I \subseteq J$ and the two ideals coincide. The quotient $R\langle x_1, \ldots, x_m\rangle/J$ therefore has the universal property of $\Lambda$, and the two algebras are isomorphic. Example. For $M$ of rank $2$ the algebra has basis $1, x, y, x \wedge y$ over $R$, with $x \wedge x = y \wedge y = 0$ and $y \wedge x = -x \wedge y$. For rank $m$ the algebra is generated by $m$ elements subject to $x_i x_j = -x_j x_i$ and $x_i^2 = 0$; it is the "squarefree" algebra on $m$ generators. Three later constructions are instances of the exterior algebra. The Clifford algebra $\mathrm{Cl}(V, q)$ is the quotient $T(V)/(x \otimes x - q(x))$ for a quadratic form $q$; the exterior algebra is the case $q = 0$, and for a nondegenerate form the associated graded algebra of $\mathrm{Cl}(V, q)$ with respect to the filtration by degree is $\Lambda(V)$, as shown. The algebra of differential forms on a smooth manifold is the exterior algebra of the cotangent space, varying smoothly over the manifold, and the exterior derivative is a graded derivation of it; this is developed with this one. The Grassmann algebra of the Berezin calculus is the exterior algebra on a finite set of odd generators, with the integral defined as a linear functional extracting the top coefficient; this is developed, also. The top exterior power $\Lambda^m M$ of a free module of rank $m$ is the determinant line, and the scalar by which $\Lambda^m f$ acts is the determinant of an endomorphism $f$. Alternating forms on $M$ are the linear functionals on $\Lambda(M)$; assembled degreewise they form the exterior algebra $\Lambda(M^*)$ of the dual when $M$ is free of finite rank, which is the algebraic model for the algebra of differential forms. The determinant, the minors, and the pairing of complementary exterior powers are treated. The exterior algebra of an $R$-module $M$ is the graded module $\Lambda(M) = \bigoplus_{n \geq 0} \Lambda^n M$, with $\Lambda^0 M = R$ and $\Lambda^1 M = M$, equipped with the wedge product that concatenates elementary wedges. It is an associative unital graded algebra, and it is graded-commutative, $uv = (-1)^{|u||v|} v u$; consequently odd elements square to zero and the even part is a commutative subalgebra, though even elements need not square to zero. $\Lambda(M)$ is the free algebra on $M$ subject to $x \wedge x = 0$, equivalently the quotient $T(M)/(x \otimes x)$ of the tensor algebra, and it is the free graded-commutative algebra on $M$. It is functorial in $M$; it converts direct sums into graded tensor products, $\Lambda(M \oplus N) \cong \Lambda(M) \hat{\otimes} \Lambda(N)$; it commutes with base change; and for a free module of rank $m$ it is free of rank $2^m$ with basis the squarefree wedges, with Poincaré series $(1+t)^m$. In characteristic $2$ graded-commutativity becomes ordinary commutativity, but $\Lambda(M)$ still differs from the symmetric algebra because $x \wedge x = 0$ remains. The exterior algebra is the $q = 0$ case of the Clifford algebra, the pointwise model of the algebra of differential forms, and the algebraic setting of the Berezin integral.The Exterior Algebra in the Wider Corpus
Summary
Summary of Notation
Symbol
Meaning
$R$
Commutative ring with identity $1 \neq 0$; the default base
$K$
Field
$M$, $N$
$R$-modules
$T(M)$
Tensor algebra $\bigoplus_{n \geq 0} M^{\otimes n}$
$\Lambda^n M$
$n$-th exterior power of $M$
$\Lambda(M) = \bigoplus_{n \geq 0} \Lambda^n M$
Exterior algebra of $M$
$\wedge$
Wedge product; $(x_1 \wedge \cdots \wedge x_p) \wedge (y_1 \wedge \cdots \wedge y_q)$ concatenates
$\vert u\vert$
Degree of a homogeneous element $u \in \Lambda^n M$
$uv = (-1)^{\vert u\vert\vert v\vert} vu$
Graded-commutativity; the Koszul sign rule
$\Lambda^{\mathrm{ev}}(M), \Lambda^{\mathrm{odd}}(M)$
Even and odd parts; the even part is a commutative subalgebra
$\Lambda(f)$
Induced graded algebra homomorphism
$\hat{\otimes}_R$
Graded tensor product, $(u \otimes v)(u' \otimes v') = (-1)^{\vert v\vert\vert u'\vert}(u u') \otimes (v v')$
$P_{\Lambda(M)}(t) = (1+t)^m$
Poincaré series for $M$ free of rank $m$
$\operatorname{Sym}(M) = T(M)/(x \otimes y - y \otimes x)$
Symmetric algebra, the commutative quotient
$\Lambda(M) = T(M)/(x \otimes x)$
Exterior algebra, the alternating quotient
$\mathrm{Cl}(V, q) = T(V)/(x \otimes x - q(x))$
Clifford algebra; the case $q = 0$ gives $\Lambda(V)$
$\Lambda^m M$
Top exterior power, the determinant line
Further Reading