Orientation defined by an ordered set of vectors.

Reversed orientation corresponds to negating the exterior product.

Geometric interpretation of graden elements in a real exterior algebra forn = 0 (signed point), 1 (directed line segment, or vector), 2 (oriented plane element), 3 (oriented volume). The exterior product ofn vectors can be visualized as anyn-dimensional shape (e.g.n-parallelotope,n-ellipsoid); with magnitude (hypervolume), andorientation defined by that of its(n − 1)-dimensional boundary and on which side the interior is.^[1][2]

In mathematics, theexterior algebra orGrassmann algebra of avector space${\displaystyle V}$ is anassociative algebra that contains${\displaystyle V,}$ which has a product, calledexterior product orwedge product and denoted with${\displaystyle \wedge }$, such that${\displaystyle v\wedge v=0}$ for every vector${\displaystyle v}$ in${\displaystyle V.}$ The exterior algebra is named afterHermann Grassmann,^[3] and the names of the product come from the "wedge" symbol${\displaystyle \wedge }$ and the fact that the product of two elements of${\displaystyle V}$ is "outside"${\displaystyle V.}$

The wedge product of${\displaystyle k}$ vectors${\displaystyle v_1\wedge v_2\wedge \cdots \wedge v_k}$ is called ablade of degree${\displaystyle k}$ or${\displaystyle k}$-blade. The wedge product was introduced originally as an algebraic construction used ingeometry to studyareas,volumes, and their higher-dimensional analogues: themagnitude of a2-blade${\displaystyle v\wedge w}$ is the area of theparallelogram defined by${\displaystyle v}$ and${\displaystyle w,}$ and, more generally, the magnitude of a${\displaystyle k}$-blade is the (hyper)volume of theparallelotope defined by the constituent vectors. Thealternating property that${\displaystyle v\wedge v=0}$ implies a skew-symmetric property that${\displaystyle v\wedge w=-w\wedge v,}$ and more generally any blade flips sign whenever two of its constituent vectors are exchanged, corresponding to a parallelotope of opposite orientation.

The full exterior algebra contains objects that are not themselves blades, butlinear combinations of blades; a sum of blades of homogeneous degree${\displaystyle k}$ is called ak-vector, while a more general sum of blades of arbitrary degree is called amultivector.^[4] Thelinear span of the${\displaystyle k}$-blades is called the${\displaystyle k}$-th exterior power of${\displaystyle V.}$ The exterior algebra is thedirect sum of the${\displaystyle k}$-th exterior powers of${\displaystyle V,}$ and this makes the exterior algebra agraded algebra.

The exterior algebra isuniversal in the sense that every equation that relates elements of${\displaystyle V}$ in the exterior algebra is also valid in every associative algebra that contains${\displaystyle V}$ and in which the square of every element of${\displaystyle V}$ is zero.

The definition of the exterior algebra can be extended for spaces built from vector spaces, such asvector fields andfunctions whosedomain is a vector space. Moreover, the field ofscalars may be any field. More generally, the exterior algebra can be defined formodules over acommutative ring. In particular, the algebra ofdifferential forms in${\displaystyle k}$ variables is an exterior algebra over the ring of thesmooth functions in${\displaystyle k}$ variables.

The two-dimensionalEuclidean vector space${\displaystyle {\mathbf{R}}^2}$ is areal vector space equipped with abasis consisting of a pair of orthogonalunit vectors$${\displaystyle {\mathbf{e}}_1=\left[\begin{array}{c}1\\ 0\end{array}\right],{\mathbf{e}}_2=\left[\begin{array}{c}0\\ 1\end{array}\right].}$$

Suppose that$${\displaystyle \mathbf{v}=\left[\begin{array}{c}a\\ b\end{array}\right]=a{\mathbf{e}}_1+b{\mathbf{e}}_2,\mathbf{w}=\left[\begin{array}{c}c\\ d\end{array}\right]=c{\mathbf{e}}_1+d{\mathbf{e}}_2}$$are a pair of given vectors in⁠${\displaystyle {\mathbf{R}}^2}$⁠, written in components. There is a unique parallelogram having${\displaystyle \mathbf{v}}$ and${\displaystyle \mathbf{w}}$ as two of its sides. Thearea of this parallelogram is given by the standarddeterminant formula:

Consider now the exterior product of${\displaystyle \mathbf{v}}$ and⁠${\displaystyle \mathbf{w}}$⁠:where the first step uses the distributive law for theexterior product, and the last uses the fact that the exterior product is analternating map, and in particular${\displaystyle {\mathbf{e}}_2\wedge {\mathbf{e}}_1=-({\mathbf{e}}_1\wedge {\mathbf{e}}_2).}$ (The fact that the exterior product is an alternating map also forces${\displaystyle {\mathbf{e}}_1\wedge {\mathbf{e}}_1={\mathbf{e}}_2\wedge {\mathbf{e}}_2=0.}$) Note that the coefficient in this last expression is precisely the determinant of the matrix[vw]. The fact that this may be positive or negative has the intuitive meaning thatv andw may be oriented in a counterclockwise or clockwise sense as the vertices of the parallelogram they define. Such an area is called thesigned area of the parallelogram: theabsolute value of the signed area is the ordinary area, and the sign determines its orientation.

The fact that this coefficient is the signed area is not an accident. In fact, it is relatively easy to see that the exterior product should be related to the signed area if one tries to axiomatize this area as an algebraic construct. In detail, ifA(v,w) denotes the signed area of the parallelogram of which the pair of vectorsv andw form two adjacent sides, then A must satisfy the following properties:

1. A(rv,sw) =rsA(v,w) for any real numbersr ands, since rescaling either of the sides rescales the area by the same amount (and reversing the direction of one of the sides reverses the orientation of the parallelogram).
2. A(v,v) = 0, since the area of thedegenerate parallelogram determined byv (i.e., aline segment) is zero.
3. A(w,v) = −A(v,w), since interchanging the roles ofv andw reverses the orientation of the parallelogram.
4. A(v +rw,w) = A(v,w) for any real numberr, since adding a multiple ofw tov affects neither the base nor the height of the parallelogram and consequently preserves its area.
5. A(e₁,e₂) = 1, since the area of the unit square is one.

With the exception of the last property, the exterior product of two vectors satisfies the same properties as the area. In a certain sense, the exterior product generalizes the final property by allowing the area of a parallelogram to be compared to that of any chosen parallelogram in a parallel plane (here, the one with sidese₁ ande₂). In other words, the exterior product provides abasis-independent formulation of area.^[5]

For vectors inR³, the exterior algebra is closely related to thecross product andtriple product. Using the standard basis{e₁,e₂,e₃}, the exterior product of a pair of vectors

${\displaystyle \mathbf{u}=u_1{\mathbf{e}}_1+u_2{\mathbf{e}}_2+u_3{\mathbf{e}}_3}$

and

${\displaystyle \mathbf{v}=v_1{\mathbf{e}}_1+v_2{\mathbf{e}}_2+v_3{\mathbf{e}}_3}$

is

${\displaystyle \mathbf{u}\wedge \mathbf{v}=(u_1{v}_2-u_2{v}_1)({\mathbf{e}}_1\wedge {\mathbf{e}}_2)+(u_3{v}_1-u_1{v}_3)({\mathbf{e}}_3\wedge {\mathbf{e}}_1)+(u_2{v}_3-u_3{v}_2)({\mathbf{e}}_2\wedge {\mathbf{e}}_3)}$

where {e₁ ∧e₂,e₃ ∧e₁,e₂ ∧e₃} is the basis for the three-dimensional space ⋀²(R³). The coefficients above are the same as those in the usual definition of thecross product of vectors in three dimensions, the only difference being that the exterior product is not an ordinary vector, but instead is abivector.

Bringing in a third vector

${\displaystyle \mathbf{w}=w_1{\mathbf{e}}_1+w_2{\mathbf{e}}_2+w_3{\mathbf{e}}_3,}$

the exterior product of three vectors is

${\displaystyle \mathbf{u}\wedge \mathbf{v}\wedge \mathbf{w}=(u_1{v}_2{w}_3+u_2{v}_3{w}_1+u_3{v}_1{w}_2-u_1{v}_3{w}_2-u_2{v}_1{w}_3-u_3{v}_2{w}_1)({\mathbf{e}}_1\wedge {\mathbf{e}}_2\wedge {\mathbf{e}}_3)}$

wheree₁ ∧e₂ ∧e₃ is the basis vector for the one-dimensional space ⋀³(R³). The scalar coefficient is thetriple product of the three vectors.

The cross product and triple product in three dimensions each admit both geometric and algebraic interpretations. The cross productu ×v can be interpreted as a vector which is perpendicular to bothu andv and whose magnitude is equal to the area of the parallelogram determined by the two vectors. It can also be interpreted as the vector consisting of theminors of the matrix with columnsu andv. The triple product ofu,v, andw is geometrically a (signed) volume. Algebraically, it is the determinant of the matrix with columnsu,v, andw. The exterior product in three dimensions allows for similar interpretations. In fact, in the presence of a positively orientedorthonormal basis, the exterior product generalizes these notions to higher dimensions.

The exterior algebra${\displaystyle \bigwedge (V)}$ of a vector space${\displaystyle V}$ over afield${\displaystyle K}$ is defined as thequotient algebra of thetensor algebraT(V), where

${\displaystyle T(V)=\underset{k=0}{\overset{\mathrm{\infty }}{⨁}}{T}^{k}V=K\oplus V\oplus (V\otimes V)\oplus (V\otimes V\otimes V)\oplus \cdots ,}$

by the two-sidedideal${\displaystyle I}$ generated by all elements of the form${\displaystyle x\otimes x}$ such that${\displaystyle x\in V}$.^[6] Symbolically,

${\displaystyle \bigwedge (V):=T(V)/I.}$

The exterior product${\displaystyle \wedge }$ of two elements of${\displaystyle \bigwedge (V)}$ is defined by

${\displaystyle \alpha \wedge \beta =\alpha \otimes \beta (\mathrm{mod}I).}$

The exterior product is by constructionalternating on elements of⁠${\displaystyle V}$⁠, which means that${\displaystyle x\wedge x=0}$ for all${\displaystyle x\in V,}$ by the above construction. It follows that the product is alsoanticommutative on elements of⁠${\displaystyle V}$⁠, for supposing that⁠${\displaystyle x,y\in V}$⁠,

${\displaystyle 0=(x+y)\wedge (x+y)=x\wedge x+x\wedge y+y\wedge x+y\wedge y=x\wedge y+y\wedge x}$

hence

${\displaystyle x\wedge y=-(y\wedge x).}$

More generally, if${\displaystyle \sigma }$ is apermutation of the integers⁠${\displaystyle [1,\dots ,k]}$⁠, and⁠${\displaystyle x_1}$⁠,⁠${\displaystyle x_2}$⁠, ...,⁠${\displaystyle x_k}$⁠ are elements of⁠${\displaystyle V}$⁠, it follows that

${\displaystyle x_{\sigma (1)}\wedge x_{\sigma (2)}\wedge \cdots \wedge x_{\sigma (k)}=\mathrm{sgn}(\sigma )x_1\wedge x_2\wedge \cdots \wedge x_k,}$

where${\displaystyle \mathrm{sgn}(\sigma )}$ is thesignature of the permutation⁠${\displaystyle \sigma }$⁠.^[7]

In particular, if${\displaystyle x_i=x_j}$ for some⁠${\displaystyle i\ne j}$⁠, then the following generalization of the alternating property also holds:

${\displaystyle x_1\wedge x_2\wedge \cdots \wedge x_k=0.}$

Together with the distributive property of the exterior product, one further generalization is that a necessary and sufficient condition for${\displaystyle \{x_1,x_2,\dots ,x_k\}}$ to be a linearly dependent set of vectors is that

${\displaystyle x_1\wedge x_2\wedge \cdots \wedge x_k=0.}$

Thekthexterior power of⁠${\displaystyle V}$⁠, denoted⁠${\displaystyle {\bigwedge }^k(V)}$⁠, is thevector subspace of⁠${\displaystyle \bigwedge (V)}$⁠spanned by elements of the form

${\displaystyle x_1\wedge x_2\wedge \cdots \wedge x_k,x_i\in V,i=1,2,\dots ,k.}$

If⁠${\displaystyle \alpha \in {\bigwedge }^k(V)}$⁠, then${\displaystyle \alpha }$ is said to be ak-vector. If, furthermore,${\displaystyle \alpha }$ can be expressed as an exterior product of${\displaystyle k}$ elements of⁠${\displaystyle V}$⁠, then${\displaystyle \alpha }$ is said to bedecomposable (or simple, by some authors; or a blade, by others). Although decomposable⁠${\displaystyle k}$⁠-vectors span⁠${\displaystyle {\bigwedge }^k(V)}$⁠, not every element of${\displaystyle {\bigwedge }^k(V)}$ is decomposable. For example, given⁠${\displaystyle {\mathbf{R}}^4}$⁠ with a basis⁠${\displaystyle \{e_1,e_2,e_3,e_4\}}$⁠, the following 2-vector is not decomposable:

${\displaystyle \alpha =e_1\wedge e_2+e_3\wedge e_4.}$

If thedimension of${\displaystyle V}$ is${\displaystyle n}$ and${\displaystyle \{e_1,\dots ,e_n\}}$ is abasis for${\displaystyle V}$, then the set

is a basis for⁠${\displaystyle {\bigwedge }^k(V)}$⁠. The reason is the following: given any exterior product of the form

${\displaystyle v_1\wedge \cdots \wedge v_k,}$

every vector${\displaystyle v_j}$ can be written as alinear combination of the basis vectors⁠${\displaystyle e_i}$⁠; using the bilinearity of the exterior product, this can be expanded to a linear combination of exterior products of those basis vectors. Any exterior product in which the same basis vector appears more than once is zero; any exterior product in which the basis vectors do not appear in the proper order can be reordered, changing the sign whenever two basis vectors change places. In general, the resulting coefficients of the basisk-vectors can be computed as theminors of thematrix that describes the vectors${\displaystyle v_j}$ in terms of the basis⁠${\displaystyle e_i}$⁠.

By counting the basis elements, the dimension of${\displaystyle {\bigwedge }^k(V)}$ is equal to abinomial coefficient:

${\displaystyle \dim {\bigwedge }^k(V)=(\genfrac{}{}{0ex}{}{n}{k}),}$

where⁠${\displaystyle n}$⁠ is the dimension of thevectors, and⁠${\displaystyle k}$⁠ is the number of vectors in the product. The binomial coefficient produces the correct result, even for exceptional cases; in particular,${\displaystyle {\bigwedge }^k(V)=\{0\}}$ for⁠${\displaystyle k>n}$⁠.

Any element of the exterior algebra can be written as a sum ofk-vectors. Hence, as a vector space the exterior algebra is adirect sum

${\displaystyle \bigwedge (V)={\bigwedge }^0(V)\oplus {\bigwedge }^1(V)\oplus {\bigwedge }^2(V)\oplus \cdots \oplus {\bigwedge }^n(V)}$

(where, by convention,⁠${\displaystyle {\bigwedge }^0(V)=K}$⁠, thefield underlying⁠${\displaystyle V}$⁠, and⁠${\displaystyle {\bigwedge }^1(V)=V}$⁠), and therefore its dimension is equal to the sum of the binomial coefficients, which is⁠${\displaystyle 2^n}$⁠.

If⁠${\displaystyle \alpha \in {\bigwedge }^k(V)}$⁠, then it is possible to express${\displaystyle \alpha }$ as a linear combination of decomposablek-vectors:

${\displaystyle \alpha ={\alpha }^{(1)}+{\alpha }^{(2)}+\cdots +{\alpha }^{(s)}}$

where each${\displaystyle {\alpha }^{(i)}}$ is decomposable, say

${\displaystyle {\alpha }^{(i)}={\alpha }_1^{(i)}\wedge \cdots \wedge {\alpha }_k^{(i)},i=1,2,\dots ,s.}$

Therank of thek-vector${\displaystyle \alpha }$ is the minimal number of decomposablek-vectors in such an expansion of⁠${\displaystyle \alpha }$⁠. This is similar to the notion oftensor rank.

Rank is particularly important in the study of 2-vectors (Sternberg 1964, §III.6) (Bryant et al. 1991). The rank of a 2-vector${\displaystyle \alpha }$ can be identified with half therank of the matrix of coefficients of${\displaystyle \alpha }$ in a basis. Thus if${\displaystyle e_i}$ is a basis for⁠${\displaystyle V}$⁠, then${\displaystyle \alpha }$ can be expressed uniquely as

${\displaystyle \alpha =\sum _{i,j}{a}_{ij}{e}_i\wedge e_j}$

where${\displaystyle a_{ij}=-a_{ji}}$ (the matrix of coefficients isskew-symmetric). The rank of the matrix${\displaystyle a_{ij}}$ is therefore even, and is twice the rank of the form${\displaystyle \alpha }$.

In characteristic 0, the 2-vector${\displaystyle \alpha }$ has rank${\displaystyle p}$ if and only if

${\displaystyle \underset{p}{\underbrace{\alpha \wedge \cdots \wedge \alpha }}\ne 0}$ and${\displaystyle \underset{p+1}{\underbrace{\alpha \wedge \cdots \wedge \alpha }}=0.}$

The exterior product of ak-vector with ap-vector is a${\displaystyle (k+p)}$-vector, once again invoking bilinearity. As a consequence, the direct sum decomposition of the preceding section

${\displaystyle \bigwedge (V)={\bigwedge }^0(V)\oplus {\bigwedge }^1(V)\oplus {\bigwedge }^2(V)\oplus \cdots \oplus {\bigwedge }^n(V)}$

gives the exterior algebra the additional structure of agraded algebra, that is

${\displaystyle {\bigwedge }^k(V)\wedge {\bigwedge }^p(V)\subset {\bigwedge }^{k+p}(V).}$

Moreover, ifK is the base field, we have

${\displaystyle {\bigwedge }^0(V)=K}$ and${\displaystyle {\bigwedge }^1(V)=V.}$

The exterior product is graded anticommutative, meaning that if${\displaystyle \alpha \in {\bigwedge }^k(V)}$ and⁠${\displaystyle \beta \in {\bigwedge }^p(V)}$⁠, then

${\displaystyle \alpha \wedge \beta =(-1{)}^{kp}\beta \wedge \alpha .}$

In addition to studying the graded structure on the exterior algebra,Bourbaki (1989) studies additional graded structures on exterior algebras, such as those on the exterior algebra of agraded module (a module that already carries its own gradation).

LetV be a vector space over the fieldK. Informally, multiplication in${\displaystyle \bigwedge (V)}$ is performed by manipulating symbols and imposing adistributive law, anassociative law, and using the identity${\displaystyle v\wedge v=0}$ forv ∈V. Formally,${\displaystyle \bigwedge (V)}$ is the "most general" algebra in which these rules hold for the multiplication, in the sense that any unital associativeK-algebra containingV with alternating multiplication onV must contain a homomorphic image of⁠${\displaystyle \bigwedge (V)}$⁠. In other words, the exterior algebra has the followinguniversal property:^[8]

To construct the most general algebra that containsV and whose multiplication is alternating onV, it is natural to start with the most general associative algebra that containsV, thetensor algebraT(V), and then enforce the alternating property by taking a suitablequotient. We thus take the two-sidedidealI inT(V) generated by all elements of the formv ⊗v forv inV, and define${\displaystyle \bigwedge (V)}$ as the quotient

${\displaystyle \bigwedge (V)=T(V)/I}$

(and use∧ as the symbol for multiplication in⁠${\displaystyle \bigwedge (V)}$⁠). It is then straightforward to show that${\displaystyle \bigwedge (V)}$ containsV and satisfies the above universal property.

As a consequence of this construction, the operation of assigning to a vector spaceV its exterior algebra${\displaystyle \bigwedge (V)}$ is afunctor from thecategory of vector spaces to the category of algebras.

Rather than defining${\displaystyle \bigwedge (V)}$ first and then identifying the exterior powers${\displaystyle {\bigwedge }^k(V)}$ as certain subspaces, one may alternatively define the spaces${\displaystyle {\bigwedge }^k(V)}$ first and then combine them to form the algebra⁠${\displaystyle \bigwedge (V)}$⁠. This approach is often used in differential geometry and is described in the next section.

Given acommutative ring${\displaystyle R}$ and an${\displaystyle R}$-module⁠${\displaystyle M}$⁠, we can define the exterior algebra${\displaystyle \bigwedge (M)}$ just as above, as a suitable quotient of the tensor algebra⁠${\displaystyle \mathrm{T}(M)}$⁠. It will satisfy the analogous universal property. Many of the properties of${\displaystyle \bigwedge (M)}$ also require that${\displaystyle M}$ be aprojective module. Where finite dimensionality is used, the properties further require that${\displaystyle M}$ befinitely generated and projective. Generalizations to the most common situations can be found inBourbaki (1989).

Exterior algebras ofvector bundles are frequently considered in geometry and topology. There are no essential differences between the algebraic properties of the exterior algebra of finite-dimensional vector bundles and those of the exterior algebra of finitely generated projective modules, by theSerre–Swan theorem. More general exterior algebras can be defined forsheaves of modules.

For a field of characteristic not 2,^[9] the exterior algebra of a vector space${\displaystyle V}$ over${\displaystyle K}$ can be canonically identified with the vector subspace of${\displaystyle \mathrm{T}(V)}$ that consists ofantisymmetric tensors. For characteristic 0 (or higher than⁠${\displaystyle \dim V}$⁠), the vector space of${\displaystyle k}$-linear antisymmetric tensors is transversal to the ideal⁠${\displaystyle I}$⁠, hence, a good choice to represent the quotient. But for nonzero characteristic, the vector space of⁠${\displaystyle K}$⁠-linear antisymmetric tensors could be not transversal to the ideal (actually, for⁠${\displaystyle k\ge \mathrm{char}K}$⁠, the vector space of${\displaystyle K}$-linear antisymmetric tensors is contained in${\displaystyle I}$); nevertheless, transversal or not, a product can be defined on this space such that the resulting algebra is isomorphic to the exterior algebra: in the first case the natural choice for the product is just the quotient product (using the available projection), in the second case, this product must be slightly modified as given below (along Arnold setting), but such that the algebra stays isomorphic with the exterior algebra, i.e. the quotient of${\displaystyle \mathrm{T}(V)}$ by the ideal${\displaystyle I}$ generated by elements of the form⁠${\displaystyle x\otimes x}$⁠. Of course, for characteristic⁠${\displaystyle 0}$⁠ (or higher than the dimension of the vector space), one or the other definition of the product could be used, as the two algebras are isomorphic (see V. I. Arnold or Kobayashi-Nomizu).

Let${\displaystyle {\mathrm{T}}^r(V)}$ be the space of homogeneous tensors of degree${\displaystyle r}$. This is spanned by decomposable tensors

${\displaystyle v_1\otimes \cdots \otimes v_r,v_i\in V.}$

Theantisymmetrization (or sometimes theskew-symmetrization) of a decomposable tensor is defined by

${\displaystyle {\mathcal{A}}^{(\mathrm{r})}(v_1\otimes \cdots \otimes v_r)=\sum _{\sigma \in {\mathfrak{S}}_r}\mathrm{sgn}(\sigma )v_{\sigma (1)}\otimes \cdots \otimes v_{\sigma (r)}}$

and, when${\displaystyle r!\ne 0}$ (for nonzero characteristic field${\displaystyle r!}$ might be 0):

${\displaystyle {\mathrm{Alt}}^{(r)}(v_1\otimes \cdots \otimes v_r)=\frac{1}{r!}{\mathcal{A}}^{(\mathrm{r})}(v_1\otimes \cdots \otimes v_r)}$

where the sum is taken over thesymmetric group of permutations on the symbols⁠${\displaystyle \{1,\dots ,r\}}$⁠. This extends by linearity and homogeneity to an operation, also denoted by${\displaystyle \mathcal{A}}$ and${\displaystyle \mathrm{A}\mathrm{l}\mathrm{t}}$, on the full tensor algebra⁠${\displaystyle \mathrm{T}(V)}$⁠.

Note that

${\displaystyle {\mathcal{A}}^{(\mathrm{r})}{\mathcal{A}}^{(\mathrm{r})}=r!{\mathcal{A}}^{(\mathrm{r})}.}$

Such that, when defined,${\displaystyle {\mathrm{Alt}}^{(r)}}$ is the projection for the exterior (quotient) algebra onto the r-homogeneous alternating tensor subspace.On the other hand, the image${\displaystyle \mathcal{A}(\mathrm{T}(V))}$ is always thealternating tensor graded subspace (not yet an algebra, as product is not yet defined), denoted⁠${\displaystyle A(V)}$⁠. This is a vector subspace of⁠${\displaystyle \mathrm{T}(V)}$⁠, and it inherits the structure of a graded vector space from that on⁠${\displaystyle \mathrm{T}(V)}$⁠. Moreover, the kernel of${\displaystyle {\mathcal{A}}^{(r)}}$ is precisely⁠${\displaystyle I^{(r)}}$⁠, the homogeneous subset of the ideal⁠${\displaystyle I}$⁠, or the kernel of${\displaystyle \mathcal{A}}$ is⁠${\displaystyle I}$⁠. When${\displaystyle \mathrm{Alt}}$ is defined,${\displaystyle A(V)}$ carries an associative graded product${\displaystyle \hat{\otimes }}$ defined by (the same as the wedge product)

${\displaystyle t\wedge s=t\hat{\otimes }s=\mathrm{Alt}(t\otimes s).}$

Assuming${\displaystyle K}$ has characteristic 0, as${\displaystyle A(V)}$ is a supplement of${\displaystyle I}$ in⁠${\displaystyle \mathrm{T}(V)}$⁠, with the above given product, there is a canonical isomorphism

${\displaystyle A(V)\cong \bigwedge (V).}$

When the characteristic of the field is nonzero,${\displaystyle \mathcal{A}}$ will do what${\displaystyle \mathrm{A}\mathrm{l}\mathrm{t}}$ did before, but the product cannot be defined as above. In such a case, isomorphism${\displaystyle A(V)\cong \bigwedge (V)}$ still holds, in spite of${\displaystyle A(V)}$ not being a supplement of the ideal⁠${\displaystyle I}$⁠, but then, the product should be modified as given below (${\displaystyle \dot{\wedge }}$ product, Arnold setting).

Finally, we always get⁠${\displaystyle A(V)}$⁠ isomorphic with⁠${\displaystyle \bigwedge (V)}$⁠, but the product could (or should) be chosen in two ways (or only one). Actually, the product could be chosen in many ways, rescaling it on homogeneous spaces as${\displaystyle c(r+p)/c(r)c(p)}$ for an arbitrary sequence${\displaystyle c(r)}$ in the field, as long as the division makes sense (this is such that the redefined product is also associative, i.e. defines an algebra on⁠${\displaystyle A(V)}$⁠). Also note, the interior product definition should be changed accordingly, in order to keep its skew derivation property.

Suppose thatV has finite dimensionn, and that a basise₁, ...,e_n ofV is given. Then any alternating tensort ∈ A^r(V) ⊂T^r(V) can be written inindex notation with theEinstein summation convention as

${\displaystyle t=t^{i_1{i}_2\cdots i_r}{\mathbf{e}}_{i_1}\otimes {\mathbf{e}}_{i_2}\otimes \cdots \otimes {\mathbf{e}}_{i_r},}$

wheret^{i₁⋅⋅⋅i_r} iscompletely antisymmetric in its indices.

The exterior product of two alternating tensorst ands of ranksr andp is given by

${\displaystyle t\hat{\otimes }s=\frac{1}{(r+p)!}\sum _{\sigma \in {\mathfrak{S}}_{r+p}}\mathrm{sgn}(\sigma )t^{i_{\sigma (1)}\cdots i_{\sigma (r)}}{s}^{i_{\sigma (r+1)}\cdots i_{\sigma (r+p)}}{\mathbf{e}}_{i_1}\otimes {\mathbf{e}}_{i_2}\otimes \cdots \otimes {\mathbf{e}}_{i_{r+p}}.}$

The components of this tensor are precisely the skew part of the components of the tensor products ⊗t, denoted by square brackets on the indices:

${\displaystyle (t\hat{\otimes }s{)}^{i_1\cdots i_{r+p}}=t^{[i_1\cdots i_r}{s}^{i_{r+1}\cdots i_{r+p}]}.}$

Theinterior product may also be described in index notation as follows. Let${\displaystyle t=t^{i_0{i}_1\cdots i_{r-1}}}$ be an antisymmetric tensor of rank⁠${\displaystyle r}$⁠. Then, forα ∈V^∗,⁠${\displaystyle {\iota }_{\alpha }t}$⁠ is an alternating tensor of rank⁠${\displaystyle r-1}$⁠, given by

${\displaystyle ({\iota }_{\alpha }t{)}^{i_1\cdots i_{r-1}}=r\sum _{j=0}^n{\alpha }_j{t}^{j{i}_1\cdots i_{r-1}}.}$

wheren is the dimension ofV.

Given two vector spacesV andX and a natural numberk, analternating operator fromV^k toX is amultilinear map

${\displaystyle f:V^k\to X}$

such that wheneverv₁, ...,v_k arelinearly dependent vectors inV, then

${\displaystyle f(v_1,\dots ,v_k)=0.}$

The map

${\displaystyle w:V^k\to {\bigwedge }^k(V),}$

which associates to${\displaystyle k}$ vectors from${\displaystyle V}$ their exterior product, i.e. their corresponding${\displaystyle k}$-vector, is also alternating. In fact, this map is the "most general" alternating operator defined on${\displaystyle V^k;}$ given any other alternating operator${\displaystyle f:V^k\to X,}$ there exists a uniquelinear map${\displaystyle \varphi :{\bigwedge }^k(V)\to X}$ with${\displaystyle f=\varphi \circ w.}$ Thisuniversal property characterizes the space of alternating operators on${\displaystyle V^k}$ and can serve as its definition.

The above discussion specializes to the case when⁠${\displaystyle X=K}$⁠, the base field. In this case an alternating multilinear function

${\displaystyle f:V^k\to K}$

is called analternating multilinear form. The set of allalternatingmultilinear forms is a vector space, as the sum of two such maps, or the product of such a map with a scalar, is again alternating. By the universal property of the exterior power, the space of alternating forms of degree${\displaystyle k}$ on${\displaystyle V}$ isnaturally isomorphic with thedual vector space⁠${\displaystyle ({\bigwedge }^k(V){)}^{\ast }}$⁠. If${\displaystyle V}$ is finite-dimensional, then the latter isnaturally isomorphic^{[clarification needed]} to⁠${\displaystyle {\bigwedge }^k\left(V^{\ast }\right)}$⁠. In particular, if${\displaystyle V}$ is${\displaystyle n}$-dimensional, the dimension of the space of alternating maps from${\displaystyle V^k}$ to${\displaystyle K}$ is thebinomial coefficient⁠${\displaystyle (\genfrac{}{}{0ex}{}{n}{k})}$⁠.

Under such identification, the exterior product takes a concrete form: it produces a new anti-symmetric map from two given ones. Supposeω :V^k →K andη :V^m →K are two anti-symmetric maps. As in the case oftensor products of multilinear maps, the number of variables of their exterior product is the sum of the numbers of their variables. Depending on the choice of identification of elements of exterior power with multilinear forms, the exterior product is defined as

${\displaystyle \omega \wedge \eta =\mathrm{Alt}(\omega \otimes \eta )}$

or as

${\displaystyle \omega \dot{\wedge }\eta =\frac{(k+m)!}{k!m!}\mathrm{Alt}(\omega \otimes \eta ),}$

where, if the characteristic of the base field${\displaystyle K}$ is 0, the alternation Alt of a multilinear map is defined to be the average of the sign-adjusted values over all thepermutations of its variables:

${\displaystyle \mathrm{Alt}(\omega )(x_1,\dots ,x_k)=\frac{1}{k!}\sum _{\sigma \in S_k}\mathrm{sgn}(\sigma )\omega (x_{\sigma (1)},\dots ,x_{\sigma (k)}).}$

When thefield${\displaystyle K}$ hasfinite characteristic, an equivalent version of the second expression without any factorials or any constants is well-defined:

${\displaystyle \omega \dot{\wedge }\eta (x_1,\dots ,x_{k+m})=\sum _{\sigma \in {\mathrm{S}\mathrm{h}}_{k,m}}\mathrm{sgn}(\sigma )\omega (x_{\sigma (1)},\dots ,x_{\sigma (k)})\eta (x_{\sigma (k+1)},\dots ,x_{\sigma (k+m)}),}$

where hereSh_k,m ⊂S_k+m is the subset of(k,m) shuffles:permutationsσ of the set{1, 2, ...,k +m} such thatσ(1) <σ(2) < ⋯ <σ(k), andσ(k + 1) <σ(k + 2) < ... <σ(k +m). As this might look very specific and fine tuned, an equivalent raw version is to sum in the above formula over permutations in left cosets ofS_k+m / (S_k ×S_m).

Suppose that${\displaystyle V}$ is finite-dimensional. If${\displaystyle V^{\ast }}$ denotes thedual space to the vector space⁠${\displaystyle V}$⁠, then for each⁠${\displaystyle \alpha \in V^{\ast }}$⁠, it is possible to define anantiderivation on the algebra⁠${\displaystyle \bigwedge (V)}$⁠,

${\displaystyle {\iota }_{\alpha }:{\bigwedge }^k(V)\to {\bigwedge }^{k-1}(V).}$

This derivation is called theinterior product with⁠${\displaystyle \alpha }$⁠, or sometimes theinsertion operator, orcontraction by⁠${\displaystyle \alpha }$⁠.

Suppose that⁠${\displaystyle w\in {\bigwedge }^k(V)}$⁠. Then${\displaystyle w}$ is a multilinear mapping of${\displaystyle V^{\ast }}$ to⁠${\displaystyle K}$⁠, so it is defined by its values on thek-foldCartesian product⁠${\displaystyle V^{\ast }\times V^{\ast }\times \cdots \times V^{\ast }}$⁠. Ifu₁,u₂, ...,u_k−1 are${\displaystyle k-1}$ elements of⁠${\displaystyle V^{\ast }}$⁠, then define

${\displaystyle ({\iota }_{\alpha }w)(u_1,u_2,\dots ,u_{k-1})=w(\alpha ,u_1,u_2,\dots ,u_{k-1}).}$

Additionally, let${\displaystyle {\iota }_{\alpha }f=0}$ whenever${\displaystyle f}$ is a pure scalar (i.e., belonging to⁠${\displaystyle {\bigwedge }^0(V)}$⁠).

The interior product satisfies the following properties:

1. For each⁠${\displaystyle k}$⁠ and each⁠${\displaystyle \alpha \in V^{\ast }}$⁠ (where by convention${\displaystyle {\mathrm{\Lambda }}^{-1}(V)=\{0\}}$),

   ${\displaystyle {\iota }_{\alpha }:{\bigwedge }^k(V)\to {\bigwedge }^{k-1}(V).}$

2. If${\displaystyle v}$ is an element of${\displaystyle V}$ (⁠${\displaystyle ={\bigwedge }^1(V)}$⁠), then⁠${\displaystyle {\iota }_{\alpha }v=\alpha (v)}$⁠ is the dual pairing between elements of${\displaystyle V}$ and elements of⁠${\displaystyle V^{\ast }}$⁠.
3. For each⁠${\displaystyle \alpha \in V^{\ast }}$⁠,${\displaystyle {\iota }_{\alpha }}$ is agraded derivation of degree −1:

   ${\displaystyle {\iota }_{\alpha }(a\wedge b)=({\iota }_{\alpha }a)\wedge b+(-1{)}^{\mathrm{deg}a}a\wedge ({\iota }_{\alpha }b).}$

These three properties are sufficient to characterize the interior product as well as define it in the general infinite-dimensional case.

Further properties of the interior product include:

• ${\displaystyle {\iota }_{\alpha }\circ {\iota }_{\alpha }=0.}$
• ${\displaystyle {\iota }_{\alpha }\circ {\iota }_{\beta }=-{\iota }_{\beta }\circ {\iota }_{\alpha }.}$

Suppose that${\displaystyle V}$ has finite dimension⁠${\displaystyle n}$⁠. Then the interior product induces a canonical isomorphism of vector spaces

${\displaystyle {\bigwedge }^k(V^{\ast })\otimes {\bigwedge }^n(V)\to {\bigwedge }^{n-k}(V)}$

by the recursive definition

${\displaystyle {\iota }_{\alpha \wedge \beta }={\iota }_{\beta }\circ {\iota }_{\alpha }.}$

In the geometrical setting, a non-zero element of the top exterior power${\displaystyle {\bigwedge }^n(V)}$ (which is a one-dimensional vector space) is sometimes called avolume form (ororientation form, although this term may sometimes lead to ambiguity). The name orientation form comes from the fact that a choice of preferred top element determines an orientation of the whole exterior algebra, since it is tantamount to fixing an ordered basis of the vector space. Relative to the preferred volume form⁠${\displaystyle \sigma }$⁠, the isomorphism is given explicitly by

${\displaystyle {\bigwedge }^k(V^{\ast })\to {\bigwedge }^{n-k}(V):\alpha \mapsto {\iota }_{\alpha }\sigma .}$

If, in addition to a volume form, the vector spaceV is equipped with aninner product identifying${\displaystyle V}$ with⁠${\displaystyle V^{\ast }}$⁠, then the resulting isomorphism is called theHodge star operator, which maps an element to itsHodge dual:

${\displaystyle \star :{\bigwedge }^k(V)\to {\bigwedge }^{n-k}(V).}$

The composition of${\displaystyle \star }$ with itself maps${\displaystyle {\bigwedge }^k(V)\to {\bigwedge }^k(V)}$ and is always a scalar multiple of the identity map. In most applications, the volume form is compatible with the inner product in the sense that it is an exterior product of anorthonormal basis of⁠${\displaystyle V}$⁠. In this case,

${\displaystyle \star \circ \star :{\bigwedge }^k(V)\to {\bigwedge }^k(V)=(-1{)}^{k(n-k)+q}\mathrm{i}\mathrm{d}}$

where id is the identity mapping, and the inner product hasmetric signature(p,q) —p pluses andq minuses.

For⁠${\displaystyle V}$⁠ a finite-dimensional space, aninner product (or apseudo-Euclidean inner product) on⁠${\displaystyle V}$⁠ defines an isomorphism of${\displaystyle V}$ with⁠${\displaystyle V^{\ast }}$⁠, and so also an isomorphism of${\displaystyle {\bigwedge }^k(V)}$ with⁠${\displaystyle ({\bigwedge }^{k}V{)}^{\ast }}$⁠. The pairing between these two spaces also takes the form of an inner product. On decomposable⁠${\displaystyle k}$⁠-vectors,

${\displaystyle ⟨v_1\wedge \cdots \wedge v_k,w_1\wedge \cdots \wedge w_k⟩=det(⟨v_i,w_j⟩),}$

the determinant of the matrix of inner products. In the special casev_i =w_i, the inner product is the square norm of thek-vector, given by the determinant of theGramian matrix(⟨v_i,v_j⟩). This is then extended bilinearly (or sesquilinearly in the complex case) to a non-degenerate inner product on${\displaystyle {\bigwedge }^k(V).}$ Ife_i,i = 1, 2, ...,n, form anorthonormal basis of⁠${\displaystyle V}$⁠, then the vectors of the form

constitute an orthonormal basis for⁠${\displaystyle {\bigwedge }^k(V)}$⁠, a statement equivalent to theCauchy–Binet formula.

With respect to the inner product, exterior multiplication and the interior product are mutually adjoint. Specifically, for⁠${\displaystyle \mathbf{v}\in {\bigwedge }^{k-1}(V)}$⁠,⁠${\displaystyle \mathbf{w}\in {\bigwedge }^k(V)}$⁠, and⁠${\displaystyle x\in V}$⁠,

${\displaystyle ⟨x\wedge \mathbf{v},\mathbf{w}⟩=⟨\mathbf{v},{\iota }_{x^{\mathrm{♭}}}\mathbf{w}⟩}$

wherex^♭ ∈V^∗ is themusical isomorphism, the linear functional defined by

${\displaystyle x^{\mathrm{♭}}(y)=⟨x,y⟩}$

for all⁠${\displaystyle y\in V}$⁠. This property completely characterizes the inner product on the exterior algebra.

Indeed, more generally for⁠${\displaystyle \mathbf{v}\in {\bigwedge }^{k-l}(V)}$⁠,⁠${\displaystyle \mathbf{w}\in {\bigwedge }^k(V)}$⁠, and⁠${\displaystyle \mathbf{x}\in {\bigwedge }^l(V)}$⁠, iteration of the above adjoint properties gives

${\displaystyle ⟨\mathbf{x}\wedge \mathbf{v},\mathbf{w}⟩=⟨\mathbf{v},{\iota }_{{\mathbf{x}}^{\mathrm{♭}}}\mathbf{w}⟩}$

where now${\displaystyle {\mathbf{x}}^{\mathrm{♭}}\in {\bigwedge }^l\left(V^{\ast }\right)\simeq ({\bigwedge }^l(V){)}^{\ast }}$ is the dual⁠${\displaystyle l}$⁠-vector defined by

${\displaystyle {\mathbf{x}}^{\mathrm{♭}}(\mathbf{y})=⟨\mathbf{x},\mathbf{y}⟩}$

for all⁠${\displaystyle \mathbf{y}\in {\bigwedge }^l(V)}$⁠.

There is a correspondence between the graded dual of the graded algebra${\displaystyle \bigwedge (V)}$ and alternating multilinear forms on⁠${\displaystyle V}$⁠. The exterior algebra (as well as thesymmetric algebra) inherits a bialgebra structure, and, indeed, aHopf algebra structure, from thetensor algebra. See the article ontensor algebras for a detailed treatment of the topic.

The exterior product of multilinear forms defined above is dual to acoproduct defined on⁠${\displaystyle \bigwedge (V)}$⁠, giving the structure of acoalgebra. Thecoproduct is a linear function⁠${\displaystyle \mathrm{\Delta }:\bigwedge (V)\to \bigwedge (V)\otimes \bigwedge (V)}$⁠, which is given by

${\displaystyle \mathrm{\Delta }(v)=1\otimes v+v\otimes 1}$

on elements⁠${\displaystyle v\in V}$⁠. The symbol${\displaystyle 1}$ stands for the unit element of the field⁠${\displaystyle K}$⁠. Recall that⁠${\displaystyle K\simeq {\bigwedge }^0(V)\subseteq \bigwedge (V)}$⁠, so that the above really does lie in⁠${\displaystyle \bigwedge (V)\otimes \bigwedge (V)}$⁠. This definition of the coproduct is lifted to the full space${\displaystyle \bigwedge (V)}$ by (linear) homomorphism. The correct form of this homomorphism is not what one might naively write, but has to be the one carefully defined in thecoalgebra article. In this case, one obtains

${\displaystyle \mathrm{\Delta }(v\wedge w)=1\otimes (v\wedge w)+v\otimes w-w\otimes v+(v\wedge w)\otimes 1.}$

Expanding this out in detail, one obtains the following expression on decomposable elements:

${\displaystyle \mathrm{\Delta }(x_1\wedge \cdots \wedge x_k)=\sum _{p=0}^k\sum _{\sigma \in Sh(p,k-p)}\mathrm{sgn}(\sigma )(x_{\sigma (1)}\wedge \cdots \wedge x_{\sigma (p)})\otimes (x_{\sigma (p+1)}\wedge \cdots \wedge x_{\sigma (k)}).}$

where the second summation is taken over all(p,k−p)-shuffles. By convention, one takes that Sh(k,0) and Sh(0,k) equals {id: {1, ...,k} → {1, ...,k}}. It is also convenient to take the pure wedge products${\displaystyle v_{\sigma (1)}\wedge \cdots \wedge v_{\sigma (p)}}$ and${\displaystyle v_{\sigma (p+1)}\wedge \cdots \wedge v_{\sigma (k)}}$to equal 1 forp = 0 andp =k, respectively (the empty product in${\displaystyle \bigwedge (V)}$). The shuffle follows directly from the first axiom of a co-algebra: the relative order of the elements${\displaystyle x_k}$ ispreserved in the riffle shuffle: the riffle shuffle merely splits the ordered sequence into two ordered sequences, one on the left, and one on the right.