Exterior algebra

The area of a parallelogram in terms of the determinant of the matrix of coordinates of two of its vertices.

The Cartesian plane ${\displaystyle \mathbb {R} ^{2}}$ is a real vector space equipped with a basis consisting of a pair of unit vectors

${\displaystyle {\mathbf {e} }_{1}={\begin{bmatrix}1\\0\end{bmatrix}},\quad {\mathbf {e} }_{2}={\begin{bmatrix}0\\1\end{bmatrix}},}$ with the orientation ${\displaystyle \mathbf {e} _{1}\times \mathbf {e} _{2}}$ and with the metric ${\displaystyle {\begin{bmatrix}1&0\\0&1\end{bmatrix}}.}$

Suppose that

${\displaystyle {\mathbf {v} }={\begin{bmatrix}a\\b\end{bmatrix}}=a{\mathbf {e} }_{1}+b{\mathbf {e} }_{2},\quad {\mathbf {w} }={\begin{bmatrix}c\\d\end{bmatrix}}=c{\mathbf {e} }_{1}+d{\mathbf {e} }_{2}}$

are a pair of given vectors in ${\displaystyle \mathbb {R} ^{2},}$ written in components. There is a unique parallelogram having v and w as two of its sides. The area of this parallelogram is given by the standard determinant formula:

${\displaystyle {\text{Area}}={\Bigl |}\det {\begin{bmatrix}{\mathbf {v} }&{\mathbf {w} }\end{bmatrix}}{\Bigr |}={\Biggl |}\det {\begin{bmatrix}a&c\\b&d\end{bmatrix}}{\Biggr |}=\left|ad-bc\right|.}$

Consider now the exterior product of v and w:

${\displaystyle {\begin{aligned}{\mathbf {v} }\wedge {\mathbf {w} }&=(a{\mathbf {e} }_{1}+b{\mathbf {e} }_{2})\wedge (c{\mathbf {e} }_{1}+d{\mathbf {e} }_{2})\\&=ac{\mathbf {e} }_{1}\wedge {\mathbf {e} }_{1}+ad{\mathbf {e} }_{1}\wedge {\mathbf {e} }_{2}+bc{\mathbf {e} }_{2}\wedge {\mathbf {e} }_{1}+bd{\mathbf {e} }_{2}\wedge {\mathbf {e} }_{2}\\&=\left(ad-bc\right){\mathbf {e} }_{1}\wedge {\mathbf {e} }_{2}\end{aligned}}}$

where the first step uses the distributive law for the exterior product, and the last uses the fact that the exterior product is alternating, 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 alternating 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 [v w]. The fact that this may be positive or negative has the intuitive meaning that v and w may be oriented in a counterclockwise or clockwise sense as the vertices of the parallelogram they define. Such an area is called the signed area of the parallelogram: the absolute 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, if A(v, w) denotes the signed area of the parallelogram of which the pair of vectors v and w form two adjacent sides, then A must satisfy the following properties:

1. A(rv, sw) = rsA(v, w) for any real numbers r and s, 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 the degenerate parallelogram determined by v (i.e., a line segment) is zero.
3. A(w, v) = −A(v, w), since interchanging the roles of v and w reverses the orientation of the parallelogram.
4. A(v + rw, w) = A(v, w) for any real number r, since adding a multiple of w to v 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.

The cross product (blue vector) in relation to the exterior product (light blue parallelogram). The length of the cross product is to the length of the parallel unit vector (red) as the size of the exterior product is to the size of the reference parallelogram (light red).

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 sides e₁ and e₂). In other words, the exterior product provides a basis-independent formulation of area.[6]

For vectors in a 3-dimensional oriented vector space with a bilinear scalar product, the exterior algebra is closely related to the cross product and triple product. Using a 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

${\displaystyle \mathbf {u} \wedge \mathbf {v} =(u_{1}v_{2}-u_{2}v_{1})(\mathbf {e} _{1}\wedge \mathbf {e} _{2})+(u_{2}v_{3}-u_{3}v_{2})(\mathbf {e} _{2}\wedge \mathbf {e} _{3})+(u_{3}v_{1}-u_{1}v_{3})(\mathbf {e} _{3}\wedge \mathbf {e} _{1}),}$

where (e₁ ∧ e₂, e₂ ∧ e₃, e₃ ∧ e₁) is a basis for the three-dimensional space ${\textstyle \bigwedge \nolimits ^{2}\left(\mathbb {R} ^{3}\right).}$ The coefficients above are the same as those in the usual definition of the cross product of vectors in three dimensions with a given orientation, the only differences being that the exterior product is not an ordinary vector, but instead is a 2-vector, and that the exterior product does not depend on the choice of orientation.

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

${\displaystyle \mathbf {u} \wedge \mathbf {v} \wedge \mathbf {w} =(u_{1}v_{2}w_{3}+u_{2}v_{3}w_{1}+u_{3}v_{1}w_{2}-u_{1}v_{3}w_{2}-u_{2}v_{1}w_{3}-u_{3}v_{2}w_{1})(\mathbf {e} _{1}\wedge \mathbf {e} _{2}\wedge \mathbf {e} _{3})}$

where e₁ ∧ e₂ ∧ e₃ is the basis vector for the one-dimensional space ${\textstyle \bigwedge \nolimits ^{3}\left(\mathbb {R} ^{3}\right).}$ The scalar coefficient is the triple product of the three vectors.

The cross product and triple product in a three dimensional Euclidean vector space each admit both geometric and algebraic interpretations, via Hodge star duality. The cross product u × v can be interpreted as a vector which is perpendicular to both u and v 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 the minors of the matrix with columns u and v. The triple product of u, v, and w is a signed scalar representing a geometric oriented volume. Algebraically, it is the determinant of the matrix with columns u, v, and w. The exterior product in three dimensions allows for similar interpretations: it, too, can be identified with oriented lines, areas, volumes, etc., that are spanned by one, two or more vectors. The exterior product generalizes these geometric notions to all vector spaces and to any number of dimensions, even in the absence of a scalar product.

In Einstein's theories of relativity, the Electromagnetic field is generally given as a differential 2-form ${\displaystyle F=dA}$ in 4-space or as the equivalent alternating tensor field ${\displaystyle F_{ij}=A_{[i,j]}=A_{[i;j]},}$ the Electromagnetic tensor. Then ${\displaystyle dF=ddA=0}$ or the equivalent Bianchi identity ${\displaystyle F_{[ij,k]}=F_{[ij;k]}=0.}$ None of this requires a metric.

Adding the Lorentz metric and an orientation provides the Hodge star operator ${\displaystyle \star }$ and thus makes it possible to define ${\displaystyle J={\star }d{\star }F}$ or the equivalent tensor divergence ${\displaystyle J^{i}=F_{,j}^{ij}=F_{;j}^{ij}}$ where ${\displaystyle F^{ij}=g^{ik}g^{jl}F_{kl}.}$

The exterior product is by construction alternating on elements of ${\displaystyle V,}$ which means that ${\textstyle x\wedge x=0}$ for all ${\displaystyle x\in V,}$ by the above construction. It follows that the product is also anticommutative 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 σ is a permutation of the integers [1, ..., k], and x₁, x₂, ..., x_k are elements of V, it follows that

${\displaystyle x_{\sigma (1)}\wedge x_{\sigma (2)}\wedge \cdots \wedge x_{\sigma (k)}=\operatorname {sgn} (\sigma )x_{1}\wedge x_{2}\wedge \cdots \wedge x_{k},}$

where sgn(σ) is the signature of the permutation σ.[8]

In particular, if x_i = x_j for some i ≠ 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 if and only if ${\displaystyle \{x_{1},x_{2},\cdots ,x_{k}\}}$ is a linearly dependent set of vectors, then

${\displaystyle x_{1}\wedge x_{2}\wedge \cdots \wedge x_{k}=0.}$

The kth exterior power of V, denoted ${\textstyle \bigwedge \nolimits ^{k}(V),}$ is the vector subspace of ${\textstyle \bigwedge (V)}$ spanned by elements of the form

${\displaystyle x_{1}\wedge x_{2}\wedge \cdots \wedge x_{k},\quad x_{i}\in V,i=1,2,\ldots ,k.}$

If ${\textstyle \alpha \in \bigwedge \nolimits ^{k}(V),}$ then α is said to be a k-vector. If, furthermore, α can be expressed as an exterior product of k elements of V, then α is said to be decomposable. Although decomposable k-vectors span ${\textstyle \bigwedge \nolimits ^{k}(V),}$ not every element of ${\textstyle \bigwedge \nolimits ^{k}(V)}$ is decomposable. For example, in ${\displaystyle \mathbb {R} ^{4},}$ the following 2-vector is not decomposable:

${\displaystyle \alpha =e_{1}\wedge e_{2}+e_{3}\wedge e_{4}.}$

(This is a symplectic form, since α ∧ α ≠ 0.[9])

If the dimension of V is n and { e₁, …, e_n } is a basis for V, then the set

${\displaystyle \{\,e_{i_{1}}\wedge e_{i_{2}}\wedge \cdots \wedge e_{i_{k}}~{\big |}~~1\leq i_{1}<i_{2}<\cdots <i_{k}\leq n\,\}}$

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

${\displaystyle v_{1}\wedge \cdots \wedge v_{k},}$

every vector v_j can be written as a linear combination of the basis vectors 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 basis k-vectors can be computed as the minors of the matrix that describes the vectors v_j in terms of the basis e_i.

By counting the basis elements, the dimension of ${\textstyle \bigwedge \nolimits ^{k}(V)}$ is equal to a binomial coefficient:

${\displaystyle \dim {\textstyle \bigwedge }^{k}(V)={\binom {n}{k}}\,.}$

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

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

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

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

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

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

where each α⁽ⁱ⁾ is decomposable, say

${\displaystyle \alpha ^{(i)}=\alpha _{1}^{(i)}\wedge \cdots \wedge \alpha _{k}^{(i)},\quad i=1,2,\ldots ,s.}$

The rank of the k-vector α is the minimal number of decomposable k-vectors in such an expansion of α. This is similar to the notion of tensor 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 α can be identified with half the rank of the matrix of coefficients of α in a basis. Thus if e_i is a basis for V, then α can be expressed uniquely as

${\displaystyle \alpha =\sum _{i,j}a_{ij}e_{i}\wedge e_{j}}$

where a_ij = −a_ji (the matrix of coefficients is skew-symmetric). The rank of the matrix a_ij is therefore even, and is twice the rank of the form α.

In characteristic 0, the 2-vector α has rank p if and only if

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

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

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

gives the exterior algebra the additional structure of a graded algebra, that is

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

Moreover, if K is the base field, we have

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

The exterior product is graded anticommutative, meaning that if ${\textstyle \alpha \in \bigwedge \nolimits ^{k}(V)}$ and ${\textstyle \beta \in \bigwedge \nolimits ^{p}(V),}$ 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 a graded module (a module that already carries its own gradation).

Let V be a vector space over the field K. Informally, multiplication in ${\textstyle \bigwedge (V)}$ is performed by manipulating symbols and imposing a distributive law, an associative law, and using the identity ${\displaystyle v\wedge v=0}$ for v ∈ V. Formally, ${\textstyle \bigwedge (V)}$ is the "most general" algebra in which these rules hold for the multiplication, in the sense that any unital associative K-algebra containing V with alternating multiplication on V must contain a homomorphic image of ${\textstyle \bigwedge (V).}$ In other words, the exterior algebra has the following universal property:[10]

Universal property of the exterior algebra

To construct the most general algebra that contains V and whose multiplication is alternating on V, it is natural to start with the most general associative algebra that contains V, the tensor algebra T(V), and then enforce the alternating property by taking a suitable quotient. We thus take the two-sided ideal I in T(V) generated by all elements of the form v ⊗ v for v in V, and define ${\textstyle \bigwedge (V)}$ as the quotient

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

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

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

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

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

Exterior algebras of vector 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 the Serre–Swan theorem. More general exterior algebras can be defined for sheaves of modules.

Suppose that V has finite dimension n, and that a basis e₁, ..., e_n of V is given. Then any alternating tensor t ∈ A^r(V) ⊂ T^r(V) can be written in index notation as

${\displaystyle t=t^{i_{1}i_{2}\cdots i_{r}}\,{\mathbf {e} }_{i_{1}}\otimes {\mathbf {e} }_{i_{2}}\otimes \cdots \otimes {\mathbf {e} }_{i_{r}},}$

where t^{i₁⋅⋅⋅i_r} is completely antisymmetric in its indices.

The exterior product of two alternating tensors t and s of ranks r and p is given by

${\displaystyle t~{\widehat {\otimes }}~s={\frac {1}{(r+p)!}}\sum _{\sigma \in {\mathfrak {S}}_{r+p}}\operatorname {sgn} (\sigma )t^{i_{\sigma (1)}\cdots i_{\sigma (r)}}s^{i_{\sigma (r+1)}\cdots i_{\sigma (r+p)}}{\mathbf {e} }_{i_{1}}\otimes {\mathbf {e} }_{i_{2}}\otimes \cdots \otimes {\mathbf {e} }_{i_{r+p}}.}$

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

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

The interior 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 r. Then, for α ∈ V^∗, i_αt is an alternating tensor of rank r − 1, given by

${\displaystyle (i_{\alpha }t)^{i_{1}\cdots i_{r-1}}=r\sum _{j=0}^{n}\alpha _{j}t^{ji_{1}\cdots i_{r-1}}.}$

where n is the dimension of V.

Given two vector spaces V and X and a natural number k, an alternating operator from V^k to X is a multilinear map

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

such that whenever v₁, ..., v_k are linearly dependent vectors in V, then

${\displaystyle f(v_{1},\ldots ,v_{k})=0.}$

The map

${\displaystyle w\colon V^{k}\to {\textstyle \bigwedge }^{\!k}(V)}$

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 ${\displaystyle f:V^{k}\rightarrow X,}$ there exists a unique linear map ${\displaystyle \phi$ :\wedge ^{k}(V)\rightarrow X} with ${\displaystyle f=\phi \circ w.}$ This universal property characterizes the space ${\displaystyle \wedge ^{k}(V)}$ and can serve as its definition.

Geometric interpretation for the exterior product of n 1-forms (ε, η, ω) to obtain an n-form ("mesh" of coordinate surfaces, here planes),[12] for n = 1, 2, 3. The "circulations" show orientation.[13][14]

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

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

is called an alternating multilinear form. The set of all alternating multilinear 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 k on V is naturally isomorphic with the dual vector space ${\textstyle {\bigl (}\bigwedge \nolimits ^{k}(V){\bigr )}^{*}.}$ If V is finite-dimensional, then the latter is naturally isomorphic to ${\textstyle \bigwedge \nolimits ^{k}\left(V^{*}\right).}$ In particular, if V is n-dimensional, the dimension of the space of alternating maps from V^k to K is the binomial coefficient ${\textstyle {\binom {n}{k}}.}$

Under such identification, the exterior product takes a concrete form: it produces a new anti-symmetric map from two given ones. Suppose ω : V^k → K and η : V^m → K are two anti-symmetric maps. As in the case of tensor 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

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

or as

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

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

${\displaystyle \operatorname {Alt} (\omega )(x_{1},\ldots ,x_{k})={\frac {1}{k!}}\sum _{\sigma \in S_{k}}\operatorname {sgn} (\sigma )\,\omega (x_{\sigma (1)},\ldots ,x_{\sigma (k)}).}$

When the field K has finite characteristic, an equivalent version of the second expression without any factorials or any constants is well-defined:

${\displaystyle {\omega \wedge \eta (x_{1},\ldots ,x_{k+m})}=\sum _{\sigma \in \mathrm {Sh} _{k,m}}\operatorname {sgn} (\sigma )\,\omega (x_{\sigma (1)},\ldots ,x_{\sigma (k)})\,\eta (x_{\sigma (k+1)},\ldots ,x_{\sigma (k+m)}),}$

where here Sh_k,m ⊂ S_k+m is the subset of (k,m) shuffles: permutations σ of the set {1, 2, ..., k + m} such that σ(1) < σ(2) < ⋯ < σ(k), and σ(k + 1) < σ(k + 2) < ⋯ < σ(k + m).

Suppose that V is finite-dimensional. If V^∗ denotes the dual space to the vector space V, then for each α ∈ V^∗, it is possible to define an antiderivation on the algebra ${\textstyle \bigwedge (V),}$

${\displaystyle i_{\alpha }:{\textstyle \bigwedge }^{k}V\rightarrow {\textstyle \bigwedge }^{k-1}V.}$

This derivation is called the interior product with α, or sometimes the insertion operator, or contraction by α.

Suppose that ${\textstyle w\in \bigwedge \nolimits ^{k}V.}$ Then w is a multilinear mapping of V^∗ to K, so it is defined by its values on the k-fold Cartesian product V^∗ × V^∗ × ... × V^∗. If u₁, u₂, ..., u_k−1 are k − 1 elements of V^∗, then define

${\displaystyle (i_{\alpha }{\mathbf {w} })(u_{1},u_{2},\ldots ,u_{k-1})={\mathbf {w} }(\alpha ,u_{1},u_{2},\ldots ,u_{k-1}).}$

Additionally, let i_αf = 0 whenever f is a pure scalar (i.e., belonging to ${\textstyle \bigwedge \nolimits ^{0}V}$).

The interior product satisfies the following properties:

1. For each k and each α ∈ V^∗,
   $${\displaystyle i_{\alpha }:{\textstyle \bigwedge }^{k}V\rightarrow {\textstyle \bigwedge }^{k-1}V.}$$

   
   (By convention, ${\textstyle \bigwedge \nolimits ^{-1}V=\{0\}.}$)
2. If v is an element of V (${\textstyle =\bigwedge \nolimits ^{1}V}$), then i_αv = α(v) is the dual pairing between elements of V and elements of V^∗.
3. For each α ∈ V^∗, i_α is a graded derivation of degree −1:
   $${\displaystyle i_{\alpha }(a\wedge b)=(i_{\alpha }a)\wedge b+(-1)^{\deg a}a\wedge (i_{\alpha }b).}$$

   

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:

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

Suppose that V has finite dimension n. Then the interior product induces a canonical isomorphism of vector spaces

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

by the recursive definition

${\displaystyle i_{\alpha \wedge \beta }=i_{\beta }\circ i_{\alpha }.}$

In the geometrical setting, a non-zero element of the top exterior power ${\textstyle \bigwedge \nolimits ^{n}(V)}$ (which is a one-dimensional vector space) is sometimes called a volume form (or orientation 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 σ, the isomorphism is given explicitly by

${\displaystyle {\textstyle \bigwedge }^{k}(V^{*})\to {\textstyle \bigwedge }^{n-k}(V):\alpha \mapsto i_{\alpha }\sigma .}$

If, in addition to a volume form, the vector space V is equipped with an inner product identifying V with V^∗, then the resulting isomorphism is called the Hodge star operator, which maps an element to its Hodge dual:

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

The composition of ${\displaystyle \star }$ with itself maps ${\textstyle \bigwedge \nolimits ^{k}(V)}$ → ${\textstyle \bigwedge \nolimits ^{k}(V)}$ 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 an orthonormal basis of V. In this case,

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

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

For V a finite-dimensional space, an inner product (or a pseudo-Euclidean inner product) on V defines an isomorphism of V with V^∗, and so also an isomorphism of ${\textstyle \bigwedge \nolimits ^{k}V}$ with ${\textstyle {\bigl (}\bigwedge \nolimits ^{k}V{\bigr )}^{*}.}$ The pairing between these two spaces also takes the form of an inner product. On decomposable k-vectors,

${\displaystyle \left\langle v_{1}\wedge \cdots \wedge v_{k},w_{1}\wedge \cdots \wedge w_{k}\right\rangle =\det {\bigl (}\langle v_{i},w_{j}\rangle {\bigr )},}$

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

${\displaystyle e_{i_{1}}\wedge \cdots \wedge e_{i_{k}},\quad i_{1}<\cdots <i_{k},}$

constitute an orthonormal basis for ${\textstyle \bigwedge \nolimits ^{k}(V).}$

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

${\displaystyle \langle x\wedge \mathbf {v} ,\mathbf {w} \rangle =\langle \mathbf {v} ,i_{x^{\flat }}\mathbf {w} \rangle }$

where x^♭ ∈ V^∗ is the musical isomorphism, the linear functional defined by

${\displaystyle x^{\flat }(y)=\langle x,y\rangle }$

for all y ∈ V. This property completely characterizes the inner product on the exterior algebra.

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

${\displaystyle \langle \mathbf {x} \wedge \mathbf {v} ,\mathbf {w} \rangle =\langle \mathbf {v} ,i_{\mathbf {x} ^{\flat }}\mathbf {w} \rangle }$

where now ${\textstyle x^{\flat }\in \bigwedge \nolimits ^{l}\left(V^{*}\right)\simeq {\bigl (}\bigwedge \nolimits ^{l}(V){\bigr )}^{*}}$ is the dual l-vector defined by

${\displaystyle \mathbf {x} ^{\flat }(\mathbf {y} )=\langle \mathbf {x} ,\mathbf {y} \rangle }$

for all ${\textstyle y\in \bigwedge \nolimits ^{l}(V).}$

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

The exterior product of multilinear forms defined above is dual to a coproduct defined on ${\textstyle \bigwedge (V),}$ giving the structure of a coalgebra. The coproduct is a linear function Δ : ${\textstyle \bigwedge (V)}$ → ${\textstyle \bigwedge (V)}$ ⊗ ${\textstyle \bigwedge (V)}$ which is given by

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

on elements v∈V. The symbol 1 stands for the unit element of the field K. Recall that K ⊂ ${\textstyle \bigwedge (V)}$, so that the above really does lie in ${\textstyle \bigwedge (V)}$ ⊗ ${\textstyle \bigwedge (V)}$. This definition of the coproduct is lifted to the full space ${\textstyle \bigwedge (V)}$ 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 the coalgebra article. In this case, one obtains

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

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

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

where the second summation is taken over all (p+1, k−p)-shuffles. The above is written with a notational trick, to keep track of the field element 1: the trick is to write ${\displaystyle x_{0}=1,}$ and this is shuffled into various locations during the expansion of the sum over shuffles. The shuffle follows directly from the first axiom of a co-algebra: the relative order of the elements ${\displaystyle x_{k}}$ is preserved 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.

Observe that the coproduct preserves the grading of the algebra. Extending to the full space ${\textstyle \bigwedge (V),}$ one has

${\displaystyle \Delta$ :{\textstyle \bigwedge }^{k}(V)\to \bigoplus _{p=0}^{k}{\textstyle \bigwedge }^{p}(V)\otimes {\textstyle \bigwedge }^{k-p}(V)}

The tensor symbol ⊗ used in this section should be understood with some caution: it is not the same tensor symbol as the one being used in the definition of the alternating product. Intuitively, it is perhaps easiest to think it as just another, but different, tensor product: it is still (bi-)linear, as tensor products should be, but it is the product that is appropriate for the definition of a bialgebra, that is, for creating the object ${\textstyle \bigwedge (V)}$ ⊗ ${\textstyle \bigwedge (V)}$. Any lingering doubt can be shaken by pondering the equalities (1 ⊗ v) ∧ (1 ⊗ w) = 1 ⊗ (v ∧ w) and (v ⊗ 1) ∧ (1 ⊗ w) = v ⊗ w, which follow from the definition of the coalgebra, as opposed to naive manipulations involving the tensor and wedge symbols. This distinction is developed in greater detail in the article on tensor algebras. Here, there is much less of a problem, in that the alternating product ∧ clearly corresponds to multiplication in the bialgebra, leaving the symbol ⊗ free for use in the definition of the bialgebra. In practice, this presents no particular problem, as long as one avoids the fatal trap of replacing alternating sums of ⊗ by the wedge symbol, with one exception. One can construct an alternating product from ⊗, with the understanding that it works in a different space. Immediately below, an example is given: the alternating product for the dual space can be given in terms of the coproduct. The construction of the bialgebra here parallels the construction in the tensor algebra article almost exactly, except for the need to correctly track the alternating signs for the exterior algebra.

In terms of the coproduct, the exterior product on the dual space is just the graded dual of the coproduct:

${\displaystyle (\alpha \wedge \beta )(x_{1}\wedge \cdots \wedge x_{k})=(\alpha \otimes \beta )\left(\Delta (x_{1}\wedge \cdots \wedge x_{k})\right)}$

where the tensor product on the right-hand side is of multilinear linear maps (extended by zero on elements of incompatible homogeneous degree: more precisely, α ∧ β = ε ∘ (α ⊗ β) ∘ Δ, where ε is the counit, as defined presently).

The counit is the homomorphism ε : ${\textstyle \bigwedge (V)}$ → K that returns the 0-graded component of its argument. The coproduct and counit, along with the exterior product, define the structure of a bialgebra on the exterior algebra.

With an antipode defined on homogeneous elements by ${\displaystyle S(x)=(-1)^{\binom {{\text{deg}}\,x\,+1}{2}}x,}$ the exterior algebra is furthermore a Hopf algebra.[15]

If ${\displaystyle 0\to U\to V\to W\to 0}$ is a short exact sequence of vector spaces, then

${\displaystyle 0\to {\textstyle \bigwedge }^{1}(U)\wedge {\textstyle \bigwedge }(V)\to {\textstyle \bigwedge }(V)\to {\textstyle \bigwedge }(W)\to 0}$

is an exact sequence of graded vector spaces,[16] as is

${\textstyle 0\to \bigwedge (U)\to \bigwedge (V).}$[17]

In particular, the exterior algebra of a direct sum is isomorphic to the tensor product of the exterior algebras:

${\displaystyle {\textstyle \bigwedge }(V\oplus W)\cong {\textstyle \bigwedge }(V)\otimes {\textstyle \bigwedge }(W).}$

This is a graded isomorphism; i.e.,

${\displaystyle {\textstyle \bigwedge }^{k}(V\oplus W)\cong \bigoplus _{p+q=k}{\textstyle \bigwedge }^{p}(V)\otimes {\textstyle \bigwedge }^{q}(W).}$

In greater generality, for a short exact sequence of vector spaces ${\textstyle 0\to U\mathrel {\overset {f}{\to }} V\mathrel {\overset {g}{\to }} W\to 0,}$ there is a natural filtration

${\displaystyle 0=F^{0}\subseteq F^{1}\subseteq \cdots \subseteq F^{k}\subseteq F^{k+1}={\textstyle \bigwedge }^{k}(V)}$

where ${\displaystyle F^{p}}$ for ${\displaystyle p\geq 1}$ is spanned by elements of the form ${\displaystyle u_{1}\wedge \ldots \wedge u_{k+1-p}\wedge v_{1}\wedge \ldots v_{p-1}}$ for ${\displaystyle u_{i}\in U}$ and ${\displaystyle v_{i}\in V.}$ The corresponding quotients admit a natural isomorphism

${\displaystyle F^{p+1}/F^{p}\cong {\textstyle \bigwedge }^{k-p}(U)\otimes {\textstyle \bigwedge }^{p}(W)}$ given by ${\displaystyle u_{1}\wedge \ldots \wedge u_{k+1-p}\wedge v_{1}\wedge \ldots \wedge v_{p-1}\mapsto u_{1}\wedge \ldots \wedge u_{k+1-p}\otimes g(v_{1})\wedge \ldots \wedge g(v_{p-1}).}$

In particular, if U is 1-dimensional then

${\displaystyle 0\to U\otimes {\textstyle \bigwedge }^{k-1}(W)\to {\textstyle \bigwedge }^{k}(V)\to {\textstyle \bigwedge }^{k}(W)\to 0}$

is exact, and if W is 1-dimensional then

${\displaystyle 0\to {\textstyle \bigwedge }^{k}(U)\to {\textstyle \bigwedge }^{k}(V)\to {\textstyle \bigwedge }^{k-1}(U)\otimes W\to 0}$

is exact.[18]

In applications to linear algebra, the exterior product provides an abstract algebraic manner for describing the determinant and the minors of a matrix. For instance, it is well known that the determinant of a square matrix is equal to the volume of the parallelotope whose sides are the columns of the matrix (with a sign to track orientation). This suggests that the determinant can be defined in terms of the exterior product of the column vectors. Likewise, the k × k minors of a matrix can be defined by looking at the exterior products of column vectors chosen k at a time. These ideas can be extended not just to matrices but to linear transformations as well: the determinant of a linear transformation is the factor by which it scales the oriented volume of any given reference parallelotope. So the determinant of a linear transformation can be defined in terms of what the transformation does to the top exterior power. The action of a transformation on the lesser exterior powers gives a basis-independent way to talk about the minors of the transformation.

Let[19] ${\displaystyle V}$ be an n-dimensional vector space over field ${\displaystyle K}$ with basis ${\displaystyle \{e_{1},\ldots ,e_{n}\}.}$

• For ${\displaystyle A\in \operatorname {End} (V),}$ define ${\textstyle \bigwedge ^{k}A\in \operatorname {End} {\bigl (}\bigwedge ^{k}V{\bigr )}}$ on simple tensors by
  $${\displaystyle {\textstyle \bigwedge }^{k}A(v_{1}\wedge \cdots \wedge v_{k})=Av_{1}\wedge \cdots \wedge Av_{k}}$$

  
  and expand the definition linearly to all tensors. More generally, we can define ${\textstyle \bigwedge ^{p}A^{k}\in \operatorname {End} {\bigl (}\bigwedge ^{p}V{\bigr )},(p\geq k)}$ on simple tensors by
  $${\displaystyle {\begin{aligned}&\left({\textstyle \bigwedge }^{p}A^{k}\right)(v_{1}\wedge \cdots \wedge v_{p})\\[10mu]&\qquad =\sum _{0\leq i_{1}<\cdots <i_{k}\leq p}v_{1}\wedge \cdots \wedge Av_{i_{1}}\wedge \cdots \wedge Av_{i_{k}}\wedge \cdots \wedge v_{p}\end{aligned}}}$$

  
  i.e. choose k components on which A would act, then sum up all results obtained from different choices. If ${\displaystyle p<k,}$ define ${\textstyle \bigwedge ^{p}A^{k}=0.}$ Since ${\textstyle \bigwedge ^{n}V}$ is 1-dimensional with basis ${\displaystyle e_{1}\wedge \cdots \wedge e_{n},}$ we can identify ${\textstyle \bigwedge ^{n}A^{k}}$ with the unique number ${\displaystyle \kappa \in K}$ satisfying
  $${\displaystyle {\textstyle \bigwedge }^{n}A^{k}(e_{1}\wedge \cdots \wedge e_{n})=\kappa (e_{1}\wedge \cdots \wedge e_{n}).}$$

  
• For ${\displaystyle \varphi \in \operatorname {End} {\bigl (}{\textstyle \bigwedge }^{p}V{\bigr )},}$ define the exterior transpose ${\textstyle \varphi ^{\mathrm {T} }\in \operatorname {End} {\bigl (}\bigwedge ^{n-p}V{\bigr )}}$ to be the unique operator satisfying ${\textstyle (\varphi ^{\mathrm {T} }\omega _{n-p})\wedge \omega _{p}=\omega _{n-p}\wedge (\varphi \omega _{p})}$ for any ${\textstyle \omega _{p}\in {\textstyle \bigwedge }^{p}V}$ and ${\textstyle \omega _{n-p}\in {\textstyle \bigwedge }^{n-p}V.}$
• For ${\displaystyle A\in \operatorname {End} (V),}$ define ${\textstyle \det A=\bigwedge ^{n}A^{n},}$ ${\textstyle \operatorname {Tr} (A)=\bigwedge ^{n}A^{1},}$ ${\textstyle \operatorname {adj} A={\bigl (}\bigwedge ^{n-1}A^{n-1}{\bigr )}^{\mathrm {T} }.}$ These are equivalent to the previous definitions.

All results obtained from other definitions of the determinant, trace and adjoint can be obtained from this definition (since these definitions are equivalent). Here are some basic properties related to these new definitions:

• ${\displaystyle (\cdot )^{\mathrm {T} }}$ is ${\displaystyle K}$-linear.
• ${\displaystyle (AB)^{\mathrm {T} }=B^{\mathrm {T} }A^{\mathrm {T} }.}$
• We have a canonical isomorphism
  $${\displaystyle {\begin{cases}\psi$$ :\operatorname {End} {\bigl (}{\textstyle \bigwedge }^{k}V{\bigr )}\cong \operatorname {End} {\bigl (}{\textstyle \bigwedge }^{n-k}V{\bigr )}\\A\mapsto A^{\mathrm {T} }\end{cases}}}

  
  However, there is no canonical isomorphism between ${\textstyle \bigwedge ^{k}V}$ and ${\textstyle \bigwedge ^{n-k}V.}$
• ${\textstyle \operatorname {Tr} {\bigl (}\bigwedge ^{k}A{\bigr )}=\bigwedge ^{n}A^{k}.}$ The entries of the transposed matrix of ${\textstyle \bigwedge ^{k}A}$ are ${\displaystyle k\times k}$-minors of ${\displaystyle A.}$
• For all ${\displaystyle k\leq n-1,p\leq k,A\in \operatorname {End} (V),}$
  $${\displaystyle \sum _{q=0}^{p}{\bigl (}{\textstyle \bigwedge }^{n-k}A^{p-q}{\bigr )}^{\mathrm {T} }{\bigl (}{\textstyle \bigwedge }^{k}A^{q}{\bigr )}={\bigl (}{\textstyle \bigwedge }^{n}A^{p}{\bigr )}\operatorname {Id} \in \operatorname {End} (V).}$$

  
  In particular,
  $${\displaystyle {\bigl (}{\textstyle \bigwedge }^{n-1}A^{p-1}{\bigr )}^{\mathrm {T} }A+{\bigl (}{\textstyle \bigwedge }^{n-1}A^{p}{\bigr )}^{\mathrm {T} }={\bigl (}{\textstyle \bigwedge }^{n}A^{p}{\bigr )}\operatorname {Id} }$$

  
  and hence
  $${\displaystyle (\operatorname {adj} A)A={\bigl (}{\textstyle \bigwedge }^{n-1}A^{n-1}{\bigr )}^{\mathrm {T} }A={\bigl (}{\textstyle \bigwedge }^{n}A^{n}{\bigr )}\operatorname {Id} =(\det A)\operatorname {Id} .}$$

  
• ${\displaystyle {\bigl (}{\textstyle \bigwedge }^{n-1}A^{p}{\bigr )}^{\mathrm {T} }=\sum _{q=0}^{p}{\bigl (}{\textstyle \bigwedge }^{n}A^{p-q}{\bigr )}(-A)^{q}=\sum _{q=0}^{p}\operatorname {Tr} {\bigl (}{\textstyle \bigwedge }^{p-q}A{\bigr )}(-A)^{q}.}$
  In particular,
  $${\displaystyle \operatorname {adj} A=\sum _{q=0}^{n-1}{\bigl (}{\textstyle \bigwedge }^{n}A^{n-q-1}{\bigr )}(-A)^{q}.}$$

  
• ${\displaystyle \operatorname {Tr} {\bigl (}{\textstyle \bigwedge }^{k}\operatorname {adj} A{\bigr )}={\textstyle \bigwedge }^{n}(\operatorname {adj} A)^{k}=(\det A)^{k-1}{\bigl (}{\textstyle \bigwedge }^{n}A^{n-k}{\bigr )}=(\det A)^{k-1}\operatorname {Tr} {\bigl (}{\textstyle \bigwedge }^{n-k}A{\bigr )}.}$
• ${\displaystyle \operatorname {Tr} \!{\Bigl (}{\bigl (}{\textstyle \bigwedge }^{n-1}A^{k}{\bigr )}^{\mathrm {T} }{\Bigr )}=(n-k){\textstyle \bigwedge }^{n}A^{p}=(n-k)\operatorname {Tr} \left({\textstyle \bigwedge }^{p}A\right).}$
• The characteristic polynomial ${\displaystyle \operatorname {ch} _{A}(t)}$ of ${\displaystyle A\in \operatorname {End} (V)}$ can be given by
  $${\displaystyle \operatorname {ch} _{A}(t)=\sum _{k=0}^{n}\operatorname {Tr} {\bigl (}{\textstyle \bigwedge }^{k}A{\bigr )}(-t)^{n-k}=\sum _{k=0}^{n}\left({\textstyle \bigwedge }^{n}A^{k}\right)(-t)^{n-k}.}$$

  
  Similarly,
  $${\displaystyle \operatorname {ch} _{\operatorname {adj} A}(t)=\sum _{k=0}^{n}\left({\textstyle \bigwedge }^{n}(\operatorname {adj} A)^{k}\right)(-t)^{n-k}=\sum _{k=0}^{n}(\det A)^{k-1}\left({\textstyle \bigwedge }^{n}A^{n-k}\right)(-t)^{n-k}}$$