Multilinear algebra

Multilinear algebra studies maps linear in each of several arguments, including determinants, tensors, and exterior products.

Index notation describes vectors, matrices and tensors by their components relative to a chosen basis. For example, matrix multiplication reads $(AB{)}_{ij}={\sum }_k{A}_{ik}{B}_{kj}$. The free indices $i,j$ identify the resulting component, while $k$ is summed. An explicit summation sign avoids depending on an implicit Einstein summation convention.

A comonoid in the monoidal category of modules over a commutative ring, with tensor ${\otimes }_k$ and unit $k$.

A linear subspace $I$ of a coalgebra satisfying the displayed conditions. These are exactly the conditions for the comultiplication and counit to descend to the quotient vector space $C/I$. An ideal that is also a coideal in a bialgebra gives a quotient bialgebra.

For unital algebras over a field $A,B$, a coalgebra $C$ measures $A$ into $B$ when a linear map $p:C\to {\mathrm{Hom}}_k(A,B)$ satisfies the displayed identity and $p(c)(1_A)=\epsilon (c)1_B$. Group-like elements therefore act by unital algebra homomorphisms over a field.

A terminal measuring coalgebra: every measuring map from $C$ factors uniquely through its universal measuring map by a coalgebra homomorphism $C\to P(A,B)$. Composition of universal measurings makes $P(A,A)$ a bialgebra, with the identity map providing its unit.

An element of a coalgebra satisfying the displayed identities. A group-like element in a Hopf algebra is invertible with inverse its antipode image.

A right comodule is an object $M$ with a coaction $\rho :M\to M\otimes C$ satisfying the coassociativity and counit laws. This definition extends from coalgebras to comonoids in a monoidal category.

A natural underlying transformation of tensor bifunctors on right comodules over a field has components ${\phi }_{M,N}(u\otimes v)=\sum u_{(0)}\otimes v_{(0)}\gamma (u_{(1)},v_{(1)})$. Recover $\gamma =(\epsilon \otimes \epsilon ){\phi }_{H,H}$; colinearity and monoidal axioms impose further equations.

For a transformation from the tensor built using $n$ to the tensor built using $m$, colinearity is $m\ast \gamma =\gamma \ast n$: $\sum m(a_{(1)},b_{(1)})\gamma (a_{(2)},b_{(2)})=\sum \gamma (a_{(1)},b_{(1)})n(a_{(2)},b_{(2)})$. Scalar maps enter this convolution product for coalgebra maps via the common algebra unit.

A comonoid morphism $f:H\to K$ induces a functor $f_{\ast }$ on right comodules, replacing $\rho$ by $(1\otimes f)\rho$ and preserving underlying objects and arrows. For bimonoids, it is strict monoidal exactly when $f$ is also a monoid morphism.

A natural transformation between corestriction functors over a field has components ${\beta }_M(u)=\sum u_{(0)}\alpha (u_{(1)})$, recovered by $\alpha =\epsilon {\beta }_H$. Monoidality forces $\alpha$ to be a unital algebra homomorphism over a field; for a Hopf algebra, $\alpha S$ supplies the inverse.

The structure map of a comodule; a right coaction satisfies $(\rho \otimes 1)\rho =(1\otimes \Delta )\rho$ with coherent parentheses, and $(1\otimes \epsilon )\rho$ is the inverse right unitor.

For a coalgebra $C$ and a unital associative algebra over a commutative ring $A$, maps $f,g:C\to A$ have convolution $f\ast g=m_A(f\otimes g){\Delta }_C$, with unit $j_A{\epsilon }_C$. The two-sided convolution inverse of $1_H$ is the antipode.

The notation $\Delta h=\sum h_{(1)}\otimes h_{(2)}$ suppresses the summation index of a comultiplication; iterated indices use coassociativity. For a comodule, write $\rho (u)=\sum u_{(0)}\otimes u_{(1)}$.

A multilinear map $F:V_1\times \cdots \times V_r\to W$ is a map between vector spaces that is linear in each argument when all other arguments are fixed. For example the evaluation $(\omega ,v)\mapsto \omega (v)$ is bilinear, and the determinant is multilinear in its columns. The universal property of a tensor product replaces this separate linearity by a single linear map from $V_1\otimes \cdots \otimes V_r$.

Multilinearity is the property of being a multilinear map: each argument enters linearly while every other argument is fixed.

A bilinear map is a map $B:V\times W\to U$ between vector spaces that is linear in each argument when the other is fixed. The nonlinear temperature advection can be written as the diagonal of a bilinear map, $\mathcal{N}(\varphi ,\chi )=-\mathbf{u}[\varphi ]\cdot \nabla \chi$, when velocity depends linearly on temperature. Unlike a bilinear form, its output need not be a scalar.

A bilinear map between normed vector spaces is jointly continuous exactly when it satisfies the displayed product bound. Continuity at zero gives a bound on a product of balls; rescale each nonzero input separately to obtain the inequality. Conversely, expand the increment as $F(x+h,y+k)-F(x,y)=F(h,y)+F(x,k)+F(h,k)$ and apply the product bound to every term. Thus continuity at zero already implies joint continuity everywhere for a bilinear map.

A map $B:V\times W\to Z$ between vector spaces is bilinear when it is linear in either argument with the other held fixed. A scalar-valued bilinear form is the case $V=W$ and $Z$ is the scalar field. Checking bilinearity is one of the algebraic steps in verifying a real inner product; being a symmetric bilinear form and having positive squared norm impose further conditions.

Let $S:W\times W\to W$ be a symmetric bilinear map on a finite-dimensional real vector space. If $S(v,v)$ is parallel to $v$ for every $v$, then it has the displayed form for one covector $V$. Write $S(e_i,e_i)={\alpha }_i{e}_i$ in a basis. Applying the condition to $e_i+e_j$ and $e_i-e_j$ excludes every component outside their span and gives $S(e_i,e_j)=({\alpha }_j{e}_i+{\alpha }_i{e}_j)/2$. Set $V(e_i)={\alpha }_i/2$ and extend by bilinearity. Dimension one follows by the same definition. The trace identity $S^a{}_{ac}=(n+1)V_c$ proves uniqueness without depending on the chosen basis.

A continuous bilinear map obeys $\Vert f(h,k)\Vert \le C\Vert h\Vert \Vert k\Vert$. Expanding $f(a+h,b+k)$ leaves the displayed linear term and remainder $f(h,k)$. For product norm $\rho =(\Vert h{\Vert }^2+\Vert k{\Vert }^2{)}^{1/2}$, the remainder has norm at most $C{\rho }^2/2$, hence is $o(\rho )$. This proves the Fréchet derivative formula.

Let $U,V$ be nonzero finite-dimensional complex vector spaces, and let $\varphi :U\otimes V\to W$ have injective restrictions when either nonzero factor is fixed. It induces a map from ${\mathbb{CP}}^{a-1}\times {\mathbb{CP}}^{b-1}$ to ${\mathbb{CP}}^{r-1}$, where $a=\dim U$, $b=\dim V$, and $r=\mathrm{rank}\varphi$. Pullback of the degree-two generator is $x+y$, since both factor restrictions are projective linear embeddings. Its $(a+b-2)$th power is $\left(\genfrac{}{}{0ex}{}{a+b-2}{a-1}\right)x^{a-1}{y}^{b-1}\ne 0$ in integral cohomology. The target must therefore have complex dimension at least $a+b-2$. Positivity of both factor dimensions is necessary; a zero factor can make the claimed inequality false.

For nonzero finite-dimensional complex vector spaces $U,V$, suppose a bilinear map never vanishes on a pair of nonzero vectors. Equivalently, each map obtained by fixing one nonzero argument is injective. If its associated map on $U\otimes V$ has image dimension $r$, projectivization gives a map to ${\mathbb{CP}}^{r-1}$ whose positive degree-two class pulls back to $x+y$. The nonzero top power $(x+y{)}^{\dim U+\dim V-2}$ in the product cohomology ring of complex projective space proves the bound. Multiplication of polynomials of bounded degree attains equality. Nonzero hypotheses matter: with a zero factor, slice conditions can be vacuous.

An alternating multilinear map vanishes whenever two arguments are equal. Swapping two arguments reverses its sign.

A multilinear map is antisymmetric when swapping two arguments negates its value. Over a field of characteristic different from $2$, this is equivalent to being an alternating multilinear map: repeated arguments give $T=-T$, hence $T=0$; conversely expand $T(\dots ,u+v,\dots ,u+v,\dots )=0$ to get the sign change. In characteristic $2$, the repeated-argument condition is stronger than the sign-change condition alone.

An alternating trilinear form is a scalar-valued map, linear in three arguments, which vanishes whenever two arguments agree. In characteristic zero, interchanging two arguments negates its value, while cyclic permutation preserves it. For a symmetric bilinear form $B$ on a Lie algebra, the tensor $B([x,y],z)$ is alternating exactly when $B$ is an invariant bilinear form on a Lie algebra.

A tensor product turns a bilinear map into a linear map. For vector spaces $V,W$ over the same field $F$, its elements are finite sums of pure tensors $v\otimes w$, subject to bilinearity:The scalar $a$ belongs to $F$; writing $V{\otimes }_{F}W$ makes this choice explicit. These rules preserve separate linearity while allowing arbitrary linear combinations of products.^[ref]

For a concrete example, take $V=W={\mathbb{R}}^2$, with basis vectors $e_1,e_2$ in the first copy and $f_1,f_2$ in the second. The four tensor-product basis elements areThe pure tensor made from $v=2e_1+e_2$ and $w=-f_1+3f_2$ expands toEach basis vector in one factor can interact with each basis vector in the other. This is why dimensions multiply; the tensor-product coefficient matrix displays these four interactions as a two-by-two grid.

The tensor product is different from the Cartesian product $V\times W$. With componentwise vector addition and scalar multiplication, the latter contains pairs and has dimension $\dim V+\dim W$. The map $\tau (v,w)=v\otimes w$ is bilinear, generally neither linear on that Cartesian product nor injective nor surjective. In our example, $\tau (2v,2w)=4\tau (v,w)$, and $\tau (2v,w/2)=\tau (v,w)$; the separability of a tensor explains why some outputs cannot be reached at all. Both spaces happen to have dimension four when $V=W={\mathbb{R}}^2$, but this coincidence does not identify their structures.

A partial inner product contracts one factor of a bipartite state against a fixed bra, leaving a vector in the other factor. For $|\Psi ⟩={\sum }_{i,j}{c}_{ij}|i⟩\otimes |j⟩$ and $|a⟩={\sum }_i{a}_i|i⟩$, the result is ${\sum }_{i,j}\overline{a_i}{c}_{ij}|j⟩$. It is linear in $|\Psi ⟩$ and conjugate-linear in $|a⟩$. For a normalized rank-one projective outcome on the first factor, its squared norm is the outcome probability; divide by that norm to obtain the conditional pure state when it is nonzero.

For two unital associative algebras over a field $k$, their tensor product carries the multiplication in the title, extended by bilinearity, with identity $1\otimes 1$. Choosing a basis $(b_j)$ of the second factor gives the basis of a module $(1\otimes b_j)$ for the resulting left $A$-module; dual basis functionals recover the coefficients uniquely.

For associative algebras of finite dimension over a field $k$, write a central tensor as $x={\sum }_j{a}_j\otimes b_j$ in a basis of $B$. Commutation with every $a\otimes 1$ forces each $a_j$ into the center of an associative algebra $Z(A)$. Reexpand in a basis of $Z(A)$ and commute with every $1\otimes b$ to force the second coefficients into $Z(B)$. The reverse inclusion follows by multiplying pure tensors.

For a bilinear map $B:V\times W\to U$, there is exactly one linear map $L:V{\otimes }_{F}W\to U$ satisfying $L(v\otimes w)=B(v,w)$. Equivalently, $B=L\circ \tau$, where $\tau (v,w)=v\otimes w$. The universal property of a tensor product concerns the vector space together with this distinguished bilinear map.^[ref]

Here is a complete calculation over the real numbers. Define the bilinear mapIts values on pairs of basis vectors determine $L$ on the four tensor-product basis elements:Consequently, for $t={\sum }_{i,j=1}^2{c}_{ij}{e}_i\otimes f_j$,For $v=(2,1)$ and $w=(-1,3)$, the direct calculation gives $B(v,w)=(-2+12+3,-1-6)=(13,-7)$. Using the expansion of $v\otimes w$ instead gives $L(v\otimes w)=-2(1,0)+6(2,-1)-(0,1)+3(1,0)=(13,-7)$. The linear map also handles a nonseparable tensor such as $t=e_1\otimes f_1+e_2\otimes f_2$: $L(t)=(2,0)$, even though no single pair has $v\otimes w=t$.

A short proof uses a tensor-product basis: assign $L(e_i\otimes f_j)=B(e_i,f_j)$ and extend by linearity. Expanding both inputs verifies $L(v\otimes w)=B(v,w)$ by bilinearity; the assigned values on a basis force uniqueness. More generally, every multilinear map $F:V_1\times \cdots \times V_r\to U$ factors uniquely through a linear map from $V_1\otimes \cdots \otimes V_r$. This lets formulas involving several inputs define linear maps without choosing coordinates; tensor contraction uses the evaluation pairing between a vector and a covector.

If $(e_1,\dots ,e_m)$ and $(f_1,\dots ,f_n)$ are bases of finite-dimensional vector spaces $V,W$ over $F$, then the $mn$ products $e_i\otimes f_j$ form a basis of $V{\otimes }_{F}W$. There is one independent coordinate for every pair of indices.^[ref]

For a proof outline, bilinearity expands each pure tensor asso these products form a spanning set. For linear independence, take the coordinate functionals $e^k,f^{\ell }$ from the two dual bases. The bilinear map $(v,w)\mapsto e^k(v)f^{\ell }(w)$ induces a linear functional ${\lambda }_{k\ell }$ on the tensor product, satisfying ${\lambda }_{k\ell }(e_i\otimes f_j)={\delta }_{ki}{\delta }_{\ell j}$. Applying it to a proposed zero linear combination selects just $c_{k\ell }$, forcing every coefficient to vanish. Thus dimensions multiply, rather than add. This finite-dimensional count includes a zero factor, whose tensor product has dimension zero.

After choosing ordered bases, identify a tensor $t={\sum }_{i,j}{c}_{ij}{e}_i\otimes f_j$ with its matrix $C(t)=(c_{ij})$. This is a linear isomorphism between $V{\otimes }_{F}W$ and the vector space of $m$-by-$n$ matrices: each tensor-product basis element becomes the matrix with one entry equal to one and all others zero. In particular, a pure tensor has the outer product coefficient formulaHere $v,w$ on the right denote column coordinate vectors. The identification depends on the chosen bases, while the tensor itself does not. Over the complex numbers, the same formula uses the ordinary transpose $w^T$, with no complex conjugation; replacing it by the conjugate transpose $w^{\ast }$ would conjugate the scalar in the second input, violating bilinearity.

In a two-factor tensor product over a field, a pure tensor is also called a separable or decomposable tensor: it can be expressed as one product $v\otimes w$. Every tensor is a finite sum of pure tensors, but a sum need not itself be a pure tensor.^[ref]

For finite-dimensional vector spaces, the tensor-product coefficient matrix gives a decisive test: a nonzero tensor is a pure tensor exactly when its matrix rank is one. Indeed, every column of $v{w}^T$ is a scalar multiple of $v$. Conversely, if all columns of a nonzero matrix are multiples of one nonzero column $v$, collect their multipliers in $w$ to obtain $C=v{w}^T$. The zero tensor is also a pure tensor, for example $0\otimes w$, but its matrix rank is zero.

In ${\mathbb{R}}^2{\otimes }_{\mathbb{R}}{\mathbb{R}}^2$, the tensoris nonseparable: its matrix rank is two and its determinant is one, whereas $\det (v{w}^T)=0$. Directly, trying $t=(a{e}_1+b{e}_2)\otimes (c{f}_1+d{f}_2)$ requires $ac=bd=1$ and $ad=bc=0$. The first two equations force all four scalars nonzero, contradicting the others. In contrast, products sharing a factor do combine: $e_1\otimes f_1+e_1\otimes f_2=e_1\otimes (f_1+f_2)$. Thus being written as two summands alone does not establish nonseparability.

The two-factor tensor rank is the smallest number of pure tensors whose sum equals $t$, with zero allowed as the empty sum. For two finite-dimensional vector spaces over the same field, it equals the matrix rank of the tensor-product coefficient matrix. A sum of $r$ outer products has column space contained in the span of their $r$ first factors, so its matrix rank is at most $r$. Conversely, choose a basis $u_1,\dots ,u_r$ for the column space, express each column in that basis, and collect the coefficients into row vectors $w_k^T$. This yields $C={\sum }_{k=1}^r{u}_k{w}_k^T$, hence a sum of $r$ pure tensors.

The pure tensor $(2,1)\otimes (-1,3)$ therefore has two-factor tensor rank one, whereas $e_1\otimes f_1+e_2\otimes f_2$ has two-factor tensor rank two. Both have two factors: the number of factors and the minimum number of product summands are different counts. Over the real numbers, the singular value decomposition supplies a useful decomposition $C={\sum }_k{\sigma }_k{u}_k{v}_k^T$, with one nonzero term per matrix rank. The two-factor rank equality does not extend to three or more factors by merely choosing one matrix layout.

The tensor algebra of a vector space $V$ is the graded algebrawith multiplication given by concatenation of tensors.

The truncated tensor algebra discards products of degree greater than $N$. Its multiplication is $(a\otimes b{)}^{(k)}={\sum }_{j=0}^k{a}^{(j)}\otimes b^{(k-j)}$. Every element with scalar part $1$ is invertible by a finite geometric series in its positive-degree part. Truncated exponential and logarithm are finite polynomials.

The nth tensor power is the iterated tensor product of modules $V^{\otimes n}=V\otimes \cdots \otimes V$.

The tensor square is the second tensor power $V^{\otimes 2}=V\otimes V$. Over a field of characteristic different from two it decomposes as the direct sum of the symmetric square and exterior square.

The symmetric algebra is the quotient of $T(V)$ by the ideal generated by $v\otimes w-w\otimes v$. It is graded by symmetric powers:

The nth symmetric power $S^{n}V$ is the quotient of $V^{\otimes n}$ that identifies tensors differing by a permutation of their factors. Equivalently, over characteristic zero it is the subspace of symmetric tensors.

In the symmetric power ${\mathrm{Sym}}^m{\mathbb{C}}^n$ of the defining unitary group representation, diagonal torus weights are indexed by nonnegative integer tuples summing to $m$, with one monomial per weight space. Torus averaging extracts a monomial from any nonzero invariant complex subspace. The complexified Lie algebra acts by $E_{ij}=x_i{\partial }_{x_j}$ on the symmetric algebra. Whenever $a_j>0$, the displayed action moves one unit of degree from $j$ to $i$ with nonzero coefficient. Finite transfers connect every degree-$m$ monomial, so the subspace is the entire representation. The case $m=0$ is the one-dimensional trivial representation. This proof needs neither the Weyl character formula nor the Weyl dimension formula.

The three-dimensional irreducible representation $V_2$ of SU(2) has weights $2,0,-2$. Thus the character of its $n$th symmetric power is ${\sum }_{a+b+c=n}{t}^{2(a-c)}$. Its weight $2\ell$ has multiplicity $\lfloor (n-|\ell |)/2\rfloor +1$ for $|\ell |\le n$. Subtracting adjacent weight multiplicities gives one copy of each $V_{2n},V_{2n-4},\dots$. The invariant subspace is one-dimensional for even $n$ and zero for odd $n$.

For eigenvalues ${\lambda }_1,\dots ,{\lambda }_d$ of a finite-order linear action, its symmetric power characters satisfy ${\sum }_{n\ge 0}{\chi }_{S^{n}V}{t}^n={\prod }_j(1-{\lambda }_{j}t{)}^{-1}$. The denominator is the alternating generating polynomial of its exterior power characters.

The symmetric square is the quotient of $V\otimes V$ by $v\otimes w-w\otimes v$, and records symmetric tensors of rank two.

For $n\ge 2$, contraction with the invariant symmetric bilinear form splits off the trivial line in the symmetric square of the defining special orthogonal group representation. Its null space is the traceless symmetric square. It contains a highest-weight vector of weight $2e_1$, and the Weyl dimension formula gives $(n+1)(2n-1)$, exactly its dimension. Complete reducibility of compact-group representations therefore makes that null space irreducible. The entire symmetric square is reducible because of the invariant line.

For $V={\mathbb{C}}^{2n}$, $n\ge 3$, the symmetric square splits into an invariant line and its irreducible traceless part of highest weight $2{\epsilon }_1$. The exterior square of the defining orthogonal representation is the adjoint module of highest weight ${\epsilon }_1+{\epsilon }_2$. For $n=2$, the exterior square splits instead into highest weights ${\epsilon }_1-{\epsilon }_2$ and ${\epsilon }_1+{\epsilon }_2$. This exception reflects that so4 Lie algebra is semisimple but not simple.

Over every field, including characteristic two, there is a natural isomorphismThe middle summand maps $v\otimes w$ to the mixed product $vw$. Monomial bases prove bijectivity, and the Leibniz rule makes the map equivariant for Lie algebra representations.

For the defining real representation of $SO(d)$, the symmetric square splits as a scalar trace plus symmetric trace-free tensors: ${\mathrm{Sym}}^2({\mathbb{R}}^d)=\mathbb{R}\oplus {\mathrm{Sym}}_0^2({\mathbb{R}}^d)$. The trace projection is $S\mapsto S-(\mathrm{tr}S)I/d$, so the second summand has dimension $d(d+1)/2-1$.

For $d\ge 3$ the trace-free summand is an irreducible real group representation. To see this, diagonalize a nonzero symmetric trace-free matrix in a nonzero invariant subspace. Two diagonal entries differ. Infinitesimal rotation in their plane yields a nonzero symmetric off-diagonal matrix; rotations carry it to every coordinate pair, and a rotation through $\pi /4$ produces a difference of two diagonal entries. These off-diagonal matrices and diagonal differences span all symmetric trace-free matrices, so the invariant subspace is the whole summand. For $d=24$, its dimension is 299, the third-level nontrivial multiplet of the lowest levels of a fully transverse ND bosonic string.

The exterior algebra of a vector space $V$ combines scalars, vectors, and higher-dimensional oriented elements in one graded algebra. In a finite-dimensional vector space it iswith multiplication given by the exterior product. Degree zero is the scalar field, degree one is $V$, and degree two is the space of bivectors. Multiplication adds degrees: combining two vectors produces a bivector, rather than another ordinary vector.^[ref]

The essential rule is $v\wedge v=0$. Repeating the same direction cannot produce a nonzero oriented area. Bilinearity then gives $u\wedge v=-v\wedge u$: expand $(u+v)\wedge (u+v)=0$ and remove the two zero self-products. The alternating rule remains part of the definition in characteristic two, where sign reversal alone would not force a self-product to vanish.

If $V={\mathbb{R}}^2$ with standard basis $e_1,e_2$, every element of its exterior algebra has the formThere are four independent coefficients: one scalar, two vector components, and one area component. In ${\mathbb{R}}^3$, the respective degree dimensions are $1,3,3,1$, giving eight components in total. More generally, the exterior power basis gives total dimension ${\sum }_{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)=2^n$.

This language is useful when direction and orientation matter together. Determinants measure the action on top-degree exterior powers; differential forms use exterior powers of a dual space to describe quantities integrated along paths, surfaces, and higher-dimensional regions. The Grassmann algebra viewpoint also makes the sign changes in anticommuting generators systematic. A single multiplication rule connects geometry and algebra.

A bivector is an element of the second exterior power ${\Lambda }^{2}V$. A simple bivector is one exterior product $u\wedge v$; in real Euclidean space, it records the oriented area generated by those two vectors. A general bivector is a sum of such elements and need not describe just one plane. Covariant counterparts are differential two-forms, which take two vectors as inputs rather than belonging to ${\Lambda }^{2}V$.

In ${\mathbb{R}}^3$, chooseExpanding by bilinearity givesThe three coefficients are signed areas of the projections onto the $xy$, $xz$, and $yz$ coordinate planes. They are three components of one bivector, not the three coordinates of either input vector.

With the usual dot product and a fixed orientation of a vector space, the cross product represents this area element by the normal vector $(6,-3,1)$. The order of the components matters: $e_1\wedge e_2$ corresponds to $e_3$, while $e_1\wedge e_3$ corresponds to $-e_2$. The Gram determinant checks the full area:Here the norm on bivectors is the one induced by the Euclidean inner product. The exterior product exists without that metric and in arbitrary dimension; the familiar cross product description relies on the additional three-dimensional geometry.

A simple bivector can be expressed as one exterior product of two vectors. Changing the pair need not change its area element: $u\wedge (v+cu)=u\wedge v$ for every scalar $c$, by bilinearity and $u\wedge u=0$. This operation shears the represented parallelogram while preserving its oriented area.

Every bivector in a real vector space of dimension at most three is simple, but this fails in dimension four. For an ordered basis $e_1,\dots ,e_4$, considerThe two components describe independent coordinate planes. Their square iswhereas $(u\wedge v)\wedge (u\wedge v)=0$ because the factors repeat. Thus this $B$ cannot be one simple bivector. Over the real numbers in dimension four, the converse also holds: a nonzero bivector with $B\wedge B=0$ is simple. The nonzero factor two is important; that criterion cannot be applied unchanged in characteristic two.

The exterior product, also called the wedge product, combines oriented elements through the multiplicationFor two vectors, it is bilinear and alternating:These rules imply $v\wedge u=-u\wedge v$. They distinguish the exterior product from both the symmetric dot product and the unconstrained tensor product.^[ref]

In an oriented orthonormal plane, write $u=(a,b)$ and $v=(c,d)$. Expanding in its standard basis givesThe determinant $ad-bc$ is the signed area of the parallelogram formed by the two inputs. Its absolute value is the ordinary area. Reversing the input order reverses the sign; parallel inputs produce zero.

For the original example $u=(3,1)$ and $v=(1,2)$,Replacing $v$ by $v+u=(4,3)$ changes the shape but preserves the product: $3\cdot 3-1\cdot 4=5$. Both parallelograms have the same oriented area element. If $v=2u$, the exterior product is zero, since $u\wedge (2u)=2(u\wedge u)=0$.

The same rule adds directions in higher dimensions. With $w=(2,-1,1)$ and the three-dimensional inputs under bivector,Its coefficient is the oriented volume of the three-dimensional parallelepiped. Repeating or linearly depending on earlier inputs collapses the volume, giving zero.

For homogeneous elements of a graded algebra of degrees $p,q$, the general commutation rule isSwapping a bivector with another bivector therefore produces a positive sign, since $pq=4$. It is the number of individual vector swaps that controls the sign, not a rule that every pair of objects must anticommute.

For a polynomial ring $A$, a degree-$p$ polynomial polyvector field is an element of ${\bigwedge }_A^p{\mathrm{Der}}_k(A)$. Its product is the exterior product. Together with the Schouten-Nijenhuis bracket these spaces form a Gerstenhaber algebra.

The left Schouten-Nijenhuis bracket on polynomial polyvector fields is the degree-minus-one Gerstenhaber bracket extending the commutator of derivations and $[D,f]=D(f)$ by graded antisymmetry and the left graded Leibniz rule. For $\Pi ={\partial }_X\wedge {\partial }_Y$, this convention gives $[h\Pi ,f]=h(f_Y{\partial }_X-f_X{\partial }_Y)$ and $[F{\partial }_X+G{\partial }_Y,h\Pi ]=(D(h)-h(F_X+G_Y))\Pi$. A right insertion convention reverses the degree-two-with-function formula.