Exterior power (${\Lambda }^{k}V$)

The $k$th exterior power ${\Lambda }^{k}V$ is the quotient vector space of the tensor power $V^{\otimes k}$ that makes every elementary tensor with two equal factors zero. More precisely, take the quotient vector space by the linear span of all such tensors. Its elementary products $v_1\wedge \cdots \wedge v_k$ turn alternating multilinear maps into linear maps: every alternating $k$-linear map on $V$ factors uniquely through ${\Lambda }^{k}V$.^[ref]

If $e_1,\dots ,e_n$ is an ordered basis, the increasing productsform a basis of ${\Lambda }^{k}V$. There is one for each $k$-element choice of basis directions, soRepeated directions vanish; products with a different order can be sorted using their sign. Coordinate determinants distinguish the remaining increasing products, establishing linear independence.^[ref]

For $V={\mathbb{R}}^3$, a basis of ${\Lambda }^{2}V$ is $e_1\wedge e_2,e_1\wedge e_3,e_2\wedge e_3$. This explains the three bivector components in the worked example. The top exterior power ${\Lambda }^{3}V$ has one basis element, $e_1\wedge e_2\wedge e_3$, while ${\Lambda }^{4}V=0$: four independent directions cannot fit into that vector space. By convention ${\Lambda }^{0}V$ is the scalar field, even when $V$ has dimension zero.

The exterior cube of a vector space $V$ over a field is its third exterior power ${\Lambda }^{3}V$. It is generated by products $u\wedge v\wedge w$, multilinear in the three vectors and zero whenever two inputs coincide. Every alternating trilinear map out of $V^3$ factors uniquely through a linear map from ${\Lambda }^{3}V$.

If $e_1,\dots ,e_n$ is a basis, the products $e_i\wedge e_j\wedge e_k$ for $i<j<k$ form a basis, so $\dim {\Lambda }^{3}V=\left(\genfrac{}{}{0ex}{}{n}{3}\right)$. In particular it vanishes for $n<3$ and is one-dimensional for $n=3$.

In an oriented three-dimensional coordinate space,Thus the exterior product packages oriented volume; swapping two vectors reverses its sign. The alternating condition, including vanishing repeated inputs, remains the definition even in characteristic two, where minus and plus signs agree.

The exterior square is the quotient of $V\otimes V$ by tensors $v\otimes v$, and records antisymmetric tensors of rank two.

The exterior square of a direct sum splits into wedges of two vectors from $V$, two from $W$, or one from each. The mixed map sends $v\otimes w$ to $(v,0)\wedge (0,w)$. A combined basis proves it is an isomorphism, and the exterior-power Lie algebra representation proves equivariance for modules. No division by $2$ is used.

For a symplectic vector space $(V,\omega )$, contraction is the map $c:{\Lambda }^{2}V\to \mathbb{C}$ defined by $c(v\wedge w)=\omega (v,w)$. It is equivariant for the symplectic Lie algebra. In dimension four its five-dimensional kernel is irreducible of highest weight ${\omega }_2$, while the invariant inverse-form bivector spans a complementary trivial Lie algebra representation.

The primitive exterior square of a symplectic vector space is the kernel of its symplectic contraction of an exterior square. In dimension four, the invariant inverse-form bivector $\Omega$ has nonzero wedge square. Choose $\mathrm{vol}=\Omega \wedge \Omega /2$; the symmetric bilinear form $\xi \wedge \eta =B(\xi ,\eta )\mathrm{vol}$ is nondegenerate on ${\Lambda }^{2}V$. The primitive subspace is ${\Omega }^{\perp }$, and therefore inherits a nondegenerate form of dimension five. The symplectic Lie algebra acts on it by infinitesimal orthogonal transformations.