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 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.