Using the orthonormal basis V = { V₀, V₁, V₂, V₃, V₄, V₅, V₆ }, here is the baseline 2-in-7 cross product, chosen arbitrarily from 480 possibilities:

	… × V0	… × V1	… × V2	… × V3	… × V4	… × V5	… × V6
V0 × …	0	+V2	−V1	+V4	−V3	+V6	−V5
V1 × …	−V2	0	+V0	+V5	−V6	−V3	+V4
V2 × …	+V1	−V0	0	−V6	−V5	+V4	+V3
V3 × …	−V4	−V5	+V6	0	+V0	+V1	−V2
V4 × …	+V3	+V6	+V5	−V0	0	−V2	−V1
V5 × …	−V6	+V3	−V4	−V1	+V2	0	+V0
V6 × …	+V5	−V4	−V3	+V2	+V1	−V0	0

The other 479 differ only by permutation of basis elements and changes of sign.

With V_n corresponding to U_n+1, there is a clear resemblance to this excerpt from the baseline 3-in-8 table:

	… × U1	… × U2	… × U3	… × U4	… × U5	… × U6	… × U7
U0 × U1 × …	0	+U3	−U2	+U5	−U4	+U7	−U6
U0 × U2 × …	−U3	0	+U1	+U6	−U7	−U4	+U5
U0 × U3 × …	+U2	−U1	0	−U7	−U6	+U5	+U4
U0 × U4 × …	−U5	−U6	+U7	0	+U1	+U2	−U3
U0 × U5 × …	+U4	+U7	+U6	−U1	0	−U3	−U2
U0 × U6 × …	−U7	+U4	−U5	−U2	+U3	0	+U1
U0 × U7 × …	+U6	−U5	−U4	+U3	+U2	−U1	0

Let W = { W₀, W₁, W₂ … W_n−1 } be an orthonormal basis of an n-dimensional vector space over the real numbers. Then there is a widely recognized definition of the cross product of n − 1 factors, and it is most succinctly defined as the determinant of a formal matrix.

Here it is for the case of three vectors in four dimensions (the "3-in-4"):

A × B × C = det

a0 a1 a2 a3 b0 b1 b2 b3 c0 c1 c2 c3 W0 W1 W2 W3

The extension to other numbers of dimensions is so obvious that there is little need to try writing the general formula. The two-factor three-dimensional cross product, indispensible to physicists, looks like this:

A × B = det

a0 a1 a2 b0 b1 b2 W0 W1 W2

With four factors in five dimensions, the following is obtained:

A × B × C × D = det

a0 a1 a2 a3 a4 b0 b1 b2 b3 b4 c0 c1 c2 c3 c4 d0 d1 d2 d3 d4 W0 W1 W2 W3 W4

Aside from the 2-in-7 and 3-in-8, mathematicians rarely define any cross product that does not fall into this determinant-based pattern. However, the next section describes a different multiplication, using multiple vector spaces, that resembles the cross product of n − 1 factors in n dimensions.

In all the discussions so far, the assumption has been that the output of cross multiplication resides in the same vector space as the inputs. If this constraint is lifted, another kind of generalization of the cross product becomes possible, namely the wedge product. Although the wedge product is often defined abstractly by certain of its behaviors, in this report we give a more concrete presentation by way of example.

Start with a vector space G₁ that has orthonormal basis { e₀, e₁, e₂, e₃ }; it helps to call vectors within G₁ univectors. For any two vectors A and B in G₁, the wedge product is symbolized A ∧ B. The wedge product, like most multiplications, is distributive over addition:

A ∧ (B + C) = (A ∧ B) + (A ∧ C)

Like cross products, the product of two vectors from G₁ is anticommutative:

A ∧ B = − B ∧ A
A ∧ A = 0

and it is linear in each factor:

(xA) ∧ (yB) = xy(A ∧ B)

Unlike the cross products, the wedge product is associative:

(A ∧ B) ∧ C = A ∧ (B ∧ C)

Thus with basis vectors:

e_i ∧ e_j = − e_j ∧ e_i
e_i ∧ e_i = 0

A key feature of the wedge product is that although e_i ∧ e_j is a valid expression, it cannot be simplified. In this regard, it resembles an expression like 3 + 4i in complex numbers. Moreover, e_i ∧ e_j is a vector that resides in the six-dimensional space G₂:

an orthonormal basis for G₂ = { e₀ ∧ e₁, e₀ ∧ e₂, e₀ ∧ e₃, e₁ ∧ e₂, e₁ ∧ e₃, e₂ ∧ e₃ }

In building the basis, it little matters whether we choose e₀ ∧ e₁ or its negative e₁ ∧ e₀; the same vector space results either way. A condensed notation for the products of G₁'s basis vectors is convenient: e_ij = e_i ∧ e_j. Of course, e_ij = − e_ji. Thus:

the same basis for G₂ = { e₀₁, e₀₂, e₀₃, e₁₂, e₁₃, e₂₃ }

An element of G₂ is called a bivector.

Multiplication is not limited to two vectors. Consider that e_i ∧ e_j ∧ e_k (more briefly e_ijk) is nonzero when i, j and k are distinct. Such a product of three univectors resides in the four-dimensional space G₃:

an orthonormal basis for G₃ = { e₀₁₂, e₀₁₃, e₀₂₃, e₁₂₃ }

Because of associativity, there is little substantive difference between the product of three univectors versus the product of one univector and one bivector. Continuing the pattern of names, elements of G₃ are trivectors.

Last in the sequence is the one-dimensional G₄:

a basis for G₄ = { e₀₁₂₃ }

While vectors in this space might be called quadrivectors, they are more likely termed pseudoscalars, because G₄ is isomorphic to scalar space, which itself could be labeled G₀ and which would have the basis { 1 }.

In total, there are five disjoint vector spaces here: G₀, G₁, G₂, G₃, and G₄. With vector A in G_a and vector B in G_b, a general result is that A ∧ B, if not zero, is in G_a+b.

In the three-dimensional case, a likely choice for the vector spaces would be:

space	an orthonormal basis	containing
F0	{ 1 }	scalars
F1	{ e0, e1, e2 }	univectors
F2	{ e12, e20, e01 }	bivectors
F3	{ e012 }	trivectors or pseudoscalars

Using the Hodge dual, we can consolidate four vector spaces into two by declaring that:

• the pseudoscalar e₀₁₂ is identical to the scalar 1
• the bivectors e₁₂, e₂₀, and e₀₁ are identical respectively to the univectors e₀, e₁, and e₂.

In a consolidation of the earlier example, G₄ would be identified with G₀, and G₃ would be identified with G₁, but G₂ would stand alone.

This consolidation is not always appropriate. Suppose the components of an F₁ vector are not merely numbers, but numbers carrying a physical unit, say meters. Then the components of an F₂ vector would be numbers carrying the physical unit square meters, and it is very difficult to establish a plausible identity between meters and square meters.