In general, the term ‘cross product’ is used for any operation denoted by the symbol ‘$\times$’, such as cartesian product, direct product, Tychonoff product, or (subsuming all of these) product in a category.
However, there is another completely different context, used in the elementary analysis of vector spaces, and that is what we discuss here. Originally isolated from the multiplication operation in quaternions as a binary operation on $\mathbb{R}^3 \simeq Im(\mathbb{H})$, the cross product now has generalisations to other arities, other dimensions, and other ground fields.
The classical cross product on the Cartesian space $\mathbb{R}^3$ is the bilinear function
given by
This operation is invariant under orthogonal transformations and there is nothing special here about the real numbers, so given any $3$-dimensional oriented inner product space $V$ over any field, we have a bilinear cross product
given, upon choosing any oriented orthonormal basis for $V$, by the formula above.
We have already, trivially, generalized the cross product to other ground fields. One way to generalise it to other dimensions is to identify characteristic features as a bilinear operation and see what operations in other dimensions have these.
In this vein, a binary cross product on any inner product space $V$ is a bilinear function
such that for all $x, y\in V$ we have
Alternation: $x \times x = 0$.
Orthgonality: $x \times y$ is orthogonal to both $x$ and $y$; that is, $x \cdot (x \times y) = (x \times y) \cdot y = 0$.
Area: ${\|x \times y\|} = {\|x\|} {\|y\|}$ if $x, y$ are orthogonal.
From these, we can prove a more general formula for ${\|x \times y\|}$:
or equivalently
(Using the polarization identity to express $x \cdot y$ in terms of $\|x\|$, $\|y\|$, and either $\|x + y\|$ or $\|x - y\|$, this is the double of Hero’s Formula for the area of a triangle.)
Conversely, using this more general area formula, we can prove both the restricted area formula and alternation, so that only orthogonality is needed as a separate axiom.
We then have over the real numbers:
These cross products exist over any base field, but as far as I know there may be additional cross products over some. (Of course, the claim that there are uncountably many cross products in $7$ dimensions should be generalised and made more precise; the space of these inner products is some algebraic variety.)
Binary cross products are closely related to normed division algebras (NDAs). Given a normed division algebra $A$, the imaginary hyperplane $Im(A)$ inherits an inner product from $A$ and gains a cross product as
Conversely, given an inner product space $V$ with a binary cross product, the orthogonal direct sum $K \oplus V$ becomes a NDA as
where $K$ is the ground field.
By Hurwitz's theorem?, the only finite-dimensional NDAs over $\mathbb{R}$ are $\mathbb{R}$ itself (the real numbers), $\mathbb{C}$ (the complex numbers), $\mathbb{H}$ (the quaternions), and $\mathbb{O}$ (the octonions). Thus the limited possibilities for binary cross products are determined by the limited possibilities for NDA structures.
Given an oriented inner product space $V$ of finite dimension $n$, we can define the signed volume of an $n$-tuple of vectors. (See also volume form.) This allows us to characterise a co-unary cross product of $n - 1$ vectors as a multilinear operation
such that
always. There is exactly one such cross product on any such $V$ (so two if we start with an unoriented inner product space).
In $3$ dimensions, this also recovers the classical cross product.
Generalizing all of the above, let a vector-valued cross product on any inner product space $V$ be a multilinear function
for some natural number $k$ (called the arity?) such that:
Alternation: $⨉(v_1,\ldots,v_k) = 0$ if $v_i = v_j$ for some $i \ne j$.
Orthgonality: $⨉(v_1,\ldots,v_k)$ is orthogonal to each $v_i$.
Area: ${\|⨉(v_1,\ldots,v_k)\|} = \prod_i {\|v_i\|}$ if the $v_i$ are mutually orthogonal.
We can again extend (3) to get the magnitude of the cross product of any $k$ vectors; its square is the determinant of the matrix whose $(i,j)$th entry is $v_i \cdot v_j$ (the Gram determinant?), and then (1) again follows.
Then for an inner product space $V$ over $\mathbb{R}$ of finite dimension $n$, we have:
Fixing a field $K$, let Vect be $Vect_K$, the symmetric monoidal category of vector spaces over $K$ (with the usual tensor product), and let $T$ be any symmetric monoidal functor from $Vect$ to itself. Note that any inner product $g\colon V \otimes V \to K$ extends to an inner product $T(g)\colon T(V) \otimes T(V) \to K$ (ignoring questions of degeneracy); similarly, any element $x$ of $V$, thought of as a linear map $x\colon K \to V$, gives rise to an element $T(x)$ of $T(V)$. The vector-valued cross products above use the identity functor for $T$, but other possible choices for $T$ are $V \mapsto V \otimes V$, $V \mapsto \Lambda^2 V$, and the constant functor $V \mapsto K$. We could also take $Vect$ to be a full subcategory of $Vect_K$ closed under the tensor product, such as $Fin Vect_K$; this may allow more possibilities for $T$ in exchange for fewer possibilities for $V$.
Given an inner-product space $V$ and a symmetric monoidal functor $T$, a $T$-valued cross product on $V$ is a multilinear function
for some natural arity $k$ such that:
Alternation: $⨉(v_1,\ldots,v_k) = 0$ if $v_i = v_j$ for some $i \ne j$.
Orthgonality: $⨉(v_1,\ldots,v_k)$ is orthogonal (in $T(V)$) to each $T(v_i)$.
Area: ${\|⨉(v_1,\ldots,v_k)\|} = \prod_i {\|v_i\|}$ if the $v_i$ are mutually orthogonal (in $V$).
Again it follows that in any case ${\|⨉(v_1,\ldots,v_k)\|}$ is a square root of the Gram determinant, and again this implies both (1) and (3).
I do not know a full list of these, but one important example is the scalar-valued binary cross product in $2$ dimensions:
Actually, this scalar-valued cross product $x \times y$ is simply the dot product $x \cdot \times{y}$, where $\times{y}$ is the unary vector-valued cross product in $2$ dimensions. (In a counterclockwise-oriented plane, it rotates a vector clockwise through a right angle.)
More generally, in any number $n \geq 2$ of dimensions, there is a multivector-valued binary cross product whose values are $(n-2)$-multivectors?; this includes the scalar-valued cross product when $n = 2$ and the classical cross product when $n = 3$, but gets more complicated for larger values of $n$. Or generalizing the scalar-valued binary cross product in a different way, the volume form on an $n$-dimensional inner-product space is a scalar-valued $n$-ary cross product. More generally still, combining the dot product with any vector-valued cross product produces a scalar-valued cross product of $1$ higher arity.
The cross product is also called ‘outer product’, and both of these terms are sometimes also used for the exterior product. In its most basic form, the exterior product of two vectors $u,v$ is a bivector $u \wedge v$. But note that this is not a bivector-valued cross product by the definition above, since it lacks orthogonality (and indeed has nothing to do with the inner product).
In $3$ dimensions, given an inner product and an orientation, we can use the Hodge dual to turn this into a vector, and this is the classical cross product once more. In $2$ dimensions, using the same structure, we can turn the bivector into a scalar; this recovers the scalar-valued binary cross product above. In general, this produces the binary $(n-2)$-vector-valued cross product.
Using only the inner product but not the orientation, we get (respectively) a pseudovector? (sometimes called an axial vector) or a pseudoscalar?; this perspective is common in geometric algebra. In general in dimension $n$, a bivector becomes an $(n-2)$-pseudo-vector, but this is not usually an simplification.
In classical applications of the cross product, often not all of the structure is needed, and the exterior product is really the fundamental concept.
If $M$ is a Riemannian manifold, then the tangent space at each point is an inner product space, so it may be possible to smoothly assign a $k$-ary cross product to these spaces. If this is done, then we can take the curl of a $(k-1)$-vector field? as follows:
This vector field is the curl of the original $(k-1)$-vector field. This justifies the notation $\Del \times X$ for the curl.
(It's important that the cross product is alternating and multilinear, so that it makes sense to apply it to a $k$-vector rather than to $k$ individual vectors.)
When $k = 2$ and $n = 3$, there is one smooth choice of cross product for each orientation of $M$, and we recover the classical notion of curl.
When $k = 1$ and $n = 2$, we may also consider the scalar-valued curl, using the scalar-valued binary cross product described above. The scalar-valued curl of a vector field $X$ is the same as the divergence of the rotated vector field $\times{X}$ (using the unary vector-valued cross product in $2$ dimensions); that is, $\Del \times X = \Del \cdot \times{X}$.
Last revised on March 28, 2018 at 17:13:04. See the history of this page for a list of all contributions to it.