【A Brief Introduction to Exterior Differentiation】2. The Power of Antisymmetry
Inner Product and Outer Product
The power of vectors (here, for the time being, meaning vectors in two- or three-dimensional space) lies in the fact that they come equipped with an inner product and an outer product (more often called the cross product, vector product, etc.). Both are operations between two vectors, and while the inner product is defined to be symmetric, the outer product is defined to be antisymmetric; both satisfy distributivity.
Following the conventions of textbooks, we denote the inner product by $\langle,\rangle$ and the outer product by $\land$. For the outer product, the more common notation is $\times$, but to avoid introducing too many symbols, we will uniformly use $\land$. Let us write vectors in terms of a basis, e.g.
$$\boldsymbol{A}=\boldsymbol{e}_{\mu}A^{\mu} \tag{1} $$
where $\boldsymbol{e}_{\mu}$ denotes a set of basis vectors, and $A^{\mu}$ are the components of the vector. Let's compute the inner and outer products of two vectors $\boldsymbol{A},\boldsymbol{B}$, namely
$$\begin{aligned}&\langle \boldsymbol{A}, \boldsymbol{B}\rangle=\langle \boldsymbol{e}_{\mu}A^{\mu}, \boldsymbol{e}_{\nu}B^{\nu}\rangle=\langle\boldsymbol{e}_{\mu},\boldsymbol{e}_{\nu}\rangle A^{\mu}A^{\nu}\\ &\boldsymbol{A}\land \boldsymbol{B}=(\boldsymbol{e}_{\mu}A^{\mu})\land (\boldsymbol{e}_{\nu}B^{\nu})=\boldsymbol{e}_{\mu}\land\boldsymbol{e}_{\nu} A^{\mu}B^{\nu} \end{aligned} \tag{2} $$more
And then? Well, there's no "and then," because we haven't yet defined $\langle\boldsymbol{e}_{\mu},\boldsymbol{e}_{\nu}\rangle$ and $\boldsymbol{e}_{\mu}\land\boldsymbol{e}_{\nu}$. In analytic geometry, the inner product is defined as follows: let $\boldsymbol{e}_{\mu}$ be a set of orthonormal basis vectors, then
$$\langle\boldsymbol{e}_{\mu},\boldsymbol{e}_{\nu}\rangle=\delta_{\mu\nu} \tag{3} $$
where $\mu=\nu$ gives $\delta_{\mu\nu}=1$, and 0 otherwise. With this, we can compute the inner product of any two vectors, and having this definition, the inner product becomes a tool for determining perpendicularity (two vectors have inner product 0), as well as a tool for computing length (the inner product of a vector with itself).
Now let's turn to the outer product. In two-dimensional space, the outer product is defined as follows: let $\boldsymbol{e}_{\mu}$ be a set of orthonormal basis vectors, then
$$\boldsymbol{e}_1\land\boldsymbol{e}_2=1 \tag{4} $$
Note that by antisymmetry we automatically get $\boldsymbol{e}_1\land\boldsymbol{e}_1=\boldsymbol{e}_2\land\boldsymbol{e}_2=0, \boldsymbol{e}_2\land\boldsymbol{e}_1=-1$, so this definition is already complete. We can then compute
$$\boldsymbol{A}\land \boldsymbol{B}=A^1 B^2 - A^2 B^1 \tag{5} $$
Here the outer product is a number, and its absolute value is exactly the area of the parallelogram spanned by $\boldsymbol{A}$ and $\boldsymbol{B}$.
In three-dimensional space, on the other hand, we define
$$\boldsymbol{e}_1\land\boldsymbol{e}_2=\boldsymbol{e}_3,\boldsymbol{e}_2\land\boldsymbol{e}_3=\boldsymbol{e}_1,\boldsymbol{e}_3\land\boldsymbol{e}_1=\boldsymbol{e}_2 \tag{6} $$
With this definition, the outer product in three-dimensional space is itself a vector, one that is perpendicular to the original two vectors, with a length equal to the area of the parallelogram spanned by those two vectors.
Looking back over this whole process, we can understand it this way: the inner and outer products are, at their core, purely algebraically defined symmetric and antisymmetric operations. As for their geometric meaning, that is something bestowed afterward, once the inner and outer products of the basis vectors have been fixed. Of course, the definitions of the inner and outer products do have a certain historical origin, but since this is not itself especially difficult, we'll set aside any study of that history here. As we can see, the definition of the inner product clearly generalizes to higher dimensions, whereas the outer product does not generalize in quite the same way. In any case, we can now see the general approach clearly: pure algebraic definition (mainly, defining the inner and outer products of the basis vectors) → seeking geometric meaning → looking back at historical origins.
The Power of Antisymmetry
Ever since we first learned arithmetic, nearly every operation we've encountered has been symmetric, i.e., satisfying $ab=ba$. Addition and multiplication of numbers are like this, and even the inner product of vectors we learn in high school remains this way. Since relatively few schools actually teach the outer product of vectors in high school, many students don't encounter non-commutative operations (i.e., $ab\neq ba$) until college — matrix multiplication, for example. Among all non-commutative operations, antisymmetric operations are a particularly special and richly meaningful kind. Not only that, but they also bring genuine convenience to computation. Consider the following example.
Consider the problem of a particle moving under a fixed gravitational center, for which we have the equation of motion
$$\ddot{\boldsymbol{x}}=-\frac{\mu\boldsymbol{x}}{|\boldsymbol{x}|^3} \tag{7} $$
Taking the outer product of both sides with $\boldsymbol{x}$, we get
$$\boldsymbol{x}\land \ddot{\boldsymbol{x}}=-\boldsymbol{x}\land \frac{\mu\boldsymbol{x}}{|\boldsymbol{x}|^3}=0 \tag{8} $$
Noting that
$$\frac{d}{dt}(\boldsymbol{x}\land \dot{\boldsymbol{x}}) = \dot{\boldsymbol{x}}\land \dot{\boldsymbol{x}}+\boldsymbol{x}\land \ddot{\boldsymbol{x}}=\boldsymbol{x}\land \ddot{\boldsymbol{x}} \tag{9} $$
the above equation implies
$$\frac{d}{dt}(\boldsymbol{x}\land \dot{\boldsymbol{x}})=0 \tag{10} $$
and therefore
$$\boldsymbol{x}\land \dot{\boldsymbol{x}}=\boldsymbol{C} \tag{11} $$
This is, in fact, exactly the conservation of angular momentum. Since it's a vector equation, writing it out in components gives three separate equations. In just a few simple steps, we've obtained three constants of integration. Tracing this back to its root, it's precisely the antisymmetry of the outer product, $a\land b = -b\land a$, that forces $a\land a=0$. This is a property of any antisymmetric quantity, and it is exactly the power of antisymmetry: it naturally makes many terms — which ought to vanish — vanish automatically.
Translated automatically with claude-sonnet-5; all equations are reproduced verbatim from the source. Copyright remains with the original author.