# Unicorns of Geometry: Getting Cross About Products

Or: Why  you should forget about the vector cross product!

So titled because while both spoken about in tales and legends, neither unicorns nor cross products actually exist. Forget what you think you know about vector geometry!

A question for the beginner physics students: consider two vectors $u = u_x \hat{x} + u_y \hat{y}+ u_z \hat{z}$ and $v = v_x\hat{x} + v_y \hat{y} + v_z \hat{z}$. The coefficients $u_i, v_i$ have units of length. You know the dot product

$u \cdot v = u_xv_x + u_yv_y + u_x v_z$,

which is a number, has units of area, and is the square of the length of one vector projected onto the other. You probably also know the other interesting operation, the cross product

$u \times v = (u_yv_z - u_zv_y)\hat{x} + (u_zv_x - u_xv_z)\hat{y} + (u_xv_y - u_yv_x)\hat{z}$

which is a vector. Its length is the magnitude of the parallelogram defined by $u$ and $v$. But its coefficients have units of area. How can it still be a vector in the same space as the other two if its coefficients have different units?

The actual answer is that the cross product is presented all wrong and simply does not exist as it stands. The hidden truth: all vectors in all spaces have a product! In actual fact the generalization is staggeringly obvious when looked at correctly, but we are sacrificing so much extra information in the standard notion of the cross product that we are actually blind to it. If you have ever been frustrated by unsatisfactory explanations about the cross product, then read on!

If you want to jump straight into the new ideas, skip the next section. Otherwise, here is a little perspective on higher-dimensional attempts at “cross products”.

It is interesting that, in any undergraduate course on mathematics or physics wherever you go, the same old story told is told about vector products. If you have been through such an institution, you might know what I’m talking about. When vectors and their operations are first introduced, the mantra is repeated over and over again: the cross product can only happen in three dimensions. Only in three. Only in three. Only in three.

Some bright spark (maybe you) may have put up your hand and asked why? Depending on your lecturer, you may have had a more or less satisfactory answer. A common one is that in three dimensions there is always exactly one direction perpendicular to a given pair of vectors, so a vector-valued product may be defined — a property that obviously fails to hold in higher-dimensional spaces.

An even brighter spark (still you?) could have then asked what about other things like the cross product, but for higher-dimensional spaces — after all, we can define a so-called “scalar cross product” on two dimensional vectors by imagining the plane embedded in 3D space and performing the product there, taking the result simply to be the magnitude along the direction “off” the plane. Answers to this much broader question vary wildly depending on the background of the lecturer.

A very good answer looks at one of the ways we compute the cross product. We stick the two vectors into the last two rows of a $3 \times 3$ matrix. Into the top row we put the basis vectors in order.

$u \times v = \left| \begin{array}{ccc} \hat{x} & \hat{y} & \hat{z} \\ u_x & u_y & u_z \\ v_x & v_y & v_z \end{array} \right|$

The lecturer says something (or doesn’t say anything) about how this doesn’t necessarily make sense since some entries are vectors and some are scalars, but we’re physics kids so we don’t really see the problem. A maths student in the class puts up their hand and asks if this is basis-independent, but we don’t really understand that either. We do know enough linear algebra to see why $u\times v = - v\times u$ because swapping rows changes the sign of a determinant. Thumbs up!

We also see why this shouldn’t really work for more than three dimensions! We would be trying to compute a determinant like

$\left| \begin{array}{cccc} \hat{x}_1 & \hat{x}_2 & \cdots & \hat{x}_n \\ u_1 & u_2 & \cdots & u_n \\ v_1 & v_2 & \cdots & v_n \end{array} \right|$

which doesn’t even exist! Then the lecturer might go on to add that there’s an obvious way to fix this; we just make a product that takes more than two vectors — $(n-1)$ of them, to be precise. Then we can define

$v_1 \times v_2 \times \cdots \times v_{n-1} = \left| \begin{array}{cccc} \hat{x}_1 & \hat{x}_2 & \cdots & \hat{x}_n \\ v_{1, 1} & v_{1, 2} & \cdots & v_{1, n} \\ v_{2, 1} & v_{2, 2} & \cdots & v_{2, n} \\ \vdots&&&\vdots \\ v_{n-1, 1} & v_{n-1, 2} & \cdots & v_{n-1, n} \end{array} \right|$

We see that this looks about right, and scratch our heads a while and ask what it measures. After all, the cross product takes an area and attaches its magnitude to the direction perpendicular to that area. Well, it’s harder to picture, but just as two independent vectors in 3D define a 2D surface with one perpendicular direction left over, (n-1) vectors in nD define an (n-1)D hypersurface…with one perpendicular direction left over. We take the “area” of this hypersurface and attach it to this remaining perpendicular direction. Win!

It’s interesting that we can go so far as this, while still skirting around what is actually happening here. For more, read on.

What are the properties of the cross product? It is bilinear:

$(u_1 + Au_2) \times v = (u_1 \times v) + A(u_2 \times v)$

and

$u \times (v_1 + Bv_2) = (u\times v_1 ) + B(u \times v_2)$

and it it is antisymmetric:

$u\times v = - v\times u$.

The geometric interpretation behind this second property is that when you are choosing a perpendicular to a little 2D plane surface (the one defined by $u$ and $v$) you can equally well chose the perpendicular to sit on either side. Perpendiculars chosen in opposite senses have opposite signs, hence the factor of $(-1)$. We standardize the choice by picking a handedness: conventionally, right-handed, though this is arbitrary.

Schematically the cross product looks like:

• Take a(n ordered) pair of vectors
• Form their (oriented) parallelogram
• Scale the (right-handed) perpendicular by the area of the parallelogram

Now, very simply, forget the last step. The real outcome of the product is the plane element, not the perpendicular. This is the fundamental object that will allow us to take “cross products” in arbitrary dimensions — we just have to let go of our desires for a vector out of it at the end. Given a handedness convention, the notion of orientation is now simply recorded by the ordering of the vectors — or, equivalently, think of little spirally arrows drawn on the surface swirling in the sense of the first vector to the second.

To further distance our minds from the cross product, let’s use a new symbol, the wedge: $\wedge$. The so-called wedge product of two vectors is simply that. It looks like:

$u\wedge v$.

It is defined to have the same properties as the cross product — bilinearity and antisymmetry. Abstractly, this is all it is. It turns out that the geometric interpretation follows — even in many dimensions. For instance:

$u \wedge v = (u_1 \hat{x}_1 + \cdots + u_n \hat{x}_n) \wedge (v_1 \hat{x}_1 + \cdots + v_n \hat{x}_n) = \sum_{i

Each term in the above measures the area of the little plane element spanned by $u_i \hat{x}_i + u_j \hat{x}_j$ and $v_i \hat{x}_i + v_j \hat{x}_j$, which is some magnitude $(u_iv_j - u_jv_i)$ times the fundamental unit-area element $\hat{x}_i \wedge \hat{x}_j$ lying along the $i-j$ plane (with ordering of $\hat{x}_i$ and $\hat{x}_j$ chosen in a right-handed sense). Altogether the linear combination $u \wedge v$ is the area element spanned by $u$ and $v$!

If $n = 3$ then this looks just like the normal cross product. However, it also works in arbitrary dimension — after all, two independent vectors always define a parallelogram in any vector space, so why shouldn’t we be able to record that information when $n \neq 3$?

But it doesn’t stop there! We can take triple wedge products

$u\wedge v \wedge w$

$\sum_{i

Which record the volume element defined by three vectors, being a linear combination of the fundamental unit-volume cubes $\hat{x}_i\wedge\hat{x}_j\wedge\hat{x}_k$. We can take quadruple wedges, quintuple, et cetera – up to the dimension of the vector space in question. Once we do this, we find we cannot take any more. For if we have

$\hat{x}_1\wedge\hat{x}_2\wedge \cdots \wedge\hat{x}_n$

called the top form, from the antisymmetry property we have

$\left(\hat{x}_1\wedge\hat{x}_2\wedge \cdots \wedge\hat{x}_n\right) \wedge \hat{x}_i = 0 \qquad \forall i = 1, 2, \ldots, n$.

We have been freely talking about linear combinations of area elements and their higher-dimensional analogues. They do indeed form a vector space. In fact, in an n-dimensional space $V$ there are exactly $\binom{n}{k}$ many independent k-dimensional hypervolume elements, and this vector space is called the kth exterior power of $V$, denoted $\Omega^k(V)$. The first exterior power $\Omega^1(V)$ is just $V$ again, and the zeroth exterior power $\Omega^0(V)$ is defined to be the scalars of $V$.

This is why we couldn’t tell the cross product apart from the wedge product in 3 dimensions, because there are the same number of independent 1-dimensional elements (vectors) as there are 2-dimensional ones (areas). This is the same reason why we are able to define an n-dimensional “cross product” of (n-1) many vectors: in general, there are as many independent (n-1)-dimensional hypervolume elements as there are 1-dimensional elements, so really we are taking some (n-1)-fold wedge product

$v_1\wedge v_2\wedge \cdots\wedge v_{n-1} = \sum_{i = 1}^n A_i \wedge_{j \neq i}$

(where $\wedge_{j\neq i}$ simply means the oriented wedge product $\hat{x}_1\wedge\cdots\wedge\hat{x}_n$ where $\hat{x}_i$ does not appear) and then claiming that we just want to associate

$\wedge_{j \neq i} \mapsto \hat{x}_i$.

Clearly this means

$\hat{x} \wedge \hat{y} \mapsto \hat{z}, \quad \hat{y} \wedge \hat{z} \mapsto \hat{x}, \quad \hat{z} \wedge \hat{x} \mapsto \hat{y}$.

Look familiar? There are no equals signs because the objects are not the same,  as they are describing very different geometric things, but the association lets us pretend that we can multiply vectors to get a vector. But that’s not all. If we run the association backwards, then we can capture the dot product as part of the wedge product as well!

What? Well, from $v = v_x \hat{x} + v_y \hat{y} + v_z \hat{z}$, let’s define

$\star(v) = v_x \hat{y}\wedge \hat{z} + v_y \hat{z}\wedge \hat{x} + v_z \hat{x}\wedge \hat{y}$

which just replaces each basis vector with the area element orthogonal to it. Then notice that

$u \wedge \star(v) = (u_x \hat{x} + u_y \hat{y} + u_z \hat{z})\wedge (v_x \hat{y}\wedge \hat{z} + v_y \hat{z}\wedge \hat{x} + v_z \hat{x}\wedge \hat{y})$

$=$

$(u_xv_x + u_yv_y + u_zv_z)\hat{x}\wedge\hat{y}\wedge\hat{z} = (u \cdot v) \hat{x}\wedge\hat{y}\wedge\hat{z}$.

Now if we choose to associate $\hat{x}\wedge\hat{y}\wedge\hat{z} \mapsto 1$ via the same $\star$ operation — which we can, as scalars and top forms are both one-dimensional, then we have the lovely formula

$u \cdot v = \star\left(u \wedge \star(v)\right)$.

Aside: If you are interested to see how the $\star$ idea interacts with higher-dimensional spaces, consider the following generalization: for every basis element $\hat{x}_{i_1}\wedge\cdots\wedge\hat{x}_{i_k}$ in $\Omega^k(V)$define its $\star$ to be the corresponding basis element $\hat{x}_{j_1} \wedge \cdots \wedge \hat{x}_{j_{n-k}}$ in $\Omega^{n-k}(V)$ where none of the “i”s or “j”s match! Of course, we have to worry about ordering, so we pick the one where the wedge of this element with its starred self is a positive multiple of the top form.

We can see now that the wedge operation right away captures nearly all of the vector geometry that is so familiar to us in physics. It truly is a bird’s-eye view of vector spaces, of shape and length and size. Wedge products are so powerful that they deserve independent study. Indeed, we can collect all the wedge products together — all the exterior powers of a space — into one big vector space, called the exterior algebra of a vector space. It is denoted $\Omega(V)$ and the wedge behaves just like a kind of multiplication.

You may think that the exterior algebra is interesting, but truly this is nothing compared to what happens when we try and include calculus! Multivariable and vector calculus, when cast in the powerful language of differential forms, produces some truly deep statements about the nature of space and measurement. Many physical theories reveal themselves to have been statements about exterior algebra all along, and some of the most cherished results in calculus all appear as special cases of one all-encompassing theorem!