The Unapologetic Mathematician

Mathematics for the interested outsider

The Cross Product and Pseudovectors

Finally we can get to something that is presented to students in multivariable calculus and physics classes as if it were a basic operation: the cross product of three-dimensional vectors. This only works out because the Hodge star defines an isomorphism from A^2(V) to V when \dim(V)=3. We define

u\times v=*(u\wedge v)

All the usual properties of the cross product are really properties of the wedge product combined with the Hodge star. Geometrically, u\times v is defined as a vector perpendicular to the plane spanned by u and v, which is exactly what the Hodge star produces. We choose which perpendicular direction by the “right-hand rule”, but this is only because we choose the basis vectors e_1, e_2, and e_3 (or as these classes often call them: \hat{\imath}, \hat{\jmath}, and \hat{k}) by the same convention, and this defines an orientation we have to stick with when we define the Hodge star. The length of the cross product is the area of the parallelogram spanned by u and v, again as expected from the Hodge star. Algebraically, the cross product is anticommutative and linear in each variable. These are properties of the wedge product, and the Hodge star — being linear — preserves them.

The biggest fib we tell students is that the value of the cross product is a vector. It certainly looks like a vector on the surface, but the problem is that it doesn’t transform like a vector. Before the advent of thinking of all these things geometrically, people thought of a vector quantity as a triple of real numbers that transform in a certain way when we change to a different orthonormal basis. This is inspired by the physical world, where there’s no magic orthonormal basis floating out somewhere to pick out coordinates. We should be able to turn our heads and translate the laws of physics to compensate exactly. These rotations form the special orthogonal group of orientation- and inner product-preserving transformations, but we can also throw in reflections to get the whole orthogonal group, of all transformations from one orthonormal basis to another.

So let’s imagine what happens to a cross product when we reflect the world. In fact, stand by a mirror and hold out your right hand in the familiar way, with your index finger along one imagined vector u, your middle finger along another vector v, and your thumb pointing in the direction of the cross product u\times v. Now look in the mirror.

The orientation has been reversed, and mirror-you is holding out its left hand! If mirror-you tried to use its version of the cross product, it would find that the cross product should go in the other direction. The cross product doesn’t behave like all the other vectors in the world, because it doesn’t reflect the same way.

Physicists to this day use the old language describing a triple of real numbers that transform like a vector under rotations, but point the wrong way under reflections. They call such a quantity a “pseudovector”. And they also have a word for a single real number that somehow mysteriously flips its sign when we apply a reflection: a “pseudoscalar”. Whenever we read about scalar, vector, pseudovector, and pseudoscalar quantities, they just mean real numbers (or triples of them) and specify how they change under certain orthogonal transformations.

But geometrically we can see exactly what’s going on. These are just the spaces A^0(V)=\mathbb{R}, A^1(V)=v, A^2(V), and A^3(V), along with their representations of the orthogonal group \mathrm{O}(V). And the “pseudo” means we’ve used the Hodge star — which depends essentially on a choice of orientation — to pretend that bivectors in A^2(V) and trivectors in A^3(V) are just like vectors in V and scalars in \mathbb{R}, respectively. And we can get away with it for a long time, until a mirror shows up.

The only essential tool from multivariable calculus or introductory physics built from the cross product that we might have need of is the “triple scalar product”, which takes three vectors u, v, and w. It calculates the cross product v\times w of two of them, and then the inner product \langle u,v\times w\rangle=\langle u,*(v\wedge w)\rangle with the third to get a scalar. But this is the coefficient of our unit cube \omega in the definition of the Hodge star:

\displaystyle\langle u,*(v\wedge w)\rangle\omega=u\wedge**(v\wedge w)=u\wedge v\wedge w

since **(v\wedge w)=(-1)^{2\cdot(3-2)}v\wedge w. That is, the triple scalar product gives the (oriented) volume of the parallelepiped spanned by u, v, and w, just as we remember from those classes. We really don’t need the cross product as a primitive operation at all, and in the long run it only leads to confusion as it identifies vectors and pseudovectors without the explicit use of the orientation-dependent Hodge star to keep us straight.

November 10, 2009 - Posted by | rants


  1. Thank you! I’ve always wanted to see a clear explanation of this online, and I think the Wikipedia articles are a little too intimidating.

    Comment by Qiaochu Yuan | November 10, 2009 | Reply

  2. […] turns out that we already know of an example of a Lie algebra: the cross product of vectors in . Indeed, take three vectors , , and and try multiplying them out in all three […]

    Pingback by Lie Algebras « The Unapologetic Mathematician | May 17, 2011 | Reply

  3. […] product on -forms, while the inner ones are the inner product on vector fields. This is exactly the cross product of vector fields on . Advertisement Eco World Content From Across The Internet. Featured on […]

    Pingback by A Hodge Star Example « The Unapologetic Mathematician | October 11, 2011 | Reply

  4. Hi there!! Nice post! Could you please explain why there is such isomorphism when $\dim V = 3$?

    Comment by Aaron | September 28, 2015 | Reply

    • In general, the Hodge star (linked above in the post) defines an isomorphism A^{k}(V)\leftrightarrow A^{n-k}(V).

      We know that A^1(V)\cong V. We also know that any antisymmetric binary operation on V must factor through A^2(V). That is, if we have f:V\times V\rightarrow X then it must factor as V\times V\rightarrow A^2(V)\rightarrow X

      In this case, we’re interested in a binary “multiplication” from V\times V to V itself, meaning we’re looking for a map A^2(V)\rightarrow V\cong A^1(V). This will be the Hodge star in the case where k=2 and n-k=1, so n=3, so this works out in three-dimensional space.

      Comment by John Armstrong | September 28, 2015 | Reply

      • Thanks for clarifying!

        Comment by Aaron | September 28, 2015 | Reply

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