# Why invent the definition of pseudovector?

According to Wikipedia:

a pseudovector (or axial vector) is a quantity that transforms like a vector under a proper rotation, but in three dimensions gains an additional sign flip under an improper rotation such as a reflection.

It seems that pseudovectors "are not real vectors". But if you think about it, every vector in $\mathbb{R}^3$ can be written as a cross product of two vectors. Let $\vec{v}_1 = (a,b,c)$, then $\vec{v}_2 = \frac{1}{\sqrt{a}}(-c,0,a), \vec{v}_3 = \frac{1}{\sqrt{a}}(-b,a,0)$ satisfy $$\vec{v}_3\times\vec{v}_2 = (a,b,c)$$ So we get that $\vec{v}_1$ is a pseudovector. But this means every vector is a pseudovector, so this definition seems empty to me in 3D.

• To quote the same wikipedia article: "The pseudovectors of $V$ form a vector space with the same dimension as $V$". It is enough to say that these spaces are isomorphic in finite-dimensioned case. – TZakrevskiy Nov 21 '17 at 9:00
• Pseudovectors are just confusing. It's better to learn about bivectors instead. See answers here and here, for example – Hans Lundmark Nov 21 '17 at 14:31
• Because you're ignoring the action. You're not looking at a vector space, you're looking at a representation of a group. – user14972 Nov 22 '17 at 21:56

You're right that $\vec{v}_1$ has an equal claim to being a vector and a pseudovector. But the conclusion is not that it is both. It is neither! Look at how you introduced it:

every vector in $\mathbb R^3$... Let $\vec{v}_1 = (a,b,c)$...

What does that mean? If you fix three real numbers $a,b,c$, for example the first three prime numbers $2,3,5$, then the quantity $(a,b,c)=(2,3,5)\in\mathbb R^3$ is neither a vector nor a pseudovector with respect to rotations in three-dimensional physical space. It is a scalar, because it doesn't transform under rotations. That scalar just happens to have an index in a three-dimensional trivial representation of the spatial symmetry group.

It is healthier to consider that vectors and pseudovectors live in separate spaces, endowed with different representations of the symmetry group, in this case $O(3)$. Both spaces are three-dimensional, so there exists an isomorphism between them (by fixing a coordinate frame). But they are not naturally isomorphic to each other, nor are they naturally isomorphic to $\mathbb R^3$.

Let's call $\vec{v}$ any vector and define $\vec v_1$ and $\vec v_2$ as its "components", meaning $\vec v=\vec v_2\times\vec v_1$. Let's say that $R:\mathbb R^3\rightarrow\mathbb R^3$ is any rotation. Then, in a not so simple way, you can verify $R(\vec v)=R(\vec v_2)\times R(\vec v_1)$. Take for example $\vec v_2$ and $\vec v_1$ in the $xy-$plane, and make a rotation around the $z-$axis. Now, let's say that $T:\mathbb R^3\rightarrow\mathbb R^3$ is a reflection, that means $T(\vec x)=-\vec x$. Then $T(\vec v_2)\times T(\vec v_1)=(-\vec v_2)\times (-\vec v_1)=\vec v_2\times\vec v_1=\vec v\neq T(\vec v)=-\vec v$. It's the "components" that don't behave well and that's why the difference.

It would be better to say that a vector transforms in a polar or axial way, than to say that it is a polar or an axial vector (a.k.a. pseudovector). The latter is however easier to say and write, so I will do that.

Just because a vector can be written as a cross product of other vectors, $\vec w = \vec u \times \vec v$, it doesn't have to be axial. If one of $\vec u$ and $\vec v$ is axial and the other polar, then $\vec w$ is polar.

You can't see on a vector whether it is polar or axial. It's a property of the type of entity the vector represents. An example is in physics where we have $\vec L = \vec r \times \vec p$ (angular momentum) and $\vec F = q \vec v \times \vec B$ (force of a charged particle moving in a magnetic field). Here the angular momentum $\vec L$ and the magnetic field $\vec B$ are axial vector entities while $\vec r$, $\vec p$, $\vec F$ and $\vec v$ are polar vector entities.

As Hans Lundmark suggests it might be better to use bivectors instead of axial vectors. Bivectors also work in other dimensions than 3. One way to represent them is as antisymmetric (skew) matrices: $$\mathbf L = \left(\begin{matrix} 0 & x p_y - y p_x & x p_z - z p_x \\ y p_x - x p_y & 0 & y p_z - z p_y \\ z p_x - x p_z & z p_y - y p_z & 0 \\ \end{matrix}\right)$$ i.e. $L_{ij} = x_i p_j - x_j p_i,$ where $(x_1, x_2, x_3) = (x, y, z)$ and $(p_1, p_2, p_3) = (p_x, p_y, p_z),$ and $$\mathbf B = \left(\begin{matrix} 0 & B_z & -B_y \\ -B_z & 0 & B_x \\ B_y & -B_x & 0 \end{matrix}\right)$$ i.e. $B_{ij} = \epsilon_{ijk} B_k,$ where $\epsilon_{ijk}$ is the Levi-Civita symbol.