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In learning about particle physics, I stumble upon an apparent paradox, so I must misunderstand something. I am reading about pseudovectors, which are vectors that do not change sign under an inversion. Cross products are examples of this. So in my mind, this distinction would in my mind imply that

$$\neg(\forall z ∈ ℝ³)(∃x,y ∈ ℝ³)(z =x × y).$$

This, on the other hand, seems ludicrous, since if any $z ∈ ℝ³$ would be expressible as a a cross product, all vectors should be, due to rotational invariance about the origin. Denying this seems to imply that some points in 3D-space are more ‘special’ than others.

Where is the fallacy here?

Example of a contradiction

We know that $\hat{z} = \hat{x} × \hat{y}$. Now for $P$ the inversion, i.e. $(x,y,z) ↦ (-x,-y,-z)$, we obtain

$$ -\hat{z} ≠ \hat{z} = \hat{x} × \hat{y} = P(\hat{x} × \hat{y}) = P(\hat{z}) = -\hat{z}.$$

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    $\begingroup$ Are you asking if every vector in $\mathbb{R}^3$ can be written as a cross product of two other vectors in $\mathbb{R}^3$? $\endgroup$
    – Cameron Williams
    Oct 8, 2020 at 20:21
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    $\begingroup$ What do you mean by an inversion? $\endgroup$
    – Servaes
    Oct 8, 2020 at 20:22
  • $\begingroup$ I'm not sure I follow your argument here. The cross product has a "handedness" which affects the sign, but if you have two cross products, you get two sign changes and the effect of the convention disappears. My understanding is that (in the cases I know about) measurable quantities are physically related in such a way that the sign convention cancels out. [The fact that the cross product behaves so much like a vector is a particular feature of three dimensions and is not true more generally]. But what does this have to do with the representation of vectors as a cross product? $\endgroup$
    – Mark Bennet
    Oct 8, 2020 at 20:27
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    $\begingroup$ If you think about the orthogonality property of the cross product, I think it will become evident that vectors in $\mathbb R^3$ are represented in this fashion. $\endgroup$
    – hardmath
    Oct 8, 2020 at 20:28
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    $\begingroup$ It seems that being a pseudovector might be a property of (bilinear) maps $V\times V\ \longrightarrow V$, not of vectors $v\in V$. $\endgroup$
    – Servaes
    Oct 8, 2020 at 20:31

3 Answers 3

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To the question in the title; yes, every vector in $\Bbb{R}^3$ is a cross product of two vectors in $\Bbb{R}^3$. The argument you sketch for this is entirely correct, and there is no fallacy in your thinking.

The fallacy is in the way you (or your source) define pseudovectors. Of course the only vector in $\Bbb{R}^3$ that is invariant under inversion is the zero vector. However, for any pair of vectors $x,y\in\Bbb{R}^3$ you have $$x\times y=(-x)\times(-y),$$ which shows that the cross product, as a map $\Bbb{R}^3\times\Bbb{R}^3\ \longrightarrow\ \Bbb{R}^3$, is invariant under precomposition by the inversion $x\ \longmapsto\ -x$. Perhaps this is what your source is getting at?

Of course this is just a very particular instance of the fact that the cross product 'plays nicely' with orthogonal transformation, which you already use in your sketch for the argument that every vector in $\Bbb{R}^3$ is a cross product. More precisely, the fact that for any orthogonal transformation $A$ you have $$A(x)\times A(y)=(\det A)A(x\times y).$$

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  • $\begingroup$ So 'a' pseudovector is actually a pair of vectors? The strange (and actually, quite infuriating) is that the text literally writes: "When applied to a vector $a$, $P$ produces a vector in the opposite direction: $P(a) = -a$. [...] Well, if $P$ changes the sign of $a$ and $b$, then $c [= a × b]$ does not change sign: $P(c) = c$." Can you imagine my confusion?? Now that I understand it, it turns out he puts to thing from different spaces on an operator that isn't even defined on the other space, so he can't possibly do that! He needs two inversions: one on $ℝ³$ and on on $ℝ³× ℝ³$, right? $\endgroup$
    – Jos van Nieuwman
    Oct 9, 2020 at 13:22
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    $\begingroup$ If you have some function $f:\ \Bbb{R}^3\ \longrightarrow\ \Bbb{R}^3$, for example an inversion, you can ask whether it leaves the cross product unchanged, i.e. whether $f(x)\times f(y)=x\times y$ for all $x,y\in\Bbb{R}^3$. Or more generally, whether it leaves some (bilinear) map $$B: \Bbb{R}^3\times \Bbb{R}^3\ \longrightarrow\ \Bbb{R}^3,$$ unchanged, i.e. whether $B(f(x),f(y))=B(x,y)$, for all $x,y\in\Bbb{R}^3$. This is not a property of pairs of vectors, but a property of a bilinear map. $\endgroup$
    – Servaes
    Oct 9, 2020 at 13:38
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    $\begingroup$ In particular, if $P$ produces a vector in the opposite direction, i.e. if $P(a)=-a$ for all $a\in\Bbb{R}^3$, then of course also $P(c)=-c$. It just isn't true that $$P(a)\times P(b)=P(a\times b).$$ What the text intends to convey, I think, is not that $P(c)=c$, but that $P(a)\times P(b)=c$. It seems that your text is very poorly worded. $\endgroup$
    – Servaes
    Oct 9, 2020 at 13:39
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    $\begingroup$ My attempt at a more 'physics-oriented' explanation would be that the construction of the cross product is a pseudovector; we can construct the cross product of any pair of vectors in $\Bbb{R}^3$, and the result of this construction is unchanged if we apply an inversion to $\Bbb{R}^3$ first, and then construct the cross product. Emphasis on the construction; the result of the construction is just a vector, but the construction itself has the desired property of a pseudovector (this is my best guess, of course). $\endgroup$
    – Servaes
    Oct 9, 2020 at 13:45
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I’d like to build on Servaes’s answer, and hope to get at the root of your question. I’m no physicist, but I believe that In physics it is very common (perhaps fundamental) to think of vectors as existing outside of any particular coordinate system. Rather, they are quantities (like the velocity of a particle) that can be measured against any given reference frame.

At the highest level, I’d say the fallacy here is in your assumption of what it means for two “vectors” $a,b$ to be equal. It goes beyond just saying that $a$ and $b$ have the same numerical coordinates (which is the sense of equality embodied in your set notation $z = x \times y$). What if changing the coordinate system results in different numerical values for $a$ and $b$? Then they aren’t truly equal in this broader sense. It’s not enough that every point in $\mathbb R^3$ can be written as the cross product of two other points.

As the change-of-basis equation in Servaes’s answer shows, any orientation-preserving rotation $A$ will have $\det A = 1$ and so will transform cross products invariantly. However, if $A$ is a reflection then $\det A = -1$ and a sign change occurs under this change of coordinates.

For this reason I can see it could be convenient to distinguish vectors that arise naturally from cross products (e.g. angular velocity) as being in a different category, evidently called pseudovectors. Then two pseudovectors that are numerically equal remain equal under any orthogonal basis change, as do two vectors that are numerically equal. But a vector and a pseudo vector that are numerically equal in one reference frame will flip signs in some change-of-coordinates and remain equal in others. Keeping the two in separate conceptual buckets prevents this ambiguity from arising.

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    $\begingroup$ I feel there is an analogy to be drawn here with the concept of covariance and contravariance in tensor analysis. However I don’t feel qualified to draw a conclusive comparison, and I hope someone else can :). $\endgroup$
    – Erick Wong
    Oct 8, 2020 at 22:22
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So I think I may have found a way to express this in the language of algebraic topology (which I am just learning). If we let $C$ be the cross product map, and $P$ the inversion again, then the following diagram commutes (I don't know how to draw an arrow from bottom left to upper right)

$\require{AMScd}$ \begin{CD} ℝ^3 × ℝ^3 @>{P}>> ℝ^3 × ℝ^3\\ @A \text{id} A A @VVCV \\ ℝ^3 × ℝ^3 @>{C}>> ℝ^3 \end{CD}

If $C$ is a covering map here, which I think it should be, then $P$ is the unique lift of $C$ across $C$. Also, we could note that $P ∈ \text{Aut}( ℝ^3 × ℝ^3 / ℝ^3)$

Is this in any way useful..? XD

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