I have some trouble with the concept of a "pseudotensor". Wikipedia distinguishes between that and a tensor density (e.g., here, where both concepts are used simultaneously) while e.g. in Eric Weisstein's Mathworld they say that

A pseudotensor is sometimes also called a tensor density.

These are two manifestly incompatible statements, and if asked, I'm inclined to believe that Wikipedia's definition of a pseudotensor, other than a synonyme for a tensor density, can not be rigorously formulated, i.e., no such thing exists. The purpose of this question is to confirm or refute this claim, mathematically. Some details follow to justify my case. "Pseudo" is used in its "Wikipedia" meaning: changing sign under inversion, whatever inversion means (discussed below).

I would appreciate it if answerers read the question in full and consider my problem specifically (and using my "language", if possible), rather than going with "stock" definitions or examples. No "looking at something in a mirror" either unless you can put that into formulas and argue it's a passive or active transform of some kind. Also, bear in mind that the "usual" sources in your country / curriculum may not be readily available to me so I would greatly appreciate it if when citing them you could also quote the relevant part.

With this out of the way, I appreciate that all of these quantities are defined through formulas their elements in given coordinate systems undergo in transformations. However what most sources neglect is distinction between passive and active transforms, and, worse, between the orientation of a vector space and an orientation of its basis. For simplicity, let us only assume linear transforms of linear spaces; this should be WLOG for the purposes of the question as stated.

So, under a passive transform, one compares the decompositions of the same quantity in two bases $E = (e_1, \ldots, e_n)$ and $F = (f_1, \ldots, f_n)$, where $f_i = S^j_{\ i} e_j$, $S = (S^k_{\ l})_{k,l=1}^n \in GL(n)$. A $k$ times covariant and $l$ times contravariant tensor density $T$ of weight $w$ transforms, by the definition I know, as

$$\tilde T^{i_1,\ldots,i_l}_{j_1,\ldots,j_k} = (\det S)^w\ S^{m_1}_{\hphantom{m_1}j_1} \ldots S^{m_k}_{\hphantom{m_k}j_k}\ (S^{-1})^{i_1}_{\hphantom{i_1}n_1} \ldots (S^{-1})^{i_l}_{\hphantom{i_l}n_l}\ T\strut^{n_1, \ldots, n_l}_{m_1, \ldots, m_k}$$ where $T\strut^\ldots_\ldots$ and $\tilde T^\ldots_\ldots$ denote the components in $E$ and in $F$, respectively. (The sign convention of $w$ may differ.)

Since we are looking at the same object in a different systems of coordinates, there is no reason the object changes, only its representation. I've been told ([1], [2]) that a typical example of a pseudoscalar in $\mathbb{R}^3$ is a triple product, $(\vec a,\vec b,\vec c) = \vec a \cdot (\vec b \times \vec c)$ since it "changes sign" under "inversion". Consider, though: in my notation, the inversion is just another $GL(n)$ matrix $S^i_{\ j} = -\delta^i_{\ j}$, and the "coordinate-less" definition of the triple product is $$s := (a,b,c) = \omega(a,b,c)$$ where $\omega$ is a certain 3-form associated with the space, known as its volume form. In a "right-handed" orthonormal basis $E$, that is, $$e_i \cdot e_j = \delta_{ij}, \qquad \omega(e_1,e_2,e_3) = +1,$$ the components of $\omega$ are $$\omega_{ijk} = \epsilon_{ijk}$$ and thus we can express $s$ as $$s = \omega_{ijk} a^i b^j c^k = \epsilon_{ijk} a^i b^j c^k.$$ If we now pick a basis $F = (-e_1,-e_2,-e_3)$ (which is left-handed), the quantity $s$ transforms as $$\tilde s = \tilde w_{ijk} \tilde a^i \tilde b^j \tilde c^k,$$ where all of $a,b,c$ are contravariant and $\omega$ is fully covariant, so $$\tilde a^i = -a^i, \quad \tilde b^j = -b^j, \quad \tilde c^k = -c^k$$ and also $$\tilde \omega_{ijk} = (-1)^3 \omega_{ijk} = -\epsilon_{ijk}$$ by the above relation. Thus $$\tilde s = -\epsilon_{ijk} (-a^i) (-b^j) (-c^k) = +\epsilon_{ijk} a^i b^j c^k = +s,$$ as expected. Of course, just looking at the same quantity (be it $s$, $\omega$, a cross product or anything else) in a different basis there is no reason the quantity should change, only perhaps its numeric description – that's what a passive transform means.

I'm not opposed to believing that $s$ is as pseudoscalar, but I'm firm in saying that a passive transform can't manifest distinction from a proper scalar. This is because although the orientation of two bases can easily differ, the orientation of the space itself is its inherent property, unaffected by what basis we choose to decompose its elements, so there are no "orientation-changing" passive transforms. In other words, while the above result shows that $s$ is consistent with being a scalar (of density zero), it does not show it could not actually be a pseudoscalar, the coordinate inversion just did not test this because no change in orientation (of the space) actually took place.

(The very same reasoning works with no significant changes if $s$ is replaced by another typical example, the vector product, as defined using the Hodge star.)

This leaves active transforms that could possibly distinguish a pseudoscalar from a proper scalar. Well, defining (for comparison) $T = -\mathbb{I}$ and introducing new (actively transformed) vectors $$A^i = T^i_j a_j = -\delta^i_{\ j} a_j = -a_i, \quad B^i = -b^i, \quad C^i = -c^i$$ and their triple product in the original vector space, $$S = \omega(A,B,C) = \epsilon_{ijk} A^i B^j C^k = (-1)^3 \epsilon_{ijk} a^i b^j c^k,$$ we indeed obtain $-s$ (in accordance with the cited sources). But how is it justified to tell anything about the properties of $s$ when $S$ is a new quantity (only defined analogously)?

What's worse, if we take an "arbitrary" active transform given by a generic $T \in GL(n)$ and recompute $S = (A,B,C)$ for the images $A, B, C$, we get $$S = \epsilon_{ijk}\ T^i_{\ p} T^j_{\ q} T^k_{\ r}\ a^p b^q c^r = \det T\ \epsilon_{pqr} a^p b^q c^r = \det T \cdot s,$$ so if we accept that $s$ is a pseudoscalar just because $S$ had a different sign for a particular choice of a $\det < 0$ matrix, we can see that the same reasoning would lead to $S$ changing its magnitude as well (making it actually behave like a scalar density, rather than a pseudoscalar, in terms of its transformation as expressed using $s$). One could argue that active transforms by $T \not\in O(3)$ are not physical, but then again – improper rotations can't be "reached" in real world any better than e.g. volume-nonpreserving ones.


2 Answers 2


The volume form $\omega$ is a legitimate rank-3 tensor. Hence $\omega(a,b,c)$ is a legitimate scalar, and indeed $\tilde{s} = +s$. When people say that the scalar triple product is a pseudoscalar, they must define it by $$ a\cdot(b \times c) := \epsilon_{ijk}a^i b^j c^k \,,$$ where $\epsilon_{ijk}$ is the Levi-Civita symbol – crucially, not the components of a legitimate rank-3 tensor.

Note that Mathworld is often flat-out wrong. In particular, in the first page you linked to, both instances of $\mathsf{A}$ should be replaced by $\mathsf{a}$.

  • $\begingroup$ Thanks for taking the time! But then, how would this ever be a scalar quantity under anything larger than $O(3)$, say, under scaling? $\endgroup$
    – The Vee
    Feb 15, 2018 at 12:57
  • $\begingroup$ (This is the same question as in my last paragraph, as keeping the epsilon is just like transforming a,b,c only under the active transform. The problem then is why in general transforms, not limited to (+/–) rotations, only the sign should be changing. Otherwise it's not a pseudoscalar but a proper scalar density. Or if one restricts oneself to $O(3)$ there is no difference between the two concepts.) $\endgroup$
    – The Vee
    Feb 15, 2018 at 13:14
  • $\begingroup$ @TheVee When people talk about the scalar triple product in this way they're most likely restricting attention to $O(3)$ tensors. However, one could easily extend the concept by allowing $\epsilon_{ijk}$ to transform under more general transformations than orthogonal ones. We would take precisely the pseudotensor transformation law, $\epsilon_{ijk} \to \mathrm{sgn}(\det S)S^l{}_{i} S^m{}_{j} S^n{}_{k} \epsilon_{lmn} = |\det S| \epsilon_{ijk}$. $\endgroup$
    – gj255
    Feb 15, 2018 at 14:28
  • $\begingroup$ Sure, I see. So one could say that taking a "pseudo" transformation law for $\epsilon$, we obtain a "pseudo" result. That's fair. The mathematical interpretation of this pseudo-3-form, as opposed to the volume form (with which it formally agrees in all positive bases), is eluding me, but then it's clear it would work this way. (+1) $\endgroup$
    – The Vee
    Feb 15, 2018 at 20:29

You've already got there, but it seems that you're having trouble taking the consequences at face value. The thing that marks $s$ as a pseudoscalar is simply the transformation property $$ S=\det(T)\cdot s, $$ as you have derived it. However, you go a bit off-track when saying that this means that the magnitude of $S$ depends on the transformation, because the set of linear transformations $\mathrm{GL}(3)$ is just too big. The set of active transformations that can be said to implement equivalent physical situations is $\mathrm{O}(3)$, and there the magnitude of $S$ doesn't change since $|\det(T)| = 1$.

Moreover, $\mathrm{O}(3)$ splits into two distinct subsets, the subgroup $\mathrm{SO}(3)$ which can be reached with rigid rotations that can be physically implemented in the real world, and its coset $(-\mathbb I)\mathrm{SO}(3)$ which requires you to re-build a separate copy of your experiment with an opposite orientation. With that in mind, then:

  • scalar quantities are those that are invariant under the full action of $\mathrm{SO}(3)$, while
  • pseudoscalar quantities are those that change their sign under the action of any active transformation drawn from the coset $(-\mathbb I)\mathrm{SO}(3)$.

The point that the coset $(-\mathbb I)\mathrm{SO}(3)$ can't be "reached" in the real world is precisely why we speak about a "mirror-world" experiment, i.e. a completely separate copy of the experiment that cannot be smoothly transformed into the original one.

  • $\begingroup$ Thank you! So for $O(n)$ there's really no difference between pseudotensors and tensor densities of (any) odd weight, and for $GL(n)$ the former isn't defined? $\endgroup$
    – The Vee
    Feb 15, 2018 at 13:25
  • $\begingroup$ There's nothing special about the $\rm O(3)$-to-$\rm SO(3)$ distinction. You can do the same with $\mathrm{GL}(n)$ over the reals - it is the disjoint union of the subgroup $\mathrm{GL}^+(n)$ with positive determinant and its coset $(-\mathbb I)\mathrm{GL}^+(n)$, and you cannot smoothly reach the latter without squishing all space into a plane at some point. $\endgroup$ Feb 15, 2018 at 13:30
  • $\begingroup$ (+1). Also (in the sense of my last sentence and your last paragraph), what justifies not saying that $O(3)$ is not "too big" either? My $SO(3)$ group preserves distances, angles and the right hand rule (or, $g_{ij}$ and $\omega_{ijk}$). I would say that breaking one is just as serious change in the model as breaking any other subset of these. $\endgroup$
    – The Vee
    Feb 15, 2018 at 13:30
  • $\begingroup$ A transformation in $(-\mathbb I)\mathrm{SO}(3)$ specifies a recipe for taking an experiment and building a separate copy of it that's entirely identical as far as internal distances and angles, but with the quirk that it has 'opposite orientation' (cannot be rotated to match the original). In what way is this unphysical? $\endgroup$ Feb 15, 2018 at 13:35
  • $\begingroup$ It's not unphysical. But I could also build an experiment that looks the same but is exactly twice as large in every direction (angles and orientation remain). And I believe that those quantities for which I had to adjust sign (to make physics work again), I also need to rescale them now by an odd power of det T (and some other quantities by even powers). That there are none which don't scale but get sign-reversed, it's always det T (density), not sgn det T (pseudo). $\endgroup$
    – The Vee
    Feb 15, 2018 at 13:44

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