Let's consider the passive transformation i.e. inversion only of the basis vectors (coordinate axes) and all other vectors remaining the same and check if the cross product is a pseudo vector.

After the inversion using a passive transformation a vector $a$ remains $a$, $b$ remains $b$, $a \times b$ remains $a \times b$ and any other vector remains the same. So in this case $a \times b$ behaves like any other vector, it remains the same geometrically. Where is the 'pseudo-vector' in this case?

Do we have to use a left-handed screw rule in the inverted coordinate system to find the cross product? Why should we change our rule on how to find the cross product?

This is very confusing to me. Can anyone please help me.


3 Answers 3


The extra minus sign which distinguishes the transformation behavior of a pseudovector from that of a vector arises in two different situations - an active transformation which sends a vector $A\rightarrow -A$, and an orientation-reversing passive transformation.

The cross-product of two vectors $A$ and $B$ is usually defined component-wise as $$C_i = \sum_{j,k=1}^3\epsilon_{ijk} A_j B_k \tag{$\star$}$$ In other words, the components of $C$ in some basis are to be computed from the components $A$ and $B$ in that basis via $(\star)$. As a result, if we make a passive transformation to a new basis $(\hat x',\hat y',\hat z')$, the components of $A\times B$ in the new basis are given by $$ \overline C_i = \sum_{j,k=1}^3\epsilon_{ijk} \overline A^i \overline B^j\tag{$\star\star$}$$

where the line denotes the components in the new basis. It's not hard to show that under a transformation $T$ which transforms the components of ordinary vectors as $A^i\mapsto \overline A^i = T^i_{\ \ j} A^j$, the components of $C$ transform as $C^i \mapsto \overline C^i =\mathrm{det}(T) T^i_{\ \ j} C^j$. This extra factor of $\mathrm{det}(T)$ means that $C$ picks up an extra minus sign under orientation-reversing transformations.

The true nature of pseudovectors can be understood by constructing the exterior algebra over $\mathbb R^3$, denoted $\bigwedge(\mathbb R^3)$. Formally, the idea is that we take our basis $\hat x,\hat y, \hat z$ and define the so-called bivectors $\hat x \wedge \hat y, \hat y \wedge \hat z$, and $ \hat z \wedge \hat x$, where $\wedge$ is understood to be antisymmetric so e.g. $\hat x \wedge \hat y = - \hat y \wedge \hat x$ and $\hat x \wedge \hat x = 0$. We further introduce the trivector $\hat x \wedge \hat y \wedge \hat z$.

The vector space of linear combinations of these objects is the exterior algebra $\bigwedge(\mathbb R^3)$. It includes: $$ a\tag{scalars}$$ $$ a\hat x + b \hat y + c \hat z \tag{vectors}$$ $$ a (\hat x\wedge\hat y) + b(\hat y \wedge \hat z) + c (\hat z \wedge \hat x) \tag{bivectors}$$ $$ a(\hat x \wedge \hat y \wedge \hat z) \tag{trivectors}$$

To standardize the language and allow for easier generalization, we call these objects $k$-blades. Scalars are $0$-blades, vectors are $1$-blades, bivectors are $2$-blades, and trivectors are $3$-blades.

There are two important features to note here. First, because $\mathbb R^3$ is a 3-dimensional vector space, there is only one linearly independent $3$-blade and there are no non-zero $k$-blades for $k>3$; the antisymmetry property of $\wedge$ means that if we repeat a vector (e.g. $\hat x \wedge \hat y \wedge \hat z \wedge \hat x$) then the result must be zero.

The second thing to note is that there are the same number of scalars as $3$-blades, and the same number of $1$-blades as $2$-blades. For a general $n$-dimensional vector spaces, there are the same number of $k$-blades as $(n-k)$-blades - namely ${n\choose{k}}=\frac{n!}{k!(n-k)!}$ of them.

For $\bigwedge(\mathbb R^3)$, this suggests that we can define a pairing between $k$-blades and $(3-k)$-blades, but to do so we need one more piece of information - we need to choose a special $3$-blade $\omega$ which is constructed by wedging together our orthonormal basis vectors. The standard choice is $\omega := \hat x \wedge \hat y \wedge \hat z$. Once we have this, we define the Hodge dual $\star$ of e.g. $\hat x$ via $$\hat x \wedge (\star \hat x) = \omega \implies \star \hat x = \hat y \wedge \hat z$$ Similarly, $$\star 1 = \hat x \wedge \hat y \wedge \hat z$$ $$\star \hat x = \hat y\wedge\hat z \qquad \star \hat y = \hat z \wedge \hat x \qquad \star \hat z = \hat x \wedge \hat y$$ $$\star (\hat y\wedge \hat z) = \hat x \qquad \star (\hat z \wedge \hat x) = \hat y \qquad \star (\hat x \wedge \hat y) = \hat z$$ $$\star (\hat x \wedge \hat y \wedge \hat z) = 1$$

This extends to all elements of $\bigwedge(\mathbb R^3)$ by linearity.

Important note: the orientation $\omega$ should be understood as part of the definition of the basis, i.e. its order; $(\hat x,\hat y,\hat z)$ is different from $(\hat y,\hat x,\hat z)$ despite using the same three unit vectors. As a result, when we change basis $(\hat x,\hat y,\hat z)\rightarrow (\hat x',\hat y',\hat z')$, it is understood that we also change the orientation $\omega = \hat x \wedge\hat y \wedge \hat z \rightarrow \hat x' \wedge \hat y' \wedge \hat z' = \omega'$.

Having defined all of this machinery, we can understand the cross product of two vectors to be the Hodge dual of their wedge product, i.e. $$A \times B := \star (A\wedge B)$$ For example, let $A = 3\hat x + \hat z$ and $B = \hat x + \hat y$. Their wedge product is $$A\wedge B = (3\hat x + \hat z)\wedge(\hat x + \hat y) = 3\underbrace{\hat x \wedge \hat x}_{=0} + 3\hat x \wedge \hat y + \hat z \wedge \hat x + \hat z \wedge \hat y$$ $$= 3\hat x \wedge \hat y + \hat z \wedge \hat x \color{red}{-} \hat y \wedge \hat z$$ Taking the hodge dual using the pairings defined above yields $$\star (A\wedge B) = 3\star(\hat x \wedge \hat y) + \star(\hat z \wedge \hat x) - \star(\hat y \wedge \hat z) = 3 \hat z + \hat y - \hat x$$ which is indeed the cross product $A\times B$ as expected.

This gives us the ability to understand the cross-product at a much deeper level. Its easy to see that under the inversion $(\hat x,\hat y,\hat z)\rightarrow (\hat x',\hat y',\hat z')=(-\hat x,-\hat y, - \hat z)$, the components of a vector (1-blade) $A$ change their sign, but the components of the 2-blade $A\wedge B$ do not; this is because, for example, $$\hat x \wedge \hat y \rightarrow \hat x' \wedge \hat y' = (-\hat x)\wedge(-\hat y) = \hat x \wedge \hat y$$ Inheriting a new orientation $\omega'=\hat x'\wedge\hat y'\wedge \hat z'$, we find that the components of $A\times B := \star(A\wedge B)$ are invariant, and thus differ from the transformation behavior of the components of a vector by a sign flip.

  • $\begingroup$ Thank you sir. It'll take some time for me understand it all. $\endgroup$
    – Kashmiri
    Commented Nov 16, 2021 at 12:54
  • $\begingroup$ You said "but the components of A×B are not - it is in this sense that it is a pseudovector" ,my first doubt is when we have done the passive transformation does the rule of cross product change in any way? Is the right hand screw rule applied when you write "but the components of A×B are not - it is in this sense that it is a pseudovector"? $\endgroup$
    – Kashmiri
    Commented Nov 16, 2021 at 12:55
  • $\begingroup$ By the rule of cross product i meant the right hand screw rule ,in which the direction of $A\times B$ is given by sweeping A to B and thumb gives direction of $A\times B$. So as a first step i would like to have an idea that when I apply the passive transformation and then I want to find $A\times B$ do I use the rule said above? If you could clarify it, then I'll study the bivector you neatly wrote. $\endgroup$
    – Kashmiri
    Commented Nov 16, 2021 at 13:33
  • $\begingroup$ If the rule of screw is same then the vector $A\times B$ is the same in inverted basis , , but the components will be inverted. But that has happened to all vectors, even to a displacement vector $r$ . Then why is $A\times B$ a pseudo vector if it behaves like any other vector? $\endgroup$
    – Kashmiri
    Commented Nov 16, 2021 at 13:46
  • $\begingroup$ The pairing mentioned is a duality. Vectors are dual to bivectors in $\mathbb{R}^3$, just as scalars are dual to the trivector. Is this correct? $\endgroup$
    – JAlex
    Commented Nov 16, 2021 at 13:50

. . . inversion only of the basis vectors (coordinate axes) . . .

$\hat x \to \hat x' = -\hat x,\, \hat y \to \hat y' = -\hat y$ and $\hat z \to \hat z' = -\hat z,$

Rotate the new coordinate axes by $\pi$ about the z-axis.

Now $\hat x' = \hat x,\, \hat y' = \hat y$ but $\hat z' = -\hat z$ so the new coordinate axes are left-handed.

If you want $a\,\hat x' \times b\,\hat y'$ to equal $ab\,\hat z'$ then you must use the left-hand.

By convention the right handed system is used by most although there are some that do not eg Why does DirectX use a left-handed coordinate system?

  • $\begingroup$ Thank you. You said "If you want $a\,\hat x' \times b\,\hat y'$ to equal $ab\,\hat z$ " did you mean $ab\,\hat z'$ ? $\endgroup$
    – Kashmiri
    Commented Nov 16, 2021 at 9:57
  • $\begingroup$ No I did mean $\hat z$. What I was pointing out is that if you use the left-hand system of coordinates but want the resulting direction of a cross product to be the same direction as that when using the right hand coordinates you need to use the left hand. $\endgroup$
    – Farcher
    Commented Nov 16, 2021 at 10:20
  • 1
    $\begingroup$ I think I'm misunderstanding something please bear with me. If I apply the right hand rule on $x'$ and $y'$ I'll get $z$ . I don't need left hand rule, it appears if I apply left handed rule by crossing $x'$ and $y'$ I'll get $z'$ not $z$. $\endgroup$
    – Kashmiri
    Commented Nov 16, 2021 at 11:00
  • $\begingroup$ @Kashmiri So sorry, my comment was incorrect an did point me at my error with your comment. I have now made the correction. $\endgroup$
    – Farcher
    Commented Nov 16, 2021 at 17:33
  • $\begingroup$ Thank you, it confused me. $\endgroup$
    – Kashmiri
    Commented Nov 18, 2021 at 8:18


$$\vec a=a_1\hat g_1+a_2\hat g_2+a_3\,\hat g_3\\ \vec b=b_1\hat g_1+b_2\hat g_2+b_3\hat g_3$$

where $~a_i~,b_i~$ are the vectors components and $~\hat g_i~$ are the basis vectors

with $~\hat g_i\cdot\hat g_j=1~,i=j~$ and $~\hat g_i\cdot\hat g_j=0~,i\ne j$

the cross product

$$\vec c=\vec a\times \vec b$$

now $~\hat g_i\mapsto -\hat g_i~$

$$\vec a\mapsto -a_1\hat g_1-a_2\hat g_2-a_3\hat g_3=-\vec a\\ \vec b\mapsto -b_1\hat g_1-b_2\hat g_2-b_3\hat g_3=-\vec b$$

the cross product

$$\vec c\mapsto -\vec a\times (-\vec b)=\vec a\times \vec b$$

coordinate axes are left -handed.

$$\hat g_1\times \hat g_3=\hat g_2\\ \hat g_2\times \hat g_1=\hat g_3\\ \hat g_3\times \hat g_2=\hat g_1$$

coordinate axes are right -handed.

$$\hat g_1\times \hat g_2=\hat g_3\\ \hat g_2\times \hat g_3=\hat g_1\\ \hat g_3\times \hat g_1=\hat g_2$$

  • 1
    $\begingroup$ That was helpful. $\endgroup$
    – Kashmiri
    Commented Nov 22, 2021 at 7:59

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