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I am confused as to the tensor nature of the cross product. Maybe I'm misunderstanding what I'm reading. But I seem to be getting conflicting messages.

user782220
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2 Answers2

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The definition of 'tensor' is often different to physicists and mathematicians. To a mathematician a tensor is a multilinear object - an element of a tensor product space. A cross product is a vector, therefore it's a tensor.

To a physicist it's particularly an object which transforms tensorially under changes of coordinates, ie, with one copy of the coordinate transformation matrix per index. A cross product doesn't transform like a vector in the standard $(x,y,z)$ basis, but instead like a thing called a pseudovector.

Pseudovectors and pseudotensors can also be represented as antisymmetric elements of a tensor product space. Cross products act like antisymmetric elements of $ \mathbb{R}^3 \otimes \mathbb{R}^3 $, which is a tensor product space, and when written in this space (with basis $(x \wedge y, y \wedge z, z \wedge x)$) they do transform tensorially. It's just when you write them as elements of $\mathbb{R}^3$ that they don't.

The coordinates of $a \times b$ are given by: $a_i b_j - a_j b_i$, which has two indices and therefore must transform with two copies of a coordinate transformation. However, when you map write pseudovectors as vectors, you map 'ij' to a single index ($x \times y \rightarrow z$) using the Hodge star operation, even though it should still transform somehow with two copies of the coordinate transformation. Hence, it does not transform 'tensorially'.

Exterior algebra clears this all up. Cross products should be written in the basis $(x \wedge y, y \wedge z, z \wedge x)$, and transform indices with $\wedge^2 R$ if coordinates transform as $R$.

See also this question.

Alex Kritchevsky
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  • Where can I read about the cross product from the mathematician's point of view as an element of $\mathbb{R} ^3 \otimes \mathbb{R}^3$. Why is the physicist's "transforms tensorially under change of coordinates" is different from the how mathematician's tensor behaves. For example isn't $e_1\otimes e_2=(a_{11}e_1' + a_{21}e_2'+ a_{31}e_3') \otimes (a_{12}e_1' + a_{22}e_2'+a_{32}e_3')$ and expanding out the same as expressing how it "transforms under coordinate change". – user782220 Oct 04 '18 at 04:26
  • Wikipedia on pseudovector has a lot to say about it. – Alex Kritchevsky Oct 04 '18 at 05:43
  • Basically:

    The problem is that $x \wedge y = x \otimes y - y \otimes x$ transforms with two copies of a coordinate change $P$ (via $P \otimes P$). But if you map it to $z$, you might try to change it with one copy.

    Similarly, the indexes of a bivector $a_{ij} e^i \wedge e^j$ should transform with two copies of $P$ (because they're tensor products, really). But if you write them as $a_k e^k$, as a vector, then you might wrongly try to transform them with one copy of $P$. The coordinate should still change like this: $(P^{-1})^i_m (P^{-1})^j_n a_{ij}$.

     – Alex Kritchevsky Oct 04 '18 at 05:49
  • Consider what would happen if you computed the cross product $\vec{v} = x \times y = z$ and then transformed it with a transformation that mapped $y \rightarrow x$ and fixes $x$ and $z$. The value $\vec{v}$ should become 0 (because $x \times x = 0$), but your transformation fixed $z$, so if you wrote $\vec{v} = z$ you would leave it unchanged.

    If you instead wrote $\vec{v} = x \wedge y$, then the transformation would map $\vec{v} \rightarrow x \wedge x = 0$.

     – Alex Kritchevsky Oct 04 '18 at 05:52
  • So what I think I understand is that the cross product as a first order tensor (a vector) doesn't work. But as a second order tensor it truly is a tensor. But I still don't understand what you mean by "A physicists' pseudotensor is a subtype of a mathematician's tensor". – user782220 Oct 04 '18 at 06:46
  • Mathematicians aren't usually as concerned with how things change under changes of coordinates. Technically $\vec{a} \times \vec{b} = (a_y b_z - a_z b_y) \hat{x} + (a_z b_x - a_x b_z) \hat{y} + (a_x b_y - a_y b_x) \hat{z}$ is a vector, which is a type of tensor. It's only when you talk about how it changes coordinates that you realize it doesn't change 'tensorially', because it 'actually' transforms as a second order tensor . In physics that means it doesn't count as a tensor, so they call it something else. – Alex Kritchevsky Oct 04 '18 at 16:05
  • See also this for bunch more discussion. – Alex Kritchevsky Oct 04 '18 at 16:06
  • So are you saying then that physicists call the cross product a first order pseudotensor (or pseudovector) because they use the first order form of the cross product rather than the second order tensor form? – user782220 Oct 04 '18 at 19:42
  • Yes, specifically because they often reserve the word 'tensor' for things that transform in the regular way (with one copy of the transformation matrix for each index). – Alex Kritchevsky Oct 04 '18 at 19:48
  • Just to be clear the cross product is a second order contravariant tensor and not a second order covariant tensor – user782220 Oct 05 '18 at 05:24
  • Note we are talking about the cross product of (polar) vectors here. That's an axial (a.k.a. pseudo-) vector. But the cross product of a polar vector with an axial vector is actually a polar vector. One such example from physics is the force on a moving charge in a magnetic field, $\vec{F} = q \vec{v} \times \vec{B}.$ Here $\vec{F}$ and $\vec{v}$ are polar vectors, but $\vec{B}$ is an axial vector. – md2perpe Oct 05 '18 at 20:44
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The cross product is a bilinear vector valued mapping hence is a tensor.

janmarqz
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  • What kind of tensor? How many indices does it have? – md2perpe Oct 02 '18 at 19:19
  • @md2perpe it is THE bilinear map $\Bbb R^3\times\Bbb R^3\to\Bbb R^3$ given by $(v,w)\mapsto v\times w$ – janmarqz Oct 02 '18 at 21:41
  • @janmarqz But isn't there suppose to be something about flipping signs incorrectly for the cross product on change of basis that don't preserve orientation? – user782220 Oct 02 '18 at 22:46
  • @user782220 tell us what definition do you want to use? – janmarqz Oct 03 '18 at 01:12
  • give an eye to https://en.m.wikipedia.org/wiki/Cross_product#Multilinear_algebra – janmarqz Oct 03 '18 at 01:18
  • @janmarqz so https://en.wikipedia.org/wiki/Pseudotensor says that a pseudotensor is a generalization of a Pseudovector. And I read that the cross product is a pseudovector. Is a pseudovector not a pseudotensor but instead a tensor so there is no contradiction? – user782220 Oct 03 '18 at 03:12