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This question is not only about "tensors under rotation" or "Lorentz tensors", which are one the examples of what I'm asking; as I will explain below, the problem is with the concept of "tensor under a given transformation".

Introduction

For starters, I know there are more abstract definitions of tensors. The definition I am accustomed to is that a $(r,s)$ tensor on a $\mathbb{R}$-vector space $V$, $\mathrm{dim}V=n$, is a multilinear map. $$T:\underbrace{V^*\times...\times V^*}_r\times\underbrace{V\times...\times V}_s\rightarrow\mathbb{R}$$ The set of all such map forms a vector space denoted by $$V^{\otimes r}\otimes V^{*\otimes s}.\tag{tensor product space}\label{tensor}$$ Without delving too much into the details which are pretty standard, under a basis change the components of an $(r,s)$ tensor obey a certain transformation law. Fix two bases and let $A=(a^i_j)\in\mathrm{GL(n,\mathbb{R})}$ and $B=A^{-1}=(b^i_j)$ be the transformation matrix and its inverse, then (using Einstein's convention on repeated indexes) $$T'^{i_1...i_r}_{j_1...j_s}=a^{i_1}_{k_1}...a^{i_r}_{k_r}b^{h_1}_{j_1}...b^{h_s}_{j_s}T^{k_1...k_r}_{h_1...h_s}\tag{1}\label{1}$$ In Physics this transformation law is sometimes used as a definition of vectors and tensors, defining them as numbers that transform in such way under a coordinate change (see e.g. this answer, contravariant vectors).

Let me rewrite \eqref{1} for a vector for future reference $$v^i=a^i_kv^k\tag{1a}\label{1a}$$

Problem

Although I don't think it is the best definition, I have nothing against this viewpoint as long as things works fine and we're really working with the same objects. Now, the problem arises with other definitions which are more specific, such as Lorentz tensors (or 4 tensors), which are defined similarly to \eqref{1} but restricting to a specific class of coordinate transformation, in this case Lorentz transformations. In the case of a 4-vector $$v'^{\mu}=\Lambda^{\mu}_{\nu}v^{\nu}\qquad\Lambda\in\mathrm{O}(1;3)\tag{2a}\label{2a}$$ In the same spirit, 3-vectors are sometimes defined as 3 real numbers $(v_i)$ that transform under rotation (also called "tensors under rotations") like this $$v'_i=R^i_j v^j\qquad R\in\mathrm{SO}(3)\tag{2b}\label{2b}$$ with the obvious generalizations to tensors. Now, \eqref{2a} and \eqref{2b} clearly seem a weaker requirement than \eqref{1a} and the same goes for the analogous version for tensors compared to \eqref{1}.

Question

Are this objects even tensors in the standard sense I described above? In case they are, on what space? The components of a vector (in the standard sense) in $\mathbb{R}^3$ satisfy \eqref{1a}, not just \eqref{2b} and the same goes for a vector in $\mathbb{R}^4$ with lorentzian metric. Also, pseudotensors would be tensors according to such definitions.

Edit (after comments)

Alright, I acknowledge using an example i.e. that of Lorentz tensors was not the best choice I could do. I'm not sure what's wrong with the previous paragraphs but it seems I'll need to expand what I mean here. Consider a vector space $V$ and forget manifolds and tangent spaces. A $(r,s)$ tensor on such space is an element of \eqref{tensor}. After a change of basis of the vector space $V$ (again, no manifold), the coordinates of a tensor transform in accordance with \eqref{1}, where $A=(a^i_j)$ is the matrix of the change of basis (a generic invertible matrix). This condition characterizes tensors.

Now, Physicists sometimes restrict the transformation law by requiring only that $A$ is a special orthogonal matrix and call object satisfying it "tensors under rotation", see \eqref{2b}. Of course tensors satisfying the transformation rule for arbitrary basis also respect this restricted condition. Vice versa, an object satisfying the restricted condition won't necessarily satisfy the condition \eqref{1} for an arbitrary change of basis (think e.g. of cross products under reflections, which do not acquire a minus sign because they are hodge duals of bivectors). So if I restrict to orthogonal linear transformations I'm not dealing only with elements of \eqref{tensor} anymore.

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  • $\begingroup$ Related. $\endgroup$ Dec 17, 2022 at 16:14
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    $\begingroup$ I think you should make yourself aware of the following: a) tensors/vectors according to (1) and (1a) are multi-linear maps that are invariant under a change of basis. b) what you see in (2a) (2b) are better called tensor fields. I.e. tensor valued functions on a manifold. To answer your question in the title: a Lorentz tensor is then a tensor field on space-time that is invariant (i.e. transforms accordingly) under Lorentz transformations. The book Space Time and Geometry by Sean Carroll explains this pretty well. Most other GR books also. $\endgroup$
    – Kurt G.
    Dec 17, 2022 at 17:18
  • $\begingroup$ A tensor field is a smooth section of a tensor bundle i.e. a map that associates to any point a tensor of a tensor bundle constructed on the tangent bundle. In my equations I'm working in a fixed tangent space and changing the basis there, so I'm working with tensors, not tensors fields here. $\endgroup$ Dec 17, 2022 at 17:25
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    $\begingroup$ I apologize if the terminology in the title is misleading or non-standard in that case, I hope that clarifies what I mean: the point is not whether we're dealing with tensors or tensor fields, rather restricting to a class of coordinate transformation which is a subset of all the possible coordinate transformation in a vector space (or for tensor fields on a manifold) and calling that object a tensor regardless. $\endgroup$ Dec 17, 2022 at 17:26
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    $\begingroup$ Related: math.stackexchange.com/questions/4713853/… $\endgroup$
    – ASlateff
    Oct 19, 2023 at 11:32

3 Answers 3

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This can be understood in the language of representation theory. Given any Lie group $G$ and any representation $V$ of $G$, we can think of the elements of $v \in V$ as "generalized tensors of type $(G, V)$" (I don't believe there's a standard term for this). We recover tensors in the usual sense by taking $G = GL_n(\mathbb{R})$ and taking $V$ to be some product of tensor products of the standard representation $\mathbb{R}^n$ and its dual, but this generalizes to taking $G$ to be, say, the orthogonal group or the Lorentz group or whatever else you want. The action map $\rho : G \to GL(V)$ is describing exactly how the "generalized tensor" transforms under the action of $G$.

Globalizing this definition requires the notion of a $G$-structure and of an associated bundle. If $M$ is an $n$-manifold and $G$ is a Lie group equipped with a map to $GL_n(\mathbb{R})$, then we can ask for what is called a reduction of the structure group to $G$; e.g. if $G = SL_n(\mathbb{R})$ this amounts to asking for an orientation, if $G = O(n)$ this amounts to asking for a Riemannian metric, etc. Given such a reduction, together with a representation $V$ of $G$, we can construct the associated bundle with fiber $V$, and sections of this bundle are "generalized tensor fields of type $(G, V)$." Probably the most famous case of this construction is the notion of a spinor field, which comes from taking $G$ to be a Spin group and $V$ to be a spin representation.

Generalized tensors are not, strictly speaking, tensors in the usual sense, both because $G$ is not required to be $GL_n(\mathbb{R})$ and because, even if it is, $V$ is not required to be a tensor product of copies of the standard representation and its dual. One can ask whether a given type of generalized tensor can be interpreted as a tensor in the usual sense, via a combination of embedding $G$ into some $GL_n(\mathbb{R})$ and compatibly embedding $V$ into some tensor representation of $GL_n(\mathbb{R})$. Such interpretations don't necessarily exist and they aren't necessarily unique if they do exist.

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  • $\begingroup$ Excellent answer. It clarifies my doubts and the spinors reference was spot on too. $\endgroup$ Dec 25, 2022 at 9:55
  • $\begingroup$ Is there any reference where "generalized tensors of type $(G,V)$" are discussed that you can advise? $\endgroup$ Dec 29, 2022 at 8:52
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    $\begingroup$ Unfortunately no. It's really too bad because it is a precise way to say what we mean when we say a tensor transforms in such-and-such a way under such-and-such change of coordinates; that is precisely describing a representation of a Lie group! $\endgroup$ Dec 29, 2022 at 19:44
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    $\begingroup$ In fact, there are several references. It's simply the usual point of view of a Cartan geometry. Given a linear representation $\rho: G \to GL(V)$ of the structure group G of a principal G-bundle (P,p,M,G) on a vector space V, a "tensor of type $(V, \rho)$ is a (smooth) map from P to V equivariant with respect to the associated left-G-action on P and the $\rho$-action on V. This notation occurs e.g. in Sharpe: Differential Geometry, Springer GTM 166 or in more recent books about Cartan Geometries, Parabolic Geometries (Cap et al) or Ricci flow (Chow et al) or in Kolar/Michor/Slovak. $\endgroup$
    – ASlateff
    Jun 6, 2023 at 19:05
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    $\begingroup$ @Mr.Feynman You're welcome. Be careful with left and right operation of G on P! Think of P as the total space of the pointwise sets of coordinatized parametrizations for the fibres, e.g. the bundle of linear frames, then G acts naturally from the right on P. The associated left action then requires to take an inverse! Mistake: math.stackexchange.com/questions/4681055/… I also had a somewhat related question: math.stackexchange.com/questions/4713853/… $\endgroup$
    – ASlateff
    Oct 19, 2023 at 11:26
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The problem is that you have to be a bit restrictive with what transformations you allow. There are many vectors in physics that aren't actually bona fide vectors with respect to the transformation (1), instead they are pseudovectors that may be not be invariant with respect to certain transformations. However, if we rule out these transformations (improper rotations) then we do in fact have something that actually transforms as a vector under the transformations that we deem as "allowed".

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    $\begingroup$ If I understand correctly, this definition includes a broader class of objects that are not tensors in the canonical sense, (e.g. pseudotensors/tensor densities). $\endgroup$ Dec 18, 2022 at 9:50
  • $\begingroup$ I'm not quite sure what you're asking. Do you want a proof that the multilinear algebra definition of a tensor implies the physicists' definition? $\endgroup$
    – K.defaoite
    Dec 18, 2022 at 17:38
  • $\begingroup$ No, I'm not asking for a proof. I'm just asking for confirmation that although the multilinear algebra definition of a tensor implies the physicists' definition, the converse is not true unless we consider all invertible transformations and not just rotations or Lorentz transformations or whatever. $\endgroup$ Dec 18, 2022 at 17:46
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    $\begingroup$ @Feynman_00 Well, I must admit I do not know how to answer that question. $\endgroup$
    – K.defaoite
    Dec 18, 2022 at 23:04
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ADDED: Today, my answer below looks not quite right. The tensors in physics are all standard tensors. But if there is a Riemannian or Lorentzian metric in the picture, then you have to keep track of it.

In particular, to write explicit formulas for the components of connection and curvature, you need to fix either coordinates or, more generally, a frame of vector fields (commonly called a moving frame), which we can denote $(e_1, \dots, e_n)$. What some physicists and mathematicians do is to choose an orthonormal frame, i.e., if $g$ is the metric tensor, then $$ g(e_i,e_j) = \delta_{ij}. $$ This simplifies many calculations. Any two orthonormal frames differ at each point by an orthogonal transformation. That is the reason for the restriction.

You can, however, use more general frames. The difference is that the metric tensor and its inverse will appear explicitly in the formulas. This happens, for example, with the Hodge star operator, which is defined in terms of the metric. On the other hand, the formula for the Hodge star operator with respect to an arbitrary frame is a lot messier than the one with respect to an orthonormal frame.

The answer below was motivated by the fact that a surprising amount of physics, notably classical mechanics, can be formulated without using any metric. But I don't believe this is the case with Maxwell's equations or presumably whatever you're interested in.

The answer above does not address spinors, but Qiaochu explains the situation there well.

EARLIER ANSWER: Your question is a good one. There are tensors that are important in physics that are not tensors in the way you define them in the introduction. They transform like standard tensors only if you restrict to transformations that preserve the volume form, i.e, transformations whose determinant is $1$ (or possibly $-1$).

The tensors are not invariant under transformations whose determinant is not $\pm 1$. Pseudovectors are an example of this. I suggest you look for expositions of Maxwell's equations using differential forms. See, for example, https://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field#Differential_forms_approach

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  • $\begingroup$ I've read your latest edit and I would say the earliest answer is more compatible with the accepted answer your previous answer; in your edit you say "The tensors in physics are all standard tensors. But if there is a Riemannian or Lorentzian metric in the picture, then you have to keep track of it" but the sole example of pseudotensors which you had mentioned before is sufficient to say it is not really the case. Of course when there is a metric you want to restrict to frames in which it looks "nice". $\endgroup$ Dec 27, 2022 at 14:02
  • $\begingroup$ As the accepted answer explains physicist do not mean general frames and then they choose to work only in "good" frames, but instead they describe the behaviour of such object only for those specific transformations, so I would say your previous answer was actually correct. $\endgroup$ Dec 27, 2022 at 14:04

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