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In physics, a tensor is defined as a multidimensional array with a special transformation law.

Therefore, a tensor of type $(r, s)$ is an geometric object

$T_{i_{1}\dots i_{s}}^{j_{1}\dots j_{r}}[\,\underline{e}\,]$

to each basis $\underline{e} = (e_1, ..., e_n)$ of an n-dimensional vector space such that the multidimensional array obeys the transformation law

$T_{i^{\prime}_{1}\dots i^{\prime}_{s}}^{j^{\prime}_{1}\dots j^{\prime}_{r}}[M\cdot \underline{e}\,]=(M^{-1})_{j_{1}}^{j^{\prime}_{1}}\cdot\dots\cdot (M^{-1})_{j_{r}}^{j^{\prime}_{r}}\cdot T_{i_{1}\dots i_{s}}^{j_{1}\dots j_{r}}\cdot M_{i_{1}^{\prime}}^{i_{1}}\cdot\dots\cdot M_{i_{s}^{\prime}}^{i_{s}}$

where $M$ is the transformation matrix.

In calculus this objects can be defined more abstract: There a tensor of type (r,s) is an element of the abstract tensor product

$T\in \underbrace{V\otimes \dots \otimes V}_{r-\text{times}}\otimes \underbrace{V^{\ast}\otimes\dots\otimes V^{\ast}}_{s-\text{times}}$

or, because of can. isomorphy, a tensor can be viewed as a multilinear function

$T:\underbrace{V^{\ast}\times\dots\times V^{\ast}}_{r-\text{times}}\times \underbrace{V\times \dots \times V}_{s-\text{times}}\to \mathbb{R}$.

Now to my question: In physics, there is also the definition of a ''pseudo-tensor'', which is an geometric object

$T_{i_{1}\dots i_{s}}^{j_{1}\dots j_{r}}[\,\underline{e}\,]$

with the the transformation law

$T_{i^{\prime}_{1}\dots i^{\prime}_{s}}^{j^{\prime}_{1}\dots j^{\prime}_{r}}[M\cdot \underline{e}\,]=\mathrm{sign}(\mathrm{det}(M))\cdot(M^{-1})_{j_{1}}^{j^{\prime}_{1}}\cdot\dots\cdot (M^{-1})_{j_{r}}^{j^{\prime}_{r}}\cdot T_{i_{1}\dots i_{s}}^{j_{1}\dots j_{r}}\cdot M_{i_{1}^{\prime}}^{i_{1}}\cdot\dots\cdot M_{i_{s}^{\prime}}^{i_{s}}$

Is there also an abstract definition for pseudo-tensors?

Thank you!

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  • $\begingroup$ Probably the tensor product with the sign representation of the matrix monoid. $\endgroup$ Commented Aug 21, 2019 at 20:04
  • $\begingroup$ Thank you for your answer..... Can you describe this in more detail? I dont really know what the sign representation and the matrix monid is..... $\endgroup$
    – B.Hueber
    Commented Aug 22, 2019 at 8:14
  • $\begingroup$ I'm not totally confident I'm correct, but it would be a vector space where acting with a matrix is the same ss multiplying by the sign of the determinant. $\endgroup$ Commented Aug 22, 2019 at 8:46

1 Answer 1

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I don't think that pseudotensors are really necessary. The pseudotensors that I know of (and those are pseudovectors) can be described as 2-tensors:

Angular momentum $\mathbf{L} = \mathbf{r} \times \mathbf{p}$ can be described as the antisymmetric 2-tensor $L_{ij} = x_i p_j - x_j p_i$ instead of $L_i = \epsilon_i{}^{jk} x_j p_k$

The magnetic field $\mathbf{B}$ defined by $\mathbf{F} = Q \mathbf{v} \times \mathbf{B}$ can likewise be described as an antisymmetric 2-tensor $B_{ij}$ such that $F_i = Q v^j B_{ij}$.

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  • $\begingroup$ Thank you for your answer..... I found a solution by myself meanwhile..... In the abstract sense, you can define pseudotensors(fields) and even more generally tensor densities (fields) as section if a tensor product of a tensor bundle with a density bundle over a manifold... But I agree with you that in practise it is no very usefull.....I was just looking for a coordinate free and abstract definition........ $\endgroup$
    – B.Hueber
    Commented Jan 14, 2020 at 13:53
  • $\begingroup$ I remember reading somewhere that the physics pseudotensors may be described as sections of a jet bundle...can't remember where.. $\endgroup$
    – R. Rankin
    Commented Sep 25, 2023 at 4:57

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