# Formal definition of a Pseudotensor

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!

• Probably the tensor product with the sign representation of the matrix monoid. Commented Aug 21, 2019 at 20:04
• 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..... Commented Aug 22, 2019 at 8:14
• 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. Commented Aug 22, 2019 at 8:46

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 $$F_i = Q v^j B_{ij}$$.

• 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........ Commented Jan 14, 2020 at 13:53
• I remember reading somewhere that the physics pseudotensors may be described as sections of a jet bundle...can't remember where.. Commented Sep 25, 2023 at 4:57