In this topic https://physics.stackexchange.com/questions/129417/what-is-pseudo-tensor one answer was the next:

*The action of parity on a tensor or pseudotensor depends on the number of indices it has (i.e. its tensor rank):*

* - Tensors of odd rank (e.g. vectors) reverse sign under parity.*

- Tensors of even rank (e.g. scalars, linear transformations, bivectors, metrics) retain their sign under parity.

- Pseudotensors of odd rank (e.g. pseudovectors) retain their sign under parity.

- Pseudotensors of even rank (e.g. pseudoscalars) reverse sign under parity.

But I don't understand one thing. Is that statement only for Euclidean three dimensions? I attempted understand it myself. And it is my thoughts. Pseudotensor is determined as:

$\hat{P}^{i_1\ldots i_q}_{\,j_1\ldots j_p} =

(-1)^A A^{i_1} {}_{k_1}\cdots A^{i_q} {}_{k_q}

B^{l_1} {}_{j_1}\cdots B^{l_p} {}_{j_p}

P^{k_1\ldots k_q}_{l_1\ldots l_p} $

where $(-1)^A = \mathrm{sign}(\det(A^{i_q} {}_{k_q})) = \pm{1}$.

Let's consider a pseudovector in Euclidean three dimensions. Then $\det(A^{i_q} {}_{k_q})$ is

\begin{pmatrix}

-1 & 0 & 0 \\

0 & -1 & 0 \\

0 & 0 & -1

\end{pmatrix}

And $(-1)^A=-1$

Let's consider a pseudovector in Euclidean three dimensions. Then $\det(A^{i_q} {}_{k_q})$ is

\begin{pmatrix}

-1 & 0 & 0 & 0\\

0 & -1 & 0 & 0\\

0 & 0 & -1 &0 \\

0 & 0 & 0 & -1

\end{pmatrix}

And $(-1)^A=1$

Let's consider a pseudovector in Minkovski space. Then $\det(A^{i_q} {}_{k_q})$ is

\begin{pmatrix}

1 & 0 & 0 & 0\\

0 & -1 & 0 & 0\\

0 & 0 & -1 &0 \\

0 & 0 & 0 & -1

\end{pmatrix}

And $(-1)^A=-1$

am I right?