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*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.

- 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?