# pseudotensor in different dimensions

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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?

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