Is wedge product a tensor or a pseudo tensor?

Question

I'm doing an exercise where $J$ is a 1-form on a manifold $M$ of dimension $N$.

The exercise ask me to calculate $J∧(*J)$ with $J=dx^0+2dx^1$ in a minkowski space with metric =(-1,1,1,1) where $*J$ is the Hodge dual of $J$. And then recalculate in a rotated basis $x^{(\alpha')}=\frac{\partial x^{\alpha'}}{\partial x^\beta}x^\beta$

I proof that $J∧(*J)=\sqrt{g}/p! J_{j_1..j_p}J^{j_1..j_p}dx^1∧..dx^N$

So forthe exercise case the wedge product would be equal to $3dx^0∧..dx^3$

Then I try to do the rotation of the basis and I get: \begin{equation*} dx^0∧..dx^n=\frac{\partial x^0}{\partial x^{\beta 0'}}..\frac{\partial x^3}{\partial x^{\beta 3'}}dx^{\beta 0'}...dx^{\beta 3'}= \end{equation*}

\begin{equation} \frac{\partial x^0}{\partial x^{\beta 0'}}..\frac{\partial x^3}{\partial x^{\beta 3'}}\epsilon^{\beta 0'...\beta 3'}dx^{0'}\wedge ...dx^{3'}= \end{equation}

\begin{equation} det(\frac{\partial x}{\partial x'})dx^{0'}\wedge ...dx^{3'}=\sqrt{|g'|}dx^{0'}\wedge ...dx^{3'} \end{equation} So $J\wedge(*J)=3\sqrt{|g'|}dx^{0'}\wedge ...dx^{3'}$

So that implies that wedge product is a pseudo form or is the Hodge dual the pseudo form?

Answer 1

In the language of differential forms in 4-dimensional space

$$J\wedge *J = \langle J,J \rangle \mathrm{d}\mathit{vol}$$

If a reflection is applied $\mathbf{r} \rightarrow -\mathbf{r}$ the volume element changes sign, as it should be, under a complete time-space reflection it does not the sign. In that respect the volume element transforms as a pseudo-scalar or as a scalar density. See in particular this post:

Is 4-volume element a scalar or a pseudoscalar in special relativity?

In n-dimensional space the result depends on the dimension $n$.