# Is wedge product a tensor or a pseudo tensor?

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: $$\begin{equation*} dx^0∧..dx^n=\frac{\partial x^0}{\partial x^{\beta 0'}}..\frac{\partial x^3}{\partial x^{\beta 3'}}dx^{\beta 0'}...dx^{\beta 3'}= \end{equation*}$$

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

$$$$det(\frac{\partial x}{\partial x'})dx^{0'}\wedge ...dx^{3'}=\sqrt{|g'|}dx^{0'}\wedge ...dx^{3'}$$$$ So $$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?

• I actually think both the Hodge dual and the wedge product lead to tensors. After all, they are operations that take forms to forms, and forms are tensors Mar 13 at 13:18
• My MSE answer math.stackexchange.com/a/4707701/171560 is relevant to this question. Might post an actual answer here later. Mar 13 at 13:20
• I Will read it thanks! Right now my teacher confirmed that the basis write as wedge product is a tensorial density Mar 13 at 13:38

$$J\wedge *J = \langle J,J \rangle \mathrm{d}\mathit{vol}$$
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:
In n-dimensional space the result depends on the dimension $$n$$.