I am taking a course in Multivariable Analysis and I am asked to do the following problem:
Show that $\ast\ast\omega = (-1)^{k(n-k)}\omega$
So I start as follows:
We know that $\displaystyle \ast\omega =\sum_I \text{sgn}(I,J)\omega_I dx^J$ then
$\ast\ast\omega=\sum_I \text{sgn}(I,J)(\sum_I \text{sgn}(I,J)\omega_I dx^J)dx^J$
but I dont know how to rearrange this to avoid the sums and get the result because I think I need to have some wedges products there, but well I dont know how to proceed, I have already read the other questions about this topic, but they are very advanced so I decided to post it with my definition.
Can you help me with this please? Thanks a lot :) in advance for your help.
My other attempt:
We can write the operator like this:
$\ast\omega= \ast [\sum_I \omega_I dx^I] = \sum_I \omega_I \ast (dx^I) = \sum_I \omega_I dx^J $
where $dx^J$ is a (n-k)-form so what it remains to be proven is the following:
$$\ast \ast \omega= \ast[ \sum_I \omega_I \ast (dx^I)] = \sum_I \omega_I \ast \ast dx^I $$.
then $(-1)^{k(n-k)}dx^I=\ast \ast dx^I$
this is $\text{sgn}(I,J)\text{sgn}(J,I)=(-1)^{k(n-k)}$
but how can I do that? and Am I right in the above arguments?
Thanks