Wedge Product as an (r+s) form

[cathalcummins]

Homework Statement

If $\omega \in \Lambda^{r}(V^*)$ and $\nu \in \Lambda^{s}(V^*)$ show that $\omega \wedge \nu \in \Lambda^{r+s}(V^*)$

Homework Equations

$\omega \wedge \nu (v_1,... ,v_r,v_{r+1},... ,v_{r+s})= \frac{1}{(r+s)!} \Sigma_{\sigma}(-1)^\sigma \omega(v_{\sigma(1)},v_{\sigma(2)},...,v_{\sigma(r)}) \nu(v_{\sigma(r+1)},v_{\sigma(r+2)},...,v_{\sigma(r+s)})$

The Attempt at a Solution

It would take really long to write out my solution thus far, I broke the sum on the right into even and odd sums. Then proceeded to show (I hope) that as $\omega$ and $\nu$ behave like $\omega$(odd interchange) $= - \omega$(no interchange) and the same for $\nu$.

Then the "odd sum" will have three negatives, one from the $(-1)^\sigma$ and two from the above relations. This leaves a net (-1) as per definition of an (r+s) form, similar reasoning is used for the even sum.

Combining these two, we have, up to a scaling factor, that this new object behaves just as an (r+s) form.

Cheers, it's mainly the permutation thing is bugging me.

 

[mighty2000]

Thank you for your post. Your solution so far seems to be on the right track. The key concept here is the alternating property of differential forms. As you mentioned, \omega and \nu behave in a certain way under odd permutations, and this allows us to show that their wedge product \omega \wedge \nu also behaves in the same way. This is because the (-1)^\sigma term in the definition of \omega \wedge \nu takes into account the behavior of \omega and \nu under permutations.

In addition, you correctly noted that the even and odd sums will have a net (-1) factor, which is what we need to show that \omega \wedge \nu is an (r+s) form. Keep working through your solution and make sure to pay attention to the alternating property. Good luck!