# wedge product with the exterior derivative of the form $\omega:= dz +x_1 \, dy_1+ x_2 \, dy_2 + \cdots + x_n \, dy_n$.

Write the coordinates on $\mathbb {R} ^{2n+1}$ as $\displaystyle{ (x_1 , y_1, x_2, y_2, \cdots ,x_n, y_n ,z)}$. Define the 1-form $\displaystyle{ \omega:= dz +x_1 \, dy_1+ x_2 \, dy_2 + \cdots + x_n \, dy_n}$.

Compute $\displaystyle{ \omega \wedge (d \omega \wedge d \omega \wedge \cdots \wedge d \omega )}$ where the wedge product is taken n times.

I first work out the simply cases $n=1,2,3$ and I guess that it must be

$\displaystyle{ \omega \wedge (d \omega \wedge d \omega \wedge \cdots \wedge d \omega ) =n dz \wedge dx_1 \wedge dy_1 \wedge dx_2 \wedge dy_2 \wedge \cdots \wedge dx_n \wedge dy_n }$

but I have no proof for the general case.

• I forgot how to do this correctly and completely. Have you tried induction on $n$? – math-visitor Jun 16 '12 at 18:53
• @math-visitor: How with induction? I tried but it was complicated with the wedge products. – passenger Jun 16 '12 at 18:56

We have $d\omega=\sum_{j=1}^ndx_j\wedge dy_j$ hence \begin{align} (d\omega)^n&=\left(\sum_{j=1}^ndx_j\wedge dy_j\right)^n\\ &=\sum_{1\leq i_1,\ldots,i_n\leq n}\bigwedge_{j=1}^n(dx_{i_j}\wedge dy_{i_j})\\ &=\sum_{\sigma\in\mathfrak S_n}\bigwedge_{j=1}^n(dx_{\sigma(j)}\wedge dy_{\sigma(j)}), \end{align} where $\mathfrak S_n$ denotes the set of the permutations of $\{1,\dots,n\}$ since in the second line the terms such that $i_k=i_j$ for $j\neq k$ vanish ($dx_k\wedge dx_k=0$).
We can see when $\sigma$ is a transposition that $\bigwedge_{j=1}^n(dx_{\sigma(j)}\wedge dy_{\sigma(j)})=\bigwedge_{j=1}^n(dx_j\wedge dy_j)$, and a permutation is a composition of transpositions, so $$(d\omega)^n=n!\bigwedge_{j=1}^n(dx_j\wedge dy_j).$$ We conclude that $$\omega\wedge (d\omega)^n=n!dz\wedge \bigwedge_{j=1}^n(dx_j\wedge dy_j)=n! \left(\bigwedge_{j=1}^ndx_j\wedge dy_j\right)\wedge dz.$$
• I think there is a small typo in the last equality since it should be $n!$. Anyway thank you very much for your anwser! It is very clear! – passenger Jun 16 '12 at 19:27
Begin with $d\omega = \sum_{i=1}^n dx_i\wedge dy_i$. Notice that wedge product of 2-forms is commutative. Wedging $d\omega$ with itself $n$ times, you get $n!$ nonzero terms, since you can only use each value of $i$ once. All these terms are the same, so you get $n!\bigwedge_{i=1}^n (dx_i\wedge dy_i)$. Wedging this with $\omega$ itself is the same was wedging with $dz$, because all other differentials are already in. Hence, $n!dz\wedge \bigwedge_{i=1}^n (dx_i\wedge dy_i)$.