# wedge product of differential form

If $\alpha$ is one form over some manifold $M$ $2n-1$ dimensional real, and $X= M\times (0,\infty)$. $r$ is the coordinate for the second factor. Define two form on $X$: $$\omega= d(r^2\alpha)$$ Then we have to calculate $\omega^n$. I am sorry if following doubt are too silly: My doubts are:

1- I think, $\omega^n:= \omega\wedge..\wedge\omega$, n times.

2- As $\omega$ is two form hence $\omega\wedge \omega \neq 0$. but for any one form $\alpha$, we must have $\alpha\wedge\alpha= 0$.[As $\alpha\wedge\alpha= c(\alpha\otimes \alpha- \alpha\otimes \alpha)$

3- What is the guarantee that $\omega^n\neq 0$, As in one form(as in 2nd part above), can we say when $\omega^n=0$ for any two form.

-

1. That depends on the convention in your textbook or notes. But I have seen that notation used, so I won't rule it out.

2. "As $\omega$ is a two-form hence $\omega\wedge\omega\neq 0$" is false. Consider the two form $\mathrm{d}x\wedge \mathrm{d}y$ on $\mathbb{R}^4 = \{(w,x,y,z)| w,x,y,z\in\mathbb{R}\}$. This two form wedged with itself is zero. What you meant is that "there exists two forms such that $\omega\wedge \omega \neq 0$", which is true.

3. There are some cases when $\omega^n = 0$ can be easily guaranteed by algebraic constraints. For starters, if the dimension of $X$ is less than $2n$, then since $\omega^n$ is a $2n$-form, it must be identically 0. This can be generalised using the rank of the differential form using the notion of form envelope.

Let $V$ be a vector space and let $\eta \in \wedge^p V$. We can consider the smallest subspace $W\subseteq V$ such that $\eta \in \wedge^p W$. It is clear that if $kp > \mathrm{dim}(W)$ that $\eta^k = 0$. It is easy to see that since every one form lives in a one dimensional sub-space, this means that for one-forms $\alpha^k = 0$ if $k > 1$.

For your specific form, you have that $$\omega = 2 r\mathrm{d}r \wedge \alpha + r^2 \mathrm{d}\alpha$$ The main constraint then is what $\mathrm{d}\alpha$ looks like. But noting that $$(\mathrm{d}r \wedge \alpha)^2 = 0$$ you can directly compute (using basically the binomial formula) that $$\omega^n = r^{2n} (\mathrm{d}\alpha)^n + 2n r^{2n-1} \mathrm{d} r \wedge \alpha \wedge (\mathrm{d}\alpha)^{n-1} ~.$$ Whether this vanish is determined by whether $(\mathrm{d}\alpha)^n$ vanishes and whether $\alpha\wedge(\mathrm{d}\alpha)^{n-1}$ vanishes.

Now, since per your edit, $M$ is $2n-1$ dimensional. Necessarily $(\mathrm{d}\alpha)^n$ which is a $2n$ form on $M$ must be zero. So you are reduced to $$\omega^n = 2n r^{2n-1} \mathrm{d}r\wedge\alpha\wedge(\mathrm{d}\alpha)^{n-1}$$

This is the best you can do in the abstract, unless more information about $\alpha$ is given. As an example, take $n = 2$ and $M = \mathbb{R}^3$ with coordinates $x,y,z$. Let $\alpha = \mathrm{d}z + x\mathrm{d}y$. Then $\mathrm{d}\alpha = \mathrm{d}x\wedge \mathrm{d}y$ and $$\omega^2 = 4 r^3 \mathrm{d}r \wedge \mathrm{d}z \wedge \mathrm{d}x\wedge \mathrm{d}y~.$$

-
Thanks for the wonderful answer... this is exactly what i need... Thanks a lot. – Junu May 29 '12 at 10:23