# $m$-th wedge product of a differential $n$-form with itself?

I have a differential 2-form $$\omega = \omega_{ij}dx^i \wedge dx^j$$ (Einstein-summation convention) on a 2-n dimensional manifold. I am trying to compute $$\omega^n = \omega \wedge \cdots \wedge \omega$$. Being a "volume" form, it should look like $$\omega = W_{1\cdots 2n}dx^1\wedge\cdots \wedge dx^{2n}$$. I guess that there is a simple formula which gives me the relation between the $$\omega_{ij}$$ and the $$W_{1 \cdots 2n}$$, which I am not able to figure out.

Edit: Here is my try for the two-form.

Claim: For $$m \leq n$$ we have $$\omega^m(v_1,\cdots,v_{2m}) = \frac{1}{2m!} \sum_{\sigma \in S_{2m}} \mathrm{sgn}(\sigma)\prod_{i =1}^m \omega(v_{\sigma(2i-1)},v_{\sigma(2i)})$$ (Where $$S_k$$ is the symmetric group of $$k$$ element)

We proceed by induction: for $$m = 1$$, it is trivial, since $$\omega(v_1,v_2) = 1/2 (\omega(v_1,v_2) - \omega(v_2,v_1))$$.

For the induction, we assume that the result holds until $$m-1$$, then \begin{align} \omega^m(v_1,\cdots,v_{2n}) & = \omega \wedge \omega^{m-1}(v_1,\cdots,v_{2m}) \\ &= \frac{1}{2m!} \sum_{\rho \in S_{2m}} \mathrm{sgn}(\rho) \omega(v_{\rho(1)},v_{\rho(2)})\omega^{m-1}(v_{\rho(3)},\cdots,v_{\rho(2m)}) \\ & = \frac{1}{2m!} \sum_{\rho \in S_{2m}} \mathrm{sgn}(\rho) \omega(v_{\rho(1)},v_{\rho(2)})\frac{1}{(2m-2)!} \sum_{\sigma \in S_{2m-2}} \mathrm{sgn}(\sigma)\prod_{i =2}^m \omega(v_{\sigma(\rho(2i-1))},v_{\sigma(\rho(2i))}) \\ & = \frac{1}{2m!} \sum_{\rho \in S_{2m}} \mathrm{sgn}(\rho) \omega(v_{\rho(1)},v_{\rho(2)})\prod_{i =2}^m \omega(v_{\rho(2i-1)},v_{\rho(2i)}) \\ & = \frac{1}{2m!} \sum_{\rho \in S_{2m}} \mathrm{sgn}(\rho)\prod_{i =1}^m \omega(v_{\rho(2i-1)},v_{\rho(2i)}). \end{align}

QED? This seems to work, but I am still not sure whether everything is correct. It seems to me that going from the third to fourth line, summing over $$S_{2m-2}$$ just "overcounts" the permutations of $$S_{2m}$$ $$(2m-2)!$$-times, am I right?

Maybe more generally, is there a simple way to see what the $$m^{th}$$ wedge product of an $$n$$-form with itself is on a smooth $$k$$-manifold ? (provided $$n \cdot m \leq k$$)

Any tips on how to compute that would be welcome.

• Your matrix $(\omega_{ij})$ is, of course, skew-symmetric. It has a normal form with $2\times 2$ blocks corresponding to the conjugate pairs of imaginary eigenvalues. Jun 23 at 0:17