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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.

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  • $\begingroup$ Please see this $\endgroup$
    – Doug
    Jun 22 at 13:05
  • $\begingroup$ 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. $\endgroup$ Jun 23 at 0:17

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