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I want to calculate $w\wedge w$ for $w=\sum a_{ij} e_i\wedge e_j$ where the sum is over all $1\leqslant i<j\leqslant 4$. Is there a neat way to do it without writing all the terms out? What about using double sums, then i get $\sum\sum a_{ij} e_{kl} e_i\wedge e_j\wedge e_k\wedge e_l$, but what about about the range of summation? can I restrict it somehow? I now what the result is supposed to be, I somehow seem to get every factor twice. I mean I can restrict the sum to $i<j,k<l$ all distinct (since expression such as $e_1\wedge e_2\wedge e_1\wedge e_4$ vanish?), but what worries me is that I keep getting both $a_{12} a_{34}e_1\wedge e_2\wedge e_3\wedge e_4$ and $a_{34} a_{12} e_3\wedge e_4\wedge e_1\wedge e_2$, which I think is not supposed to be

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  • $\begingroup$ A basic property of exterior product is that $w\land w=0$ for all $w$. $\endgroup$ – Berci Aug 16 '14 at 23:06
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    $\begingroup$ @Berci his $w$ is a two-form. $\endgroup$ – James S. Cook Aug 17 '14 at 2:38
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It might just be easier define $a_{ji} := -a_{ij}$ for $i < j$ and write $w = \tfrac{1}{2}\sum_{i,j} a_{ij} e^i \wedge e^j$, so that $$ w \wedge w = \frac{1}{4} \sum_{i,j,k,l} a_{ij}a_{kl} e^i \wedge e^j \wedge e^k \wedge e^l = \sum_{i<j<k<l} \frac{1}{4}\left(\sum_{\pi \in S_4} (-1)^\pi a_{\pi(i)\pi(j)}a_{\pi(k)\pi(l)} \right)e^i \wedge e^j \wedge e^k \wedge e^l. $$ This convention, then, really ought to take care of any worries about double-counting.

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