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

  • $\begingroup$ A basic property of exterior product is that $w\land w=0$ for all $w$. $\endgroup$ – Berci Aug 16 '14 at 23:06
  • 2
    $\begingroup$ @Berci his $w$ is a two-form. $\endgroup$ – James S. Cook Aug 17 '14 at 2:38

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.