Let $n\geq4 $ and $ \omega \in \Lambda^2(\mathbb R^n)^* $ an alternating 2-linear form. How can one show that for $v_1,\ldots,v_4 \in \mathbb R^n$:

$$ \omega \land \omega(v_1,\ldots, v_4)=2 \omega(v_1,v_2)\omega(v_3,v_4)-2\omega(v_1,v_3)\omega(v_2,v_4)+2\omega(v_1,v_4)\omega(v_2,v_3).$$

Now I thought to try something by using the alternator, but still I cant solve this. I got something like $$\sum sign(\sigma)\omega(v_{\sigma(1)},v_{\sigma(2)})\omega(v_{\sigma(3)},v_{\sigma(4)})+\sum sign(\sigma)\omega(v_{\sigma(1)},v_{\sigma(2)})\omega(v_{\sigma(3)},v_{\sigma(4)})+\sum sign(\sigma)\omega(v_{\sigma(1)},v_{\sigma(2)})\omega(v_{\sigma(3)},v_{\sigma(4)}) .$$ I am not sure about my idea. Any help or a hint is much appreciated. Thanks!

  • $\begingroup$ How did you define the wedge product of two multilinear alternating maps? $\endgroup$ – levap Jan 28 '17 at 14:25
  • $\begingroup$ We defined it via Determinante. But I may use the alternator aswell $\endgroup$ – user409387 Jan 28 '17 at 14:49

You might find it useful to prove the following formula for the wedge product:

$$ (\omega \wedge \eta)(v_1,v_2,v_3,v_4) = \sum_{\sigma \in S_{2,2}} \operatorname{sign}(\sigma) \omega(v_{\sigma(1)}, v_{\sigma(2)})\eta (v_{\sigma(3)},v_{\sigma(4)}) $$

where $S_{2,2}$ denotes the $(2,2)$-shuffles which are the permutations $\sigma$ of $\{1,2,3,4\}$ that satisfy $\sigma(1) < \sigma(2)$ and $\sigma(3) < \sigma(4)$. The advantage of this formula (which can be proved by a combinatorical argument and generalizes for higher order forms) is that it involves working with only $|S_{2,2}| = {4 \choose 2} = 6$ permutations and not $4! = 24$ permutations.

The relevant permutations in your case are

$$ \operatorname{id}, (23), (243), (123), (1243), (13)(24) $$

and so

$$ (\omega \wedge \omega)(v_1,v_2,v_3,v_4) = \omega(v_1,v_2)\omega(v_3,v_4) - \omega(v_1,v_3)\omega(v_2,v_4) + \omega(v_1,v_4)\omega(v_2,v_3) \\ + \omega(v_2,v_3)\omega(v_1,v_4) - \omega(v_2,v_4)\omega(v_1,v_3) + \omega(v_3,v_4)\omega(v_1,v_2) \\ = 2\omega(v_1,v_2)\omega(v_3,v_4) - 2\omega(v_1,v_3)\omega(v_2,v_4) + 2\omega(v_2,v_3)\omega(v_1,v_4).$$

Alternatively, you can brute-force your way through all $4! = 24$ permutations and use the fact that $\omega$ is alternating to get the same expression. More explicitly, you have

$$ (\omega \wedge \eta)(v_1,v_2,v_3,v_4) = \frac{4!}{2!\cdot 2!} \operatorname{Alt}(\omega \otimes \eta)(v_{\sigma(1)}, v_{\sigma(2)}, v_{\sigma(3)}, v_{\sigma(4)}) \\ = \frac{1}{2! \cdot 2!} \sum_{\sigma \in S_{4}} \operatorname{sign}(\sigma) \omega(v_{\sigma(1)}, v_{\sigma(2)})\eta (v_{\sigma(3)},v_{\sigma(4)}). $$

I have already computed the right hand side for six permutations. Now show that each of the six permutations above comes with three permutations such that result in the same value. For example, the permutations

$$ \operatorname{id}, (12)(34), (12), (34) $$

fix $\{ 1,2 \}$ and $\{ 3, 4 \}$ and give us

$$ \omega(v_1,v_2)\omega(v_3,v_4) + \omega(v_2,v_1)\omega(v_4,v_3) - \omega(v_2,v_1)\omega(v_3,v_4) - \omega(v_1,v_2)\omega(v_4,v_3) = 4 \omega(v_1,v_2) \omega(v_3, v_4) $$

and similarly for each of the five other permutations.

  • $\begingroup$ Thank you for the nice answer. I tried to solve the problem with my idea, but I do not get it. It would be great, if you could show me, how its done. Cheers $\endgroup$ – user409387 Jan 28 '17 at 20:44
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    $\begingroup$ @JohnDoe: I've added some details, enjoy. $\endgroup$ – levap Jan 29 '17 at 0:37
  • $\begingroup$ Thank you. The only problem remaining is, that I do not understand what the fix{1,2} {3,4} mean. I tried the other permutations, but I still don´t get it. I tried for example the permutations (13), (23)(14), (14),(23), but then I do not get any of the above results. Sorry for stupid questions, I am totaly new to this thematic and did not understand it yet. Thanks $\endgroup$ – user409387 Jan 29 '17 at 10:37
  • $\begingroup$ Let's say we are interested only in the permutations in the sum such that $\omega$ eats only the first two variables $v_1,v_2$ (in some order) and then (necessarily) $\eta$ eats the second variables $v_3,v_4$ (in some order). There are four such permutations - $\operatorname{id}, (12)(34), (12),(34)$ and each result in the same summand $\omega(v_1,v_2)\eta(v_3,v_4)$. Now do the same for the $6$ other choices of pairs of variables. $\endgroup$ – levap Jan 29 '17 at 18:33
  • $\begingroup$ Thanks. So I just have to choose fix{1,2} {3,4}, fix{1,3} {2,4}, fix{1,4} {2,3}, fix{3,4}{1,2}, fix{2,4}{1,2}, fix{2,3}{1,4}. Then ALWAYS use the said permutation id, (12)(34),(12),(34). Compute it and then I should get the above result. Is this right? $\endgroup$ – user409387 Feb 1 '17 at 10:17

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