Let $S_{12}$, $S_{13}$, $S_{14}$, $S_{23}$, $S_{24}$, $S_{34}$ be complex homogeneous polynomials in 4 variables satisfying the Plücker relation : $$S_{12}S_{34}-S_{13}S_{24}+S_{14}S_{23}=0 .$$ Suppose also that they don't have any common non trivial zero. Let $d_{ij}=\deg(S_{ij})$. Let $d$ be the integer $$d=d_{12}+d_{34}=d_{13}+d_{24}=d_{14}+d_{23}.$$ By an indirect way (I used this to construct vector bundles on $\mathbb{P}_3$), I find that $d$ should be a divisor of $d_{12}d_{13}d_{14}$. I would like to know if this is correct, and if it is true what is a direct proof.
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$\begingroup$ I wonder if there is a direct proof of the particular case $d_{12}=d_{13}=d_{14}=d_{23}=d_{34}=d_{24}=2k+1$. Equivalently, is there a Pythagorean 6-tuple of odd degree complex homogeneous polynomials in 4 variables without nontrivial common zero? For real polynomials the latter question is answered easily. Cf. mathoverflow.net/questions/62820/pythagorean-5-tuples. $\endgroup$– Mikhail SkopenkovAug 16, 2016 at 15:46
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$\begingroup$ I would be very grateful to you if you share any text with the proof that d | d12 d13 d14 or with your motivation for studying polynomials satisfying the Pluecker relation as well, whenever such a text appears. $\endgroup$– Mikhail SkopenkovAug 17, 2016 at 18:36
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1$\begingroup$ I can explain where this appears. Using the $S_{ij}$ you build an antisymmetric 4x4-matrix of polynomials. The Plücker relation implies that it is of rank 2 at every point of $\mathbb{P}^3$. It is the matrix of a morphism of rank 2 $E\to F$ of rank 4 vector bundles on $\mathbb{P}^3$, which are direct sums of line bundles. By computing the second Chern classes of the kernel and cokernel (which are integers), I find that $d$ should divide $d_{12}d_{13}d_{14}$. $\endgroup$– HephaistosAug 18, 2016 at 6:57
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$\begingroup$ Thanks. The motivation for the question is very interesting as well. $\endgroup$– Mikhail SkopenkovAug 18, 2016 at 7:34
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