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Let $D'$ be the set of degree $4$ smooth surfaces in $\mathbb{P}^3$, where for each $S\in D'$ there are 6 hyperplanes $H_1, \dots, H_6\subset\mathbb{P}^3$ in general positions, such that $S$ passes through all the intersection points of these planes, i. e. $(H_i\cap H_j\cap H_k)\in S$ for all $1\leq i < j < k \leq 6$. Let $D=\overline{D'}$, the closure of $D'$.

I think for any quadric surface $Q=z_1z_4-z_2z_3$, i.e. $\mathbb{P}^1\times\mathbb{P}^1$, that $Q^2\notin D$, where $Q^2$ is the scheme defined by $(z_1z_4-z_2z_3)^2=0$. But I don't have a rigorous proof, and hope to give $D$ a more tractable description.

Here I call an element in $D$ a Darboux surface because this is a generalization of Darboux curves (degree 4 Darboux curves are called Luroth quartics, which has been studied intensively). I think this is a classical subject, but didn't find any reference about it, so I guess maybe it has some other names? Thanks a lot!

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    $\begingroup$ This a very nice question! I wonder if a similar statement holds in $\mathbb P^2$ - if you take five lines, $10$ intersection points of them and consider quartics that contain these $10$ points, is it true that double conic is not in the Zariski closure of the space of such quartics? $\endgroup$ Mar 1, 2013 at 21:57
  • $\begingroup$ @Dmitri: If you fix the lines, you just get a ${\bf P}^4$ of quartic curves, which contains no double conic. Do you want the lines to vary too? $\endgroup$ Mar 1, 2013 at 23:01
  • $\begingroup$ Noam, sure I want lines to vary too, so this becomes YQ's question in dimension one less. $\endgroup$ Mar 1, 2013 at 23:33
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    $\begingroup$ @Dmitri, thank you for your interest, and your statement is true. Please refer to Proposition 4.1, $\href{arxiv.org/pdf/0911.2101.pdf}{On singular L¨uroth quartics}$, by G. Ottaviani - E. Sernesi. The proof relies on the descriptions of points in the two components of the hypersurface of Luroth quartics. $\endgroup$
    – JYQ
    Mar 2, 2013 at 16:38
  • $\begingroup$ @Noam, if we allow lines to vary too, we have a degree 54 hypersurface in $\mathbb{P}^{14}$ by taking the closure. $\endgroup$
    – JYQ
    Mar 2, 2013 at 17:02

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