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A generic quintic threefold has 2875 lines. Is there an example of a quintic threefold explicitly defined by a polynomial that can be proven to have exactly 2875 distinct lines?

I am particularly interested in an answer in the following form. Oguiso-Yu (Section 2, arXiv:1504.05011) classified the automorphism groups of smooth quintic threefolds into 22 types. I would like to know one (or more, if possible) quintic threefold $X$ from their list such that $X$ has 2875 distinct lines and its automorphism group $Aut(X)\subset PGL_5(\mathbb{C})$ acts faithfully on those lines, i.e. the map from $Aut(X)$ into the symmetric group of the 2875 lines is injective.

I apologize for my ignorance in algebraic geometry as my background is in topology. I am aware of this math.stackexchange post (Lines on a Quintic Threefold) asking for a quintic threefold which can be shown to contain only finitely many lines. But it doesn't answer my question. In a comment, Nefertiti mentioned that Macaulay2 has a command command Fano(k,I) which will calculate the ideal of the Fano variety of k-planes inside the projective variety defined by the homogenous ideal $I$. I tried this approach for the quintic threefold $x^4y+y^4z+z^4w+w^4u+u^4x=0$ on my personal computer. The program got killed automatically after an hour. I guess my computer ran out of memory for this task. I am new to stackexchange and don't have enough reputation to comment on that post.

Thank you very much for any input.

Weiyan Chen
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  • Upon proving that some of the Galois groups of enumerative problems, among them the 2875 lines on the smooth quintic 3-fold in ${\Bbb P}^4$ is the full symmetric group, Harris writes "In particular, in every case in which current theory had failed to discern any intrinsic structure on the set of solutions -- it is proved here -- there is in fact none." – Jan-Magnus Økland Jul 19 '24 at 12:23
  • @Jan-MagnusØkland Thanks. How is that remark relevant here? – Weiyan Chen Jul 19 '24 at 15:20
  • Probably isn’t, since the 2875 lines are of codim two, unlike for the cubic surfaces where the 27 lines are codim one – Jan-Magnus Økland Jul 20 '24 at 10:05
  • I tried to run the same command on your examples and I don't think it's running out of memory but maybe gives up on some sort of convergence since it complains about running numeric algorithms. Maybe there's a way to check specific $\mathbb{Q}$-examples by looking at their reductions to sufficiently large $\mathbb{Z}/p$? (Also, hi!) – ronno Jul 23 '24 at 13:24
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    Actually I got Macaulay to finish the computation for the second polynomial and it claims to produce 4005 solutions, so something is wrong (if the number of lines is $>2875$ then the variety of lines should be positive dimensional). One of the generators of the ideal produced by Fano(1, I) is of the form $\text{monomial} - 10^{16} \cdot \text{monomial}$, so I'm guessing that's causing numeric issues. – ronno Jul 23 '24 at 18:46
  • @ronno Special quintics, like the Fermat, can have infinite lines. Maybe then the Fano command might malfunction like that? Big integers like the ones you see shouldn’t be a problem – Jan-Magnus Økland Jul 24 '24 at 16:46
  • @Jan-MagnusØkland Yes, I am aware that the variety of lines can be positive dimensional but I also verified that the ideal Macaulay produces defines something $0$-dimensional. I wouldn't be suspicious if there were big numbers everywhere but the ideal had dozens of generators (necessarily, to be $0$ dimensional in the Grasmmannian) and that was the only coefficient other than $\pm 1$. I'm guessing the $p - \text{large} \cdot q$ term is supposed to just be $q$, and because of this perturbation it's turning some of the solutions into multiple. – ronno Jul 24 '24 at 17:07
  • However, I am a beginner at using Macaulay so it's possible that I'm misusing/misunderstanding something. – ronno Jul 24 '24 at 17:07
  • @ronno Hi! Thanks! I just realized that the second example $x^4y+y^5+z^5+w^5+u^5=0$ does have infinitely many lines. For example, it contains any line $[s:\zeta s :zt:wt:ut]$ where $\zeta^4=-1$ and $z^5+w^5+u^5=0$. That seems to be a 1-dimensional family of lines. So that's not an example to look at. Sorry about that. I will edit the question. I still want to know about the first example though, or any example of a generic quintic threefold. I also tried to compute the Fano(1,I) for the first example over $Z/pZ$ where $p=2,3,5,7$. My computer does not produce any results after hours. – Weiyan Chen Jul 24 '24 at 21:15

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