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Let $H$ be an irreducible hypersurface in $\mathbb P^n$ of large-ish degree, say 14. This question is about subvarieties $V$ of $H$ such that

  • $V$ has codimension 1 in $H$ (i.e. $V$ has dimension $n-2$),
  • $V$ has degree 3, and
  • $V$ is contained in a quadratic hypersurface in $\mathbb P^n$.

Can $H$ have infinitely many such subvarieties? If $H$ has infinitely many such subvarieties, what does that say about $H$?

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    $\begingroup$ Are these subvarieties all iterated cones on cubic normal curves? $\endgroup$
    – Will Sawin
    Commented Oct 11, 2023 at 15:04
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    $\begingroup$ @WillSawin: it could also be an iterated cone over a cubic scroll. $\endgroup$
    – Sasha
    Commented Oct 11, 2023 at 16:02
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    $\begingroup$ @Sasha . . . or it could be a linearly degenerate variety that is a cubic hypersurface in a hyperplane (and the quadratic hypersurface is a reducible quadratic hypersurface). $\endgroup$
    – Jason Starr
    Commented Oct 11, 2023 at 17:41
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    $\begingroup$ @JasonStarr: right, and this covers all possibilities. $\endgroup$
    – Sasha
    Commented Oct 11, 2023 at 17:50
  • $\begingroup$ @Sasha would you consider posting that as an answer (with an epsilon more detail)? This trichotomy would already be enough for the application I have in mind. $\endgroup$
    – Simon L Rydin Myerson
    Commented Oct 12, 2023 at 20:13

1 Answer 1

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If the linear span of $V$ has dimension $n-1$, then $V$ is a cubic hypersurface in a hyperplane. Otherwise, $V$ is a variety of minimal degree, hence it is a cone over a linear section of $\mathbb{P}^1 \times \mathbb{P}^2$, see Corollary 1.12 in [Saint-Donat, B. Projective models of $K-3$ surfaces. Amer. J. Math. 96 (1974), 602--639].

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  • $\begingroup$ If I remember MO protocol correctly I should wait a day or so before accepting this as the answer. But it gives all the information I need, I think - in particular about the codimension of the singular locus of the quadratic hypersurface, which is very restricted in my application. $\endgroup$
    – Simon L Rydin Myerson
    Commented Oct 14, 2023 at 22:42
  • $\begingroup$ Having looked at that reference the other results given there are very explicit and helpful, so much so I will accept the answer now. $\endgroup$
    – Simon L Rydin Myerson
    Commented Oct 15, 2023 at 11:16

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