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Let $X, Y\in \mathbb{P}^n$ be two singular Fano complete intersections of the same multidegree $(d_1,…,d_r)$.

If we assume there is an isomorphism $f\colon X\rightarrow Y$ are there any assumptions so that we can conclude that $f$ is induced by an action of the Automorphism group of $\mathbb{P}^n$, $\operatorname{PGL}(n+1)$?

I was reading the "Algebraic Hypersurfaces" note by J. Kollár (https://www.ams.org/journals/bull/2019-56-04/S0273-0979-2019-01663-2/S0273-0979-2019-01663-2.pdf) and I was wondering if there was a generalisation of Theorem $30$ for complete intersections.

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    $\begingroup$ Yes, at least if $\dim X\geq 3$ and $X$ is smooth. Then the embedding $X\hookrightarrow \mathbb{P}^n$ is defined by the generator of $\operatorname{Pic(X)} $, and therefore canonical. This is well-known, the question would be better appropriate at MSE. $\endgroup$
    – abx
    Commented Nov 22, 2020 at 14:09
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    $\begingroup$ @abx Thank you for your comment. I realize that my original question was not too specific, as I am mostly interested in singular Fano complete intersections. I have edited the question to reflect that. $\endgroup$
    – Theodoros Papazachariou
    Commented Nov 23, 2020 at 15:06
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    $\begingroup$ In fact the argument works provided $\dim(X)\geq 3$, with no smoothness assumptions. This is SGA 2, Exp. 12, Cor. 3.7. $\endgroup$
    – abx
    Commented Nov 23, 2020 at 15:48
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    $\begingroup$ @abx Your answer is excellent, but I would disagree about the appropriateness of this question for MO. It may be well-known to algebraic geometers, but likely not to those in adjacent (or even more distant) fields. So I think this is fine for MO, and do not feel that it should be closed. (And abx, maybe you could turn your two comments into an answer.) $\endgroup$
    – Joe Silverman
    Commented Nov 23, 2020 at 15:57
  • $\begingroup$ @Joe Silverman: You are right, especially for the singular case which is not so well-known. I will write my comment as an answer. $\endgroup$
    – abx
    Commented Nov 23, 2020 at 17:13

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This always holds for $\dim(X)\geq 3$. The point is that in this case the Picard group of $X$ is cyclic, generated by the line bundle $\mathscr{O}_X(1):= \mathscr{O}_{\mathbb{P}^n}(1)_{|X}$; this is SGA 2, Exp. 12, Cor. 3.7. Therefore any isomorphism $f:X\rightarrow Y$ induces an isomorphism $f^*\mathscr{O}_Y(1)\cong \mathscr{O}_X(1)$. Since the embedding in $\mathbb{P}^n$ is given by the global sections of this line bundle, this implies that $f$ is the restriction of an automorphism of $\mathbb{P}^n$.

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  • $\begingroup$ What about $\dim X=2$, say complete intersection of two quadrics in $\mathbb P^4$? $\endgroup$
    – Jesus Martinez Garcia
    Commented Nov 27, 2020 at 14:03
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    $\begingroup$ In your particular example this will still work, because $\mathscr{O}_X(1)=K_X^{-1}$ is anticanonical, hence preserved by any automorphism. However this would be false for a $(2,3)$ complete intersection in $\mathbb{P}^4$, or a $(2,2,2)$ in $\mathbb{P}^5$, which may have non-projective automorphisms. $\endgroup$
    – abx
    Commented Nov 27, 2020 at 14:34
  • $\begingroup$ Thank you @abx. When you say ‘false’ you mean definitely false or may be false because the (anti-)canonical argument doesn’t apply? $\endgroup$
    – Jesus Martinez Garcia
    Commented Nov 28, 2020 at 18:30
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    $\begingroup$ I mean definitely false. For instance the automorphism group may be infinite, while the group of projective automorphisms is always finite. $\endgroup$
    – abx
    Commented Nov 28, 2020 at 19:55

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